Electrochemistry - Cell Potential

[nothing123]

Hi,

Just having a little trouble understanding the concept of standard cell potential...what is it exactly? I understand that it is compared to a reference potential of 0 for the reaction 2H+ + 2e- -> H2(g). But what does it mean when a given reaction has positive potential (other than the fact that it's spontaneous)? For example, if a given reaction (A -> A2+ + 2e-) has a cell potential of 2V, does that mean the electrons and positive ions in the reaction actually generate a potential of 2 volts (at least relative to the reference potential)?

One of the questions I encountered asked something like this: "Given the following 2 reactions, which one generates more electrons for the circuit when both reactions run to completion (assume standard conditions and same starting concentrations)"?

A --> A2+ + 2e-, E* = 1.00
B --> B2+ + 2e-. E* = 2.00

Apparently, they generate the same number of electrons...how exactly? If one has a higher voltage, wouldn't it give more charges and thus electrons? Further, I tried to look at this mathematically:
deltaG* = -nfE* and
deltaG = deltaG* + RTlnK.
Then at equilibrium,
0 = -nFE* + RTlnK so
nfE* = RTlnK

Well certainly if E* is larger, then K would be larger meaning more products are favored at equilibrium and in turn more electrons, no?

Thanks for any help, much appreciated.

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[cwfergus]

I dont know if its that simple, but look at the two reactions, each releases 2 electrons....

One releases more voltage, but they release the same amount of electrons, two

 

[nfg05]

One thing I think you may be missing from your understanding is that a cell potential implies that you have two half reactions: a reduction half reaction AND an oxidation half reaction.

For example, A -> A2+ + 2e- cannot have a "cell potential". As it is written, the standard E value for that half reaction should be referred to as a standard oxidation potential. By the way, standard oxidation/reduction/cell potentials imply standard concentrations (1 M) for all substances and standard temp (25 degrees Celsius) and I think standard pressure of 1 atm as well but you really shouldn't have to worry about a nonstandard pressure in cases like this for the MCAT. If conditions weren't standard and you wanted to find the cell potential, you'd need the Nernst equation, along with the reduction potentials and oxidation potentials of the involved species at the nonstandard temperature and the (nonstandard) concentrations of the species.

The potentials you see listed in tables are typically given as standard reduction potentials, which would be A2+ + 2e- -> A for that particular substance. Of course, the magnitude of the oxidation and reduction potentials are equal, but their signs are opposite. For example,

A2+ + 2e- -> A standard E of 0.76 V
A -> A2+ + 2e- standard E of -0.76 V

The standard hydrogen electrode does indeed give you some reference potentials, but they are standard reduction and standard oxidation potentials, not exactly a point of reference to compare to when looking at a standard cell potential. The standard reduction AND oxidation potentials (reversing the reduction reaction and negating the standard reduction potential of 0 still gives you 0) of hydrogen are arbitrarily set at zero to give you a means of comparing how much other substances want to be reduced/oxidized compared to hydrogen.

Soooo how's a cell potential different from the half reaction oxidation and reduction potentials? Well, let's say you're given the following info:

A2+ + 2e- -> A standard E = 1.5 V
B+ + e- -> B standard E = -0.3 V

A2+ has the higher standard reduction potential, so it will be A2+ will be reduced to A and B will be oxidized to B+.

Back to what the cell potential is, its
Ecell = Ecathode - Eanode.
From an ox red cat, oxidation always happens at the anode and reduction at the cathode, so

Ecell = (1.5 V) - (-0.3 V) = 1.8 V. This cell potential is positive, so it's a galvanic cell. Notice how you sub in the reduction potentials for Ecathode and Eanode, even though the oxidation reaction is what's going on at the anode. You could use the equation as

Ecell = Ecathode + Eanode, where Eanode is the oxidation potential if that's how you prefer.

Since the tables usually give reduction potentials, I like to stick with Ecell = Ecathode - Eanode and plug in the values I'm given.

So this turned out way longer than I expected, and it doesn't really address the example question you provided. It's pretty much as cwfergus said. Since each species yields 2 electrons when oxidized (presumably they are being oxidized since the question says they are generating electrons for the circuit) and the starting concentration of each is equal (1 M due to standard conditions assumption), when the reaction is run to completion (i.e. all the starting A and B is oxidized) an equal number of electrons is liberated in each case, regardless of the difference in their standard oxidation potentials.

 

[gopher22]

If a given reaction (A -> A2+ + 2e-) has a cell potential of 2V, does that mean the electrons and positive ions in the reaction actually generate a potential of 2 volts (at least relative to the reference potential)?

Not quite. That's only a half-cell potential. If there's another reaction (B -> B2 + 2e-) with cell potential 1.5V, that means element B wants to hang onto electrons slightly more than element A (or vice-versa, I'm not entirely confident at the moment, but lets assume for the sake of this example). Thus, you get a meager 0.5V when you use those two elements in a battery.

One of the questions I encountered asked something like this: "Given the following 2 reactions, which one generates more electrons for the circuit when both reactions run to completion (assume standard conditions and same starting concentrations)"?

A --> A2+ + 2e-, E* = 1.00
B --> B2+ + 2e-. E* = 2.00

Apparently, they generate the same number of electrons...how exactly?

Those are both the same type of reaction. Atoms forming a diatom will share 2 e- so two are freed

If one has a higher voltage, wouldn't it give more charges and thus electrons? Further, I tried to look at this mathematically:
deltaG* = -nfE* and
deltaG = deltaG* + RTlnK.
Then at equilibrium,
0 = -nFE* + RTlnK so
nfE* = RTlnK

Well certainly if E* is larger, then K would be larger meaning more products are favored at equilibrium and in turn more electrons, no?

Thanks for any help, much appreciated.

Regarding this last part, not sure I understand your perception of this concept. Hope the above helps.

 

[nothing123]

Wow guys, thanks for all your help so far! That definitely cleared a few things up for me. I have a couple more questions though. The thing that confused me about the electrons is, yes although both will have 2 electrons moving for every redox reaction that occurs, wouldn't the one with the *higher oxidation potential* (and not cell potential, thanks for that clarification matt0586) have a higher tendency to lose electrons? So, if the standard reduction and oxidation potentials is determined from putting a metal into solution with it's ion (correct me if I'm wrong about this, just read it up on the internet), wouldn't more electrons want to get liberated? I guess what I'm getting at is for every redox reaction, there still will be a equilibrium point and thus going to completion won't really mean all the metal at the anode disappearing. Unless, that is the purpose of the salt bridge - to prevent charge separation and essentially restore the cell potential...so that ultimately completion does mean all the metal disappearing.

 

[nfg05]

My interpretation of "both reactions run to completion" is that all of species A or B initially present is oxidized, not that each reaction is allowed to proceed until it reaches equilibrium. If it did mean that each reaction is allowed to proceed until reaching equilibrium, I would agree with what you are saying based on the Nernst equation:

Ecell = standardEcell - (RT/nF)ln(Q)
Applying this at equilibrium, Ecell = 0 and Q = Keq, so
standardEcell = (RT/nF)ln(Keq)

For simplicity, assume the other electrode of each cell is the standard hydrogen electrode, so that substance A/B will be oxidized at the anode and hydrogen will be reduced at the cathode.

The half reactions are:

Cell A
A(s) -> A2+(aq) + 2e- standardE = 1.00 V
2H+ + 2e- -> H2(g) standardE = 0.00 V

Cell B
B(s) -> B2+(aq) + 2e- standardE = 2.00 V
2H+ + 2e- -> H2(g) standardE = 0.00 V

The overall reactions are:
A(s) + 2H+ -> A2+(aq) + H2(g)
B(s) + 2H+ -> B2+(aq) + H2(g)

Using standardEcell = standardEcathode - standardEanode and plugging in only standard reduction potentials:

standardEcellA = (0 V) - (-1.00 V) = 1.00 V
standardEcellB = (0 V) - (-2.00 V) = 2.00 V

Since standardEcell = (RT/nF)ln(Keq) at equilibrium and R, T, n, and F are the same for both A and B, Keq is larger for B since standardEcellB is larger than standardEcellA. This means that more of B oxidizes to B2+ and that B supplies more electrons. As you were saying, this makes sense since the substance with the higher oxidation potential has a higher tendency to lose electrons.

However, going back to the overall equation for A:
A(s) + 2H+ -> A2+(aq) + H2(g)

When they say "both reactions run to completion," I take that to mean they are driving the reaction to completion somehow, probably via Le Chatlier's principle. Continuous removal of A2+ (perhaps via precipitation) or H2(g) would drive the reaction to completion (until there was no more of substance A or B left to oxidize). So in the case of both reactions running to completion, all of substance A initially present and all of substance B initially present is oxidized, which would provide the same number of electrons in each case.

 

[nothing123]

Well, your interpretation is certainly correct seeing as it agreed with the answer! Based on that interpretation, wouldn't the salt bridge act to drive it to completion? I mean, it's acting as a conduit for ions to neutralize each other and thus preventing charge buildup so the potential doesn't run out (and the electrons can continue flowing), right?

Thanks again.