## 14.37

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

Curtis Tam 1J
Posts: 105
Joined: Thu Jul 13, 2017 3:00 am

### 14.37

Determine the potential of each of the following cells:
a) Pt(s) / H2(g,1.0 bar) / HCl(aq, 0.075 M) // HCl(aq, 1.0M) / H2(g, 1.0 bar) /Pt(s)

In the solutions manual, they give the balanced equation as:

2H+ (aq, 1.0M) + H2(g, 1 atm) ----> 2H+ (aq, 0.075M) + H2 (g, 1 atm)

If you go on to calculate E(cell) you get:

Ecell= -0.025693V/2 * ln[(0.075)^2(1 bar)]/[(1)^2(1 bar)] = 0.067V

My question is if you were to write out the balanced equation such that you cancel the H2 gas on both sides and set the stoichiometric cofficients of the H+ to 1, would that still be okay? The answer comes out to 0.033V if this method is used. In the solutions manual, they sometimes simplify the balanced equations and sometimes don't.

Gurshaan Nagra 2F
Posts: 49
Joined: Thu Jul 27, 2017 3:01 am

### Re: 14.37

Yeah I believe that should be fine, just make sure they are able to cancel like you did.

Curtis Tam 1J
Posts: 105
Joined: Thu Jul 13, 2017 3:00 am

### Re: 14.37

Oh wait never mind. If I simplified the balanced equation to stoichiometric coefficients of 1, then the value of n in the Nernst equation changes such that I get the same answer either way. Problem solved :)

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