## Possible Solution Error on 6N.1 part b

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

Tiffany Chao 2H
Posts: 117
Joined: Fri Aug 09, 2019 12:17 am

### Possible Solution Error on 6N.1 part b

6N.1.b) Calculate the equilibrium constants for the following reactions: In^3+ (aq) + U^3+ (aq) <--> In^2+ (aq) + U^4+ (aq).

My answer differs from the solutions manual. I got 107 instead of the textbook that got 1000 for K. However, I don't understand how they got 2 electrons for the Indium half reaction. I got 1 e- and proceeded with my calculation from there. Is anyone getting 107 as well?

Thanks

Hiba Alnajjar_2C
Posts: 108
Joined: Fri Aug 09, 2019 12:17 am

### Re: Possible Solution Error on 6N.1 part b

Hi Tiffany,
I got the same answer as you! I agree with your point about n being equal to 1, particularly since the oxidation number only changes by 1.

Amy Xiao 1I
Posts: 101
Joined: Sat Jul 20, 2019 12:15 am

### Re: Possible Solution Error on 6N.1 part b

I agree, after asking a UA, we both agreed that there was an error in the solutions because the number of electrons only changes by one in the reduction.

AniP_2D
Posts: 95
Joined: Sat Aug 17, 2019 12:17 am

### Re: Possible Solution Error on 6N.1 part b

I also keep getting 107, so there probably is an error.

Angela Patel 2J
Posts: 110
Joined: Sat Aug 24, 2019 12:17 am

### Re: Possible Solution Error on 6N.1 part b

Yeah I think the oxidation number only changes by 1 for the indium so that is probably where the error is.

AKhanna_3H
Posts: 104
Joined: Wed Sep 18, 2019 12:19 am

### Re: Possible Solution Error on 6N.1 part b

My TA confirmed that it is an error in the solution manual because there should only be 1 e-, so your answer is correct!

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