(b)The idea/concept of oxidation numbers uses/applies the following simple guideline rules: Guidelines /rules applied in assigning oxidation number 1.Oxidation number of combined Oxygen is always -2 except in peroxides (Na2O2/H2O2) where its Oxidation number is -1
5 2.Oxidation number of combined Hydrogen is always +1except in Hydrides (NaH/KH) where its Oxidation number is -1 3.All atoms and molecules of elements have oxidation number 0 (zero) Atom Oxidation number Molecule Oxidation number Na 0 Cl2 0 O 0 O2 0 H 0 H2 0 Al 0 N2 0 Ne 0 O3 0 K 0 P3 0 Cu 0 S8 0 4.All combined metals and non-metals have oxidation numbers equal to their valency /oxidation state e.g. Metal/non-metal ion Valency Oxidation state Oxidation number Fe2+ 2 -2 -2 Fe3+ 3 -3 -3 Cu2+ 2 -2 -2 Cu+ 1 +1 +1 Cl- 1 -1 -1 O2- 2 -2 -2 Na+ 1 +1 +1 Al3+ 3 +3 +3 P3- 3 -3 -3 Pb2+ 2 +2 +2 5.Sum of oxidation numbers of atoms of elements making a compound is equal zero(0) e.g. Using this rule ,an unknown oxidation number of an atom in a compound can be determined as below: a) CuSO4 has- -one atom of Cu with oxidation number +2( refer to Rule 4) -one atom of S with oxidation number +6 ( refer to Rule 4) -six atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in CuSO4 = (+2 + +6 + (-2 x 6)) = 0 b) H2SO4 has- -two atom of H each with oxidation number +1( refer to Rule 2) -one atom of S with oxidation number +6 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in H2SO4 = (+2 + +6 + (-2 x 4)) = 0
6 c) KMnO4 has- -one atom of K with oxidation number +1( refer to Rule 4) -one atom of Mn with oxidation number +7 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in KMnO4 = (+1 + +7 + (-2 x 4)) = 0 Determine the oxidation number of: I.Nitrogen in; -NO => x + -2 = 0 thus x = 0 – (-2) = + 2 The chemical name of this compound is thus Nitrogen(II)oxide -NO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Nitrogen(IV)oxide -N2O => 2x + -2 = 0 thus 2x = 0 – (-2) = +2/2= +1 The chemical name of this compound is thus Nitrogen(I)oxide II. Sulphur in; -SO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Sulphur(IV)oxide -SO3 => x + (-2 x3)= 0 thus x = 0 – (-6) = + 6 The chemical name of this compound is thus Sulphur(VI)oxide -H2SO4 = ((+1 x 2) + x + (-2 x 4)) thus x= 0-( +2 +-8) =+6 The chemical name of this compound is thus Sulphuric(VI)acid -H2SO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Sulphuric(IV)acid III.


Metal/non-metal ion Valency Oxidation state Oxidation number Fe2+ 2 -2 -2 Fe3+ 3 -3 -3 Cu2+ 2 -2 -2 Cu+ 1 +1 +1 Cl- 1 -1 -1 O2- 2 -2 -2 Na+ 1 +1 +1 Al3+ 3 +3 +3 P3- 3 -3 -3 Pb2+ 2 +2 +2 5.Sum of oxidation numbers of atoms of elements making a compound is equal zero(0) e.g. Using this rule ,an unknown oxidation number of an atom in a compound can be determined as below: a) CuSO4 has- -one atom of Cu with oxidation number +2( refer to Rule 4) -one atom of S with oxidation number +6 ( refer to Rule 4) -six atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in CuSO4 = (+2 + +6 + (-2 x 6)) = 0 b) H2SO4 has- -two atom of H each with oxidation number +1( refer to Rule 2) -one atom of S with oxidation number +6 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in H2SO4 = (+2 + +6 + (-2 x 4)) = 0
6 c) KMnO4 has- -one atom of K with oxidation number +1( refer to Rule 4) -one atom of Mn with oxidation number +7 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in KMnO4 = (+1 + +7 + (-2 x 4)) = 0 Determine the oxidation number of: I.Nitrogen in; -NO => x + -2 = 0 thus x = 0 – (-2) = + 2 The chemical name of this compound is thus Nitrogen(II)oxide -NO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Nitrogen(IV)oxide -N2O => 2x + -2 = 0 thus 2x = 0 – (-2) = +2/2= +1 The chemical name of this compound is thus Nitrogen(I)oxide II. Sulphur in; -SO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Sulphur(IV)oxide -SO3 => x + (-2 x3)= 0 thus x = 0 – (-6) = + 6 The chemical name of this compound is thus Sulphur(VI)oxide -H2SO4 = ((+1 x 2) + x + (-2 x 4)) thus x= 0-( +2 +-8) =+6 The chemical name of this compound is thus Sulphuric(VI)acid -H2SO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Sulphuric(IV)acid III. Carbon in; -CO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus carbon(IV)oxide -CO => x + -2 = 0 thus x = 0 – -2 = + 2 The chemical name of this compound is thus carbon(II)oxide -H2CO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Carbonic(IV)acid IV.Manganese in; -MnO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Manganese(IV)oxide -KMnO4 = ((+1 + x + (-2 x 4)) thus x= 0-( +1 +-8) =+7 The chemical name of this compound is thus Potassium manganate(VII) V.Chromium in; - Cr2O3 => 2x + (-2 x 3)= 0 thus 2x = 0 – (-6) = +6 / 2= +3 The chemical name of this compound is thus Chromium(III)oxide -K2Cr2O7 => (+1 x 2) + 2x + (-2 x7)= 0 thus 2x = 0 – +2 +-14 = +12 / 2= +6 The chemical name of this compound is thus Potassium dichromate(VI) -K2CrO4 => (+1 x 2) + x + (-2 x4)= 0 thus 2x = 0 – +2 +-8 = +12 / 2= +6
7 The chemical name of this compound is thus Potassium chromate(VI) 6.The sum of the oxidation numbers of atoms of elements making a charged radical/complex ion is equal to its charge.


Using this rule ,an unknown oxidation number of an atom in a compound can be determined as below: a) CuSO4 has- -one atom of Cu with oxidation number +2( refer to Rule 4) -one atom of S with oxidation number +6 ( refer to Rule 4) -six atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in CuSO4 = (+2 + +6 + (-2 x 6)) = 0 b) H2SO4 has- -two atom of H each with oxidation number +1( refer to Rule 2) -one atom of S with oxidation number +6 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in H2SO4 = (+2 + +6 + (-2 x 4)) = 0
6 c) KMnO4 has- -one atom of K with oxidation number +1( refer to Rule 4) -one atom of Mn with oxidation number +7 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in KMnO4 = (+1 + +7 + (-2 x 4)) = 0 Determine the oxidation number of: I.Nitrogen in; -NO => x + -2 = 0 thus x = 0 – (-2) = + 2 The chemical name of this compound is thus Nitrogen(II)oxide -NO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Nitrogen(IV)oxide -N2O => 2x + -2 = 0 thus 2x = 0 – (-2) = +2/2= +1 The chemical name of this compound is thus Nitrogen(I)oxide II. Sulphur in; -SO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Sulphur(IV)oxide -SO3 => x + (-2 x3)= 0 thus x = 0 – (-6) = + 6 The chemical name of this compound is thus Sulphur(VI)oxide -H2SO4 = ((+1 x 2) + x + (-2 x 4)) thus x= 0-( +2 +-8) =+6 The chemical name of this compound is thus Sulphuric(VI)acid -H2SO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Sulphuric(IV)acid III. Carbon in; -CO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus carbon(IV)oxide -CO => x + -2 = 0 thus x = 0 – -2 = + 2 The chemical name of this compound is thus carbon(II)oxide -H2CO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Carbonic(IV)acid IV.Manganese in; -MnO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Manganese(IV)oxide -KMnO4 = ((+1 + x + (-2 x 4)) thus x= 0-( +1 +-8) =+7 The chemical name of this compound is thus Potassium manganate(VII) V.Chromium in; - Cr2O3 => 2x + (-2 x 3)= 0 thus 2x = 0 – (-6) = +6 / 2= +3 The chemical name of this compound is thus Chromium(III)oxide -K2Cr2O7 => (+1 x 2) + 2x + (-2 x7)= 0 thus 2x = 0 – +2 +-14 = +12 / 2= +6 The chemical name of this compound is thus Potassium dichromate(VI) -K2CrO4 => (+1 x 2) + x + (-2 x4)= 0 thus 2x = 0 – +2 +-8 = +12 / 2= +6
7 The chemical name of this compound is thus Potassium chromate(VI) 6.The sum of the oxidation numbers of atoms of elements making a charged radical/complex ion is equal to its charge. Using this rule ,the oxidation number of unknown atom of an element in a charged radical/complex ion can be determined as in the examples below; a) SO42- has- -one atom of S with oxidation number +6( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in SO42- = ( +6 + (-2 x 4)) = -2 The chemical name of this radical is thus sulphate(VI) ion b) NO3- has- -one atom of N with oxidation number +4( refer to Rule 4) -three atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in NO3- = ( +4 + (-2 x 3)) = -1 The chemical name of this radical is thus nitrate(IV) ion.


Sulphur in; -SO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Sulphur(IV)oxide -SO3 => x + (-2 x3)= 0 thus x = 0 – (-6) = + 6 The chemical name of this compound is thus Sulphur(VI)oxide -H2SO4 = ((+1 x 2) + x + (-2 x 4)) thus x= 0-( +2 +-8) =+6 The chemical name of this compound is thus Sulphuric(VI)acid -H2SO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Sulphuric(IV)acid III. Carbon in; -CO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus carbon(IV)oxide -CO => x + -2 = 0 thus x = 0 – -2 = + 2 The chemical name of this compound is thus carbon(II)oxide -H2CO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Carbonic(IV)acid IV.Manganese in; -MnO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Manganese(IV)oxide -KMnO4 = ((+1 + x + (-2 x 4)) thus x= 0-( +1 +-8) =+7 The chemical name of this compound is thus Potassium manganate(VII) V.Chromium in; - Cr2O3 => 2x + (-2 x 3)= 0 thus 2x = 0 – (-6) = +6 / 2= +3 The chemical name of this compound is thus Chromium(III)oxide -K2Cr2O7 => (+1 x 2) + 2x + (-2 x7)= 0 thus 2x = 0 – +2 +-14 = +12 / 2= +6 The chemical name of this compound is thus Potassium dichromate(VI) -K2CrO4 => (+1 x 2) + x + (-2 x4)= 0 thus 2x = 0 – +2 +-8 = +12 / 2= +6
7 The chemical name of this compound is thus Potassium chromate(VI) 6.The sum of the oxidation numbers of atoms of elements making a charged radical/complex ion is equal to its charge. Using this rule ,the oxidation number of unknown atom of an element in a charged radical/complex ion can be determined as in the examples below; a) SO42- has- -one atom of S with oxidation number +6( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in SO42- = ( +6 + (-2 x 4)) = -2 The chemical name of this radical is thus sulphate(VI) ion b) NO3- has- -one atom of N with oxidation number +4( refer to Rule 4) -three atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in NO3- = ( +4 + (-2 x 3)) = -1 The chemical name of this radical is thus nitrate(IV) ion. Determine the oxidation number of: I.Nitrogen in; -NO2- => x + (-2 x2)= -1 thus x = -1 – (-4) = + 3 The chemical name of this compound/ion/radical is thus Nitrate(III)ion II.


Carbon in; -CO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus carbon(IV)oxide -CO => x + -2 = 0 thus x = 0 – -2 = + 2 The chemical name of this compound is thus carbon(II)oxide -H2CO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Carbonic(IV)acid IV.Manganese in; -MnO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Manganese(IV)oxide -KMnO4 = ((+1 + x + (-2 x 4)) thus x= 0-( +1 +-8) =+7 The chemical name of this compound is thus Potassium manganate(VII) V.Chromium in; - Cr2O3 => 2x + (-2 x 3)= 0 thus 2x = 0 – (-6) = +6 / 2= +3 The chemical name of this compound is thus Chromium(III)oxide -K2Cr2O7 => (+1 x 2) + 2x + (-2 x7)= 0 thus 2x = 0 – +2 +-14 = +12 / 2= +6 The chemical name of this compound is thus Potassium dichromate(VI) -K2CrO4 => (+1 x 2) + x + (-2 x4)= 0 thus 2x = 0 – +2 +-8 = +12 / 2= +6
7 The chemical name of this compound is thus Potassium chromate(VI) 6.The sum of the oxidation numbers of atoms of elements making a charged radical/complex ion is equal to its charge. Using this rule ,the oxidation number of unknown atom of an element in a charged radical/complex ion can be determined as in the examples below; a) SO42- has- -one atom of S with oxidation number +6( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in SO42- = ( +6 + (-2 x 4)) = -2 The chemical name of this radical is thus sulphate(VI) ion b) NO3- has- -one atom of N with oxidation number +4( refer to Rule 4) -three atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in NO3- = ( +4 + (-2 x 3)) = -1 The chemical name of this radical is thus nitrate(IV) ion. Determine the oxidation number of: I.Nitrogen in; -NO2- => x + (-2 x2)= -1 thus x = -1 – (-4) = + 3 The chemical name of this compound/ion/radical is thus Nitrate(III)ion II. Sulphur in; -SO32- => x + (-2 x3)= -2 thus x = -2 – (-6) = + 4 The chemical name of this compound/ion/radical is thus Sulphate(IV)ion III.


Using this rule ,the oxidation number of unknown atom of an element in a charged radical/complex ion can be determined as in the examples below; a) SO42- has- -one atom of S with oxidation number +6( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in SO42- = ( +6 + (-2 x 4)) = -2 The chemical name of this radical is thus sulphate(VI) ion b) NO3- has- -one atom of N with oxidation number +4( refer to Rule 4) -three atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in NO3- = ( +4 + (-2 x 3)) = -1 The chemical name of this radical is thus nitrate(IV) ion. Determine the oxidation number of: I.Nitrogen in; -NO2- => x + (-2 x2)= -1 thus x = -1 – (-4) = + 3 The chemical name of this compound/ion/radical is thus Nitrate(III)ion II. Sulphur in; -SO32- => x + (-2 x3)= -2 thus x = -2 – (-6) = + 4 The chemical name of this compound/ion/radical is thus Sulphate(IV)ion III. Carbon in; -CO32- = x + (-2 x 3) = -2 thus x = -2 – (-6) = + 4 The chemical name of this compound/ion/radical is thus Carbonate(IV)ion IV.Manganese in; -MnO4 - = x + (-2 x 4)= -1 thus x= -1-(-2 +-8) =+7 The chemical name of this compound/ion/radical is thus manganate(VII) ion V.Chromium in -Cr2O72- => 2x + (-2 x7)= -2 thus 2x = -2 – +2 +-14 = +12 / 2= +6 The chemical name of this compound/ion//radical is thus dichromate(VI) ion -CrO42- => x + (-2 x4)= -2 thus x = -2 + (-2 x 4) = +6 The chemical name of this compound/ion//radical is thus chromate(VI) ion (c)Using the concept/idea of oxidation numbers as increase and decrease in oxidation numbers , the oxidizing and reducing species/agents can be determined as in the following examples; (i) Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) Oxidation numbers -> +2 0 +2 0 Oxidizing species/agents =>Cu2+ ;its oxidation number decrease from+2 to 0 in Cu(s)
8 Reducing species/agents => Zn2+ ;its oxidation number increase from 0 to +2 in Zn(s) (ii) 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2 (l) Oxidation numbers -> -1 0 -1 0 Oxidizing agent =>Cl2(g) ;its oxidation number decrease from 0 to-1 in 2Cl- (aq) Reducing agents => Zn2+ ;its oxidation number increase from -1 to 0 in Zn(s) (iii) Br2 (l) + Zn(s) -> Zn2+ (aq) + 2Br- (aq) Oxidation numbers -> 0 0 +2 -1 Oxidizing agent => Br2 (l) ;its oxidation number decrease from 0 to-1 in 2Br-(aq) Reducing agents => Zn(s) ;its oxidation number increase from 0 to +2 in Zn2+ (iv) 2HCl (aq) + Mg(s) -> MgCl2 (aq) + H2 (g) Oxidation numbers -> 2 (+1 -1) 0 +2 2(-1) 0 Oxidizing agent => H+ in HCl;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Mg(s) ;its oxidation number increase from 0 to +2 in Mg2+ (v) 2H2O (l) + 2Na(s) -> 2NaOH (aq) + H2 (g) Oxidation numbers -> +1 -2 0 +1 -2 +1 0 Oxidizing agent => H+ in H2O;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Na(s) ;its oxidation number increase from 0 to +1 in Na+ (vi) 5Fe2+ (aq) + 8H+ (aq) + MnO4- -> 5Fe3+ (aq) + Mn2+ (aq) + 4H2O (l) +2 +1 +7 -2 +3 +2 +1 -2 Oxidizing agent => Mn in MnO4- ;its oxidation number decrease from +7to+2 in Mn2+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (vii) 6Fe2+ (aq) + 14H+ (aq) + Cr2O72-(aq) -> 6Fe3+ (aq) + Cr3+ (aq) + 7H2O (l) +2 +1 +6 -2 +3 +3 +1 -2 Oxidizing agent: Cr in Cr2O72- ;its oxidation number decrease from +6 to+3 in Cr3+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (viii) 2Fe2+ (aq) + 2H+ (aq) + H2O2(aq) -> 2Fe3+ (aq) + 2H2O (l) +2 +1 +1 -1 +3 +1 -2 Oxidizing agent: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (ix) Cr2O72-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Cr3+ (aq) + 2H2O (l) + 5O2(g) +6 -2 +1 +1 -1 +3 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Cr in Cr2O72- its oxidation number decrease from +6 to +3 in Cr3+ Reducing agents
9 O in H2O2;its oxidation number increase from -1 to O in O2(g) O in Cr2O72- its oxidation number increase from -2 to O in O2(g) (x) 2MnO4-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Mn2+ (aq) + 8H2O (l) + 5O2(g) +7 -2 +1 +1 -1 +2 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Mn in MnO4- its oxidation number decrease from +7 to +2 in Mn2+ Reducing agents O in H2O2;its oxidation number increase from -1 to O in O2(g) O in MnO4- its oxidation number increase from -2 to O in O2(g) (ii)ELECTROCHEMICAL (VOLTAIC) CELL
10 1.


Determine the oxidation number of: I.Nitrogen in; -NO2- => x + (-2 x2)= -1 thus x = -1 – (-4) = + 3 The chemical name of this compound/ion/radical is thus Nitrate(III)ion II. Sulphur in; -SO32- => x + (-2 x3)= -2 thus x = -2 – (-6) = + 4 The chemical name of this compound/ion/radical is thus Sulphate(IV)ion III. Carbon in; -CO32- = x + (-2 x 3) = -2 thus x = -2 – (-6) = + 4 The chemical name of this compound/ion/radical is thus Carbonate(IV)ion IV.Manganese in; -MnO4 - = x + (-2 x 4)= -1 thus x= -1-(-2 +-8) =+7 The chemical name of this compound/ion/radical is thus manganate(VII) ion V.Chromium in -Cr2O72- => 2x + (-2 x7)= -2 thus 2x = -2 – +2 +-14 = +12 / 2= +6 The chemical name of this compound/ion//radical is thus dichromate(VI) ion -CrO42- => x + (-2 x4)= -2 thus x = -2 + (-2 x 4) = +6 The chemical name of this compound/ion//radical is thus chromate(VI) ion (c)Using the concept/idea of oxidation numbers as increase and decrease in oxidation numbers , the oxidizing and reducing species/agents can be determined as in the following examples; (i) Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) Oxidation numbers -> +2 0 +2 0 Oxidizing species/agents =>Cu2+ ;its oxidation number decrease from+2 to 0 in Cu(s)
8 Reducing species/agents => Zn2+ ;its oxidation number increase from 0 to +2 in Zn(s) (ii) 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2 (l) Oxidation numbers -> -1 0 -1 0 Oxidizing agent =>Cl2(g) ;its oxidation number decrease from 0 to-1 in 2Cl- (aq) Reducing agents => Zn2+ ;its oxidation number increase from -1 to 0 in Zn(s) (iii) Br2 (l) + Zn(s) -> Zn2+ (aq) + 2Br- (aq) Oxidation numbers -> 0 0 +2 -1 Oxidizing agent => Br2 (l) ;its oxidation number decrease from 0 to-1 in 2Br-(aq) Reducing agents => Zn(s) ;its oxidation number increase from 0 to +2 in Zn2+ (iv) 2HCl (aq) + Mg(s) -> MgCl2 (aq) + H2 (g) Oxidation numbers -> 2 (+1 -1) 0 +2 2(-1) 0 Oxidizing agent => H+ in HCl;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Mg(s) ;its oxidation number increase from 0 to +2 in Mg2+ (v) 2H2O (l) + 2Na(s) -> 2NaOH (aq) + H2 (g) Oxidation numbers -> +1 -2 0 +1 -2 +1 0 Oxidizing agent => H+ in H2O;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Na(s) ;its oxidation number increase from 0 to +1 in Na+ (vi) 5Fe2+ (aq) + 8H+ (aq) + MnO4- -> 5Fe3+ (aq) + Mn2+ (aq) + 4H2O (l) +2 +1 +7 -2 +3 +2 +1 -2 Oxidizing agent => Mn in MnO4- ;its oxidation number decrease from +7to+2 in Mn2+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (vii) 6Fe2+ (aq) + 14H+ (aq) + Cr2O72-(aq) -> 6Fe3+ (aq) + Cr3+ (aq) + 7H2O (l) +2 +1 +6 -2 +3 +3 +1 -2 Oxidizing agent: Cr in Cr2O72- ;its oxidation number decrease from +6 to+3 in Cr3+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (viii) 2Fe2+ (aq) + 2H+ (aq) + H2O2(aq) -> 2Fe3+ (aq) + 2H2O (l) +2 +1 +1 -1 +3 +1 -2 Oxidizing agent: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (ix) Cr2O72-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Cr3+ (aq) + 2H2O (l) + 5O2(g) +6 -2 +1 +1 -1 +3 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Cr in Cr2O72- its oxidation number decrease from +6 to +3 in Cr3+ Reducing agents
9 O in H2O2;its oxidation number increase from -1 to O in O2(g) O in Cr2O72- its oxidation number increase from -2 to O in O2(g) (x) 2MnO4-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Mn2+ (aq) + 8H2O (l) + 5O2(g) +7 -2 +1 +1 -1 +2 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Mn in MnO4- its oxidation number decrease from +7 to +2 in Mn2+ Reducing agents O in H2O2;its oxidation number increase from -1 to O in O2(g) O in MnO4- its oxidation number increase from -2 to O in O2(g) (ii)ELECTROCHEMICAL (VOLTAIC) CELL
10 1. When a metal rod/plate is put in a solution of its own salt, some of the metal ionizes and dissolve into the solution i.e.


Sulphur in; -SO32- => x + (-2 x3)= -2 thus x = -2 – (-6) = + 4 The chemical name of this compound/ion/radical is thus Sulphate(IV)ion III. Carbon in; -CO32- = x + (-2 x 3) = -2 thus x = -2 – (-6) = + 4 The chemical name of this compound/ion/radical is thus Carbonate(IV)ion IV.Manganese in; -MnO4 - = x + (-2 x 4)= -1 thus x= -1-(-2 +-8) =+7 The chemical name of this compound/ion/radical is thus manganate(VII) ion V.Chromium in -Cr2O72- => 2x + (-2 x7)= -2 thus 2x = -2 – +2 +-14 = +12 / 2= +6 The chemical name of this compound/ion//radical is thus dichromate(VI) ion -CrO42- => x + (-2 x4)= -2 thus x = -2 + (-2 x 4) = +6 The chemical name of this compound/ion//radical is thus chromate(VI) ion (c)Using the concept/idea of oxidation numbers as increase and decrease in oxidation numbers , the oxidizing and reducing species/agents can be determined as in the following examples; (i) Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) Oxidation numbers -> +2 0 +2 0 Oxidizing species/agents =>Cu2+ ;its oxidation number decrease from+2 to 0 in Cu(s)
8 Reducing species/agents => Zn2+ ;its oxidation number increase from 0 to +2 in Zn(s) (ii) 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2 (l) Oxidation numbers -> -1 0 -1 0 Oxidizing agent =>Cl2(g) ;its oxidation number decrease from 0 to-1 in 2Cl- (aq) Reducing agents => Zn2+ ;its oxidation number increase from -1 to 0 in Zn(s) (iii) Br2 (l) + Zn(s) -> Zn2+ (aq) + 2Br- (aq) Oxidation numbers -> 0 0 +2 -1 Oxidizing agent => Br2 (l) ;its oxidation number decrease from 0 to-1 in 2Br-(aq) Reducing agents => Zn(s) ;its oxidation number increase from 0 to +2 in Zn2+ (iv) 2HCl (aq) + Mg(s) -> MgCl2 (aq) + H2 (g) Oxidation numbers -> 2 (+1 -1) 0 +2 2(-1) 0 Oxidizing agent => H+ in HCl;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Mg(s) ;its oxidation number increase from 0 to +2 in Mg2+ (v) 2H2O (l) + 2Na(s) -> 2NaOH (aq) + H2 (g) Oxidation numbers -> +1 -2 0 +1 -2 +1 0 Oxidizing agent => H+ in H2O;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Na(s) ;its oxidation number increase from 0 to +1 in Na+ (vi) 5Fe2+ (aq) + 8H+ (aq) + MnO4- -> 5Fe3+ (aq) + Mn2+ (aq) + 4H2O (l) +2 +1 +7 -2 +3 +2 +1 -2 Oxidizing agent => Mn in MnO4- ;its oxidation number decrease from +7to+2 in Mn2+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (vii) 6Fe2+ (aq) + 14H+ (aq) + Cr2O72-(aq) -> 6Fe3+ (aq) + Cr3+ (aq) + 7H2O (l) +2 +1 +6 -2 +3 +3 +1 -2 Oxidizing agent: Cr in Cr2O72- ;its oxidation number decrease from +6 to+3 in Cr3+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (viii) 2Fe2+ (aq) + 2H+ (aq) + H2O2(aq) -> 2Fe3+ (aq) + 2H2O (l) +2 +1 +1 -1 +3 +1 -2 Oxidizing agent: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (ix) Cr2O72-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Cr3+ (aq) + 2H2O (l) + 5O2(g) +6 -2 +1 +1 -1 +3 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Cr in Cr2O72- its oxidation number decrease from +6 to +3 in Cr3+ Reducing agents
9 O in H2O2;its oxidation number increase from -1 to O in O2(g) O in Cr2O72- its oxidation number increase from -2 to O in O2(g) (x) 2MnO4-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Mn2+ (aq) + 8H2O (l) + 5O2(g) +7 -2 +1 +1 -1 +2 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Mn in MnO4- its oxidation number decrease from +7 to +2 in Mn2+ Reducing agents O in H2O2;its oxidation number increase from -1 to O in O2(g) O in MnO4- its oxidation number increase from -2 to O in O2(g) (ii)ELECTROCHEMICAL (VOLTAIC) CELL
10 1. When a metal rod/plate is put in a solution of its own salt, some of the metal ionizes and dissolve into the solution i.e. M(s) -> M+(aq) + e ( monovalent metal) M(s) -> M2+(aq) + 2e ( divalent metal) M(s) -> M3+(aq) + 3e ( Trivalent metal) The ions move into the solution leaving electrons on the surface of the metal rod/plate.


Carbon in; -CO32- = x + (-2 x 3) = -2 thus x = -2 – (-6) = + 4 The chemical name of this compound/ion/radical is thus Carbonate(IV)ion IV.Manganese in; -MnO4 - = x + (-2 x 4)= -1 thus x= -1-(-2 +-8) =+7 The chemical name of this compound/ion/radical is thus manganate(VII) ion V.Chromium in -Cr2O72- => 2x + (-2 x7)= -2 thus 2x = -2 – +2 +-14 = +12 / 2= +6 The chemical name of this compound/ion//radical is thus dichromate(VI) ion -CrO42- => x + (-2 x4)= -2 thus x = -2 + (-2 x 4) = +6 The chemical name of this compound/ion//radical is thus chromate(VI) ion (c)Using the concept/idea of oxidation numbers as increase and decrease in oxidation numbers , the oxidizing and reducing species/agents can be determined as in the following examples; (i) Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) Oxidation numbers -> +2 0 +2 0 Oxidizing species/agents =>Cu2+ ;its oxidation number decrease from+2 to 0 in Cu(s)
8 Reducing species/agents => Zn2+ ;its oxidation number increase from 0 to +2 in Zn(s) (ii) 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2 (l) Oxidation numbers -> -1 0 -1 0 Oxidizing agent =>Cl2(g) ;its oxidation number decrease from 0 to-1 in 2Cl- (aq) Reducing agents => Zn2+ ;its oxidation number increase from -1 to 0 in Zn(s) (iii) Br2 (l) + Zn(s) -> Zn2+ (aq) + 2Br- (aq) Oxidation numbers -> 0 0 +2 -1 Oxidizing agent => Br2 (l) ;its oxidation number decrease from 0 to-1 in 2Br-(aq) Reducing agents => Zn(s) ;its oxidation number increase from 0 to +2 in Zn2+ (iv) 2HCl (aq) + Mg(s) -> MgCl2 (aq) + H2 (g) Oxidation numbers -> 2 (+1 -1) 0 +2 2(-1) 0 Oxidizing agent => H+ in HCl;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Mg(s) ;its oxidation number increase from 0 to +2 in Mg2+ (v) 2H2O (l) + 2Na(s) -> 2NaOH (aq) + H2 (g) Oxidation numbers -> +1 -2 0 +1 -2 +1 0 Oxidizing agent => H+ in H2O;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Na(s) ;its oxidation number increase from 0 to +1 in Na+ (vi) 5Fe2+ (aq) + 8H+ (aq) + MnO4- -> 5Fe3+ (aq) + Mn2+ (aq) + 4H2O (l) +2 +1 +7 -2 +3 +2 +1 -2 Oxidizing agent => Mn in MnO4- ;its oxidation number decrease from +7to+2 in Mn2+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (vii) 6Fe2+ (aq) + 14H+ (aq) + Cr2O72-(aq) -> 6Fe3+ (aq) + Cr3+ (aq) + 7H2O (l) +2 +1 +6 -2 +3 +3 +1 -2 Oxidizing agent: Cr in Cr2O72- ;its oxidation number decrease from +6 to+3 in Cr3+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (viii) 2Fe2+ (aq) + 2H+ (aq) + H2O2(aq) -> 2Fe3+ (aq) + 2H2O (l) +2 +1 +1 -1 +3 +1 -2 Oxidizing agent: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (ix) Cr2O72-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Cr3+ (aq) + 2H2O (l) + 5O2(g) +6 -2 +1 +1 -1 +3 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Cr in Cr2O72- its oxidation number decrease from +6 to +3 in Cr3+ Reducing agents
9 O in H2O2;its oxidation number increase from -1 to O in O2(g) O in Cr2O72- its oxidation number increase from -2 to O in O2(g) (x) 2MnO4-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Mn2+ (aq) + 8H2O (l) + 5O2(g) +7 -2 +1 +1 -1 +2 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Mn in MnO4- its oxidation number decrease from +7 to +2 in Mn2+ Reducing agents O in H2O2;its oxidation number increase from -1 to O in O2(g) O in MnO4- its oxidation number increase from -2 to O in O2(g) (ii)ELECTROCHEMICAL (VOLTAIC) CELL
10 1. When a metal rod/plate is put in a solution of its own salt, some of the metal ionizes and dissolve into the solution i.e. M(s) -> M+(aq) + e ( monovalent metal) M(s) -> M2+(aq) + 2e ( divalent metal) M(s) -> M3+(aq) + 3e ( Trivalent metal) The ions move into the solution leaving electrons on the surface of the metal rod/plate. 2.The metal rod becomes therefore negatively charged while its own solution positively charged.


When a metal rod/plate is put in a solution of its own salt, some of the metal ionizes and dissolve into the solution i.e. M(s) -> M+(aq) + e ( monovalent metal) M(s) -> M2+(aq) + 2e ( divalent metal) M(s) -> M3+(aq) + 3e ( Trivalent metal) The ions move into the solution leaving electrons on the surface of the metal rod/plate. 2.The metal rod becomes therefore negatively charged while its own solution positively charged. As the positive charges of the solution increase, some of them recombine with the electrons to form back the metal atoms M+(aq) + e -> M(s) ( monovalent metal) M2+(aq) + 2e -> M(s) (divalent metal) M3+(aq) + 3e -> M(s) (Trivalent metal) 3. When a metal rod/plate is put in a solution of its own salt, it constitutes/forms a half-cell. The tendency of metals to ionize differ from one metal to the other. The difference can be measured by connecting two half cells to form an electrochemical/voltaic cell as in the below procedure: To set up an electrochemical /voltaic cell To compare the relative tendency of metals to ionize Place 50cm3 of 1M Zinc(II) sulphate(VI) in 100cm3 beaker. Put a clean zinc rod/plate into the solution. Place 50cm3 of 1M Copper(II) sulphate(VI) in another 100cm3 beaker. Put a clean copper rod/plate of equal area (length x width) with Zinc into the solution. Connect/join the two metals(to a voltmeter) using connecting wires. Dip a folded filter paper into a solution of Potassium nitrate(V) or sodium(I) chloride(I) until it soaks.


Put a clean copper rod/plate of equal area (length x width) with Zinc into the solution. Connect/join the two metals(to a voltmeter) using connecting wires. Dip a folded filter paper into a solution of Potassium nitrate(V) or sodium(I) chloride(I) until it soaks. Use the folded soaked filter paper to connect/join the two solutions in the two beakers.The whole set up should be as below Repeat the above procedure by replacing: (i)Zinc half cell with Magnesium rod/plate/ribbon dipped in 50cm3 of IM magnesium (II) sulphate(VI) solution V
11 (ii)Zinc half cell with Silver rod/plate/coin dipped in 50cm3 of IM silver(I) nitrate(V) solution (iii)Copper half cell with Iron rod/plate/spoon dipped in 50cm3 of IM Iron (II) sulphate(VI) solution Record the observations in the table below Changes on the 1st metal rod (A) Changes on the 2nd metal rod (B) Changes on the 1st solution (A(aq)) Changes on the 2nd solution (B(aq)) Voltage/voltmeter reading(Volts) Using Zn/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Zinc(II)sulphate (VI)colour remain colourless Blue Copper (II)sulphate (VI)colour fades.


Connect/join the two metals(to a voltmeter) using connecting wires. Dip a folded filter paper into a solution of Potassium nitrate(V) or sodium(I) chloride(I) until it soaks. Use the folded soaked filter paper to connect/join the two solutions in the two beakers.The whole set up should be as below Repeat the above procedure by replacing: (i)Zinc half cell with Magnesium rod/plate/ribbon dipped in 50cm3 of IM magnesium (II) sulphate(VI) solution V
11 (ii)Zinc half cell with Silver rod/plate/coin dipped in 50cm3 of IM silver(I) nitrate(V) solution (iii)Copper half cell with Iron rod/plate/spoon dipped in 50cm3 of IM Iron (II) sulphate(VI) solution Record the observations in the table below Changes on the 1st metal rod (A) Changes on the 2nd metal rod (B) Changes on the 1st solution (A(aq)) Changes on the 2nd solution (B(aq)) Voltage/voltmeter reading(Volts) Using Zn/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Zinc(II)sulphate (VI)colour remain colourless Blue Copper (II)sulphate (VI)colour fades. Brown solid/residue/ deposit 0.8 (Theoretical value=1.10V) Using Mg/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Magnesium(II) sulphate(VI) colour remain colourless Blue Copper (II)sulphate (VI)colour fades Brown solid/residue/ deposit 1.5 (Theoretical value=2.04V) Using Ag/Cu half cell -The rod increase in size /mass /deposited -silver coin/ rod /plate increase in size /mass/ deposited Blue Copper (II)sulphate (VI)colour remains Silver(I)nitrate (V)colour remain colourless 0.20 (Theoretical value=0.46V) Using Fe/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Iron(II)sulphate (VI)colour becomes more green Blue Copper (II)sulphate (VI)colour fades.Brown solid/residue/ deposit 0.60 (Theoretical value=0.78V) From the above observations ,it can be deduced that: (i)in the Zn/Cu half-cell the; -Zinc rod/plate ionizes /dissolves faster than the copper rod/plate to form Zn2+
12 Ionic equation Zn(s) -> Zn2+(aq) + 2e -blue copper ions in the Copper (II)sulphate solution gains the donated electrons to form brown copper metal/atoms Ionic equation Cu2+(aq) + 2e -> Cu(s) This reaction shows /imply the Zinc rod has a higher tendency to ionize than copper.The Zinc rod has a higher net accumulation of electrons and is more negative compared to the copper rod which has lower accumulation of electrons.


Dip a folded filter paper into a solution of Potassium nitrate(V) or sodium(I) chloride(I) until it soaks. Use the folded soaked filter paper to connect/join the two solutions in the two beakers.The whole set up should be as below Repeat the above procedure by replacing: (i)Zinc half cell with Magnesium rod/plate/ribbon dipped in 50cm3 of IM magnesium (II) sulphate(VI) solution V
11 (ii)Zinc half cell with Silver rod/plate/coin dipped in 50cm3 of IM silver(I) nitrate(V) solution (iii)Copper half cell with Iron rod/plate/spoon dipped in 50cm3 of IM Iron (II) sulphate(VI) solution Record the observations in the table below Changes on the 1st metal rod (A) Changes on the 2nd metal rod (B) Changes on the 1st solution (A(aq)) Changes on the 2nd solution (B(aq)) Voltage/voltmeter reading(Volts) Using Zn/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Zinc(II)sulphate (VI)colour remain colourless Blue Copper (II)sulphate (VI)colour fades. Brown solid/residue/ deposit 0.8 (Theoretical value=1.10V) Using Mg/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Magnesium(II) sulphate(VI) colour remain colourless Blue Copper (II)sulphate (VI)colour fades Brown solid/residue/ deposit 1.5 (Theoretical value=2.04V) Using Ag/Cu half cell -The rod increase in size /mass /deposited -silver coin/ rod /plate increase in size /mass/ deposited Blue Copper (II)sulphate (VI)colour remains Silver(I)nitrate (V)colour remain colourless 0.20 (Theoretical value=0.46V) Using Fe/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Iron(II)sulphate (VI)colour becomes more green Blue Copper (II)sulphate (VI)colour fades.Brown solid/residue/ deposit 0.60 (Theoretical value=0.78V) From the above observations ,it can be deduced that: (i)in the Zn/Cu half-cell the; -Zinc rod/plate ionizes /dissolves faster than the copper rod/plate to form Zn2+
12 Ionic equation Zn(s) -> Zn2+(aq) + 2e -blue copper ions in the Copper (II)sulphate solution gains the donated electrons to form brown copper metal/atoms Ionic equation Cu2+(aq) + 2e -> Cu(s) This reaction shows /imply the Zinc rod has a higher tendency to ionize than copper.The Zinc rod has a higher net accumulation of electrons and is more negative compared to the copper rod which has lower accumulation of electrons. The copper rod is therefore relatively more positive with respect to Zinc rod.


Use the folded soaked filter paper to connect/join the two solutions in the two beakers.The whole set up should be as below Repeat the above procedure by replacing: (i)Zinc half cell with Magnesium rod/plate/ribbon dipped in 50cm3 of IM magnesium (II) sulphate(VI) solution V
11 (ii)Zinc half cell with Silver rod/plate/coin dipped in 50cm3 of IM silver(I) nitrate(V) solution (iii)Copper half cell with Iron rod/plate/spoon dipped in 50cm3 of IM Iron (II) sulphate(VI) solution Record the observations in the table below Changes on the 1st metal rod (A) Changes on the 2nd metal rod (B) Changes on the 1st solution (A(aq)) Changes on the 2nd solution (B(aq)) Voltage/voltmeter reading(Volts) Using Zn/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Zinc(II)sulphate (VI)colour remain colourless Blue Copper (II)sulphate (VI)colour fades. Brown solid/residue/ deposit 0.8 (Theoretical value=1.10V) Using Mg/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Magnesium(II) sulphate(VI) colour remain colourless Blue Copper (II)sulphate (VI)colour fades Brown solid/residue/ deposit 1.5 (Theoretical value=2.04V) Using Ag/Cu half cell -The rod increase in size /mass /deposited -silver coin/ rod /plate increase in size /mass/ deposited Blue Copper (II)sulphate (VI)colour remains Silver(I)nitrate (V)colour remain colourless 0.20 (Theoretical value=0.46V) Using Fe/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Iron(II)sulphate (VI)colour becomes more green Blue Copper (II)sulphate (VI)colour fades.Brown solid/residue/ deposit 0.60 (Theoretical value=0.78V) From the above observations ,it can be deduced that: (i)in the Zn/Cu half-cell the; -Zinc rod/plate ionizes /dissolves faster than the copper rod/plate to form Zn2+
12 Ionic equation Zn(s) -> Zn2+(aq) + 2e -blue copper ions in the Copper (II)sulphate solution gains the donated electrons to form brown copper metal/atoms Ionic equation Cu2+(aq) + 2e -> Cu(s) This reaction shows /imply the Zinc rod has a higher tendency to ionize than copper.The Zinc rod has a higher net accumulation of electrons and is more negative compared to the copper rod which has lower accumulation of electrons. The copper rod is therefore relatively more positive with respect to Zinc rod. When the two half cells are connected , electrons therefore flow from the negative Zinc rod through the external wire to be gained by copper ions.


Brown solid/residue/ deposit 0.8 (Theoretical value=1.10V) Using Mg/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Magnesium(II) sulphate(VI) colour remain colourless Blue Copper (II)sulphate (VI)colour fades Brown solid/residue/ deposit 1.5 (Theoretical value=2.04V) Using Ag/Cu half cell -The rod increase in size /mass /deposited -silver coin/ rod /plate increase in size /mass/ deposited Blue Copper (II)sulphate (VI)colour remains Silver(I)nitrate (V)colour remain colourless 0.20 (Theoretical value=0.46V) Using Fe/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Iron(II)sulphate (VI)colour becomes more green Blue Copper (II)sulphate (VI)colour fades.Brown solid/residue/ deposit 0.60 (Theoretical value=0.78V) From the above observations ,it can be deduced that: (i)in the Zn/Cu half-cell the; -Zinc rod/plate ionizes /dissolves faster than the copper rod/plate to form Zn2+
12 Ionic equation Zn(s) -> Zn2+(aq) + 2e -blue copper ions in the Copper (II)sulphate solution gains the donated electrons to form brown copper metal/atoms Ionic equation Cu2+(aq) + 2e -> Cu(s) This reaction shows /imply the Zinc rod has a higher tendency to ionize than copper.The Zinc rod has a higher net accumulation of electrons and is more negative compared to the copper rod which has lower accumulation of electrons. The copper rod is therefore relatively more positive with respect to Zinc rod. When the two half cells are connected , electrons therefore flow from the negative Zinc rod through the external wire to be gained by copper ions. This means a net accumulation/increase of Zn2+ positive ions on the negative half cell and a net decrease in Cu2+ positive ions on the positive half cell. The purpose of the salt bridge therefore is: (i)complete the circuit (ii)maintain balance of charges /ions on both half cells.


When the two half cells are connected , electrons therefore flow from the negative Zinc rod through the external wire to be gained by copper ions. This means a net accumulation/increase of Zn2+ positive ions on the negative half cell and a net decrease in Cu2+ positive ions on the positive half cell. The purpose of the salt bridge therefore is: (i)complete the circuit (ii)maintain balance of charges /ions on both half cells. For the negative half cell the NO3- /Cl- from salt bridge decrease/neutralise the increased positive(Zn2+) ion. For the positive half cell the Na+ / K+ from salt bridge increase the decreased positive(Cu2+) ion. The voltmeter should theoretically register/read a 1.10Volts as a measure of the electromotive force (e.m.f) of the cell .Practically the voltage reading is lowered because the connecting wires have some resistance to be overcomed. A combination of two half cells that can generate an electric current from a redox reaction is called a voltaic/electrochemical cell. By convention a voltaic/electrochemical cell is represented; M(s) / M2+(aq) // N2+ (aq) / N(s) (metal rod of M)(solution ofM)(solution ofN)(metal rod ofN) Note; a)(i)Metal M must be the one higher in the reactivity series. (ii)It forms the negative terminal of the cell. (iii)It must diagrammatically be drawn first on the left hand side when illustrating the voltaic/electrochemical cell. b)(i)Metal N must be the one lower in the reactivity series. (ii)It forms the positive terminal of the cell. (iii)It must diagrammatically be drawn second/after/ right hand side when illustrating the voltaic/electrochemical cell. Illustration of the voltaic/electrochemical cell. (i)Zn/Cu cell
13 1. Zinc rod ionizes /dissolves to form Zn2+ ions at the negative terminal Zn(s) -> Zn2+(aq) + 2e 2. Copper ions in solution gain the donated electrons to form copper atoms/metal Cu2+(aq) + 2e -> Cu(s) 3.Overall redox equation Cu2+(aq) + Zn(s) -> Zn2+(aq) + Cu(s) 4.cell representation.


(i)Zn/Cu cell
13 1. Zinc rod ionizes /dissolves to form Zn2+ ions at the negative terminal Zn(s) -> Zn2+(aq) + 2e 2. Copper ions in solution gain the donated electrons to form copper atoms/metal Cu2+(aq) + 2e -> Cu(s) 3.Overall redox equation Cu2+(aq) + Zn(s) -> Zn2+(aq) + Cu(s) 4.cell representation. Zn(s) / 1M, Zn2+(aq) // 1M,Cu2+(aq) / Cu(s) E0 = +1.10 V 5.cell diagram (ii)Mg/Cu cell 1. Magnesium rod ionizes /dissolves to form Mg2+ ions at the negative terminal Mg(s) -> Mg2+(aq) + 2e 2. Copper ions in solution gain the donated electrons to form copper atoms/metal Cu2+(aq) + 2e -> Cu(s) 3.Overall redox equation Cu2+(aq) + Mg(s) -> Mg2+(aq) + Cu(s) 4.cell representation. Mg(s) / 1M, Mg2+(aq) // 1M,Cu2+(aq) / Cu(s) E0 = +2.04 V 5.cell diagram. Voltmeter
14 (iii)Fe/Cu cell 1. Magnesium rod ionizes /dissolves to form Mg2+ ions at the negative terminal Fe(s) -> Fe2+(aq) + 2e 2. Copper ions in solution gain the donated electrons to form copper atoms/metal Cu2+(aq) + 2e -> Cu(s) 3.Overall redox equation Cu2+(aq) + Fe(s) -> Fe2+(aq) + Cu(s) 4.cell representation. Fe(s) / 1M, Fe2+(aq) // 1M,Cu2+(aq) / Cu(s) E0 = +0.78 V 5.cell diagram. Fe Fe++ V
15 (iv)Ag/Cu cell 1. Copper rod ionizes /dissolves to form Cu2+ ions at the negative terminal Cu(s) -> Cu2+(aq) + 2e 2.


Fe(s) / 1M, Fe2+(aq) // 1M,Cu2+(aq) / Cu(s) E0 = +0.78 V 5.cell diagram. Fe Fe++ V
15 (iv)Ag/Cu cell 1. Copper rod ionizes /dissolves to form Cu2+ ions at the negative terminal Cu(s) -> Cu2+(aq) + 2e 2. Silver ions in solution gain the donated electrons to form silver atoms/metal 2Ag+(aq) + 2e -> 2Ag(s) 3.Overall redox equation 2Ag+(aq) + Cu(s) -> Cu2+(aq) + 2Ag(s) 4.cell representation. Cu(s) / 1M, Cu2+(aq) // 1M,2Ag+(aq) / 2Ag(s) E0 = +0.46 V 5.cell diagram. Standard electrode potential (Eᶿ) The standard electrode potential (Eᶿ) is obtained if the hydrogen half cell is used as reference. The standard electrode potential (Eᶿ) consist of inert platinum electrode immersed/dipped in 1M solution of (sulphuric(VI) acid) H+ ions. Hydrogen gas is bubbled on the platinum electrodes at: (i)a temperature of 25oC (ii)atmospheric pressure of 101300Pa/101300Nm-2/1atm/760mmHg/76cmHg (iii)a concentration of 1M(1moledm-3) of sulphuric(VI) acid/ H+ ions and 1M(1moledm-3) of the other half cell. Hydrogen is adsorbed onto the surface of the platinum. An equilibrium/balance exist between the adsorbed layer of molecular hydrogen and H+ ions in solution to form a half cell.


Hydrogen gas is bubbled on the platinum electrodes at: (i)a temperature of 25oC (ii)atmospheric pressure of 101300Pa/101300Nm-2/1atm/760mmHg/76cmHg (iii)a concentration of 1M(1moledm-3) of sulphuric(VI) acid/ H+ ions and 1M(1moledm-3) of the other half cell. Hydrogen is adsorbed onto the surface of the platinum. An equilibrium/balance exist between the adsorbed layer of molecular hydrogen and H+ ions in solution to form a half cell. ½ H2 (g) ==== H+ (aq) + e The half cell representation is: Pt,½ H2 (g) / H+ (aq), 1M Cu Ag V Cu++ Ag+ Cu(s) + 2Ag+ (aq) Cu2+(aq) + 2Ag(s)
16 The standard electrode potential (Eᶿ) is thus defined as the potential difference for a cell comprising of a particular element in contact with1M solution of its own ions and the standard hydrogen electrode. If the other electrode has a higher/greater tendency to lose electrons than the hydrogen electrode, the electrode is therefore negative with respect to hydrogen electrode and its electrode potential has negative (Eᶿ) values. If the other electrode has a lower/lesser tendency to lose electrons than the hydrogen electrode, the electrode is therefore positive with respect to hydrogen electrode and its electrode potential has positive (Eᶿ) values.


½ H2 (g) ==== H+ (aq) + e The half cell representation is: Pt,½ H2 (g) / H+ (aq), 1M Cu Ag V Cu++ Ag+ Cu(s) + 2Ag+ (aq) Cu2+(aq) + 2Ag(s)
16 The standard electrode potential (Eᶿ) is thus defined as the potential difference for a cell comprising of a particular element in contact with1M solution of its own ions and the standard hydrogen electrode. If the other electrode has a higher/greater tendency to lose electrons than the hydrogen electrode, the electrode is therefore negative with respect to hydrogen electrode and its electrode potential has negative (Eᶿ) values. If the other electrode has a lower/lesser tendency to lose electrons than the hydrogen electrode, the electrode is therefore positive with respect to hydrogen electrode and its electrode potential has positive (Eᶿ) values. Table showing the standard electrode potential (Eᶿ) of some reactions Reaction (Eᶿ) values in volts F2 (g)+ 2e -> 2F- (aq) +2.87 H2 O2 (aq)+ H+ (aq) +2e -> H2 O (l) +1.77 Mn O4- (aq)+ 4H+ (aq) +3e -> MnO2 (s) +H2 O (l) +1.70 2HClO (aq)+ 2H+ (aq) +2e -> Cl2 (aq) +2H2 O (l) +1.59 Mn O4- (aq)+ 4H+ (aq) +5e -> Mn2+ (aq) +H2 O (l) +1.51 Cl2 (g)+ 2e -> 2Cl- (aq) +1.36 Mn O2 (s)+ 4H+ (aq) +2e -> Mn2+ (aq) +2H2 O (l) +1.23 Br2 (aq)+ 2e -> 2Br- (aq) +1.09 NO3- (aq)+ 2H+ (aq) + e -> NO2 (g) + H2 O (l) +0.80 Ag+ (aq) + e -> Ag(s) +0.80 Fe3+ (aq) + e -> Fe2+ (aq) +0.77 2H+ (aq)+ O2 (g) -> H2 O2 (aq) +0.68 I2 (aq)+ 2e -> 2I- (aq) +0.54 Cu2+ (aq) + 2e -> Cu(s) +0.34 2H+ (aq) + 2e -> H2(g) +0.00 Pb2+ (aq) + 2e -> Pb(s) -0.13 Fe2+ (aq) + 2e -> Fe(s) -0.44 Zn2+ (aq) + 2e -> Zn(s) -0.77 Al3+ (aq) + 3e -> Al(s) -1.66 Mg2+ (aq) + 2e -> Mg(s) -2.37 Na+ (aq) + e -> Na(s) -2.71 K+ (aq) + e -> K(s) -2.92 Note: (i)Eᶿ values generally show the possibility/feasibility of a reduction process/oxidizing strength.


(ii)The element/species in the half cell with the highest negative Eᶿ value easily gain / acquire electrons. It is thus the strongest oxidizing agent and its reduction process is highly possible/feasible. The element/species in the half cell with the lowest positive Eᶿ value easily donate / lose electrons. 17 It is thus the strongest reducing agent and its reduction process is the least possible/feasible. (iii)The overall redox reaction is possible/feasible is it has a positive (+) Eᶿ. If the overall redox reaction is not possible/ not feasible/ forced, it has a negative (-) Eᶿ Calculation examples on Eᶿ Calculate the Eᶿ value of a cell made of: a)Zn and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram) Zn2+ (aq) + 2e ->Zn(s) Eᶿ = -0.77V(lower Eᶿ/ Left Hand Side diagram) Zn(s) ->Zn2+ (aq) + 2e Eᶿ = +0.77(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿoxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.34 – (- 0.77) = +1.10V Overall redox equation: Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) Eᶿ = +1.10V Overall conventional cell representation: Zn(s) / Zn2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) Eᶿ = +1.10V Overall conventional cell diagram: Voltmeter(1.10V) 1M Cu2+ (aq) 1M Zn2+ (aq) Zn2+
18 Zinc and copper reaction has a positive(+) overall Eᶿ therefore is possible/feasible and thus Zinc can displace/reduce Copper solution.


17 It is thus the strongest reducing agent and its reduction process is the least possible/feasible. (iii)The overall redox reaction is possible/feasible is it has a positive (+) Eᶿ. If the overall redox reaction is not possible/ not feasible/ forced, it has a negative (-) Eᶿ Calculation examples on Eᶿ Calculate the Eᶿ value of a cell made of: a)Zn and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram) Zn2+ (aq) + 2e ->Zn(s) Eᶿ = -0.77V(lower Eᶿ/ Left Hand Side diagram) Zn(s) ->Zn2+ (aq) + 2e Eᶿ = +0.77(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿoxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.34 – (- 0.77) = +1.10V Overall redox equation: Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) Eᶿ = +1.10V Overall conventional cell representation: Zn(s) / Zn2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) Eᶿ = +1.10V Overall conventional cell diagram: Voltmeter(1.10V) 1M Cu2+ (aq) 1M Zn2+ (aq) Zn2+
18 Zinc and copper reaction has a positive(+) overall Eᶿ therefore is possible/feasible and thus Zinc can displace/reduce Copper solution. b)Mg and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram) Mg2+ (aq) + 2e ->Mg(s) Eᶿ = -2.37V(lower Eᶿ/ Left Hand Side diagram) Mg(s) ->Mg2+ (aq) + 2e Eᶿ = +2.37(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.34 – (- 2.37) = +2.71V Overall redox equation: Cu2+ (aq) + Mg(s) -> Mg2+ (aq) + Cu(s) Eᶿ = +2.71V Overall conventional cell representation: Mg(s) / Mg2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) Eᶿ = +2.71V c)Ag and Pb From the table above: 2Ag+ (aq) + 2e -> 2Ag(s) Eᶿ = +0.80V(higher Eᶿ /Right Hand Side diagram) Pb2+ (aq) + 2e ->Pb(s) Eᶿ = -0.13V(lower Eᶿ/ Left Hand Side diagram) Pb(s) ->Pb2+ (aq) + 2e Eᶿ = +0.13(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.80 – (- 0.13) = +0.93V Overall redox equation: 2Ag+ (aq) + Pb(s) -> Pb2+ (aq) + 2Ag(s) Eᶿ = +0.93V Overall conventional cell representation: Pb(s) / Pb2+ (aq) 1M, // 1M,2Ag+ (aq) / Ag(s) Eᶿ = +0.93V d)Chlorine and Bromine From the table above: 2e + Cl2(g) ->2Cl- (aq) Eᶿ = +1.36V(higher Eᶿ /Right Hand Side diagram) 2e + Br2(aq) ->2Br- (aq) Eᶿ = +0.13V(lower Eᶿ/ Left Hand Side diagram) 2Br- (aq) -> Br2(aq) + 2e Eᶿ = -0.13(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = - 0.13 – (- 1.36) = +1.23V Overall redox equation:
19 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2(aq) Eᶿ = +1.23V Overall conventional cell representation: Cl2(g) / 2Cl- (aq) 1M, // 1M, 2Br- (aq) / Br2(aq) Eᶿ = +1.23V Chlorine displaces bromine from bromine water.


(iii)The overall redox reaction is possible/feasible is it has a positive (+) Eᶿ. If the overall redox reaction is not possible/ not feasible/ forced, it has a negative (-) Eᶿ Calculation examples on Eᶿ Calculate the Eᶿ value of a cell made of: a)Zn and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram) Zn2+ (aq) + 2e ->Zn(s) Eᶿ = -0.77V(lower Eᶿ/ Left Hand Side diagram) Zn(s) ->Zn2+ (aq) + 2e Eᶿ = +0.77(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿoxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.34 – (- 0.77) = +1.10V Overall redox equation: Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) Eᶿ = +1.10V Overall conventional cell representation: Zn(s) / Zn2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) Eᶿ = +1.10V Overall conventional cell diagram: Voltmeter(1.10V) 1M Cu2+ (aq) 1M Zn2+ (aq) Zn2+
18 Zinc and copper reaction has a positive(+) overall Eᶿ therefore is possible/feasible and thus Zinc can displace/reduce Copper solution. b)Mg and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram) Mg2+ (aq) + 2e ->Mg(s) Eᶿ = -2.37V(lower Eᶿ/ Left Hand Side diagram) Mg(s) ->Mg2+ (aq) + 2e Eᶿ = +2.37(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.34 – (- 2.37) = +2.71V Overall redox equation: Cu2+ (aq) + Mg(s) -> Mg2+ (aq) + Cu(s) Eᶿ = +2.71V Overall conventional cell representation: Mg(s) / Mg2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) Eᶿ = +2.71V c)Ag and Pb From the table above: 2Ag+ (aq) + 2e -> 2Ag(s) Eᶿ = +0.80V(higher Eᶿ /Right Hand Side diagram) Pb2+ (aq) + 2e ->Pb(s) Eᶿ = -0.13V(lower Eᶿ/ Left Hand Side diagram) Pb(s) ->Pb2+ (aq) + 2e Eᶿ = +0.13(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.80 – (- 0.13) = +0.93V Overall redox equation: 2Ag+ (aq) + Pb(s) -> Pb2+ (aq) + 2Ag(s) Eᶿ = +0.93V Overall conventional cell representation: Pb(s) / Pb2+ (aq) 1M, // 1M,2Ag+ (aq) / Ag(s) Eᶿ = +0.93V d)Chlorine and Bromine From the table above: 2e + Cl2(g) ->2Cl- (aq) Eᶿ = +1.36V(higher Eᶿ /Right Hand Side diagram) 2e + Br2(aq) ->2Br- (aq) Eᶿ = +0.13V(lower Eᶿ/ Left Hand Side diagram) 2Br- (aq) -> Br2(aq) + 2e Eᶿ = -0.13(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = - 0.13 – (- 1.36) = +1.23V Overall redox equation:
19 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2(aq) Eᶿ = +1.23V Overall conventional cell representation: Cl2(g) / 2Cl- (aq) 1M, // 1M, 2Br- (aq) / Br2(aq) Eᶿ = +1.23V Chlorine displaces bromine from bromine water. When chlorine gas is thus bubbled in bromine water, the pale green colour fades as displacement takes place and a brown solution containing dissolved bromine liquid is formed.


If the overall redox reaction is not possible/ not feasible/ forced, it has a negative (-) Eᶿ Calculation examples on Eᶿ Calculate the Eᶿ value of a cell made of: a)Zn and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram) Zn2+ (aq) + 2e ->Zn(s) Eᶿ = -0.77V(lower Eᶿ/ Left Hand Side diagram) Zn(s) ->Zn2+ (aq) + 2e Eᶿ = +0.77(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿoxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.34 – (- 0.77) = +1.10V Overall redox equation: Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) Eᶿ = +1.10V Overall conventional cell representation: Zn(s) / Zn2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) Eᶿ = +1.10V Overall conventional cell diagram: Voltmeter(1.10V) 1M Cu2+ (aq) 1M Zn2+ (aq) Zn2+
18 Zinc and copper reaction has a positive(+) overall Eᶿ therefore is possible/feasible and thus Zinc can displace/reduce Copper solution. b)Mg and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram) Mg2+ (aq) + 2e ->Mg(s) Eᶿ = -2.37V(lower Eᶿ/ Left Hand Side diagram) Mg(s) ->Mg2+ (aq) + 2e Eᶿ = +2.37(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.34 – (- 2.37) = +2.71V Overall redox equation: Cu2+ (aq) + Mg(s) -> Mg2+ (aq) + Cu(s) Eᶿ = +2.71V Overall conventional cell representation: Mg(s) / Mg2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) Eᶿ = +2.71V c)Ag and Pb From the table above: 2Ag+ (aq) + 2e -> 2Ag(s) Eᶿ = +0.80V(higher Eᶿ /Right Hand Side diagram) Pb2+ (aq) + 2e ->Pb(s) Eᶿ = -0.13V(lower Eᶿ/ Left Hand Side diagram) Pb(s) ->Pb2+ (aq) + 2e Eᶿ = +0.13(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.80 – (- 0.13) = +0.93V Overall redox equation: 2Ag+ (aq) + Pb(s) -> Pb2+ (aq) + 2Ag(s) Eᶿ = +0.93V Overall conventional cell representation: Pb(s) / Pb2+ (aq) 1M, // 1M,2Ag+ (aq) / Ag(s) Eᶿ = +0.93V d)Chlorine and Bromine From the table above: 2e + Cl2(g) ->2Cl- (aq) Eᶿ = +1.36V(higher Eᶿ /Right Hand Side diagram) 2e + Br2(aq) ->2Br- (aq) Eᶿ = +0.13V(lower Eᶿ/ Left Hand Side diagram) 2Br- (aq) -> Br2(aq) + 2e Eᶿ = -0.13(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = - 0.13 – (- 1.36) = +1.23V Overall redox equation:
19 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2(aq) Eᶿ = +1.23V Overall conventional cell representation: Cl2(g) / 2Cl- (aq) 1M, // 1M, 2Br- (aq) / Br2(aq) Eᶿ = +1.23V Chlorine displaces bromine from bromine water. When chlorine gas is thus bubbled in bromine water, the pale green colour fades as displacement takes place and a brown solution containing dissolved bromine liquid is formed. This reaction is feasible /possible because the overall redox reaction has a positive Eᶿ value.


b)Mg and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram) Mg2+ (aq) + 2e ->Mg(s) Eᶿ = -2.37V(lower Eᶿ/ Left Hand Side diagram) Mg(s) ->Mg2+ (aq) + 2e Eᶿ = +2.37(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.34 – (- 2.37) = +2.71V Overall redox equation: Cu2+ (aq) + Mg(s) -> Mg2+ (aq) + Cu(s) Eᶿ = +2.71V Overall conventional cell representation: Mg(s) / Mg2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) Eᶿ = +2.71V c)Ag and Pb From the table above: 2Ag+ (aq) + 2e -> 2Ag(s) Eᶿ = +0.80V(higher Eᶿ /Right Hand Side diagram) Pb2+ (aq) + 2e ->Pb(s) Eᶿ = -0.13V(lower Eᶿ/ Left Hand Side diagram) Pb(s) ->Pb2+ (aq) + 2e Eᶿ = +0.13(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.80 – (- 0.13) = +0.93V Overall redox equation: 2Ag+ (aq) + Pb(s) -> Pb2+ (aq) + 2Ag(s) Eᶿ = +0.93V Overall conventional cell representation: Pb(s) / Pb2+ (aq) 1M, // 1M,2Ag+ (aq) / Ag(s) Eᶿ = +0.93V d)Chlorine and Bromine From the table above: 2e + Cl2(g) ->2Cl- (aq) Eᶿ = +1.36V(higher Eᶿ /Right Hand Side diagram) 2e + Br2(aq) ->2Br- (aq) Eᶿ = +0.13V(lower Eᶿ/ Left Hand Side diagram) 2Br- (aq) -> Br2(aq) + 2e Eᶿ = -0.13(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = - 0.13 – (- 1.36) = +1.23V Overall redox equation:
19 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2(aq) Eᶿ = +1.23V Overall conventional cell representation: Cl2(g) / 2Cl- (aq) 1M, // 1M, 2Br- (aq) / Br2(aq) Eᶿ = +1.23V Chlorine displaces bromine from bromine water. When chlorine gas is thus bubbled in bromine water, the pale green colour fades as displacement takes place and a brown solution containing dissolved bromine liquid is formed. This reaction is feasible /possible because the overall redox reaction has a positive Eᶿ value. e)Strongest oxidizing agent and the strongest reducing agent.


When chlorine gas is thus bubbled in bromine water, the pale green colour fades as displacement takes place and a brown solution containing dissolved bromine liquid is formed. This reaction is feasible /possible because the overall redox reaction has a positive Eᶿ value. e)Strongest oxidizing agent and the strongest reducing agent. From the table above: 2e + F2(g) ->2F- (aq) Eᶿ = +2.87V(highest Eᶿ /strongest oxidizing agent) 2e + 2K+ (aq) ->2K (aq) Eᶿ = -2.92V(lowest Eᶿ/ strongest reducing agent) 2K (aq) -> 2K+ (aq) + 2e Eᶿ = +2.92V (reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = +2.87 – (-2.92) = +5.79V Overall redox equation: F2(g) + 2K(s) -> 2F- (aq) + 2K+ (aq) Eᶿ = +5.79V Overall conventional cell representation: 2K(s) / 2K+ (aq),1M, // 1M, 2F- (aq) / F2(g) Eᶿ = +5.79V The redox reactions in an electrochemical/voltaic is commercially applied to make the: (a)Dry /primary/Laclanche cell. (b)Wet /secondary /accumulators. (a)Dry/primary/Laclanche cell Examine a used dry cell. Note the positive and the negative terminal of the cell. Carefully using a knife cut a cross section from one terminal to the other. The dry cell consist of a Zinc can containing a graphite rod at the centre surrounded by a paste of; -Ammonium chloride -Zinc chloride -powdered manganese (IV) oxide mixed with Carbon.


Note the positive and the negative terminal of the cell. Carefully using a knife cut a cross section from one terminal to the other. The dry cell consist of a Zinc can containing a graphite rod at the centre surrounded by a paste of; -Ammonium chloride -Zinc chloride -powdered manganese (IV) oxide mixed with Carbon. Zinc acts/serve as the negative terminal where it ionizes/dissociates:
20 Zn(s) -> Zn2+(aq) + 2e Ammonium ions in ammonium chloride serve as the positive terminal where it is converted to ammonia gas and hydrogen gas. 2NH4+(aq) + 2e -> 2NH3(g) + H2(g) Ammonia forms a complex salt / compound /(Zn(NH3) 4)2+ (aq) / tetramminezinc(II) complex with the Zinc chloride in the paste. Manganese (IV) oxide oxidizes the hydrogen produced at the electrodes to water preventing any bubbles from coating the carbon terminal which would reduce the efficiency of the cell. Ammonium chloride is used as paste because the solid does not conduct electricity because the ions are fused/not mobile. Since the reactants are used up, the dry /primary /Laclanche cell cannot provide continous supply of electricity.The process of restoring the reactants is called recharging. b)Wet/Secondary/Accumulators 1.


Ammonium chloride is used as paste because the solid does not conduct electricity because the ions are fused/not mobile. Since the reactants are used up, the dry /primary /Laclanche cell cannot provide continous supply of electricity.The process of restoring the reactants is called recharging. b)Wet/Secondary/Accumulators 1. Wet/Secondary/Accumulators are rechargeable unlike dry /primary /Laclanche cells.Wet/Secondary/Accumulators are made up of: (i)Lead plate that forms the negative terminal (ii)Lead(IV) oxide that forms the positive terminal 2.The two electrodes are dipped in concentrated sulphuric(VI) acid of a relative density 1.2/1.3 3.At the negative terminal,lead ionizes /dissolves; Pb(s) -> Pb2+ + 2e 4.At the positive terminal,
21 (i) Lead(IV) oxide reacts with the hydrogen ions in sulphuric(VI)acid to form Pb2+ (aq) ions; PbO2(s) + 4H+(aq) + 2e -> Pb2+ (aq) + H2O(l) (ii) Pb2+ (aq) ions formed instantly react with sulphate (VI) ions/ SO42- (aq) from sulphuric (VI)acid to form insoluble Lead(II) sulphate (VI). Pb2+ (aq) + SO42- (aq) -> PbSO4(s) 5.The overall cell reaction is called discharging PbO2(s) +Pb(s) + 4H+(aq) + 2SO42- (aq)-> 2PbSO4(s) + 2H2O(l) Eᶿ = +2.0V 6.The insoluble Lead(II) sulphate (VI) formed should not be left for long since fine Lead(II) sulphate (VI) will change to a course non-reversible and inactive form making the cell less efficient. As the battery discharges ,lead and lead(IV)oxide are depleted/finished/reduced and the concentration of sulphuric(VI)acid decreases. 7.


Pb2+ (aq) + SO42- (aq) -> PbSO4(s) 5.The overall cell reaction is called discharging PbO2(s) +Pb(s) + 4H+(aq) + 2SO42- (aq)-> 2PbSO4(s) + 2H2O(l) Eᶿ = +2.0V 6.The insoluble Lead(II) sulphate (VI) formed should not be left for long since fine Lead(II) sulphate (VI) will change to a course non-reversible and inactive form making the cell less efficient. As the battery discharges ,lead and lead(IV)oxide are depleted/finished/reduced and the concentration of sulphuric(VI)acid decreases. 7. During recharging, the electrode reaction is reversed as below: 2PbSO4(s) + 2H2O(l) ->PbO2(s) +Pb(s) + 4H+(aq) + 2SO42- (aq) 8. A car battery has six Lead-acid cells making a total of 12 volts. (iii)ELECTROLYSIS (ELECTROLYTIC CELL) 1.Electrolysis is defined simply as the decomposition of a compound by an electric current/electricity. A compound that is decomposed by an electric current is called an electrolyte. Some electrolytes are weak while others are strong. 2.Strong electrolytes are those that are fully ionized/dissociated into (many) ions. Common strong electrolytes include: (i)all mineral acids (ii)all strong alkalis/sodium hydroxide/potassium hydroxide. (iii)all soluble salts
22 3.Weak electrolytes are those that are partially/partly ionized/dissociated into (few) ions. Common weak electrolytes include: (i)all organic acids (ii)all bases except sodium hydroxide/potassium hydroxide. (iii)Water 4. A compound that is not decomposed by an electric current is called nonelectrolyte. Non-electrolytes are those compounds /substances that exist as molecules and thus cannot ionize/dissociate into(any) ions . Common non-electrolytes include: (i) most organic solvents (e.g.


A compound that is not decomposed by an electric current is called nonelectrolyte. Non-electrolytes are those compounds /substances that exist as molecules and thus cannot ionize/dissociate into(any) ions . Common non-electrolytes include: (i) most organic solvents (e.g. petrol/paraffin/benzene/methylbenzene/ethanol) (ii)all hydrocarbons(alkanes /alkenes/alkynes) (iii)Chemicals of life(e.g. proteins, carbohydrates, lipids, starch, sugar) 5. An electrolytes in solid state have fused /joined ions and therefore do not conduct electricity but the ions (cations and anions) are free and mobile in molten and aqueous (solution, dissolved in water) state. 6.During electrolysis, the free ions are attracted to the electrodes. An electrode is a rod through which current enter and leave the electrolyte during electrolysis. An electrode that does not influence/alter the products of electrolysis is called an inert electrode. Common inert electrodes include: (i)Platinum (ii)Carbon graphite Platinum is not usually used in a school laboratory because it is very expensive. Carbon graphite is easily/readily and cheaply available (from used dry cells). 7.The positive electrode is called Anode.The anode is the electrode through which current enter the electrolyte/electrons leave the electrolyte 8.The negative electrode is called Cathode. The cathode is the electrode through which current leave the electrolyte / electrons enter the electrolyte 9. During the electrolysis, free anions are attracted to the anode where they lose /donate electrons to form neutral atoms/molecules. i.e. M(l) -> M+(l) + e (for cations from molten electrolytes) M(s) -> M+(aq) + e (for cations from electrolytes in aqueous state / solution / dissolved in water) The neutral atoms /molecules form the products of electrolysis at the anode. This is called discharge at anode
23 10. During electrolysis, free cations are attracted to the cathode where they gain /accept/acquire electrons to form neutral atoms/molecules.


M(l) -> M+(l) + e (for cations from molten electrolytes) M(s) -> M+(aq) + e (for cations from electrolytes in aqueous state / solution / dissolved in water) The neutral atoms /molecules form the products of electrolysis at the anode. This is called discharge at anode
23 10. During electrolysis, free cations are attracted to the cathode where they gain /accept/acquire electrons to form neutral atoms/molecules. X+ (aq) + 2e -> X(s) (for cations from electrolytes in aqueous state / solution / dissolved in water) 2X+ (l) + 2e -> X (l) (for cations from molten electrolytes) The neutral atoms /molecules form the products of electrolysis at the cathode. This is called discharge at cathode. 11. The below set up shows an electrolytic cell. BatteryAnode(+)Cathode(-)ElectrolyteSimple set up of electrolytic cellGaseous product at anodeGaseous product at cathode
24 12. For a compound /salt containing only two ion/binary salt the products of electrolysis in an electrolytic cell can be determined as in the below examples: a)To determine the products of electrolysis of molten Lead(II)chloride (i)Decomposition of electrolyte into free ions; PbCl2 (l) -> Pb 2+(l) + 2Cl-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); Pb 2+(l) + 2e -> Pb (l) (Cation / Pb 2+ gains / accepts / acquires electrons to form free atom)
25 (iii)At the anode/positive electrode(+); 2Cl-(l) -> Cl2 (g) + 2e (Anion / Cl- donate/lose electrons to form free atom then a gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid lead metal. II.At the anode pale green chlorine gas.


BatteryAnode(+)Cathode(-)ElectrolyteSimple set up of electrolytic cellGaseous product at anodeGaseous product at cathode
24 12. For a compound /salt containing only two ion/binary salt the products of electrolysis in an electrolytic cell can be determined as in the below examples: a)To determine the products of electrolysis of molten Lead(II)chloride (i)Decomposition of electrolyte into free ions; PbCl2 (l) -> Pb 2+(l) + 2Cl-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); Pb 2+(l) + 2e -> Pb (l) (Cation / Pb 2+ gains / accepts / acquires electrons to form free atom)
25 (iii)At the anode/positive electrode(+); 2Cl-(l) -> Cl2 (g) + 2e (Anion / Cl- donate/lose electrons to form free atom then a gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid lead metal. II.At the anode pale green chlorine gas. b)To determine the products of electrolysis of molten Zinc bromide (i)Decomposition of electrolyte into free ions; ZnBr2 (l) -> Zn 2+(l) + 2Br-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); Zn 2+(l) + 2e -> Zn(l) (Cation / Zn2+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 2Br-(l) -> Br2 (g) + 2e (Anion / Br- donate/lose electrons to form free atom then a liquid molecule which change to gas on heating) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid Zinc metal. II.At the anode red bromine liquid / red/brown bromine gas.


II.At the anode pale green chlorine gas. b)To determine the products of electrolysis of molten Zinc bromide (i)Decomposition of electrolyte into free ions; ZnBr2 (l) -> Zn 2+(l) + 2Br-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); Zn 2+(l) + 2e -> Zn(l) (Cation / Zn2+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 2Br-(l) -> Br2 (g) + 2e (Anion / Br- donate/lose electrons to form free atom then a liquid molecule which change to gas on heating) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid Zinc metal. II.At the anode red bromine liquid / red/brown bromine gas. c)To determine the products of electrolysis of molten sodium chloride (i)Decomposition of electrolyte into free ions; NaCl (l) -> Na +(l) + Cl-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); 2Na+(l) + 2e -> Na (l) (Cation / Na+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 2Cl-(l) -> Cl2 (g) + 2e (Anion / Cl- donate/lose electrons to form free atom then a gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid sodium metal. II.At the anode pale green chlorine gas.


II.At the anode red bromine liquid / red/brown bromine gas. c)To determine the products of electrolysis of molten sodium chloride (i)Decomposition of electrolyte into free ions; NaCl (l) -> Na +(l) + Cl-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); 2Na+(l) + 2e -> Na (l) (Cation / Na+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 2Cl-(l) -> Cl2 (g) + 2e (Anion / Cl- donate/lose electrons to form free atom then a gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid sodium metal. II.At the anode pale green chlorine gas. 26 d)To determine the products of electrolysis of molten Aluminium (III)oxide (i)Decomposition of electrolyte into free ions; Al2O3 (l) -> 2Al 3+(l) + 3O2-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); 4Al 3+ (l) + 12e -> 4Al (l) (Cation / Al 3+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 6O2-(l) -> 3O2 (g) + 12e (Anion /6O2- donate/lose 12 electrons to form free atom then three gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid aluminium metal. II.At the anode colourless gas that relights/rekindles glowing splint. 13. For a compound /salt mixture containing many ions in an electrolytic cell, the discharge of ions in the cell depend on the following factors: a) Position of cations and anions in the electrochemical series 1. Most electropositive cations require more energy to reduce (gain electrons) and thus not readily discharged. The higher elements /metals in the electrochemical series the less easily/readily it is discharged at the cathode in the electrolytic cell.


For a compound /salt mixture containing many ions in an electrolytic cell, the discharge of ions in the cell depend on the following factors: a) Position of cations and anions in the electrochemical series 1. Most electropositive cations require more energy to reduce (gain electrons) and thus not readily discharged. The higher elements /metals in the electrochemical series the less easily/readily it is discharged at the cathode in the electrolytic cell. Table I showing the relative ease of discharge of cations in an electrolytic cell K+(aq) + e -> K(s) (least readily/easily discharged) Na+(aq) + e -> Na(s) Ca2+(aq) + 2e -> Ca(s) Mg2+(aq) + 2e -> Mg(s) Al3+(aq) + 3e -> Al(s) Zn2+(aq) + 2e -> Zn(s) Fe2+(aq) + 2e -> Fe(s) Pb2+(aq) + 2e -> Pb(s) 2H+(aq) + 2e -> H2(g) (hydrogen is usually “metallic”) Cu2+(aq) + 2e -> Cu(s) Hg2+(aq) + 2e -> Hg(s) Ag+(aq) + e -> Ag(s) (most readily/easily discharged) 2.The OH- ion is the most readily/easily discharged anion . All the other anionic radicals(SO42- ,SO32- ,CO32- ,HSO4- ,HCO3- ,NO3- ,PO43-)are not/never discharged. The ease of discharge of halogen ions increase down the group. Table II showing the relative ease of discharge of anions in an electrolytic cell
27 4OH- (aq) -> 2H2O(l) + O2 (g) + 4e (most readily/easily discharged) 2 I-(aq) -> I2(aq) + 2e 2 Br-(aq) -> Br2(aq) + 2e 2 Cl-(aq) -> Cl2(aq) + 2e 2 F-(aq) -> F2(aq) + 2e SO42- ,SO32- ,CO32- ,HSO4- ,HCO3- ,NO3- ,PO43- not/never/rarely discharged.


All the other anionic radicals(SO42- ,SO32- ,CO32- ,HSO4- ,HCO3- ,NO3- ,PO43-)are not/never discharged. The ease of discharge of halogen ions increase down the group. Table II showing the relative ease of discharge of anions in an electrolytic cell
27 4OH- (aq) -> 2H2O(l) + O2 (g) + 4e (most readily/easily discharged) 2 I-(aq) -> I2(aq) + 2e 2 Br-(aq) -> Br2(aq) + 2e 2 Cl-(aq) -> Cl2(aq) + 2e 2 F-(aq) -> F2(aq) + 2e SO42- ,SO32- ,CO32- ,HSO4- ,HCO3- ,NO3- ,PO43- not/never/rarely discharged. 3.(a)When two or more cations are attracted to the cathode, the ion lower in the electrochemical series is discharged instead of that which is higher as per the table I above. This is called selective/preferential discharge at cathode. (b)When two or more anions are attracted to the anode, the ion higher in the electrochemical series is discharged instead of that which is lower as per the table I above. This is called selective/preferential discharge at anode. 4.The following experiments shows the influence /effect of selective/preferential discharge on the products of electrolysis: (i)Electrolysis of acidified water/dilute sulphuric(VI) acid Fill the Hoffmann voltameter with dilute sulphuric(VI) acid. Connect the Hoffmann voltameter to a d.c. electric supply. Note the observations at each electrode. Electrolytic cell set up during electrolysis of acidified water/dilute sulphuric(VI) acid BatteryAnode(+)Cathode(-)ElectrolyteSimple set up of electrolytic cellGaseous product at anodeGaseous product at cathode Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq)
28 H2 SO4(aq) -> SO42-(aq) + 2H+(aq) II.


Electrolytic cell set up during electrolysis of acidified water/dilute sulphuric(VI) acid BatteryAnode(+)Cathode(-)ElectrolyteSimple set up of electrolytic cellGaseous product at anodeGaseous product at cathode Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq)
28 H2 SO4(aq) -> SO42-(aq) + 2H+(aq) II. Name the ions in acidified water that are attracted/move to: Cathode- H+(aq) from either sulphuric(VI) acid (H2 SO4) or water (H2O) Anode- SO42-(aq) from sulphuric (VI) acid (H2 SO4) and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) Anode 4OH- (aq) -> 2H2O(l) + O2 (g) + 4e (4OH- ions selectively discharged instead of SO42- ions at the anode) IV. Name the products of electrolysis of acidified water. Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound Anode-Oxygen gas (colourless gas that relights /rekindles glowing splint) V. Explain the difference in volume of products at the cathode and anode. The four(4) electrons donated/lost by OH- ions to form 1 molecule/1volume/1mole of oxygen (O2)gas at the anode are gained/acquired/accepted by the four H+(aq) ions to form 2 molecule/2volume/2mole of Hydrogen (H2)gas at the cathode. The volume of Oxygen gas at the anode is thus a half the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is thus a twice the volume of Oxygen produced at the anode. VI. Why is electrolysis of dilute sulphuric(VI) acid called “electrolysis of (acidified) water”?


The volume of Oxygen gas at the anode is thus a half the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is thus a twice the volume of Oxygen produced at the anode. VI. Why is electrolysis of dilute sulphuric(VI) acid called “electrolysis of (acidified) water”? The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. This implies/means that water in the electrolyte is being decomposed into hydrogen and Oxygen gases. The electrolysis of dilute sulphuric acid is therefore called “electrolysis of acidified water.” VI. Explain the changes in concentration of the electrolyte during electrolysis of acidified water” The concentration of dilute sulphuric (VI) acid increases. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape. The concentration /mole of acid present in a given volume of solution thus continue increasing/rising. (ii)Electrolysis of Magnesium sulphate(VI) solution Fill the Hoffmann voltameter with dilute sulphuric(VI) acid. Connect the Hoffmann voltameter to a d.c. electric supply. Note the observations at each electrode. Answer the following questions:
29 I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) Mg SO4(aq) -> SO42-(aq) + Mg2+(aq) II. Name the ions in Magnesium sulphate(VI) solution that are attracted/move to: Cathode- Mg2+(aq) from Magnesium sulphate(VI) solution (Mg SO4) and H+(aq) from water (H2O) Anode- SO42-(aq) from Magnesium sulphate(VI) solution (Mg SO4) and OH- (aq) from water (H2O) III.


Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) Mg SO4(aq) -> SO42-(aq) + Mg2+(aq) II. Name the ions in Magnesium sulphate(VI) solution that are attracted/move to: Cathode- Mg2+(aq) from Magnesium sulphate(VI) solution (Mg SO4) and H+(aq) from water (H2O) Anode- SO42-(aq) from Magnesium sulphate(VI) solution (Mg SO4) and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) H+ ions selectively discharged instead of Mg2+ ions at the cathode) Anode 4OH- (aq) -> 2H2O(l) + O2 (g) + 4e (4OH- ions selectively discharged instead of SO42- ions at the anode) IV. Name the products of electrolysis of Magnesium sulphate(VI) solution Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound Anode-Oxygen gas (colourless gas that relights /rekindles glowing splint) V. Explain the difference in volume of products at the cathode and anode. The four(4) electrons donated/lost by OH- ions to form 1 molecule/1volume/1mole of oxygen (O2)gas at the anode are gained/acquired/accepted by the four H+(aq) ions to form 2 molecule/2volume/2mole of Hydrogen (H2)gas at the cathode. The volume of Oxygen gas at the anode is thus a half the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is thus a twice the volume of Oxygen produced at the anode. VI. Explain the changes in concentration of the electrolyte during electrolysis of Magnesium sulphate(VI) solution The concentration of dilute Magnesium sulphate(VI) solution increases. The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water.


VI. Explain the changes in concentration of the electrolyte during electrolysis of Magnesium sulphate(VI) solution The concentration of dilute Magnesium sulphate(VI) solution increases. The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products. The concentration /mole of acid present in a given volume of Magnesium sulphate(VI) solution thus continue increasing/rising. 30 The set – up below was used during the electrolysis of aqueous magnesium sulphate using inert electrodes. Name a suitable pair of electrodes for this experiment Identify the ions and cations in the solution On the diagram label the cathode Write ionic equations for the reactions that took place at the anode. Explain the change that occurred to the concentration of magnesium sulphate solution during the experience. During the electrolysis a current of 2 amperes was passed through the solution for 4 hours. Calculate the volume of the gas produced at the anode.(1 faraday 96500 coulombs and volume of a gas at room temperature is 24000cm3) One of the uses of electrolysis is electroplating What is meant by electroplating? Give tow reasons why electroplating is necessary. b) Concentration of the electrolytes 1.High concentrations of cations and/or anions at the electrodes block the ion/s that is likely to be discharged at the electrode. This is called over voltage. A concentrated solution therefore produces different products of electrolysis from a dilute one. 2. The following experiments show the influence/effect of concentration of electrolyte on the products of electrolysis. (i)Electrolysis of dilute and concentrated(brine)sodium chloride solution
31 I. Dissolve about 0.5 g of pure sodium chloride crystals in 100cm3 of water. Place the solution in an electrolytic cell. Note the observations at each electrode for 10 minutes. Transfer the set up into a fume chamber/open and continue to make observations for a further 10 minute. Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) NaCl(aq) -> Cl-(aq) + Na+(aq) II.


Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) NaCl(aq) -> Cl-(aq) + Na+(aq) II. Name the ions in sodium chloride solution that are attracted/move to: Cathode- Na+(aq) from Sodium chloride solution (NaCl) and H+(aq) from water (H2O) Anode- Cl-(aq) from sodiumchloride solution (NaCl) and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) H+ ions selectively discharged instead of Na+ ions at the cathode) Anode 4OH- (aq) -> 2H2O(l) + O2 (g) + 4e (4OH- ions selectively discharged instead of Cl- ions at the anode) IV. Name the products of electrolysis of dilute sodium chloride solution Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound Anode-Oxygen gas (colourless gas that relights /rekindles glowing splint) V. Explain the difference in volume of products at the cathode and anode. Four(4) electrons donated/lost by OH- ions to form 1 molecule/1volume/1mole of oxygen (O2)gas at the anode are gained/acquired/accepted by four H+(aq) ions to form 2 molecule/2volume/2mole of Hydrogen (H2)gas at the cathode. The volume of Oxygen gas at the anode is half the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is twice the volume of Oxygen produced at the anode. VI. Explain the changes in concentration of the electrolyte during electrolysis of sodium chloride solution The concentration of dilute sodium chloride solution increases. The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products. The concentration /moles of salt present in a given volume of sodium chloride solution continue increasing/rising. 32 II.


Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products. The concentration /moles of salt present in a given volume of sodium chloride solution continue increasing/rising. 32 II. Dissolve about 20 g of pure sodium chloride crystals in 100cm3 of water. Place the solution in an electrolytic cell. Note the observations continuously at each electrode for 30 minutes in a fume chamber/open. Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) NaCl(aq) -> Cl-(aq) + Na+(aq) II. Name the ions in sodium chloride solution that are attracted/move to: Cathode- Na+(aq) from Sodium chloride solution (NaCl) and H+(aq) from water (H2O) Anode- Cl-(aq) from sodium chloride solution (NaCl) and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 2H+(aq) + 2e -> H2(g) H+ ions selectively discharged instead of Na+ ions at the cathode) Anode 2Cl- (aq) -> Cl2(g) + 4e (Cl- ions with a higher concentration block the discharge of OH- ions at the anode) IV. Name the products of electrolysis of concentrated sodium chloride solution/brine Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound Anode-Chlorine gas(pale green gas that bleaches damp/moist/wet litmus papers) V. Explain the difference in volume of products at the cathode and anode. Two (2) electrons donated/lost by Cl- ions to form 1 molecule/1volume/1mole of Chlorine (Cl2)gas at the anode are gained/acquired/accepted by two H+(aq) ions to form 1 molecule/1volume/1mole of Hydrogen (H2)gas at the cathode. The volume of Chlorine gas at the anode is equal to the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is equal to the volume of Chlorine produced at the anode. VI.


Two (2) electrons donated/lost by Cl- ions to form 1 molecule/1volume/1mole of Chlorine (Cl2)gas at the anode are gained/acquired/accepted by two H+(aq) ions to form 1 molecule/1volume/1mole of Hydrogen (H2)gas at the cathode. The volume of Chlorine gas at the anode is equal to the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is equal to the volume of Chlorine produced at the anode. VI. Explain the changes in concentration of the electrolyte during electrolysis of concentrated sodium chloride solution/brine The concentration of concentrated sodium chloride solution/brine increases. The ratio of Cl2 (g): H2 (g) is 1:1 as they are combined in water. Water in the electrolyte is decomposed into only Hydrogen gas that escapes as products at cathode. 33 The concentration /moles of OH- (aq) and Na+ ion (as NaOH) present in a given volume of electrolyte continue increasing/rising. This makes the electrolyte strongly alkaline with high pH. As the electrolysis of brine continues the concentration of Cl- ions decrease and oxygen gas start being liberated at anode. The electrolyte pH is thus lowered and the concentration of brine starts again increasing. (ii)Electrolysis of dilute and concentrated Hydrochloric acid solution I. Prepare about 50cm3 of 0.05 M of dilute Hydrochloric acid in 100cm3 solution. Place the solution in an electrolytic cell. Note the observations at each electrode for 10 minutes. Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) HCl(aq) -> Cl-(aq) + H+(aq) II. Name the ions in dilute Hydrochloric acid solution that are attracted/move to: Cathode- H+(aq) from dilute Hydrochloric acid (HCl) and H+(aq) from water (H2O) Anode- Cl-(aq) from dilute Hydrochloric acid (HCl) and OH- (aq) from water (H2O) III.


Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) HCl(aq) -> Cl-(aq) + H+(aq) II. Name the ions in dilute Hydrochloric acid solution that are attracted/move to: Cathode- H+(aq) from dilute Hydrochloric acid (HCl) and H+(aq) from water (H2O) Anode- Cl-(aq) from dilute Hydrochloric acid (HCl) and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) H+ ions selectively discharged instead of Na+ ions at the cathode) Anode 4OH- (aq) -> H2O(l) +O2+ 4e (4OH- ions selectively discharged instead of Cl- ions at the anode) IV. Name the products of electrolysis of dilute Hydrochloric acid Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound Anode-Oxygen gas (colourless gas that relights /rekindles glowing splint) V. Explain the difference in volume of products at the cathode and anode. Four(4) electrons donated/lost by OH- ions to form 1 molecule/1volume/1mole of oxygen (O2)gas at the anode are gained/acquired/accepted by four H+(aq) ions to form 2 molecule/2volume/2mole of Hydrogen (H2)gas at the cathode. The volume of Oxygen gas at the anode is half the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is twice the volume of Oxygen produced at the anode. 34 VI. Explain the changes in concentration of the electrolyte during electrolysis of dilute Hydrochloric acid The concentration of dilute Hydrochloric acid increases. The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products.


Explain the changes in concentration of the electrolyte during electrolysis of dilute Hydrochloric acid The concentration of dilute Hydrochloric acid increases. The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products. The concentration /moles of HCl present in a given volume of dilute Hydrochloric acid continue increasing/rising. II. Prepare about 50cm3 of 2M of Hydrochloric acid in 100cm3 solution. Place the solution in an electrolytic cell. Note the observations at each electrode for 30 minutes CautionThis experiment should be done in the open/fume chamber. Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) HCl(aq) -> Cl-(aq) + H+(aq) II. Name the ions in 2M Hydrochloric acid solution that are attracted/move to: Cathode- H+(aq) from dilute Hydrochloric acid (HCl) and H+(aq) from water (H2O) Anode- Cl-(aq) from dilute Hydrochloric acid (HCl) and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) H+ ions selectively discharged instead of Na+ ions at the cathode) Anode 2Cl- (aq) -> Cl2+ 2e (OH- ions concentration is low.Cl- ions concentration is higher at the anode thus cause over voltage/block discharge of OH- ions) IV. Name the products of electrolysis of 2M Hydrochloric acid Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound Anode-Chlorine gas (Pale green gas that bleaches blue/red moist/wet/damp litmus papers) V. Explain the difference in volume of products at the cathode and anode.


Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) H+ ions selectively discharged instead of Na+ ions at the cathode) Anode 2Cl- (aq) -> Cl2+ 2e (OH- ions concentration is low.Cl- ions concentration is higher at the anode thus cause over voltage/block discharge of OH- ions) IV. Name the products of electrolysis of 2M Hydrochloric acid Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound Anode-Chlorine gas (Pale green gas that bleaches blue/red moist/wet/damp litmus papers) V. Explain the difference in volume of products at the cathode and anode. Two(2) electrons donated/lost by Cl- ions to form 1 molecule/1volume/1mole of Chlorine (Cl2)gas at the anode are gained/acquired/accepted by two H+(aq) ions to form 1 molecule/1volume/1mole of Hydrogen (H2)gas at the cathode. The volume of Chlorine gas at the anode is equal to the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is twice the volume of Chlorine produced at the anode. 35 VI. Explain the changes in concentration of the electrolyte during electrolysis of 2M Hydrochloric acid The concentration of Hydrochloric acid decreases. The ratio of H2 (g): Cl2 (g) is 1:1 as they are combined in Hydrochloric acid. Water in the electrolyte is decomposed only into Hydrogen gas that escapes as products at the cathode. There is a net accumulation of excess OH- (aq) ions in solution. This makes the electrolyte strongly alkaline with high pH. c) Nature of electrodes used in the electrolytic cell Inert electrodes (carbon-graphite and platinum) do not alter the expected products of electrolysis in an electrolytic cell. If another/different electrode is used in the electrolytic cell it alters/influences/changes the expected products of electrolysis. The examples below illustrate the influence of the nature of electrode on the products of electrolysis: (i)Electrolysis of copper(II) sulphate(VI) solution I.