Michael S. answered • 13d
B.S. in Chemistry, Indiana University; Organic Chem Teaching Intern
Answer: the second choice, [Zn2+] = 0.25 M and [Fe2+] = 1 M because Q < 1. It gives E = 0.338 V, the highest of the four.
Write Q down before you read the reasons, because two of these four options describe their own numbers incorrectly. For Fe2+(aq) + Zn(s) -> Fe(s) + Zn2+(aq), pure solids are left out of Q, so Zn(s) and Fe(s) do not appear and only the two ions matter:
Q = [Zn2+] / [Fe2+]
Zn2+ is a product (the zinc metal is what gets oxidized) and Fe2+ is a reactant. That single line is the whole problem.
The fast route, with no arithmetic at all. The Nernst equation subtracts a term proportional to log Q, so a bigger Q always means a smaller E. The highest potential therefore belongs to the smallest Q, and you are simply hunting for the option with the smallest [Zn2+]/[Fe2+] ratio. Those ratios are 2, 0.25, 0.5 and 4, so the second choice wins before you touch a calculator.
Now the trap, which is the real point of the question. Each option comes with a stated reason, and you have to check the reason against the numbers instead of trusting it:
Option Q = [Zn2+]/[Fe2+] Reason given Reason true? E (V) [Zn2+]=2M, [Fe2+]=1M 2 / 1 = 2.00 Q < 1 NO 0.311 [Zn2+]=0.25M, [Fe2+]=1M 0.25 / 1 = 0.25 Q < 1 YES 0.338 [Zn2+]=1M, [Fe2+]=2M 1 / 2 = 0.50 Q > 1 NO 0.329 [Zn2+]=2M, [Fe2+]=0.5M 2 / 0.5 = 4.00 Q > 1 YES 0.302
Two options state a Q relation that is false about their own concentrations. The first one claims Q < 1 while its numbers give Q = 2. The third claims Q > 1 while its numbers give Q = 0.5, which is why it is the nastiest distractor in the set: it does raise the potential above 0.32 V, so a student who only checks the direction and never checks the arithmetic will accept it, and it still is not the highest. The fourth option is the only other one whose reason is honest, and it points the wrong way, because Q > 1 lowers E.
The arithmetic for the winner. Both half reactions move two electrons (Zn -> Zn2+ + 2e- and Fe2+ + 2e- -> Fe), so n = 2:
E = E° - (0.0592 / n) log Q n = 2 electrons E = 0.32 - (0.0592 / 2) log(0.25) E = 0.32 - (0.0296)(-0.602) E = 0.32 + 0.018 E = 0.338 V
Running the same calculation on all four, plus the standard state for reference:
0.338 V [Zn2+]=0.25 M, [Fe2+]=1 M Q = 0.25 <-- highest, the answer 0.329 V [Zn2+]=1 M, [Fe2+]=2 M Q = 0.50 0.320 V both ions at 1 M Q = 1.00 (standard conditions) 0.311 V [Zn2+]=2 M, [Fe2+]=1 M Q = 2.00 0.302 V [Zn2+]=2 M, [Fe2+]=0.5 M Q = 4.00 lowest
One thing worth taking away beyond this question: concentration is a weak lever on cell potential. Going from Q = 4 down to Q = 0.25 is a sixteenfold change in the concentration ratio, and it buys you only 36 mV on a 320 mV cell, about 11 percent. The log and the division by n are what flatten it. If you wanted to double this cell to 0.64 V by concentration alone you would need log Q = -10.8, that is Q near 1.5e-11, which is not something you set up with a graduated cylinder. Choosing a different pair of metals moves E by volts; diluting a beaker moves it by millivolts.
The qualitative check that costs nothing. Read the Nernst result as Le Chatelier applied to a cell. Cell potential measures how far the reaction still has to run, so anything that pushes the reaction forward raises E. Removing a product does that, and Zn2+ is the product, so dropping it to 0.25 M is exactly the right move. Adding reactant would work too, which is why the third option is not crazy, just smaller. Piling up the product, as the first and fourth options do, drains the cell toward equilibrium, and at equilibrium Q = K and E = 0. That is also what a dead battery is: not an empty one, but one that has reached Q = K.
