question_id stringlengths 8 35 | subject stringclasses 3
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values | question stringlengths 17 24.5k | options stringlengths 2 4.26k | correct_option stringclasses 6
values | answer stringclasses 460
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1lgrl0m11 | chemistry | electrochemistry | batteries,-fuel-cells-and-corrosion | <p>For lead storage battery pick the correct statements</p>
<p>A. During charging of battery, $$\mathrm{PbSO}_{4}$$ on anode is converted into $$\mathrm{PbO}_{2}$$</p>
<p>B. During charging of battery, $$\mathrm{PbSO}_{4}$$ on cathode is converted into $$\mathrm{PbO}_{2}$$</p>
<p>C. Lead storage battery consists of gri... | [{"identifier": "A", "content": "A, B, D only"}, {"identifier": "B", "content": "B, C only"}, {"identifier": "C", "content": "B, C, D only"}, {"identifier": "D", "content": "B, D only"}] | ["D"] | null | <p>Let's review each statement:</p>
<p>A. During charging of battery, PbSO₄ on anode is converted into PbO₂.</p>
<ul>
<li>This statement is incorrect. During charging, PbSO₄ on the anode (lead plate) is converted to Pb, not PbO₂.</li>
</ul>
<p>B. During charging of battery, PbSO₄ on cathode is converted into PbO₂.<... | mcq | jee-main-2023-online-12th-april-morning-shift | 1,957 |
jaoe38c1lscrpjrl | chemistry | electrochemistry | batteries,-fuel-cells-and-corrosion | <p>Which of the following statements is not correct about rusting of iron?</p> | [{"identifier": "A", "content": "Rusting of iron is envisaged as setting up of electrochemical cell on the surface of iron object.\n"}, {"identifier": "B", "content": "Dissolved acidic oxides $$\\mathrm{SO}_2, \\mathrm{NO}_2$$ in water act as catalyst in the process of rusting.\n"}, {"identifier": "C", "content": "Coat... | ["C"] | null | <p>The statement that is not correct about the rusting of iron is Option C. Let's examine each option one by one:</p>
<p>Option A: Rusting of iron is envisaged as setting up of an electrochemical cell on the surface of iron object. This statement is correct. Rusting of iron is an electrochemical process that occurs wh... | mcq | jee-main-2024-online-27th-january-evening-shift | 1,958 |
luxzq6ft | chemistry | electrochemistry | batteries,-fuel-cells-and-corrosion | <p>Match List I with List II</p>
<p><style type="text/css">
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overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;bo... | [{"identifier": "A", "content": "A-III, B-I, C-IV, D-II\n"}, {"identifier": "B", "content": "A-I, B-II, C-III, D-IV\n"}, {"identifier": "C", "content": "A-IV, B-III, C-I, D-II\n"}, {"identifier": "D", "content": "A-II, B-III, C-IV, D-I"}] | ["C"] | null | <p>To find the correct match between List I (Cell) and List II (Use/Property/Reaction), let's first understand each item and its corresponding match:</p>
<p>List I (Cell):</p>
<ul>
<li><strong>Leclanché cell</strong>: This is a primary cell (non-rechargeable) widely used in flashlights and other portable devices. It'... | mcq | jee-main-2024-online-9th-april-evening-shift | 1,959 |
lv2ertkj | chemistry | electrochemistry | batteries,-fuel-cells-and-corrosion | <p>Fuel cell, using hydrogen and oxygen as fuels,</p>
<p>A. has been used in spaceship</p>
<p>B. has as efficiency of $$40 \%$$ to produce electricity</p>
<p>C. uses aluminum as catalysts</p>
<p>D. is eco-friendly</p>
<p>E. is actually a type of Galvanic cell only</p>
<p>Choose the correct answer from the options given... | [{"identifier": "A", "content": "A, D, E only\n"}, {"identifier": "B", "content": "A, B, C only\n"}, {"identifier": "C", "content": "A, B, D only\n"}, {"identifier": "D", "content": "A, B, D, E only"}] | ["A"] | null | <p>Fuel cells produce electricity with an efficiency of about $$70 \%$$.</p>
<p>Fuel cells are pollution free, thus, eco-friendly.</p>
<p>Fuel cells are type of Galvanic cells only.</p> | mcq | jee-main-2024-online-4th-april-evening-shift | 1,960 |
CiFEmbp3pMNxIYZE | chemistry | electrochemistry | conductance-and-electrolysis | Conductivity (Seimen’s S) is directly proportional to area of the vessel and the concentration
of the solution in it and is inversely proportional to the length of the vessel then, then constant of proportionality is expressed in : | [{"identifier": "A", "content": "Sm mol<sup>-1</sup>"}, {"identifier": "B", "content": "Sm<sup>2 </sup> mol<sup>-1</sup>"}, {"identifier": "C", "content": "S<sup>-2</sup>m<sup>2</sup> mol"}, {"identifier": "D", "content": "S<sup>2</sup>m<sup>2</sup> mol<sup>-2</sup>"}] | ["B"] | null | Given $$S \propto {{area\, \times \,conc} \over \ell } = {{\kappa {m^2}mol} \over {m \times {m^3}}}$$
<br><br>$$\therefore$$ $$\,\,\,\,\kappa = S{m^2}mo{l^{ - 1}}$$ | mcq | aieee-2002 | 1,962 |
SzA3psjoCBYySpI0 | chemistry | electrochemistry | conductance-and-electrolysis | The limiting molar conductivities Λ° for NaCl, KBr and KCl are 126, 152 and 150 S cm<sup>2</sup> mol<sup>-1</sup> respectively. The Λ° for NaBr is | [{"identifier": "A", "content": "128 S cm<sup>2</sup> mol<sup>-1</sup>"}, {"identifier": "B", "content": "278 S cm<sup>2</sup> mol<sup>-1</sup>"}, {"identifier": "C", "content": "176 S cm<sup>2</sup> mol<sup>-1</sup>"}, {"identifier": "D", "content": "302 S cm<sup>2</sup> mol<sup>-1</sup>"}] | ["A"] | null | $${A^ \circ }Nacl = {\lambda ^ \circ }N{a^ + } + \lambda C{l^ - }\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>$${A^ \circ }KBr = {\lambda ^ \circ }{K^ + } + \lambda Br\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
<br><br>$${A^ \circ }KCl = {\lambda ^ \circ }{K^ + } + \lambda C{l^ - }\,\,\,\,\,\,\,\,\,... | mcq | aieee-2004 | 1,964 |
bYbJGByTiEe5k0xZ | chemistry | electrochemistry | conductance-and-electrolysis | Aluminium oxide may be electrolysed at 1000<sup>o</sup>C to furnish aluminium metal (Atomic
mass = 27 amu; 1 Faraday = 96,500 Coulombs). The cathode reaction is Al<sup>3+</sup> + 3e<sup>-</sup> $$\to$$ Al<sup>o</sup> To prepare 5.12 kg of aluminium metal by this method would require | [{"identifier": "A", "content": "5.49 $$\\times$$ 10<sup>7</sup> C of electricity "}, {"identifier": "B", "content": "1.83 $$\\times$$ 10<sup>7</sup> C of electricity "}, {"identifier": "C", "content": "5.49 $$\\times$$ 10<sup>4</sup> C of electricity "}, {"identifier": "D", "content": "5.49 $$\\times$$ 10<sup>1</sup> ... | ["A"] | null | $$1$$ mole of $${e^ - } = 1F = 96500\,C$$
<br><br>$$27g$$ of $$Al$$ is deposited by $$3 \times 96500\,C$$
<br><br>$$5120$$ $$g$$ of $$Al$$ will be deposited by
<br><br>$$ = {{3 \times 96500 \times 5120} \over {27}}$$
<br><br>$$ = 5.49 \times {10^7}C$$ | mcq | aieee-2005 | 1,965 |
TgPxnqNHi2MYD4Ah | chemistry | electrochemistry | conductance-and-electrolysis | The highest electrical conductivity of the following aqueous solutions is of : | [{"identifier": "A", "content": "0.1 M acetic acid"}, {"identifier": "B", "content": "0.1 M chloroacetic acid"}, {"identifier": "C", "content": "0.1 M fluoroacetic acid "}, {"identifier": "D", "content": "0.1 M difluoroacetic acid "}] | ["D"] | null | Thus difluoro acetic acid being strongest acid will furnish maximum number of ions showing highest electrical conductivity. The decreasing acidic strength of the carboxylic acids given is difluoro acetic acid $$>$$ fluoro acetic acid $$>$$ chloro acitic acid $$>$$ acetic acid. | mcq | aieee-2005 | 1,966 |
gne2KSHrilaUfc6x | chemistry | electrochemistry | conductance-and-electrolysis | <div style="overflow-x: auto;">
<style type="text/css">
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.tg th{font-family:Arial, sans-serif;font-size:14px;fo... | [{"identifier": "A", "content": "517.2 "}, {"identifier": "B", "content": "552.7 "}, {"identifier": "C", "content": "390.7"}, {"identifier": "D", "content": "217.5 "}] | ["C"] | null | $$A_{HCl}^\infty = 426.2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>$$A_{AcONa}^\infty = 91.0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
<br><br>$$A_{NaCl}^\infty = 126.5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right)$$
<br><br>$$A_{AcOH}^\infty = \left(... | mcq | aieee-2005 | 1,967 |
Mq9Iv1ngH7Opdlng | chemistry | electrochemistry | conductance-and-electrolysis | The molar conductivities $$ \wedge _{NaOAc}^o$$ and $$ \wedge _{HCl}^o$$ and at infinite dilution in water at 25<sup>o</sup>C are 91.0 and 426.2 Scm<sup>2</sup>/mol respectively. To calculate $$ \wedge _{HOAc}^o$$ , the additional value required is | [{"identifier": "A", "content": "$$ \\wedge _{{H_2}O}^o$$"}, {"identifier": "B", "content": "$$ \\wedge _{KCl}^o$$ "}, {"identifier": "C", "content": "$$ \\wedge _{NaOH}^o$$"}, {"identifier": "D", "content": "$$ \\wedge _{NaCl}^o$$"}] | ["D"] | null | $$\Lambda _{C{H_3}COOH}^o$$ is given by the following equation
<br><br>$$\Lambda _{C{H_3}COOH}^o = \left( {\Lambda _{C{H_3}COONa}^o + \Lambda _{HCl}^o} \right) - \left( {\Lambda _{NaCl}^o} \right)$$
<br><br>Hence $$\Lambda _{NaCl}^ \circ $$ is required. | mcq | aieee-2006 | 1,968 |
OLMbCGTlkXvrtGM3 | chemistry | electrochemistry | conductance-and-electrolysis | Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100$$\Omega $$.
The conductivity of this solution is 1.29 S m<sup>–1</sup>. Resistance of the same cell when filled with 0.2 M of
the same solution is 520 $$\Omega $$, The molar conductivity of 0.02 M solution of the el... | [{"identifier": "A", "content": "124 $$\\times$$ 10<sup>\u20134 </sup> S m<sup>2 </sup> mol<sup>\u20131</sup>"}, {"identifier": "B", "content": "1240 $$\\times$$ 10<sup>\u20134 </sup> S m<sup>2 </sup> mol<sup>\u20131</sup>"}, {"identifier": "C", "content": "1.24 $$\\times$$ 10<sup>\u20134 </sup> S m<sup>2 </sup> mol<su... | ["D"] | null | $$R = 100\Omega ,\kappa = {1 \over R}\left( {{l \over a}} \right),{l \over a}$$ (cell constant)
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = 1.29 \times 100{m^{ - 1}}$$
<br><br>Given, $$R = 520\Omega ,C = 0.2M,\,\mu $$ (molar conductivity) $$=$$ ?
<br><br>$$\mu = \kappa ... | mcq | aieee-2006 | 1,969 |
82P6cYNxCChvG7DX | chemistry | electrochemistry | conductance-and-electrolysis | The equivalent conductances of two strong electrolytes at infinite dilution in H<sub>2</sub>O (where ions move
freely through a solution) at 25<sup>o</sup>C are given below: <br/>
$$ \wedge _{C{H_3}COONa}^o$$ = 91.0 S cm<sup>2</sup>/equiv<br/>
$$ \wedge _{HCl}^o$$ = 426.2 S cm<sup>2</sup>/equiv<br/>
What additional inf... | [{"identifier": "A", "content": "$$ \\wedge ^o$$ of chloroacetic acid (C/CH<sub>2</sub>COOH)"}, {"identifier": "B", "content": "$$ \\wedge ^o$$ of NaCl"}, {"identifier": "C", "content": "$$ \\wedge ^o$$ of CH<sub>3</sub>COOK"}, {"identifier": "D", "content": "The limiting equivalent conductance of $${H^ + }( \\wedge _{... | ["B"] | null | <b>NOTE :</b> According to Kohlrausch's law, molar conductivity of weak electrolyte acetic acid $$\left( {C{H_3}COOH} \right)$$ can be calculated as follows:
<br><br>$${\Lambda ^o}_{C{H_3}COOH} = \left( {{\Lambda ^o}_{C{H_3}COONa} + {\Lambda ^o}_{HCl}} \right) - {\Lambda ^o}_{NaCl}$$
<br><br>$$\therefore$$ Value of $$... | mcq | aieee-2007 | 1,970 |
GENLBQdzkzTSXUCQ | chemistry | electrochemistry | conductance-and-electrolysis | Resistance of 0.2 M solution of an electrolyte is 50 $$\Omega$$. The specific conductance of the solution is 1.4 S m<sup>-1</sup>. The resistance of 0.5 M solution of the same electrolyte is 280 $$\Omega$$. The molar conductivity of 0.5 M solution of the electrolyte in S m<sup>2</sup> mol<sup>-1</sup> is : | [{"identifier": "A", "content": "5 \u00d7 10<sup>3</sup>"}, {"identifier": "B", "content": "5 \u00d7 10<sup>2</sup>"}, {"identifier": "C", "content": "5 \u00d7 10<sup>-4</sup>"}, {"identifier": "D", "content": "5 \u00d7 10<sup>-3</sup>"}] | ["C"] | null | Given for $$0.2$$ $$M$$ solution
<br><br>$$R = 50\Omega $$
<br><br>$$\kappa = 1.45\,S\,{m^{ - 1}} = 1.4 \times {10^{ - 2}}\,S\,c{m^{ - 1}}$$
<br><br>Now, $$R = \rho {\ell \over a} = {1 \over \kappa } \times {\ell \over a}$$
<br><br>$$ \Rightarrow {\ell \over a} = R \times \kappa = 50 \times 1.4 \times {10^{ - 2}}$... | mcq | jee-main-2014-offline | 1,971 |
wQZhoL2I8GFK6YPP | chemistry | electrochemistry | conductance-and-electrolysis | Two Faraday of electricity is passed through a solution of CuSO<sub>4</sub>. The mass of copper deposited at the
cathode is: (at. mass of Cu = 63.5 amu) | [{"identifier": "A", "content": "63.5 g"}, {"identifier": "B", "content": "2 g"}, {"identifier": "C", "content": "127 g"}, {"identifier": "D", "content": "0 g"}] | ["A"] | null | $$C{u^{2 + }} + 2{e^ - }\buildrel \, \over
\longrightarrow Cu$$
<br><br>$$2F\,\,i.e.\,\,2 \times 96500\,C$$ deposit $$Cu=1$$ mol $$=63.5$$ $$g$$ | mcq | jee-main-2015-offline | 1,973 |
carxJYhGrMbmiDcNAGpd7 | chemistry | electrochemistry | conductance-and-electrolysis | Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing 0.2 Faraday electricity through an aqueous solution of potassium succinate, the total volume of gases (at both cathode and anode) at STP (1 atm and 273 K) is : | [{"identifier": "A", "content": "2.24 L"}, {"identifier": "B", "content": "4.48 L"}, {"identifier": "C", "content": "6.72 L "}, {"identifier": "D", "content": "8.96 L "}] | ["D"] | null | <p>The reaction is</p>
<p>$$2C{H_3}COOK + 2{H_2}O\buildrel {Electrolysis} \over
\longrightarrow C{H_3} - C{H_3} + 2C{O_2} + {H_2} + 2KOH$$</p>
<p>At the anode (Oxidation):</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l3l0or4y/3b6eb7ab-114c-466f-bd28-046bc2bfb140/6a1d6420-dbd9-11ec-adf6-01b3c... | mcq | jee-main-2016-online-10th-april-morning-slot | 1,974 |
qOEpVdldNK7Ce1uJ | chemistry | electrochemistry | conductance-and-electrolysis | How long (approximate) should water be electrolysed by passing through 100 amperes current so that the
oxygen released can completely burn 27.66 g of diborane?<br/>
(Atomic weight of B = 10.8 u) | [{"identifier": "A", "content": "1.6 hours"}, {"identifier": "B", "content": "6.4 hours"}, {"identifier": "C", "content": "0.8 hours"}, {"identifier": "D", "content": "3.2 hours"}] | ["D"] | null | Required reaction :
<br><br>B<sub>2</sub>H<sub>6</sub> + 3O<sub>2</sub> $$ \to $$ B<sub>2</sub> O<sub>3</sub> + 3 H<sub>2</sub> O
<br><br>Here molar mass of B<sub>2</sub>H<sub>6</sub> =10.8 $$ \times $$ 2 + 6 = 27.6 gm
<br><br>Given weight of B<sub>2</sub>H<sub>6</sub> = 27.66 g
<br><br>$$\therefore\,\,\,\,$$No of mo... | mcq | jee-main-2018-offline | 1,975 |
Z89d2iucGgQrFVsuxdnOP | chemistry | electrochemistry | conductance-and-electrolysis | When an electric currents passed through acidified water, 112 mL of hydrogen gas at N.T.P. was collected at the cathode in 965 seconds. The current passed, in ampere, is : | [{"identifier": "A", "content": "1.0"}, {"identifier": "B", "content": "0.5"}, {"identifier": "C", "content": "0.1"}, {"identifier": "D", "content": "2.0"}] | ["A"] | null | Reaction at cathode :
<br><br>2H<sup>+</sup> + 2e<sup>$$-$$</sup> $$ \to $$ H<sub>2</sub>
<br><br>We know,
<br><br>$$\omega $$ = zIt = $${{EIt} \over {96500}}$$
<br><<br>no. of moles of H<sub>2</sub> = $${{112} \over {22400}}$$
<br><br>$$\therefore\,\,\,$$ mass (w) of H<sub>2</sub> = $${{112} \over {22400}}$$ $$ \... | mcq | jee-main-2018-online-15th-april-morning-slot | 1,976 |
jSrsVc414XRCcl9cmjHvy | chemistry | electrochemistry | conductance-and-electrolysis | When 9.65 ampere current was passed for 1.0 hour into nitrobenzene in acidic medium, the amount of p-aminophenol produced is : | [{"identifier": "A", "content": "9.81 g"}, {"identifier": "B", "content": "10.9 g"}, {"identifier": "C", "content": "98.1 g"}, {"identifier": "D", "content": "109.0 g"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263906/exam_images/ajivxv9rvutgjeo28wkf.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2018 (Online) 16th April Morning Slot Chemistry - Electrochemistry Question 142 English Explanation">
<br>Mole... | mcq | jee-main-2018-online-16th-april-morning-slot | 1,977 |
QiJQ3TTC2PU4GT8FGk3rsa0w2w9jwv8740b | chemistry | electrochemistry | conductance-and-electrolysis | Consider the statements S1 and S2
<br/><br>S1 : Conductivity always increases with decrease in the concentration of electrolyte.
<br/><br>S2 : Molar conductivity always increases with decrease in the concentration of electrolyte.
<br/><br>The correct option among the following is :</br></br></br> | [{"identifier": "A", "content": "Both S1 and S2 are wrong"}, {"identifier": "B", "content": "S1 is correct and S2 is wrong"}, {"identifier": "C", "content": "Both S1 and S2 are correct"}, {"identifier": "D", "content": "S1 is wrong and S2 is correct "}] | ["D"] | null | We know conductivity (k) = $${G \over V}$$
<br><br>V = volume
<br><br>When concentration decreases volume increases and when volume increases then conductivity (k) decreases.
<br><br>So, we can say S1 is incorrect.
<br><br>We know that,
<br><br>$${\lambda _m} = {k \over c}$$
<br><br>where
<br><br>$$\lambda $$<sub>m</su... | mcq | jee-main-2019-online-10th-april-morning-slot | 1,978 |
wrFltoovRg58a2S6yc3rsa0w2w9jx0znlcf | chemistry | electrochemistry | conductance-and-electrolysis | Which one of the following graphs between molar conductivity ($${\Lambda _m}$$) versus $$\sqrt C $$ is correct ? | [{"identifier": "A", "content": "<picture><source media=\"(max-width: 320px)\" srcset=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266178/exam_images/mkbhkv3fhy2kd2gk2bf7.webp\"><img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264447/exam_images/eq4pkvyvlbjuf7fi7dhk.webp\" style=\"max-width... | ["A"] | null | The graph is drawn using following equation,
<br><br>$${\Lambda _m} = \Lambda _m^\infty - b\sqrt c $$
<br><br>As the size of K<sup>+</sup> is higher than the size of Na<sup>+</sup>, then the hydration radii of aqueous Na<sup>+</sup> will be more than the aqueous K<sup>+</sup>. Therefore the ionic mobility of Na<sup>+<... | mcq | jee-main-2019-online-10th-april-evening-slot | 1,979 |
95BpaQdmmggnxsGyQHSyQ | chemistry | electrochemistry | conductance-and-electrolysis | $$ \wedge _m^ \circ $$ for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm<sup>2</sup> mol<sup>–1</sup>, respectively. If the conductivity of 0.001 M HA is-
<br/><br/>5 $$ \times $$ 10<sup>–5</sup> S cm<sup>–1</sup>, degree of dissociation of HA is - | [{"identifier": "A", "content": "0.50"}, {"identifier": "B", "content": "0.125"}, {"identifier": "C", "content": "0.25"}, {"identifier": "D", "content": "0.75"}] | ["B"] | null | $$ \wedge _m^ \circ $$ (HA) = $$ \wedge _m^ \circ $$ (HCl) + $$ \wedge _m^ \circ $$ (N<i>a</i>A) $$-$$ $$ \wedge _m^ \circ $$ (N<i>a</i>Cl)
<br><br>= 425.9 + 100.5 $$-$$ 126.4
<br><br>= 400 S cm<sup>2</sup> . mol<sup>$$-$$1</sup>
<br><br> $$ \wedge _m^ c $$ = $${{K \times 1000} \over M}$$
<br><br... | mcq | jee-main-2019-online-12th-january-evening-slot | 1,980 |
mSgr7enqEwoMXtcYZZlWu | chemistry | electrochemistry | conductance-and-electrolysis | A solution of Ni(NO<sub>3</sub>)<sub>2</sub> is electrolysed between
platinum electrodes using 0.1 Faraday
electricity. How many mole of Ni will be
deposited at the cathode? | [{"identifier": "A", "content": "0.10"}, {"identifier": "B", "content": "0.15"}, {"identifier": "C", "content": "0.20"}, {"identifier": "D", "content": "0.05"}] | ["D"] | null | Cathode reaction :
<br><br>Ni<sup>+2</sup> + 2e<sup>-</sup> $$ \to $$ Ni(s)
<br><br>$$ \therefore $$ From 2 mole of electrons 1 mole of Ni is deposited at the cathode.
<br><br>So from 0.1 F or 0.1 mole of electrons $${1 \over 2} \times 0.1$$ = 0.05 mole of Ni is deposited at the cathode. | mcq | jee-main-2019-online-9th-april-evening-slot | 1,981 |
1Me5RhDnbbjMhiM2gVjgy2xukfuqlla6 | chemistry | electrochemistry | conductance-and-electrolysis | Potassium chlorate is prepared by the
electrolysis of KCl in basic solution
<br/><br/>6OH<sup>-</sup> + Cl<sup>-</sup> $$ \to $$ ClO<sub>3</sub><sup>-</sup> + 3H<sub>2</sub>O + 6e<sup>-</sup>
<br/><br/>If only 60% of the current is utilized in the
reaction, the time (rounded to the nearest hour)
required to produce 10 ... | [] | null | 11 | For synthesis of 1 mole of ClO<sub>3</sub><sup>-</sup> , 6F of charge
is required.
<br><br>$$ \therefore $$ To synthesise
$${{10} \over {122}}$$ moles of KClO<sub>3</sub>,
<br><br>Charge required = $${{10} \over {122}} \times 6$$ F
<br><br>$${{10} \over {122}} \times 6 = {{2 \times t(hr) \times 3600} \over {96500}} \t... | integer | jee-main-2020-online-6th-september-morning-slot | 1,982 |
FguWwE6LXaiCXr0XXHjgy2xukfox5xhz | chemistry | electrochemistry | conductance-and-electrolysis | The variation of molar conductivity with concentration of an electrolyte (X) in aqueous solution
is shown in the given figure.
<img src="data:image/png;base64,UklGRrAKAABXRUJQVlA4IKQKAACQVwCdASrgATwBPm02mEkkIqKhIPeJGIANiWlu8p91Z/0C4QKmttAutQ0v0K/ov4beAn9m/J3z78R3kb2m9ONmL+/fy3+x/6n+qet/+G/oX43+o/qQ/G74CPTn9i/kf9K/WT0Z/... | [{"identifier": "A", "content": "HCl"}, {"identifier": "B", "content": "CH<sub>3</sub>COOH "}, {"identifier": "C", "content": "NaCl"}, {"identifier": "D", "content": "KNO<sub>3</sub>"}] | ["B"] | null | The electrolyte (X) must be weak electrolyte as such type of variation is always for weak electrolyte. So
<br><br>X is CH<sub>3</sub>COOH. | mcq | jee-main-2020-online-5th-september-evening-slot | 1,983 |
M1cCgSXdzD4c3SuurH7k9k2k5idsa5s | chemistry | electrochemistry | conductance-and-electrolysis | 108 g of silver (molar mass 108 g mol<sup>–1</sup>) is
deposited at cathode from AgNO<sub>3</sub>(aq) solution
by a certain quantity of electricity. The
volume (in L) of oxygen gas produced at
273 K and 1 bar pressure from water by the
same quantity of electricity is _______. | [] | null | 5.66to5.68 | Cathode : Ag<sup>+</sup>(aq) + e<sup>-</sup> $$ \to $$ Ag(s)
<br><br>Moles of Ag deposited = $${{108} \over {108}}$$ = 1 mole
<br><br>Anode : 2H<sub>2</sub>O $$ \to $$ O<sub>2</sub> + 4H<sup>+</sup> + 4e<sup>-</sup>
<br><br>Here we have to find volume of O<sub>2</sub> evolved.
<br><br>Equivalance of Ag = Equivalance of... | integer | jee-main-2020-online-9th-january-morning-slot | 1,987 |
cuQCtLDotzQNDXNsFG7k9k2k5epdscq | chemistry | electrochemistry | conductance-and-electrolysis | The equation that is incorrect is : | [{"identifier": "A", "content": "$${\\left( {\\Lambda _m^0} \\right)_{KCl}} - {\\left( {\\Lambda _m^0} \\right)_{NaCl}} = {\\left( {\\Lambda _m^0} \\right)_{KBr}} - {\\left( {\\Lambda _m^0} \\right)_{NaBr}}$$"}, {"identifier": "B", "content": "$${\\left( {\\Lambda _m^0} \\right)_{NaBr}} - {\\left( {\\Lambda _m^0} \\rig... | ["B"] | null | Left hand side :
<br>$${\left( {\Lambda _m^0} \right)_{NaBr}} - {\left( {\Lambda _m^0} \right)_{NaI}}$$ = $${\left( {\Lambda _m^0} \right)_{Br^-}} - {\left( {\Lambda _m^0} \right)_{I^-}}$$ ....(1)
<br><br>Right hand side :
<br><br>$${\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{NaBr}}$$ = $${\... | mcq | jee-main-2020-online-7th-january-evening-slot | 1,988 |
tQNO5pcBpymZ5olrc41kluebp7v | chemistry | electrochemistry | conductance-and-electrolysis | Consider the following reaction<br/><br/>$$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{ + 2}} + 4{H_2}O,{E^o} = 1.51V$$.<br/><br/>The quantity of electricity required in Faraday to reduce five moles of $$MnO_4^ - $$ is ___________. (Integer answer) | [] | null | 25 | $$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{ + 2}} + 4{H_2}O,{E^o} = 1.51V$$
<br><br>1 mole of MnO<sub>4</sub><sup>-</sup> required 5 moles of
electrons or 5 F electricity.
<br><br>$$ \therefore $$ 5 moles of MnO<sub>4</sub><sup>-</sup> required 25 F electricity. | integer | jee-main-2021-online-26th-february-morning-slot | 1,989 |
lctM0Ts8eS5zEynETd1kmiuth60 | chemistry | electrochemistry | conductance-and-electrolysis | A 5.0 m mol dm<sup>$$-$$3</sup> aqueous solution of KCl has a conductance of 0.55 mS when measured in a cell of cell constant 1.3 cm<sup>$$-$$1</sup>. The molar conductivity of this solution is ___________ mSm<sup>2</sup> mol<sup>$$-$$1</sup>. (Round off to the Nearest Integer). | [] | null | 14 | Conductance = $${{Conductivity} \over {Cell\,cons\tan t}}$$<br><br>$$ \therefore $$ Conductivity = 0.55 $$\times$$ 10<sup>$$-$$3</sup> $$\times$$ 1.3 S cm<sup>$$-$$1</sup><br><br>Molar conductivity = $${{Conductivity\,(S\,c{m^{ - 1}}) \times 1000} \over {Molarity\,(mol/L)}}$$<br><br>$$ = {{0.55 \times {{10}^{ - 3}} \ti... | integer | jee-main-2021-online-16th-march-evening-shift | 1,990 |
0W9EF8d5zfw2Sn6pue1kmkjbfcy | chemistry | electrochemistry | conductance-and-electrolysis | A KCl solution of conductivity 0.14 S m<sup>$$-$$1</sup> shows a resistance of 4.19$$\Omega$$ in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops to 1.03$$\Omega$$. The conductivity of the HCl solution is ____________ $$\times$$ 10<sup>$$-$$2</sup> S m<sup>$$-$$1</sup>. (Round ... | [] | null | 57 | For KCl solution,<br><br>$$R = \left( {{1 \over K}} \right)\left( {{l \over A}} \right) \Rightarrow {l \over A} = R \times K = 4.19 \times 0.14$$<br><br>= 0.58<br><br>For HCl solution,<br><br>$$R = \left( {{1 \over K}} \right)\left( {{l \over A}} \right)$$<br><br>$$ \Rightarrow K = {{(l/A)} \over R} = {{0.58} \over {1.... | integer | jee-main-2021-online-17th-march-evening-shift | 1,991 |
1krrlotqp | chemistry | electrochemistry | conductance-and-electrolysis | Potassium chlorate is prepared by electrolysis of KCl in basic solution as shown by following equation.<br/><br/>6OH<sup>$$-$$</sup> + Cl<sup>$$-$$</sup> $$\to$$ ClO<sub>3</sub><sup>$$-$$</sup> + 3H<sub>2</sub>O + 6e<sup>$$-$$</sup><br/><br/>A current of xA has to be passed for 10h to produce 10.0g of potassium chlorat... | [] | null | 1 | $$W = {E \over F} \times I \times t$$<br><br>$$10 = {{122.6} \over {96500 \times 6}} \times x \times 10 \times 3600$$<br><br>$$x = 1.311$$<br><br>Ans. (1) | integer | jee-main-2021-online-20th-july-evening-shift | 1,992 |
1krz25rki | chemistry | electrochemistry | conductance-and-electrolysis | The conductivity of a weak acid HA of concentration 0.001 mol L<sup>$$-$$1</sup> is 2.0 $$\times$$ 10<sup>$$-$$5</sup> S cm<sup>$$-$$1</sup>. If $$\Lambda _m^o$$(HA) = 190 S cm<sup>2</sup> mol<sup>$$-$$1</sup>, the ionization constant (K<sub>a</sub>) of HA is equal to ______________ $$\times$$ 10<sup>$$-$$6</sup>. (Rou... | [] | null | 12 | $$\Lambda _m^{} = 1000 \times {\kappa \over M}$$<br><br>$$ = 1000 \times {{2 \times {{10}^{ - 5}}} \over {0.001}} = 20$$ S cm<sup>2</sup> mol<sup>$$-$$1</sup><br><br>$$ \Rightarrow \alpha = {{\Lambda _m^{}} \over {\Lambda _m^\infty }} = {{20} \over {190}} = \left( {{2 \over {19}}} \right)$$<br><br>HA $$\rightleftharp... | integer | jee-main-2021-online-27th-july-morning-shift | 1,993 |
1ktb4bcnb | chemistry | electrochemistry | conductance-and-electrolysis | Given below are two statements :<br/><br/>Statement I : The limiting molar conductivity of KCl (strong electrolyte) is higher compared to that of CH<sub>3</sub>COOH (weak electrolyte).<br/><br/>Statement II : Molar conductivity decreases with decrease in concentration of electrolyte.<br/><br/>In the light of the above ... | [{"identifier": "A", "content": "Statement I is true but Statement II is false."}, {"identifier": "B", "content": "Statement I is false but Statement II is true."}, {"identifier": "C", "content": "Both Statement I and Statement II are true."}, {"identifier": "D", "content": "Both Statement I and Statement II are false.... | ["D"] | null | <table class="tg">
<thead>
<tr>
<th class="tg-c3ow">Ion</th>
<th class="tg-c3ow">$${H^ + }$$</th>
<th class="tg-c3ow">$${K^ + }$$</th>
<th class="tg-c3ow">$$C{l^ - }$$</th>
<th class="tg-baqh">$$C{H_3}CO{O^ - }$$</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-c3ow">$$\Lambda _{m\,Sc{m^2}/m... | mcq | jee-main-2021-online-26th-august-morning-shift | 1,994 |
1ktjvcwac | chemistry | electrochemistry | conductance-and-electrolysis | Match List - I with List - II<br/><br/><table>
<thead>
<tr>
<th></th>
<th>List - I<br/>(Parmeter)</th>
<th></th>
<th>List - II<br/>(Unit)</th>
</tr>
</thead>
<tbody>
<tr>
<td>(a)</td>
<td>Cell constant</td>
<td>(i)</td>
<td>$$S\,c{m^2}mo{l^{ - 1}}$$</td>
</tr>
<tr>
<td>(b)</td>
<td>Molar conductivity</td>
<td>(ii)</td>... | [{"identifier": "A", "content": "(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)"}, {"identifier": "B", "content": "(a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)"}, {"identifier": "C", "content": "(a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)"}, {"identifier": "D", "content": "(a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)"}] | ["A"] | null | Cell constant = $$\left( {{l \over A}} \right)$$ $$\Rightarrow$$ Units = m<sup>$$-$$1</sup><br><br>Molar conductivity ($$\Lambda $$<sub>m</sub>) $$\Rightarrow$$ Units = Sm<sup>2</sup> mole<sup>$$-$$1</sup><br><br>Conductivity (K) $$\Rightarrow$$ Units = S m<sup>$$-$$1</sup><br><br>Degree of dissociation ($$\alpha$$ $$\... | mcq | jee-main-2021-online-31st-august-evening-shift | 1,996 |
1l54a24wt | chemistry | electrochemistry | conductance-and-electrolysis | <p>A dilute solution of sulphuric acid is electrolysed using a current of 0.10 A for 2 hours to produce hydrogen and oxygen gas. The total volume of gases produced a STP is _____________ cm<sup>3</sup>. (Nearest integer)</p>
<p>[Given : Faraday constant F = 96500 C mol<sup>$$-$$1</sup> at STP, molar volume of an ideal ... | [] | null | 127 | $2 \mathrm{~F}$ produces $=\frac{3}{2}$ mole of gas
<br/><br/>
$0.10 \times 2 \times 3600$ coulomb produces
<br/><br/>
$$
\begin{aligned}
& =\frac{\frac{3}{2} \times 0.1 \times 2 \times 3600}{2 \times 96500} \\\\
& =0.0056 \text { moles of gas }
\end{aligned}
$$<br/><br/>
Volume of gas produced $=0.0056 \times 22.7 \ma... | integer | jee-main-2022-online-29th-june-morning-shift | 1,998 |
1l56bki0j | chemistry | electrochemistry | conductance-and-electrolysis | <p>The quantity of electricity in Faraday needed to reduce 1 mol of Cr<sub>2</sub>O$$_7^{2 - }$$ to Cr<sup>3+</sup> is ____________.</p> | [] | null | 6 | $$
\overset{+6}{\mathrm{Cr}_2} \mathrm{O}_7^{2-} \longrightarrow 2 \mathrm{Cr}^{3+}
$$<br/><br/>
$\because$ Each $\mathrm{Cr}$ is converting from $+6$ to $+3$<br/><br/>
$\therefore 6$ faradays of charge is required | integer | jee-main-2022-online-28th-june-morning-shift | 1,999 |
1l57t17ol | chemistry | electrochemistry | conductance-and-electrolysis | <p>The limiting molar conductivities of NaI, NaNO<sub>3</sub> and AgNO<sub>3</sub> are 12.7, 12.0 and 13.3 mS m<sup>2</sup> mol<sup>$$-$$1</sup>, respectively (all at 25$$^\circ$$C). The limiting molar conductivity of AgI at this temperature is ____________ mS m<sup>2</sup> mol<sup>$$-$$1</sup>.</p> | [] | null | 14 | Given<br/><br/>
(1) $\lambda_m^{\infty}(\mathrm{NaI})=12.7 \,\mathrm{mS} \mathrm{m}^2 \mathrm{~mol}^{-1}$<br/><br/>
(2) $\lambda_{\mathrm{m}}^{\infty}\left(\mathrm{NaNO}_3\right)=12.0 \,\mathrm{mS} \mathrm{m}^2 \mathrm{~mol}^{-1}$<br/><br/>
(3) $\lambda_{\mathrm{m}}^{\infty}\left(\mathrm{AgNO}_3\right)=13.3 \,\mathrm{m... | integer | jee-main-2022-online-27th-june-morning-shift | 2,000 |
1l59rlods | chemistry | electrochemistry | conductance-and-electrolysis | <p>A solution of Fe<sub>2</sub>(SO<sub>4</sub>)<sub>3</sub> is electrolyzed for 'x' min with a current of 1.5 A to deposit 0.3482 g of Fe. The value of x is ___________. [nearest integer]</p>
<p>Given : 1 F = 96500 C mol<sup>$$-$$1</sup></p>
<p>Atomic mass of Fe = 56 g mol<sup>$$-$$1</sup></p> | [] | null | 20 | $\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe}$
<br/><br/>
$$
\text {Moles of Fe deposited }=\frac{0.3482}{56}=6.2 \times 10^{-3}
$$
<br/><br/>
For 1 mole $\mathrm{Fe}$, charge required is $3 \mathrm{~F}$
<br/><br/>
For $6.2 \times 10^{-3}$ mole $\mathrm{Fe}$, charge required is $3 \times 6.2 \times 10^{-3}... | integer | jee-main-2022-online-25th-june-evening-shift | 2,001 |
1l5bdz9jj | chemistry | electrochemistry | conductance-and-electrolysis | <p>The resistance of a conductivity cell containing 0.01 M KCl solution at 298 K is 1750 $$\Omega$$. If the conductivity of 0.01 M KCl solution at 298 K is 0.152 $$\times$$ 10<sup>$$-$$3</sup> S cm<sup>$$-$$1</sup>, then the cell constant of the conductivity cell is ____________ $$\times$$ 10<sup>$$-$$3</sup> cm<sup>$$... | [] | null | 266 | Molarity of $\mathrm{KCl}$ solution $=0.1 ~\mathrm{M}$
<br/><br/>
$$
\begin{array}{ll}
\text { Resistance } & =1750 ~\mathrm{ohm} \\\\
\text { Conductivity } & =0.152 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^{-1} \\\\
\text { Conductivity } & =\frac{\text { Cell constant }}{\text { Resistance }} \\\\
\therefore \text { ... | integer | jee-main-2022-online-24th-june-evening-shift | 2,002 |
1l6f6mp0f | chemistry | electrochemistry | conductance-and-electrolysis | <p>The molar conductivity of a conductivity cell filled with 10 moles of 20 mL NaCl solution is $${\Lambda _{m1}}$$ and that of 20 moles another identical cell heaving 80 mL NaCl solution is $${\Lambda _{m2}}$$. The conductivities exhibited by these two cells are same. The relationship between $${\Lambda _{m2}}$$ and $... | [{"identifier": "A", "content": "$${\\Lambda _{m2}}$$ = 2$${\\Lambda _{m1}}$$"}, {"identifier": "B", "content": "$${\\Lambda _{m2}}$$ = $${\\Lambda _{m1}}$$ / 2"}, {"identifier": "C", "content": "$${\\Lambda _{m2}}$$ = $${\\Lambda _{m1}}$$"}, {"identifier": "D", "content": "$${\\Lambda _{m2}}$$ = 4$${\\Lambda _{m1}}$$"... | ["A"] | null | $$\Lambda_{\mathrm{m}_{1}}=\frac{\mathrm{k}_{1} \times 1000}{\mathrm{M}_{1}}=\frac{\mathrm{k} \times 1000}{\frac{10}{0.02}}$$
<br/><br/>
$$
\Lambda_{\mathrm{m}_{2}}=\frac{\mathrm{k}_{2} \times 1000}{\frac{20}{0.08}}
$$
<br/><br/>
It is given that $$\mathrm{k}_{1}=\mathrm{k}_{2}$$
<br/><br/>
$$
\mathrm{k}_{1}=\frac{\Lam... | mcq | jee-main-2022-online-25th-july-evening-shift | 2,003 |
1l6gskady | chemistry | electrochemistry | conductance-and-electrolysis | <p>The amount of charge in $$\mathrm{F}$$ (Faraday) required to obtain one mole of iron from $$\mathrm{Fe}_{3} \mathrm{O}_{4}$$ is ___________. (Nearest Integer)</p> | [] | null | 3 | For $$\mathrm{Fe}_{3} \mathrm{O}_{4}$$,
<br/><br/>
$$
x=\frac{+8}{3}
$$
<br/><br/>
where x is oxidation state of Fe.
<br/><br/>
$$
\mathrm{Fe}_{3} \mathrm{O}_{4}+8 \mathrm{H}^{+}+8 \mathrm{e}^{-} \longrightarrow 3 \mathrm{Fe}+4 \mathrm{H}_{2} \mathrm{O}
$$
<br/><br/>
Charge required $$=\frac{8}{3} \times \mathrm{F}=\fr... | integer | jee-main-2022-online-26th-july-morning-shift | 2,004 |
1l6p7pwiy | chemistry | electrochemistry | conductance-and-electrolysis | <p>Resistance of a conductivity cell (cell constant $$129 \mathrm{~m}^{-1}$$) filled with $$74.5 \,\mathrm{ppm}$$ solution of $$\mathrm{KCl}$$ is $$100 \,\Omega$$ (labelled as solution 1). When the same cell is filled with $$\mathrm{KCl}$$ solution of $$149 \,\mathrm{ppm}$$, the resistance is $$50 \,\Omega$$ (labelled ... | [] | null | 1000 | $$\frac{l}{A}=129 \mathrm{~m}^{-1}$$
<br/><br/>
$$\mathrm{KCl}$$ solution $$1 \Rightarrow 74.5 \,\mathrm{ppm}, \mathrm{R}_{1}=100 \Omega$$
<br/><br/>
$$\mathrm{KCl}$$ solution $$2 \Rightarrow 149 \,\mathrm{ppm}, \mathrm{R}_{2}=50 \Omega$$<br/><br/>
$$
\begin{aligned}
&\text { Here, } \frac{p p m_{1}}{p p m_{2}}=\frac{M... | integer | jee-main-2022-online-29th-july-morning-shift | 2,006 |
1ldo2veix | chemistry | electrochemistry | conductance-and-electrolysis | <p>Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R)</p>
<p>Assertion (A) : An aqueous solution of $$\mathrm{KOH}$$ when used for volumetric analysis, its concentration should be checked before the use.</p>
<p>Reason (R) : On aging, $$\mathrm{KOH}$$ solution absor... | [{"identifier": "A", "content": "(A) is correct but (R) is not correct"}, {"identifier": "B", "content": "(A) is not correct but (R) is correct"}, {"identifier": "C", "content": "Both (A) and (R) are correct but (R) is not the correct explanation of (A)"}, {"identifier": "D", "content": "Both (A) and (R) are correct an... | ["D"] | null | <p>In volumetric analysis, the concentration of a solution is a crucial factor in determining the accuracy of the results. In the case of an aqueous solution of KOH, the concentration can change over time due to the absorption of atmospheric CO<sub>2</sub>. This occurs because KOH is a basic (alkaline) solution and rea... | mcq | jee-main-2023-online-1st-february-evening-shift | 2,007 |
1ldo38i30 | chemistry | electrochemistry | conductance-and-electrolysis | <p>$$1 \times 10^{-5} ~\mathrm{M} ~\mathrm{AgNO}_{3}$$ is added to $$1 \mathrm{~L}$$ of saturated solution of $$\mathrm{AgBr}$$. The conductivity of this solution at $$298 \mathrm{~K}$$ is _____________ $$\times 10^{-8} \mathrm{~S} \mathrm{~m}^{-1}$$.</p>
<p>[Given : $$\mathrm{K}_{\mathrm{SP}}(\mathrm{AgBr})=4.9 \times... | [] | null | 13039 | $$
\begin{aligned}
& \operatorname{AgBr}(\mathrm{S}) \rightleftharpoons \underset{\left(10^{-5}+\mathrm{x}\right)}{\rightleftharpoons \mathrm{Ag}^{+}}(\mathrm{aq})+\mathrm{Br}_{\mathrm{x}}^{-}(\mathrm{aq}) \\\\
& x\left(x+10^{-5}\right)=4.9 \times 10^{-13} \\\\
& x \simeq 4.9 \times 10^{-8} \mathrm{M} \\\\
& \lambda_{\... | integer | jee-main-2023-online-1st-february-evening-shift | 2,008 |
ldo9z26e | chemistry | electrochemistry | conductance-and-electrolysis | The resistivity of a $0.8 \mathrm{M}$ solution of an electrolyte is $5 \times 10^{-3} \Omega~ \mathrm{cm}$. <br/><br/>Its molar conductivity is _________ $\times 10^{4}~ \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$. (Nearest integer) | [] | null | 25 | $$
\begin{aligned}
\text { Molar conductivity } & =\frac{\mathrm{k} \times 1000}{\mathrm{C}} \\\\
& =\frac{\frac{1}{5 \times 10^{-3}} \times 1000}{0.8} \\\\
& =\frac{10^6}{4}=0.25 \times 10^6 \\\\
& =25 \times 10^4 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}
\end{aligned}
$$ | integer | jee-main-2023-online-31st-january-evening-shift | 2,009 |
1ldpp3x3f | chemistry | electrochemistry | conductance-and-electrolysis | <p>Which one of the following statements is correct for electrolysis of brine solution?</p> | [{"identifier": "A", "content": "$$\\mathrm{O}_{2}$$ is formed at cathode"}, {"identifier": "B", "content": "$$\\mathrm{H}_{2}$$ is formed at anode"}, {"identifier": "C", "content": "$$\\mathrm{Cl}_{2}$$ is formed at cathode"}, {"identifier": "D", "content": "$$\\mathrm{OH}^{-}$$ is formed at cathode"}] | ["D"] | null | <b>Cathode :</b> $\mathrm{H}_2 \mathrm{O}_{(l)}+e^{-} \longrightarrow \frac{1}{2} \mathrm{H}_{2(g)}+\mathrm{OH}_{(a q)}^{-}$
<br/><br/><b>Anode</b> $: \mathrm{Cl}_{(a q)}^{-} \longrightarrow \frac{1}{2} \mathrm{Cl}_{2(g)}+e^{-}$ | mcq | jee-main-2023-online-31st-january-morning-shift | 2,010 |
1ldst9rcb | chemistry | electrochemistry | conductance-and-electrolysis | <p>Following figure shows dependence of molar conductance of two electrolytes on concentration. $$\Lambda \mathop m\limits^o $$ is the limiting molar conductivity.</p>
<p><img src="data:image/png;base64,UklGRtYPAABXRUJQVlA4IMoPAACQ8wCdASoAA4oCP4G82WY2LqynITBJQsAwCWlu4XJ0mmNwvj6TxgDAv853eO2l29/w+cKDmbqL//9exI+Wduy9uKl4F... | [] | null | 2 | (A) $\Lambda_{\mathrm{m}}^{\circ}$ for 'A' cannot be obtained by extra polation.
<br/><br/>
(C) At infinite dilution, value of degree of dissociation approaches one.
<br/><br/>
$\therefore \mathrm{A}$ and $\mathrm{C}$ are incorrect | integer | jee-main-2023-online-29th-january-morning-shift | 2,011 |
1ldwun2ko | chemistry | electrochemistry | conductance-and-electrolysis | <p>Choose the correct representation of conductometric titration of benzoic acid vs sodium hydroxide.</p> | [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/1ldxiqesd/495bd5c7-250a-432e-8902-5c8dce5564ad/8e19a0d0-a8b3-11ed-a3d7-5b81580a6b38/file-1ldxiqese.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/1ldxiqesd/495bd5c7-250a-432e-8902-5c8dce5564ad/8e1... | ["C"] | null | $\underset{\text { (wA) }}{\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}}+\underset{\text { (SB) }}{\mathrm{NaOH}} \longrightarrow \underset{\text { (Salt) }}{\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONa}}+\mathrm{H}_2 \mathrm{O}$<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le4ek5vm/62e07637-edce-4be8... | mcq | jee-main-2023-online-24th-january-evening-shift | 2,012 |
1lgq5aavp | chemistry | electrochemistry | conductance-and-electrolysis | <p>A metal surface of $$100 \mathrm{~cm}^{2}$$ area has to be coated with nickel layer of thickness $$0.001 \mathrm{~mm}$$. A current of $$2 \mathrm{~A}$$ was passed through a solution of $$\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}$$ for '$$\mathrm{x}$$' seconds to coat the desired layer. The value of $$\mathrm{x}$$ ... | [] | null | 161 | Using the Faraday's law of electrolysis, we can directly relate the amount of substance deposited (in this case, the nickel layer) with the electric charge passed through the electrolyte.
<br/><br/>
The formula for Faraday's law of electrolysis is:
<br/><br/>
$$W = z \times i \times t$$
<br/><br/>
where W is the amoun... | integer | jee-main-2023-online-13th-april-morning-shift | 2,014 |
1lgvvgt74 | chemistry | electrochemistry | conductance-and-electrolysis | <p>The specific conductance of $$0.0025 ~\mathrm{M}$$ acetic acid is $$5 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}$$ at a certain temperature. The dissociation constant of acetic acid is __________ $$\times ~10^{-7}$$ (Nearest integer)</p>
<p>Consider limiting molar conductivity of $$\mathrm{CH}_{3} \mathrm{COOH}$$ ... | [] | null | 66 | <p>Given that the specific conductance, $k$, is $5 \times 10^{-5} ~S~cm^{-1}$ and the concentration, $C$, is $0.0025~M$, we can find the molar conductivity, $\lambda_m$, as follows:</p>
<p>$$\lambda_m = \frac{k}{C} \times 1000 = \frac{5 \times 10^{-5} \times 10^3}{0.0025} = \frac{5 \times 10^{-2}}{2.5 \times 10^{-3}} =... | integer | jee-main-2023-online-10th-april-evening-shift | 2,015 |
lsapbkde | chemistry | electrochemistry | conductance-and-electrolysis | The amount of electricity in Coulomb required for the oxidation of $1 \mathrm{~mol}$ of $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$ is __________ $\times 10^5 \mathrm{C}$. | [] | null | 2 | $\begin{aligned} & 2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{O}_2+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \\\\ & \frac{\mathrm{W}}{\mathrm{E}}=\frac{\mathrm{Q}}{96500} \\\\ & \text { mole } \times \text { n-factor }=\frac{\mathrm{Q}}{96500}\end{aligned}$
<br/><br/>$\begin{aligned} & 1 \times 2=\frac{Q}{96500} \\\\ & Q=2 ... | integer | jee-main-2024-online-1st-february-evening-shift | 2,016 |
jaoe38c1lsc6c6mx | chemistry | electrochemistry | conductance-and-electrolysis | <p>The mass of silver (Molar mass of $$\mathrm{Ag}: 108 \mathrm{~gmol}^{-1}$$ ) displaced by a quantity of electricity which displaces $$5600 \mathrm{~mL}$$ of $$\mathrm{O}_2$$ at S.T.P. will be ______ g.</p> | [] | null | 108 | <p>First, we need to determine the amount of $$\mathrm{O}_2$$ (oxygen gas) in moles that is displaced by the quantity of electricity mentioned. To do this, we'll use the molar volume of a gas at Standard Temperature and Pressure (S.T.P.), which is approximately $$22.4 \mathrm{~L/mol}$$ (or $$22400 \mathrm{~mL/mol}$$).<... | integer | jee-main-2024-online-27th-january-morning-shift | 2,017 |
jaoe38c1lsda3o5g | chemistry | electrochemistry | conductance-and-electrolysis | The values of conductivity of some materials at $298.15 \mathrm{~K}^{-1} ~\text{in} ~\mathrm{Sm}^{-1}$ are $2.1 \times 10^3$,<br/><br/> $1.0 \times 10^{-16}, 1.2 \times 10,3.91,1.5 \times 10^{-2}, 1 \times 10^{-7}, 1.0 \times 10^3$.<br/><br/> The number of conductors among the materials is _____________. | [] | null | 4 | <p>Conductivity (S m$$^{-1}$$)</p>
<p>$$\left.\begin{array}{l}
2.1 \times 10^3 \\
1.2 \times 10 \\
3.91 \\
1 \times 10^3
\end{array}\right\} \text { conductors at } 298.15 \mathrm{~K}$$</p>
<p>$$1 \times 10^{-16} \text { Insulator at } 298.15 \mathrm{~K}$$</p>
<p>$$\left.\begin{array}{l}
1.5 \times 10^{-2} \\
1 \times ... | integer | jee-main-2024-online-31st-january-evening-shift | 2,018 |
jaoe38c1lse8dput | chemistry | electrochemistry | conductance-and-electrolysis | <p>Number of alkanes obtained on electrolysis of a mixture of $$\mathrm{CH}_3 \mathrm{COONa}$$ and $$\mathrm{C}_2 \mathrm{H}_5 \mathrm{COONa}$$ is ________.</p> | [] | null | 3 | <p>$$\mathrm{CH}_3 \mathrm{COONa} \rightarrow \dot{\mathrm{C}} \mathrm{H}_3$$</p>
<p>$$\mathrm{C}_2 \mathrm{H}_5 \mathrm{COONa} \rightarrow \dot{\mathrm{C}}_2 \mathrm{H}_5$$</p>
<p>$$2 \dot{\mathrm{C}}_2 \mathrm{H}_5 \rightarrow \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3$$</p>
<p>$$2 \dot{\mathrm{CH}}_3 \r... | integer | jee-main-2024-online-31st-january-morning-shift | 2,020 |
jaoe38c1lse8fc6l | chemistry | electrochemistry | conductance-and-electrolysis | <p>One Faraday of electricity liberates $$x \times 10^{-1}$$ gram atom of copper from copper sulphate. $$x$$ is ________.</p> | [] | null | 5 | <p>To find the value of <i>x</i> when one Faraday of electricity liberates <i>x</i> times $10^{-1}$ gram atom of copper from copper sulphate, we need to understand how electricity interacts with copper ions in solution. The key reaction is:</p><p>$$\mathrm{Cu}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}$$</p><p>Thi... | integer | jee-main-2024-online-31st-january-morning-shift | 2,021 |
jaoe38c1lsfk10zs | chemistry | electrochemistry | conductance-and-electrolysis | <p>The mass of zinc produced by the electrolysis of zine sulphate solution with a steady current of $$0.015 \mathrm{~A}$$ for 15 minutes is _________ $$\times 10^{-4} \mathrm{~g}$$.</p>
<p>(Atomic mass of zinc $$=65.4 \mathrm{~amu}$$)</p> | [] | null | 46 | <p>$$\begin{aligned} & \mathrm{Zn}^{+2}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} \\ & \mathrm{W}=\mathrm{Z} \times \mathrm{i} \times \mathrm{t} \\ & =\frac{65.4}{2 \times 96500} \times 0.015 \times 15 \times 60 \\ & =45.75 \times 10^{-4} \mathrm{gm}\end{aligned}$$</p> | integer | jee-main-2024-online-29th-january-morning-shift | 2,022 |
luxzqiw7 | chemistry | electrochemistry | conductance-and-electrolysis | <p>Which out of the following is a correct equation to show change in molar conductivity with respect to concentration for a weak electrolyte, if the symbols carry their usual meaning :</p> | [{"identifier": "A", "content": "$$\\Lambda_{\\mathrm{m}}-\\Lambda_{\\mathrm{m}}^{\\circ}+\\mathrm{AC}^{\\frac{1}{2}}=0$$"}, {"identifier": "B", "content": "$$\\Lambda_{\\mathrm{m}}^2 \\mathrm{C}+\\mathrm{K}_{\\mathrm{a}} \\Lambda_{\\mathrm{m}}^{\\mathrm{o}^2}-\\mathrm{K}_{\\mathrm{a}} \\Lambda_{\\mathrm{m}} \\Lambda_{... | ["D"] | null |
<p>The dissociation of a weak electrolyte ($\mathrm{HA}$) in solution can be represented as:</p>
<p>$ \mathrm{HA}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq}) + \mathrm{A}^{-}(\mathrm{aq}) $</p>
<p>The degree of dissociation ($\alpha$) for a weak electrolyte is given by:</p>
<p>$ \alpha = \frac{\Lamb... | mcq | jee-main-2024-online-9th-april-evening-shift | 2,025 |
lv2erz0j | chemistry | electrochemistry | conductance-and-electrolysis | <p>For a strong electrolyte, a plot of molar conductivity against (concentration) $${ }^{1 / 2}$$ is a straight line, with a negative slope, the correct unit for the slope is</p> | [{"identifier": "A", "content": "$$\\mathrm{S} \\mathrm{cm}^2 \\mathrm{~mol}^{-3 / 2} \\mathrm{~L}$$\n"}, {"identifier": "B", "content": "$$\\mathrm{S} \\mathrm{cm}{ }^2 \\mathrm{~mol}^{-3 / 2} \\mathrm{~L}^{-1 / 2}$$\n"}, {"identifier": "C", "content": "$$\\mathrm{S} \\mathrm{cm}{ }^2 \\mathrm{~mol}^{-1} \\mathrm{~L}^... | ["D"] | null | <p>The molar conductivity ($\Lambda_m$) of a strong electrolyte depends on the concentration ($c$) according to Kohlrausch's law, which can be mathematically expressed as $\Lambda_m = \Lambda_m^0 - k\sqrt{c}$, where $\Lambda_m^0$ is the molar conductivity at infinite dilution, and $k$ is a constant. The graph of molar ... | mcq | jee-main-2024-online-4th-april-evening-shift | 2,027 |
lv3xmbaj | chemistry | electrochemistry | conductance-and-electrolysis | <p>Given below are two statements :</p>
<p>Statement (I) : Fusion of $$\mathrm{MnO}_2$$ with $$\mathrm{KOH}$$ and an oxidising agent gives dark green $$\mathrm{K}_2 \mathrm{MnO}_4$$.</p>
<p>Statement (II) : Manganate ion on electrolytic oxidation in alkaline medium gives permanganate ion.</p>
<p>In the light of the abo... | [{"identifier": "A", "content": "Statement I is false but Statement II is true\n"}, {"identifier": "B", "content": "Statement I is true but Statement II is false\n"}, {"identifier": "C", "content": "Both Statement I and Statement II are false\n"}, {"identifier": "D", "content": "Both Statement I and Statement II are tr... | ["D"] | null | <p>Statement I is indeed true. When $$\mathrm{MnO}_2$$ (manganese dioxide) is fused with $$\mathrm{KOH}$$ (potassium hydroxide) and an oxidising agent such as $$\mathrm{KNO}_3$$ (potassium nitrate), it forms dark green potassium manganate ($$\mathrm{K}_2\mathrm{MnO}_4$$). The reaction can be represented as follows:</p>... | mcq | jee-main-2024-online-8th-april-evening-shift | 2,028 |
lv7v49ug | chemistry | electrochemistry | conductance-and-electrolysis | <p>Molar ionic conductivities of divalent cation and anion are $$57 \mathrm{~S~cm}^2 \mathrm{~mol}^{-1}$$ and $$73 \mathrm{~S~cm}^2 \mathrm{~mol}^{-1}$$ respectively. The molar conductivity of solution of an electrolyte with the above cation and anion will be:</p> | [{"identifier": "A", "content": "$$187 \\mathrm{~S} \\mathrm{~cm}^2 \\mathrm{~mol}^{-1}$$\n"}, {"identifier": "B", "content": "$$260 \\mathrm{~S} \\mathrm{~cm}^2 \\mathrm{~mol}^{-1}$$\n"}, {"identifier": "C", "content": "$$65 \\mathrm{~S} \\mathrm{~cm}^2 \\mathrm{~mol}^{-1}$$\n"}, {"identifier": "D", "content": "$$130 ... | ["D"] | null | <p>The compound with divalent cation $$\left(\mathrm{A}^{2+}\right)$$ and anion $$\left(B^{2-}\right)$$ will be $$A B$$.</p>
<p>Molar conductivity of its solution will be</p>
<p>$$57+73=130 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$$</p> | mcq | jee-main-2024-online-5th-april-morning-shift | 2,029 |
lv9s2pv4 | chemistry | electrochemistry | conductance-and-electrolysis | <p>The quantity of silver deposited when one coulomb charge is passed through $$\mathrm{AgNO}_3$$ solution :</p> | [{"identifier": "A", "content": "$$0.1 \\mathrm{~g}$$ atom of silver\n"}, {"identifier": "B", "content": "1 chemical equivalent of silver\n"}, {"identifier": "C", "content": "$$1 \\mathrm{~g}$$ of silver\n"}, {"identifier": "D", "content": "1 electrochemical equivalent of silver"}] | ["D"] | null | <p>To determine the quantity of silver deposited when one coulomb of charge is passed through $$\mathrm{AgNO}_3$$ solution, we need to refer to Faraday's laws of electrolysis, especially to the concept of the electrochemical equivalent. The electrochemical equivalent (ECE) of a substance is the amount of the substance ... | mcq | jee-main-2024-online-5th-april-evening-shift | 2,030 |
lvc58eru | chemistry | electrochemistry | conductance-and-electrolysis | <p>A conductivity cell with two electrodes (dark side) are half filled with infinitely dilute aqueous solution of a weak electrolyte. If volume is doubled by adding more water at constant temperature, the molar conductivity of the cell will -</p>
<p><img src="data:image/png;base64,UklGRlgrAABXRUJQVlA4IEwrAACw2wGdASoAA9... | [{"identifier": "A", "content": "depend upon type of electrolyte"}, {"identifier": "B", "content": "increase sharply"}, {"identifier": "C", "content": "decrease sharply"}, {"identifier": "D", "content": "remain same or can not be measured accurately"}] | ["D"] | null | <p>The molar conductivity, denoted as $$\wedge_m$$, is calculated using the formula:</p>
<p>$$\wedge_m = \frac{\mathrm{K} \times 1000}{\mathrm{M}}$$</p>
<p>When the volume of the solution is doubled by adding more water at a constant temperature, the conductivity, $$\mathrm{K}$$, will be reduced to half, while the mo... | mcq | jee-main-2024-online-6th-april-morning-shift | 2,031 |
yTjIAY8xgsNZVa9N | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | For the following cell with hydrogen electrodes at two different pressure p<sub>1</sub> and p<sub>2</sub>. What will be the emf for the given cell :
<br/><br/>$$\eqalign{
& Pt({H_2})|{H^ + }(aq)|Pt({H_2}) \cr
& \,\,\,\,\,{p_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,1M\,\,\,\,\,\,\,\,\,\,\,\,{p_2} \cr} $$ | [{"identifier": "A", "content": "$${{RT} \\over F}{\\log _e}{{{P_1}} \\over {{P_2}}}$$ "}, {"identifier": "B", "content": "$${{RT} \\over 2F}{\\log _e}{{{P_1}} \\over {{P_2}}}$$"}, {"identifier": "C", "content": "$${{RT} \\over F}{\\log _e}{{{P_2}} \\over {{P_1}}}$$"}, {"identifier": "D", "content": "none of these"}] | ["B"] | null | Oxidation half cell : -
<br><br>$${H_2}\left( g \right)\buildrel \, \over
\longrightarrow 2{H^ + }\left( {1M} \right) + 2{e^ - }\,\,\,...{P_1}$$
<br><br>Reduction half cell
<br><br>$$2{H^ + }\left( {1M} \right) + 2{e^ - }\buildrel \, \over
\longrightarrow {H_2}\left( g \right)\,\,\,...{P_2}$$
<br><br>The net cell re... | mcq | aieee-2002 | 2,032 |
xp34ovlDlQwqPFEa | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | EMF of a cell in terms of reduction potential of its left and right electrodes is : | [{"identifier": "A", "content": "E = E<sub>left</sub> - E<sub>right</sub>"}, {"identifier": "B", "content": "E = E<sub>left</sub> + E<sub>right</sub>"}, {"identifier": "C", "content": "E = E<sub>right</sub> - E<sub>left</sub>"}, {"identifier": "D", "content": "E = -(E<sub>right</sub> + E<sub>left</sub>)"}] | ["C"] | null | $${E_{cell}} = \,\,$$ Reduction potential of cathode (right)
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$-$$ Reduction potential of anode (left)
<br><br>$$ = {E_{right}} - {E_{left}}.$$ | mcq | aieee-2002 | 2,033 |
PGFWnz6ubDo31DLQ | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | When the sample of copper with zinc impurity is to be purified by electrolysis, the appropriate
electrodes are : | [{"identifier": "A", "content": "cathode = pure zinc, anode = pure copper"}, {"identifier": "B", "content": "cathode = impure sample, anode = pure copper"}, {"identifier": "C", "content": "cathode = impure zinc, anode = impure sample"}, {"identifier": "D", "content": "cathode = pure copper, anode = impure sample"}] | ["D"] | null | Pure metal always deposits at cathode. | mcq | aieee-2002 | 2,035 |
lkr1y78y | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | For a cell given below
<br/><br/><img src="data:image/png;base64,UklGRpYPAABXRUJQVlA4IIoPAACwYwCdASoAA5wAPm00mEekIyKhJjXp4IANiWlu/B25GutQx/1T/zXaz/qv7v/Z+uK4w/2XkgvS/fL+R67/6rvj4AX5H/Xf9tvLIAvq538WpH4h9gDzA/2nho/bv+d7An83/ynox6JPsPgVEnAQAdr/XnTtf686dr/XnS651jSgh9efgWHpu7X60Dtf686dr/ICywgjJL3dR8gfYuVALxJqLrMQOoh9Sep75qo... | [{"identifier": "A", "content": "$x+2 y$\n"}, {"identifier": "B", "content": "$2 x+y$"}, {"identifier": "C", "content": "$y-x$"}, {"identifier": "D", "content": "$y-2 x$"}] | ["C"] | null | <p>The cell notation indicates that the left side (Ag|Ag+) is the anode (oxidation occurs), and the right side (Cu²⁺|Cu) is the cathode (reduction occurs). In a galvanic (voltaic) cell, electrons flow from anode to cathode.</p>
<p>The standard cell potential (E° cell) is the difference in potential between the cathode ... | mcq | aieee-2002 | 2,036 |
SoHEmoxCiUyAtlFp | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | For the redox reaction Zn(s) + Cu<sup>2+</sup>(0.1 M) $$\to$$ Zn<sup>2+</sup>(1M) + Cu(s) taking place in a cell, $$E_{cell}^o$$ is 1.10 volt. E<sub>cell</sub> for the cell will be ($$2.303{{RT} \over F}$$ = 0.0591) | [{"identifier": "A", "content": "1.80 volt"}, {"identifier": "B", "content": "1.07 volt"}, {"identifier": "C", "content": "0.82 volt"}, {"identifier": "D", "content": "2.14 volt"}] | ["B"] | null | $${E_{cell}} = {E^ \circ }_{cell} + {{0.059} \over n}\log {{\left[ {C{u^{ + 2}}} \right]} \over {\left[ {Z{n^{ + 2}}} \right]}}$$
<br><br>$$ = 1.10 + {{0.059} \over 2}\log \left[ {0.1} \right]$$
<br><br>$$ = 1.10 - 0.0295$$
<br><br>$$ = 1.07V$$ | mcq | aieee-2003 | 2,038 |
rXhD04mECCIdYnJU | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | Standard reduction electrode potentials of three metals A,B&C are respectively +0.5 V, -3.0 V & -1.2 V. The
reducing, powers of these metals are | [{"identifier": "A", "content": "A > B > C"}, {"identifier": "B", "content": "C > B > A"}, {"identifier": "C", "content": "A > C > B"}, {"identifier": "D", "content": "B > C > A"}] | ["D"] | null | $$\matrix{
A & B & C \cr
{ + 0.5C} & { - 3.0V} & { - 1.2V} \cr
} $$
<br><br><b>NOTE :</b> The higher the negative value of reduction potential, the more is the reducing power.
<br><br>Hence $$B > C > A.$$ | mcq | aieee-2003 | 2,039 |
JMgJztXuwIK7ythF | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | Consider the following E<sup>o</sup> values <br/>
<br>$$E_{F{e^{3 + }}/F{e^{2 + }}}^o$$ = 0.77 V;<br/>
<br>$$E_{S{n^{2 + }}/S{n}}^o$$ = -0.14 V<br/>
<br>Under standard conditions the potential for the reaction <br/>
<br>Sn(s) + 2Fe<sup>3+</sup>(aq) $$\to$$ 2Fe<sup>2+</sup>(aq) + Sn<sup>2+</sup>(aq) is :</br></br></br><... | [{"identifier": "A", "content": "1.68 V"}, {"identifier": "B", "content": "0.63 V "}, {"identifier": "C", "content": "0.91 V "}, {"identifier": "D", "content": "1.40 V "}] | ["C"] | null | $$F{e^{3 + }} + {e^ - } \to F{e^{2 + }}\Delta {G^ \circ }$$
<br><br>$$\,\,\,\,\,\,\,\,\,\, = - 1 \times F \times 0.77$$
<br><br>$$S{n^{2 + }} + 2{e^ - } \to Sn\left( s \right)\Delta {G^ \circ }$$
<br><br>$$ = - 2 \times F\left( { - 0.14} \right)$$
<br><br>$$Sn\left( s \right) + 2F{e^{3 + }}\left( {aq} \right)$$
<br>... | mcq | aieee-2004 | 2,040 |
5M8pT6y7VOoRf7rT | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | The standard e.m.f of a cell, involving one electron change is found to be 0.591 V at 25<sup>o</sup>C.
The equilibrium constant of the reaction is (F = 96,500 C mol<sup>-1</sup>: R = 8.314 JK<sup>-1</sup> mol<sup>-1</sup>) | [{"identifier": "A", "content": "1.0 $$\\times$$ 10<sup>1</sup>"}, {"identifier": "B", "content": "1.0 $$\\times$$ 10<sup>30</sup>"}, {"identifier": "C", "content": "1.0 $$\\times$$ 10<sup>10</sup>"}, {"identifier": "D", "content": "1.0 $$\\times$$ 10<sup>5</sup>"}] | ["C"] | null | $$E_{cell}^o = E_{cell}^o - {{0.059} \over n}\log \,{K_c}$$
<br><br>or $$\,\,\,\,\,\,$$ $$0 = 0.591 - {{0.0591} \over 1}\log {K_c}$$
<br><br>or $$\,\,\,\,\,\,$$ $$\log \,{K_c} = {{0.591} \over {0.0591}} = 10$$
<br><br>or $$\,\,\,\,\,\,$$ $${K_c} = 1 \times {10^{10}}$$ | mcq | aieee-2004 | 2,041 |
xjjIlFLTU5Mx54S4 | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | In a cell that utilises the reaction Zn(s) + 2H<sup>+</sup>
(aq) $$\to$$ Zn2+(aq) + H2(g) addition of H<sub>2</sub>SO<sub>4</sub> to cathode compartment, will | [{"identifier": "A", "content": "lower the E and shift equilibrium to the left "}, {"identifier": "B", "content": "increases the E and shift equilibrium to the left"}, {"identifier": "C", "content": "increase the E and shift equilibrium to the right "}, {"identifier": "D", "content": "Lower the E and shift equilibrium ... | ["C"] | null | $$Zn\left( s \right) + 2{H^ + } + \left( {aq} \right)\,\,\rightleftharpoons\,\,Z{n^{2 + }}\left( {aq} \right) + {H_2}\left( g \right)$$
<br><br>$${E_{cell}} = E_{cell}^ \circ - {{0.059} \over 2}\log {{\left[ {Z{n^{2 + }}} \right]\left[ {{H_2}} \right]} \over {{{\left[ {{H^ + }} \right]}^2}}}$$
<br><br>Addition of $${... | mcq | aieee-2004 | 2,042 |
0eHrZRrqDZ2WoZfL | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | The $$E_{{M^{3 + }}/{M^{2 + }}}^o$$ values for Cr, Mn, Fe and Co are – 0.41, +1.57, + 0.77 and +1.97 V
respectively. For which one of these metals the change in oxidation state form +2 to +3 is
easiest? | [{"identifier": "A", "content": "Fe"}, {"identifier": "B", "content": "Mn"}, {"identifier": "C", "content": "Cr"}, {"identifier": "D", "content": "Co"}] | ["C"] | null | The given values show that $$Cr$$ has maximum oxidation potential, therefore its oxidation will be easiest. (Change the sign to get the oxidation values) | mcq | aieee-2004 | 2,043 |
lapmeFUNf98aZaDl | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | For a spontaneous reaction the ∆G , equilibrium constant (K) and $$E_{cell}^o$$ will be respectively | [{"identifier": "A", "content": "-ve, >1, +ve"}, {"identifier": "B", "content": "+ve, >1, -ve "}, {"identifier": "C", "content": "-ve, <1, -ve "}, {"identifier": "D", "content": "-ve, >1, -ve "}] | ["A"] | null | <b>NOTE :</b> For spontaneous reaction $$\Delta G$$ should be negative. Equilibrium constant should be more than one
<br><br>$$\left( {\Delta G = - 2.303\,RT\,\log {K_c},\,\,} \right.$$
<br><br>If $${K_c} = 1\,\,$$ then $$\,\,\,\Delta G = 0;\,\,$$
<br><br>If $${K_c} < 1$$ then $$\left. {\Delta G = + ve} \right).\... | mcq | aieee-2005 | 2,044 |
D6we0EIVC1ykVqoS | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | The cell, Zn | Zn<sup>2+</sup> (1M) || Cu<sup>2+</sup> (1M) | Cu($$E_{cell}^o$$ = 1.10V) was allowed to be completely discharged at 298 K. The relative concentration of Zn<sup>2+</sup> to Cu<sup>2+</sup> $$\left[ {{{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}} \right]$$ is | [{"identifier": "A", "content": "antilog (24.08) "}, {"identifier": "B", "content": "37.3"}, {"identifier": "C", "content": "10<sup>37.3</sup>"}, {"identifier": "D", "content": "9.65 $$\\times$$ 10<sup>4</sup>\n "}] | ["C"] | null | $${E_{cell}} = 0;\,\,$$ when cell is completely discharged.
<br><br>$${E_{cell}} = {E^ \circ }_{cell} - {{0.059} \over 2}\log \left( {{{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}} \right)$$
<br><br>or$$\,\,\,\,0 - 1.1 - {{0.059} \over 2}\log \left( {{{\left[ {Z{n^{2 + }}} \right]} \over {\lef... | mcq | aieee-2007 | 2,046 |
uSB2XOq3OHyFIxIT | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | Given $$E_{C{r^{3 + }}/Cr}^o$$ = -0.72 V; $$E_{Fe^{2+}/Fe}^o$$ = -0.42V, The potential for the cell Cr | Cr<sup>3+</sup> (0.1M) || Fe<sup>2+</sup> (0.01 M) | Fe is | [{"identifier": "A", "content": "0.26 V"}, {"identifier": "B", "content": "0.399 V"}, {"identifier": "C", "content": "\u22120.339 V"}, {"identifier": "D", "content": "\u22120.26 V"}] | ["A"] | null | From the given representation of the cell, $${E_{cell}}$$ can be found as follows.
<br><br>$${E_{cell}} = E_{F{e^{2 + }}/Fe}^ \circ - E_{C{r^{3 + }}/Cr}^ \circ - {{0.059} \over 6}\log {{{{\left[ {C{r^{3 + }}} \right]}^2}} \over {{{\left[ {F{e^{2 + }}} \right]}^3}}}$$ $$\left[ {} \right.$$ Nernst - Equ. $$\left. {} \... | mcq | aieee-2008 | 2,047 |
Z1oO3ob5geBBpWEe | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | Given : $$E_{F{e^{3 + }}/Fe}^o$$ = -0.036V; $$E_{F{e^{2 + }}/Fe}^o$$ = -0.439 V<br/>
The value of standard electrode potential for the change, <br/>
Fe<sup>3+</sup> (aq) + e<sup>-</sup> $$\to$$ Fe<sup>2+</sup> (aq) will be | [{"identifier": "A", "content": "-0.072 V"}, {"identifier": "B", "content": "0.385 V"}, {"identifier": "C", "content": "0.770 V "}, {"identifier": "D", "content": "0.270"}] | ["C"] | null | Given
<br><br>$$F{e^{3 + }} + 3{e^ - } \to Fe,\,\,{E^ \circ }_{F{e^{3 + }}/Fe}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\, = - 0.036V\,\,\,...\left( i \right)$$
<br><br>$$F{e^{2 + }} + 2{e^ - } \to Fe,\,\,{E^ \circ }_{F{e^{2 + }}/Fe}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\, = - 0.439V\,\,\,...\left( {ii} \right)$$
<br><br>we have ... | mcq | aieee-2009 | 2,048 |
ZBWJikte7Qop5gW2 | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | The correct order of $$E_{{M^{2 + }}/M}^o$$ values with negative sign for the four successive elements Cr, Mn, Fe
and Co is : | [{"identifier": "A", "content": "Mn > Cr > Fe > Co"}, {"identifier": "B", "content": "Cr > Fe > Mn > Co"}, {"identifier": "C", "content": "Fe > Mn > Cr > Co"}, {"identifier": "D", "content": "Cr > Mn > Fe > Co"}] | ["A"] | null | The value of $$E_{{M^{2 + }}/M}^ \circ $$ for given metal ions are
<br><br>$$E_{M{n^{2 + }}/Mn}^ \circ = - 1.18\,V,\,\,$$
<br><br>$$E_{C{r^{2 + }}/Cr}^ \circ = - 0.9V,$$
<br><br>$$E_{F{e^{2 + }}/Fe}^ \circ = - 0.44\,V$$ and
<br><br>$$E_{C{o^{2 + }}/Co}^ \circ = - 0.28\,V.$$
<p>So when we consider these valu... | mcq | aieee-2010 | 2,049 |
LWCyjCSw4AjHEKIr | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | The reduction potential of hydrogen half cell will be negative if : | [{"identifier": "A", "content": "p(H<sub>2</sub>) = 1 atm and [H<sup>+</sup>] = 1.0 M"}, {"identifier": "B", "content": "p(H<sub>2</sub>) = 1 atm and [H<sup>+</sup>] = 2.0 M"}, {"identifier": "C", "content": "p(H<sub>2</sub>) = 2 atm and [H<sup>+</sup>] =1.0 M"}, {"identifier": "D", "content": "p(H<sub>2</sub>) = 2 atm... | ["C"] | null | $${H^ + } + {e^ - }\buildrel \, \over
\longrightarrow {1 \over 2}{H_2};$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ E = {E^ \circ } - {{0.059} \over 1}\log {{P_{{H_2}}^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}} \over {\left[ {{H^ + }} \right]}}$$
<br><br>Now if... | mcq | aieee-2011 | 2,051 |
4fZCIsWjm9yNWayS | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | The standard reduction potentials for Zn<sup>2+</sup>/ Zn, Ni<sup>2+</sup>/ Ni, and Fe<sup>2+</sup>/ Fe are –0.76, –0.23 and –0.44 V respectively. The reaction <br/><br/>X + Y<sup>2+</sup> $$\to$$ X<sup>2+</sup> + Y will be spontaneous when : | [{"identifier": "A", "content": "X = Ni, Y = Fe"}, {"identifier": "B", "content": "X = Ni, Y = Zn"}, {"identifier": "C", "content": "X = Fe, Y = Zn"}, {"identifier": "D", "content": "X = Zn, Y = Ni"}] | ["D"] | null | For a spontaneous reaction $$\Delta G$$ must be $$-ve$$
<br><br>Since $$\Delta G = - nF{E^ \circ }$$
<br><br>Hence for $$\Delta G$$ to be $$-ve$$ $$\Delta {E^ \circ }$$ has to be positive.
<br><br>Which is possible when $$X = Zn,Y = Ni$$
<br><br>$$Zn + N{i^{ + + }}\buildrel \, \over
\longrightarrow Z{n^{ + + }} + ... | mcq | aieee-2012 | 2,052 |
l3DCdJawiACbkXOv | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | Given<br/>
<br>$$E_{C{r^{2 + }}/Cr}^o$$ = -0.74 V; $$E_{MnO_4^ - /M{n^{2 + }}}^o$$ = 1.51 V<br/>
<br>$$E_{C{r_2}O_7^{2 - }/C{r^{3 + }}}^o$$ = 1.33 V; $$E_{Cl/C{l^ - }}^o$$ = 1.36 V<br/>
<br>Based on the data given above, strongest oxidising agent will be : </br></br></br> | [{"identifier": "A", "content": "Cr<sup>3+</sup>"}, {"identifier": "B", "content": "Mn<sup>2+</sup>"}, {"identifier": "C", "content": "$$MnO_4^ - $$ "}, {"identifier": "D", "content": "Cl<sup>-</sup>"}] | ["C"] | null | <p>In electrochemistry, the strongest oxidizing agent will be the one with the highest standard electrode potential (E°), because a higher E° value means a greater tendency to gain electrons, i.e., get reduced. An oxidizing agent gains electrons and in doing so, oxidizes another species.</p>
<p>From the provided data, ... | mcq | jee-main-2013-offline | 2,053 |
1xD2pQDXx8XvRJBw | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | Given below are the half-cell reactions:<br/>
<br>Mn<sup>2+</sup> + 2e<sup>-</sup> $$\to$$ Mn; E<sup>o</sup> = -1.18 V<br/>
<br>2(Mn<sup>3+</sup> + e<sup>-</sup> $$\to$$ Mn<sup>2+</sup>); E<sup>o</sup> = +1.51 V<br/>
<br>The E<sup>o</sup> for 3Mn<sup>2+</sup> $$\to$$ Mn + 2Mn<sup>3+</sup> will be :</br></br></br> | [{"identifier": "A", "content": "\u2013 0.33 V; the reaction will not occur"}, {"identifier": "B", "content": "\u2013 0.33 V; the reaction will occur"}, {"identifier": "C", "content": "\u2013 2.69 V; the reaction will not occur "}, {"identifier": "D", "content": "\u2013 2.69 V; the reaction will occur"}] | ["C"] | null | (a)$$\,\,\,\,\,$$ $$M{n^{2 + }} + 2{e^ - } \to Mn;\,\,{E^ \circ } = - 1.18V;\,...\left( i \right)$$
<br><br>(b) $$\,\,\,\,\,$$ $$M{n^{3 + }} + e \to M{n^{2 + }};\,\,{E^ \circ } = - 1.51V;\,...\left( {ii} \right)$$
<br><br>Now multiplying equation $$(ii)$$ by two and subtracting from equation $$(i)$$
<br><br>$$3M{n^{2... | mcq | jee-main-2014-offline | 2,054 |
6FwE436nRDqSrw7iaxjfz | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | What will occur if a block of copper metal is dropped into a beaker containing a
solution of 1M ZnSO<sub>4</sub>? | [{"identifier": "A", "content": "The copper metal will dissolve and zinc metal will be deposited."}, {"identifier": "B", "content": "The copper metal will dissolve with evolution of hydrogen gas."}, {"identifier": "C", "content": "The copper metal will dissolve with evolution of oxygen gas."}, {"identifier": "D", "cont... | ["D"] | null | In the reaction of copper with zinc ions, zinc is more active (easily oxidised) than copper; thus, no reaction will take place. | mcq | jee-main-2016-online-9th-april-morning-slot | 2,055 |
glD0k4y2Jz1hSVqm | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | Given<br/>
$$E_{C{l_2}/C{l^ - }}^o$$ = 1.36 V, $$E_{C{r^{3 + }}/Cr}^o$$ = - 0.74 V<br/>
$$E_{C{r_2}{O_7}^{2 - }/C{r^{3 + }}}^o$$ = 1.33 V, $$E_{Mn{O_4}^ - /Mn ^{2+}}^o$$ = 1.51 V<br/>
Among the following, the strongest reducing agent is : | [{"identifier": "A", "content": "Mn<sup>2+</sup>"}, {"identifier": "B", "content": "Cr<sup>3+</sup>\n"}, {"identifier": "C", "content": "Cl<sup>\u2013</sup>"}, {"identifier": "D", "content": "Cr"}] | ["D"] | null | $$E_{C{l_2}/C{l^ - }}^o$$ = 1.36 V, $$E_{C{r^{3 + }}/Cr}^o$$ = - 0.74 V<br>
$$E_{C{r_2}{O_7}^{2 - }/C{r^{3 + }}}^o$$ = 1.33 V, $$E_{Mn{O_4}^ - /Mn ^{2+}}^o$$ = 1.51 V
<br><br>More negative the E° value of the species, more stronger is the reducing agent. Since Cr<sup>3+</sup> is
having least reducing potential, so Cr w... | mcq | jee-main-2017-offline | 2,056 |
3wm6goYMCAXUXRu8cIpOQ | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | What is the standard reduction potential (E<sup>o</sup>) for Fe<sup>3+</sup> $$ \to $$ Fe ?
<br/>Given that :
<br/>Fe<sup>2+</sup> + 2e<sup>$$-$$</sup> $$ \to $$ Fe; $$E_{F{e^{2 + }}/Fe}^o$$ = $$-$$0.47 V
<br/>Fe<sup>3+</sup> + e<sup>$$-$$</sup> $$ \to $$ Fe<sup>2+</sup>; $$E_{F{e^{3 + }}/F{e^{2 + }}}^o$$ = +0.77 V | [{"identifier": "A", "content": "$$-$$ 0.057 V"}, {"identifier": "B", "content": "+ 0.057 V"}, {"identifier": "C", "content": "+ 0.30 V"}, {"identifier": "D", "content": "$$-$$ 0.30 V"}] | ["A"] | null | For the given reaction :
<br><br>Fe<sup>3+</sup> + e<sup>$$-$$</sup> $$ \to $$ Fe<sup>2+</sup>; $$E_{F{e^{3 + }}/F{e^{2 + }}}^o$$ = +0.77 V
<br><br>$$\Delta $$G<sup>o</sup> = -nFE<sup>o</sup>
<br><br>$$ \Rightarrow $$ $$\Delta G_1^o = - \left( 1 \right)F\left( {0.77} \right)$$ = -0.77F
<br><br>For the following reac... | mcq | jee-main-2017-online-8th-april-morning-slot | 2,057 |
c5du9JDLuSgRuPLqXGgPG | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | Consider the following standard electrode potentials (E<sup>o</sup> in volts) in aqueous solution :
<br/><br/><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{font-family:Arial, sans-serif;font-size:14px;padding:10px 5px;border-style:solid;border-width:1px;overflow:hidden;word-break:norm... | [{"identifier": "A", "content": "T1<sup>+</sup> is more stable than A1<sup>3+</sup>"}, {"identifier": "B", "content": "A1<sup>+</sup> is more stable than A1<sup>3+</sup> "}, {"identifier": "C", "content": "T1 <sup>+</sup> is more stable than A1<sup>+</sup>"}, {"identifier": "D", "content": "T1<sup>3+</sup> is more stab... | ["C"] | null | <p>The standard electrode potential E<sup>0</sup> for M<sup>+</sup>/M become negative for Tl which indicates that Tl<sup>+</sup> is more stable than Al<sup>+</sup>.</p>
<p>This can also be be explained by inert pair effect. The atoms of this group have an outer electronic configuration of s<sup>2</sup>p<sup>1</sup>. Th... | mcq | jee-main-2017-online-8th-april-morning-slot | 2,058 |
rJ5MeXS6Q6ilnTsB6bQyr | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | To find the standard potential of M<sup>3+</sup>/M electrode,the following cell is constituted : Pt/M/M<sup>3+</sup>(0.001 mol L<sup>−1</sup> )/Ag<sup>+</sup>(0.01 mol L<sup>−1</sup> )/Ag
<br/><br/>The emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction M3<sup>+ </sup> + 3e<sup>... | [{"identifier": "A", "content": "0.38 Volt"}, {"identifier": "B", "content": "0.32 Volt "}, {"identifier": "C", "content": "1.28 Volt"}, {"identifier": "D", "content": "0.66 Volt"}] | ["B"] | null | <p>According to Nernst equation, for the cell reaction</p>
<p>M(s) + 3Ag<sup>+</sup>(aq) $$\to$$ M<sup>3+</sup>(aq) + 3Ag(s)</p>
<p>$${E_{cell}} = E_{cell}^o - {{0.059} \over n}\log {{[{M^{3 + }}]} \over {{{[A{g^ + }]}^3}}}$$</p>
<p>Substituting the values, we get</p>
<p>$$0.421 = E_{cell}^o - {{0.059} \over 3}\log {{0... | mcq | jee-main-2017-online-9th-april-morning-slot | 2,059 |
Fu8Gl8B3LaApCw8cUnlNU | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the reaction
<br/><br>Zn(s) + Cu<sup>2+</sup> (aq) $$\rightleftharpoons$$ Zn<sup>2+</sup>(aq) + Cu(s)
<br/><br>at 300 K is approximately,
<br/><br>(R = 8 JK<sup>$$-$$1</sup>mol<sup>$$-$$1</sup>, F = 96000 C mol<sup>$$-$$... | [{"identifier": "A", "content": "e<sup>$$-$$80</sup>"}, {"identifier": "B", "content": "e<sup>$$-$$160</sup>"}, {"identifier": "C", "content": "e<sup>320</sup>"}, {"identifier": "D", "content": "e<sup>160</sup>"}] | ["D"] | null | $$\Delta $$G<sup>o</sup> = $$-$$ RT lnk = $$-$$nFE<sup>o</sup><sub>cell</sub>
<br><br>lnk = $${{n \times F \times {E^o}} \over {R \times T}} = {{2 \times 96000 \times 2} \over {8 \times 300}}$$
<br><br>lnk = 160
<br><br>k = e<sup>160</sup> | mcq | jee-main-2019-online-9th-january-evening-slot | 2,060 |
XSCjtZy1GFblWZclEF6pD | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | The standard Gibbs energy for the given cell
reaction in kJ mol<sup>–1</sup> at 298 K is :<br/>
<br>Zn(s) + Cu<sup>2+</sup> (aq) $$ \to $$ Zn<sup>2+</sup> (aq) + Cu (s),<br/>
<br>E° = 2 V at 298 K<br/>
<br>(Faraday's constant, F = 96000 C mol<sup>–1</sup>)</br></br></br> | [{"identifier": "A", "content": "384"}, {"identifier": "B", "content": "\u2013192"}, {"identifier": "C", "content": "\u2013384"}, {"identifier": "D", "content": "192"}] | ["C"] | null | Here Zn is losing two electrons and Cu is gaining two electrons. So only two electrons are involved in the reaction.
<br><br>$$ \therefore $$ n = 2
<br><br>$$\Delta $$G<sup>o</sup> = - nFE<sup>o</sup>
<br><br>= -2 $$ \times $$ 96000 $$ \times $$ 2
<br><br>= -384 kJ/mol | mcq | jee-main-2019-online-9th-april-morning-slot | 2,062 |
XgMMmhs2mwwqfhqjxPgjn | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | Calculate the standard cell potential in (V) of the
cell in which following reaction takes place :
<br/><br/>Fe<sup>2+</sup>(aq) + Ag<sup>+</sup>(aq) $$ \to $$ Fe<sup>3+</sup>(aq) + Ag (s)
<br/><br/>Given that<br/>
<br>$$E_{A{g^ + }/Ag}^o = xV$$<br/>
<br>$$E_{Fe^{2+ }/Fe}^o = yV$$<br/>
<br>$$E_{Fe^{3+ }/Fe}^o = zV$$</b... | [{"identifier": "A", "content": "x + 2y - 3z"}, {"identifier": "B", "content": "x - z"}, {"identifier": "C", "content": "x - y"}, {"identifier": "D", "content": "x + y - z"}] | ["A"] | null | Standard emf,
<br><br>$${E^0} = E_{A{g^ + }|Ag}^0 - E_{F{e^{3 + }}|F{e^{2 + }}}^0$$ ..............(1)
<br><br>Given $$E_{Fe^{2+ }/Fe}^o = yV$$
<br><br>$$ \therefore $$ Fe<sup>2+</sup> + 2e<sup>-</sup> $$ \to $$ Fe ..........(2)
<br><br>$${E^0}$$ = y and $$\Delta {G^0}$$ = -2Fy
<br><br>Also given $$E_{Fe^{3+ }/Fe}^o = ... | mcq | jee-main-2019-online-8th-april-evening-slot | 2,063 |
8A5KDTJeCT1IRGt6pCPIf | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | Given that $${E^\Theta }_{{O_2}/{H_2}O} = 1.23\,V$$ ;<br/>
<br>$${E^\Theta }_{{S_2}O_8^{2 - }/SO_4^{2 - }} = 2.05\,V$$<br/>
<br>$${E^\Theta }_{B{r_2}/B{r^ - }} = 1.09\,V$$<br/>
<br>$${E^\Theta }_{A{u^{3 + }}/Au} = 1.4\,V$$<br/>
<br/>The strongest oxidizing agent is :</br></br></br> | [{"identifier": "A", "content": "O<sub>2</sub>"}, {"identifier": "B", "content": "Au<sup>3+</sup>"}, {"identifier": "C", "content": "Br<sub>2</sub>"}, {"identifier": "D", "content": "$${S_2}O_8^{2 - }$$"}] | ["D"] | null | Which electrode have higher value of standard reduction potential (SRP), that electrode will be strongest oxidizing agent.
<br><br>Tendency to gain electrone is called standard reduction potential. When tendency to gain electron is more then that electrode will have more oxidizing power.
<br><br>Here given SRP's are :
... | mcq | jee-main-2019-online-8th-april-morning-slot | 2,064 |
SmPXPFiVHnfPsRt3ygw24 | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | Given the equilibrium constant:
<br/><br/>K<sub>C</sub> of the reaction :
<br/><br/>Cu(s) + 2Ag<sup>+</sup> (aq) $$ \to $$ Cu<sup>2+</sup> (aq) + 2Ag(s) is <br/><br/>10 $$ \times $$ 10<sup>15</sup>, calculate the E$$_{cell}^0$$ of this reaciton at 298 K <br/><br/>[2.303 $${{RT} \over F}$$ at 298 K = 0.059V] | [{"identifier": "A", "content": "0.4736 mV"}, {"identifier": "B", "content": "0.04736 V"}, {"identifier": "C", "content": "0.4736 V"}, {"identifier": "D", "content": "0.04736 mV"}] | ["C"] | null | We know,
<br><br>$$\Delta $$G<sup>o</sup> = -RTln(K<sub>C</sub>) ....(1)
<br><br>Also $$\Delta $$G<sup>o</sup> = -nF$$E_{cell}^o$$ ....(2)
<br><br>$$ \therefore $$ -nF$$E_{cell}^o$$ = -RTln(K<sub>C</sub>)
<br><br>$$ \Rightarrow $$ $$E_{cell}^o$$ = $${{RT} \over {nF}}\ln \left( {{K_C}} \right)$$
<br><br>= $$2.303{{RT} \... | mcq | jee-main-2019-online-11th-january-evening-slot | 2,065 |
UAZ845NByQR9VWArvbOrp | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | For the cell Zn(s) |Zn<sup>2+</sup> (aq)| |M<sup>x+</sup> (aq)| M(s), different half cells and their standard electrode potentials are given below :
<br/><br/><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{font-family:Arial, sans-serif;font-size:14px;padding:10px 5px;border-style:soli... | [{"identifier": "A", "content": "Ag<sup>+</sup>/Ag"}, {"identifier": "B", "content": "Fe<sup>3+</sup>/Fe<sup>2+</sup>"}, {"identifier": "C", "content": "Au<sup>3+</sup>/Au"}, {"identifier": "D", "content": "Fe<sup>2+</sup>/Fe"}] | ["A"] | null | Zn(s) |Zn<sup>2+</sup> (aq)| |M<sup>x+</sup> (aq)| M(s)
<br>---------------------------------------
<br> Anode Cathode
<br><br> E<sup>o</sup><sub>cell</sub> = E<sup>o</sup><sub>cathode</sub> – E<... | mcq | jee-main-2019-online-11th-january-morning-slot | 2,066 |
swzda7kwskEWBbGApD9uc | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | In the cell
<br/><br/>Pt$$\left| {\left( s \right)} \right|$$H<sub>2</sub>(g, 1 bar)$$\left| {HCl\left( {aq} \right)} \right|$$AgCl$$\left| {\left( s \right)} \right|$$Ag(s)|Pt(s)
<br/><br/>the cell potential is 0.92 V when a 10<sup>–6</sup> molal HCl solution is used. The standard electrode potential of (AgCl/ AgCl... | [{"identifier": "A", "content": "0.94 V"}, {"identifier": "B", "content": "0.40 V"}, {"identifier": "C", "content": "0.76 V"}, {"identifier": "D", "content": "0.20 V"}] | ["D"] | null | Anode : H<sub>2</sub>(g) $$ \to $$ 2H<sup>+</sup>(aq) + 2e<sup>-</sup>
<br><br>Cathode : AgCl(s) + e<sup>-</sup> $$ \to $$ Ag(s) + Cl<sup>-</sup>(aq)
<br>---------------------------------------------------------------
<br> H<sub>2</sub>(g) + 2AgCl(s... | mcq | jee-main-2019-online-10th-january-evening-slot | 2,067 |
5BVvstOicz7u7lLOaVjIi | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | Consider the following reduction processes :
<br/>Zn<sup>2+</sup> + 2e<sup>–</sup> $$ \to $$ Zn(s) ; E<sup>o</sup> = – 0.76 V
<br/>Ca<sup>2+</sup> + 2e<sup>–</sup> $$ \to $$ Ca(s); E<sup>o</sup> = –2.87 V
<br/>Mg<sup>2+</sup> + 2e<sup>–</sup> $$ \to $$ Mg(s) ; E<sup>o</sup> = – 2.36 V
<br/>Ni<sup>2</sup> + 2e<sup>–</su... | [{"identifier": "A", "content": "Ca < Mg < Zn < Ni "}, {"identifier": "B", "content": "Ni < Zn < Mg < Ca"}, {"identifier": "C", "content": "Zn < Mg < Ni < Ca"}, {"identifier": "D", "content": "Ca < Zn < Mg < Ni "}] | ["B"] | null | Higher the oxidation potential better will be reducing power. | mcq | jee-main-2019-online-10th-january-morning-slot | 2,068 |
lHYliJBzUSWdRBukUqjgy2xukeyhxqfv | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | For the disproportionation reaction
<br/>2Cu<sup>+</sup>(aq) ⇌ Cu(s) + Cu<sup>2+</sup>(aq) at 298 K. ln K
<br/>(where K is the equilibrium constant) is
<br/>___________ × 10<sup>–1</sup>.
<br/>Given :
<br/>($$E_{C{u^{2 + }}/C{u^ + }}^0 = 0.16V$$
<br/>$$E_{C{u^ + }/Cu}^0 = 0.52V$$
<br/>$${{RT} \over F} = 0.025$$) | [] | null | 144 | $$E_{cell}^0$$ = $$E_{C{u^ + }/Cu}^0$$ - $$E_{C{u^{2 + }}/C{u^ + }}^0$$
<br><br>= 0.52 – 0.16
<br><br>= 0.36 V
<br><br>At equilibrium, E<sub>cell</sub> = 0
<br><br>$$E_{cell}^0$$ = $${{RT} \over {nF}}$$ln K
<br><br>$$ \Rightarrow $$ ln K = $${{E_{cell}^0 \times nF} \over {RT}}$$
<br><br>= $${{0.36 \times 1} \over {0.02... | integer | jee-main-2020-online-2nd-september-evening-slot | 2,071 |
anp5e0kXk90TGWLSeijgy2xukg3eyne6 | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | For the given cell :
<br/><br/>Cu(s) | Cu<sup>2+</sup>(C<sub>1</sub>M) || Cu<sup>2+</sup>(C<sub>2</sub>M) | Cu(s)
<br/><br/>change in Gibbs energy ($$\Delta $$G) is negative, if : | [{"identifier": "A", "content": "C<sub>2</sub> = $$\\sqrt 2 $$C<sub>1</sub>"}, {"identifier": "B", "content": "C<sub>2</sub> = $${{{C_1}} \\over {\\sqrt 2 }}$$"}, {"identifier": "C", "content": "C<sub>1</sub> = 2C<sub>2</sub>"}, {"identifier": "D", "content": "C<sub>1</sub> = C<sub>2</sub>"}] | ["A"] | null | Given $$\Delta $$G < 0
<br><br>$$ \therefore $$ -nFE<sub>cell</sub> < 0
<br><br>$$ \Rightarrow $$ E<sub>cell</sub> > 0
<br><br>We know, E<sub>cell</sub> = $$E_{cell}^0$$ - $${{RT} \over {2F}}\ln \left( {{{{C_1}} \over {{C_2}}}} \right)$$
<br><br> = 0 - $${{RT} \over {2F}}\ln \left( {{{{C_1}} \over {{C_2}}}} \r... | mcq | jee-main-2020-online-6th-september-evening-slot | 2,072 |
gyOVIW5hXWdjpDXcrcjgy2xukfjj3uxw | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | An oxidation-reduction reaction in which<br/> 3 electrons are transferred has a $$\Delta $$Gº of 17.37 kJ mol<sup>–1</sup> at
<br/>25 <sup>o</sup>C. The value of E<sup>o</sup>
<br/>cell (in V) is ______ × 10<sup>–2</sup>.
<br/>(1 F = 96,500 C mol<sup>–1</sup>) | [] | null | -6 | $$\Delta $$Gº = 17.37 kJ ; n = 3
<br><br>$$\Delta $$Gº = -nF$$E_{cell}^o$$
<br><br>$$ \Rightarrow $$ $$E_{cell}^o$$ = $$ - {{17.37 \times 1000} \over {3 \times 96500}}$$
<br><br>= -0.06 = -6.00 $$ \times $$ 10<sup>-2</sup> | integer | jee-main-2020-online-5th-september-morning-slot | 2,073 |
FsZpKUQHnwDafSgIlwjgy2xukf9kgojo | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | <img src="data:image/png;base64,UklGRpAVAABXRUJQVlA4IIQVAAAwlQCdASr0AWIBP4G+1WS2MCunIrGqasAwCWlu/CIUNi9sXmNwur5H/y/bn/w/tP9MfMj88k/uJ2HXnX/jO9ngHef/3w82XZSAD+wvmm/i+Z/IF32p4P6+fUh+9f+v1jv+oN7vJp9ebV4Dt7nzaN55Xpq7yafXm1eaGBmk8jCeRhObwAaGhqYBvxuH9Z0V5tXmikXAEzSeRg+KzmTel3/LNhB9V+TfMYdAJR8ybZ0V5tXmQ4ggTNNnNZOVjsADyht4yxad... | [{"identifier": "A", "content": "If E<sub>ext</sub> < 1.1 V, Zn dissolves at anode and Cu\ndeposits at cathode"}, {"identifier": "B", "content": "If E<sub>ext</sub> = 1.1 V, no flow of e<sup>\u2013</sup> or current\noccurs"}, {"identifier": "C", "content": "If E<sub>ext</sub> > 1.1 V, e<sup>\u2013</sup> flows fro... | ["D"] | null | $$E_{cell}^0$$ = $$E_{C{u^{2 + }}|Cu}^0$$ - $$E_{Z{n^{2 + }}|Zn}^0$$
<br><br>= 0.34 – (–0.76)
<br><br>= 1.10 V
<br><br>If E<sub>ext </sub>< 1.1 V then Zn dissolves at anode and
copper deposits at Cathode.
<br><br>If E<sub>ext</sub> > 1.1V then Zn deposited at zinc
electrodes and Cu deposits at Cu electrode. | mcq | jee-main-2020-online-4th-september-morning-slot | 2,074 |
tBHIIQYj0GjIVURT0gjgy2xukewfeme4 | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | The Gibbs change (in J) for the given reaction at
<br/>[Cu<sup>2+</sup>] = [Sn<sup>2+</sup>] = 1 M and 298K is :
<br/><br/>Cu(s) + Sn<sup>2+</sup>(aq.) $$ \to $$ Cu<sup>2+</sup>(aq.) + Sn(s);
<br/><br/>($$E_{S{n^{2 + }}|Sn}^0 = - 0.16\,V$$,
<br/>$$E_{C{u^{2 + }}|Cu}^0 = 0.34\,V$$)
<br/>Take F = 96500 C mol<sup>–1</s... | [] | null | 96500 | $$\Delta $$G = $$\Delta $$G<sup>o</sup> + RTln $$\left[ {{{S{n^{ + 2}}} \over {C{u^{ + 2}}}}} \right]$$
<br><br>= –2 × 96500 [(–0.16) – 0.34] + RT$$\left[ {{1 \over 1}} \right]$$
<br><br>= 96500 J | integer | jee-main-2020-online-2nd-september-morning-slot | 2,076 |
7rxUdnAfy0uIrZrtPb7k9k2k5hm4ozw | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | For an electrochemical cell<br/>
<br>Sn(s) | Sn<sup>2+</sup> (aq,1M)||Pb<sup>2+</sup> (aq,1M)|Pb(s)<br/>
<br>the ratio $${{\left[ {S{n^{2 + }}} \right]} \over {\left[ {P{b^{2 + }}} \right]}}$$ when this cell attains
equilibrium is _________.
<br/><br/>
(Given $$E_{S{n^{2 + }}|Sn}^0 = - 0.14V$$,<br/>
<br>$$E_{P{b^{2 + ... | [] | null | 2.13TO2.16 | Cell reaction is :
<br><br>Sn(s) + Pb<sup>+2</sup>(aq) $$ \to $$ Sn<sup>+2</sup>(aq) + Pb(s)
<br><br>Apply Nernst equation :
<br><br>E<sub>cell</sub> = $$E_{cell}^0$$ - $${{0.06} \over 2}\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$$ ....(1)
<br><br>$$ \Rightarrow $$ $$E_{cell}^0$$ = -0.13... | integer | jee-main-2020-online-8th-january-evening-slot | 2,077 |
mvRWFFO3zxDINChkNj7k9k2k5h74b00 | chemistry | electrochemistry | electrochemical-series,-cell-and-their-emf | What would be the electrode potential for the given half cell reaction at pH = 5?
______.
<br/><br>2H<sub>2</sub>O $$ \to $$ O<sub>2</sub> + 4H<sup>$$ \oplus $$</sup> + 4e<sup>–</sup> ;
$$E_{red}^0$$ = 1.23 V
<br/><br>(R = 8.314 J mol<sup>–1</sup> K<sup>–1</sup> ; Temp = 298 k; <br/><br>oxygen under std. atm. pressure ... | [] | null | 1.52TO1.53 | E = E<sub>0</sub> - $${{0.0591} \over 4}\log {\left[ {{H^ + }} \right]^4}$$
<br><br>$$ \Rightarrow $$ E = 1.23 + 0.0591 × pH
<br><br>$$ \Rightarrow $$ E = 1.23 + 0.0591 × (5)
<br><br>$$ \Rightarrow $$ E = 1.52 | integer | jee-main-2020-online-8th-january-morning-slot | 2,078 |
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