## 6N.7b

$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$

Ryan Chang 1C
Posts: 105
Joined: Sat Aug 24, 2019 12:17 am

### 6N.7b

Calculate Ecell for each of the following concentration cells:
(b) Pt(s)|H2(g, 1 bar)|H1(aq, pH 5 4.0)||H1(aq, pH 5 3.0)|H2(g, 1 bar)|Pt(s)

When I wrote the half reactions, I got (2H+) + (2e-) -> H2 and H2 -> (2H+) + (2e-)
I would assume that n=2 based on the half reactions, but the answer key says that n=1. Why is that?

Chem_Mod
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### Re: 6N.7b

In the answer key, they simplified the reaction to H+ (pH 3) --> H+ (pH 4). This is because H2 (g) is on the reactants and products side and can be cancelled out, and the rest of the reaction can be halved. This results in only 1 electron being transferred. You can also just keep your redox reaction as is--n will be 2 and the value of Q will be different--but you will still get to the same answer.

Hailey Kim 4G
Posts: 110
Joined: Sat Jul 20, 2019 12:16 am

### Re: 6N.7b

If you use n=2, make sure you square the concentrations of your products and reactants for your value of Q (due to the coefficient of 2 in the H+). E should equal 0.06 V.

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