## 6N.1b

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

Alicia Lin 2F
Posts: 83
Joined: Wed Sep 18, 2019 12:17 am

### 6N.1b

In the answer key they say that the reduction half reaction involves two electrons transferred and therefore n would be equal to 2. However, shouldn't the reaction only be involving 1 electron transfer since it is going from In3+ to In2+ which would also make lnK=4.67 and K=107?

nicolely2F
Posts: 149
Joined: Sat Sep 14, 2019 12:17 am

### Re: 6N.1b

Yeah, it's a typo. In "Corrections to Solutions Manual" it's corrected to 1 electron. The final answer's K approximately equal to 100.

Return to “Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)”

### Who is online

Users browsing this forum: No registered users and 1 guest