Question 14.41 [ENDORSED]

[Kevin Tabibian 1A]

In question 14.41 part be you are given the concentration cell of H2, why is n=1 if the half reactions are are H2+2e -> 2H+. Shouldn't the n=2?

[Andy Nguyen 1A]

This is what I did:
Cathode: 2H+ + 2e- --> H2
Anode: H2 --> 2H+ + 2e-
Overall: 2H+ + 2e- + H2 --> 2H+ + 2e- + H2
If you cancel the H2 out you get:
2H+ + 2e- --> 2H+ + 2e-
So I guess you are supposed to factor out the 2 to get only 1 electron being transferred. I think H2 cancels out because they are the same in both half reactions while the H+ are different concentrations in both half reactions.

[Chem_Mod]

One obtains the same answer using n = 1 or 2 as long as the Q expression is self-consistent.

In other words, if you use 2H⁺ (pH = 3) --> 2H⁺ (pH = 4), then Q = ([H⁺ (pH = 4)]/[H⁺ (pH = 3)])². Here, n = 2.

If you solve using H⁺ (pH = 3) --> H⁺ (pH = 4) and Q = [H⁺ (pH = 4)]/[H⁺ (pH = 3)] then n = 1.

You should get the same answer for both approaches.

See attached detailed answer.

A lot of great questions this quarter!