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oYVD8MSW8NE-014|The oxidation occurs in one compartment and the reduction in another compartment, like an electrochemical cell.
oYVD8MSW8NE-015|The compartment where the oxidation occurs is inside the mitochondria of your cell.
oYVD8MSW8NE-019|Now in solution, the hydrophobic ends associate and the hydrophilic ends are out towards the water.
oYVD8MSW8NE-021|Inside the mitochondria, the oxidation of glucose occurs in the absence of oxygen. It's a formal oxidation.
oYVD8MSW8NE-022|I remove a pair of electrons.
oYVD8MSW8NE-027|These are large molecules-- I've drawn them as ovals here-- that span the membrane.
oYVD8MSW8NE-028|They go from one side of the membrane to the other.
oYVD8MSW8NE-029|So small molecules take our products of our oxidation, our electrons and our protons, to the membrane spanning protein.
oYVD8MSW8NE-033|So as an electrochemical cell, you might call this the cathode of the electrochemical cell.
oYVD8MSW8NE-034|This membrane spanning protein acts as a cathode where the reduction occurs.
oYVD8MSW8NE-036|So what you have is a pH gradient across the mitochondria.
sFbI9fAOsKE-000|Let's look at a chemical reaction that's at equilibrium and then we disturb it.
sFbI9fAOsKE-001|So which is true at time 1 for the chemical reaction A and B go to C in the aqueous phase?
sFbI9fAOsKE-002|And you have equilibrium and then a perturbation of one of the concentrations.
sFbI9fAOsKE-003|Which is true?
sFbI9fAOsKE-010|We're looking at a reaction that was at equilibrium.
sFbI9fAOsKE-011|Macroscopically, all the concentrations had stopped changing.
sFbI9fAOsKE-012|And then a perturbation occurred, and it looks like that perturbation was to increase the concentration of B.
sFbI9fAOsKE-015|So in this case, Q would be less than K.
sFbI9fAOsKE-016|What do we expect would happen here?
sFbI9fAOsKE-017|Well, we've increased a reactant concentration.
sFbI9fAOsKE-018|We expect the equilibrium to shift towards products.
sFbI9fAOsKE-019|So that numerator in the reaction quotient would get larger until Q is equal to K again and equilibrium is re-established.
sFbI9fAOsKE-020|But just after I add that reactant, Q is less than K, and the correct answer here is C.
MOyr35CZxF8-000|When a proton is transferred in a chemical reaction from one compound to another, that's called an acid base reaction.
MOyr35CZxF8-001|The compound that gives up the proton is the acid, the compound that accepts the proton is the base.
MOyr35CZxF8-002|Electrons can also be transferred in chemical reactions from one compound to another.
MOyr35CZxF8-003|I've written possible electron transfer reactions here.
MOyr35CZxF8-004|When electrons are transferred, we call them reduction oxidation reactions or redox reactions.
MOyr35CZxF8-005|Here's copper ions and zinc metal or zinc ions and copper metal.
MOyr35CZxF8-006|The question is, where do the electrons prefer to reside?
MOyr35CZxF8-007|That is, will zinc transfer a pair of its electrons to copper and make copper metal?
MOyr35CZxF8-008|If it did, that would be a reduction.
MOyr35CZxF8-010|Zinc would be oxidized in that process.
MOyr35CZxF8-011|Its oxidation number would go from zero, increasing to plus two.
MOyr35CZxF8-015|So in this process, copper ions act as an oxidizer.
MOyr35CZxF8-018|Once you donate a proton as an acid, you become a base.
MOyr35CZxF8-019|Same thing here.
MOyr35CZxF8-020|Once you oxidize by accepting electrons, you have the potential to be a reducer and donate electrons.
MOyr35CZxF8-021|We can look at the zinc as the reducer here.
MOyr35CZxF8-022|It's adding electrons, donating electrons to the copper to form copper metal.
MOyr35CZxF8-023|In the process, it becomes oxidized.
MOyr35CZxF8-025|It would accept electrons from the copper, itself would be reduced, while the copper is being oxidized.
MOyr35CZxF8-026|Now this reaction, as I've written, is spontaneous in this direction.
MOyr35CZxF8-027|That is, the standard state free energy difference is less than zero.
MOyr35CZxF8-029|That standard state free energy difference less than zero indicates a k, an equilibrium constant, greater than one.
MOyr35CZxF8-030|And in this case, it's a large equilibrium constant.
MOyr35CZxF8-031|It strongly favors the products.
MOyr35CZxF8-032|And we can see that happen.
MOyr35CZxF8-039|And where does the copper metal end up?
MOyr35CZxF8-042|On this side, there's no reaction.
MOyr35CZxF8-043|This represents the reverse reaction-- zinc ions and copper metal.
MOyr35CZxF8-045|Now we can look at half reactions.
MOyr35CZxF8-046|We can split up the chemical reaction into two half reactions involving electrons.
MOyr35CZxF8-049|That happens, in this case, simultaneously with a reduction of copper ions.
MOyr35CZxF8-050|The copper oxidation number goes from plus two to zero.
MOyr35CZxF8-051|That's a reduction in oxidation number.
MOyr35CZxF8-052|We break these up into what are called half reactions to help us study redox reactions.
MOyr35CZxF8-054|This will help us study redox chemistry as a whole.
aDm54rB3l2I-000|Let's look at these properties of ionization energy and electron affinity in more detail.
aDm54rB3l2I-003|That is, if you add an electron to an element, it generally accepts that element, goes to a more stable state releasing energy.
aDm54rB3l2I-004|So electrons are generally accepted.
aDm54rB3l2I-005|Free electron and an atom, the most stable state is that electron attached to the atom.
aDm54rB3l2I-011|You can see neon, at the edge of the periodic table, is in the noble gases, a relatively unreactive element.
aDm54rB3l2I-012|And it's so unreactive even-- won't even pick up free electrons.
aDm54rB3l2I-013|Sodium and, excuse me, nitrogen is a half filled o orbital.
aDm54rB3l2I-014|And the half filled p orbital is relatively stable.
aDm54rB3l2I-015|It has three parallel spins.
aDm54rB3l2I-016|When it accepts another electron, that electron has to go in anti-parallel and spin pair with one of the electrons there.
aDm54rB3l2I-017|And that's just enough perturbation to make that electron affinity not so favorable.
aDm54rB3l2I-018|Now, the ionization energies are all positive.
aDm54rB3l2I-019|It always requires energy to pull electrons away from atoms.
aDm54rB3l2I-021|Carbon, nitrogen, oxygen, fluorine.
OTQOaH4nAto-000|Let's look at the molecular orbitals and the bonding in butadiene.
OTQOaH4nAto-002|Each carbon is sp2 hybridized.
OTQOaH4nAto-005|So you know that each carbon has to have four total bonds.
OTQOaH4nAto-006|So this carbon here at this vertex has one, two, three, so there must be a hydrogen here, four.
OTQOaH4nAto-007|So shorthand notation for writing butadiene looks like this.
OTQOaH4nAto-008|But with these sp2 hybridized carbons, we'll have p orbitals left over on each carbon.
OTQOaH4nAto-012|So here's the carbon.
OTQOaH4nAto-013|Now I have each carbon and notice 120 degree bond angles.
OTQOaH4nAto-014|That's the sp2 hybridized.
OTQOaH4nAto-015|sp2 leaves behind a p orbital and here they are above and below the plane of the molecule.
OTQOaH4nAto-016|So a p orbital left over on each carbon.
OTQOaH4nAto-017|I can form from these one, two, three, four atomic orbitals four molecular orbitals.
OTQOaH4nAto-018|And now this is going to be interesting because the orbital that I form, rather than just spanning two atoms, can span four atoms.
OTQOaH4nAto-019|So I'll form a molecular orbital that looks like this.
OTQOaH4nAto-020|A long, extended, delocalized orbital.
OTQOaH4nAto-021|And delocalized I mean it's over several nuclei.
OTQOaH4nAto-022|The electron can delocalize.
OTQOaH4nAto-023|There's a probability of finding it all the way along the length of the molecule.
OTQOaH4nAto-024|So a second possibility is to have molecular orbitals that look like this.
OTQOaH4nAto-025|Here's another possibility where you have a node now in the middle.
OTQOaH4nAto-026|That's a higher energy.
OTQOaH4nAto-027|Remember, more nodes, higher energy.
OTQOaH4nAto-028|And I can continue to a two node molecular orbital and a three node molecular orbital.
OTQOaH4nAto-029|As you go up in number of nodes, of course you go up in energy.
OTQOaH4nAto-030|And we're going to scale these as pi and pi star.
OTQOaH4nAto-032|So we'll just take the upper half as antibonding and the lower half as bonding.
OTQOaH4nAto-033|Now this is interesting.