## 14.41(b)

Maria1E
Posts: 61
Joined: Sat Jul 22, 2017 3:01 am

### 14.41(b)

Calculate E for each of the following concentration cells:
b) Pt(s) I H2 (g, 1 bar) I H+ (aq, pH = 4.0) II H+ (aq, pH = 3.0) I H2 (g, 1 bar) I Pt(s)

The solution manual states that n = 1. Can someone please explain why this is? I thought that the 2 half-reactions were 2H+ + 2e --> H2 and H2 --> 2H+ + 2e. After you balance the H's, wouldn't n = 2?

Posts: 31
Joined: Sat Jul 22, 2017 3:00 am

### Re: 14.41(b)

Hey! I don't have the solutions manual so I can't see that it says n=1, but I did find a way to solve it using the same set up as you did. Using the Nernst equation, I got $(\frac{0.0592}{2})log(\frac{pH3.0}{pH4.0})$ . pH is a measure of H+, and we can use logs to find the H+ concentration. $pH= -log[H^{+}]$ so $[H^{+}]=10^{-pH}$. You also need to remember that for reaction quotients, the coefficients are raised to the power, so our final equation is $E=(\frac{0.0592}{2})log(\frac{(10^{-3})^{2}}{(10^{-4})^{2}}$ giving an answer rounding to +0.06V.

Hope this is of some help!