## 6N.9

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

Ellis Song 4I
Posts: 102
Joined: Thu Jul 11, 2019 12:17 am

### 6N.9

A tin electrode in 0.015 m Sn(NO3)2(aq) is connected to a hydrogen electrode in which the pressure of H2 is 1.0 bar. If the cell potential is 0.061 V at 25 C, what is the pH of the electrolyte at the hydrogen electrode?

Ruby Tang 2J
Posts: 102
Joined: Fri Aug 30, 2019 12:15 am

### Re: 6N.9

(1) Consider what the two electrodes are. In this case it should be hydrogen (given in the problem), and tin (from Sn(NO3)2))
(2) Figure out which one is going to be at the cathode and which will be at the anode. The standard reduction potential of H+ (2H+ + 2e- --> H2) is 0 by definition. The oxidation number of tin in Sn(NO3)2 is +2, so we are looking for the standard reduction potential of Sn2+ + 2e- --> Sn, which is -0.14 V. The hydrogen electrode must be the cathode, because we want Ecathode - Eanode to be positive, and 0 - (-0.14) = +0.14 V
(3) Use the Nernst equation (E = Enaught - (RT/nF)lnQ) to find Q, which is the reaction quotient.
(4) Write out the overall redox reaction to find the expression for Q. In this case, it's Sn(s) + 2H+(aq) --> Sn2+(aq) + H2(g). We will not consider Sn(s) in the reaction quotient, since it is a solid and therefore the concentration is unlikely to change significantly. Therefore, Q = [Sn2+]*P(H2)/[H+]^2.
(5). Solve for [H+] using the value for Q you calculated in step 3 and the concentrations given to you in the problem: [Sn2+] = 0.015 M, and P(H2) is 1 bar.
(6) Use pH = -log[H+] to get the pH.

Hope this helps!

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