Edmund Zhi 2B
Posts: 118
Joined: Sat Jul 20, 2019 12:16 am

In this question, we are given M(s)|M2+(aq)||Cu^2+(aq)|Cu(s) as the cell diagram and are asked to find E(M2+/M) given E(cell) = -0.689.
Now normally, we would consider the right side to be the cathode, right? So in that case, we setup the equation to as
E(cell) = E(cathode) - E(anode)
-0.689 = 0.337 - E(anode)
E(anode) = 1.026

However, the answer key gives us a value of -0.349. To get this, we would have to consider copper as the ANODE. Why is this?

Alex Chen 2L
Posts: 86
Joined: Wed Sep 18, 2019 12:21 am

Re: 6M.1: reading the cell diagram

I'm going to agree with you and say that the book made a mistake. Seems like they accidentally flipped the sign for E cell during calculations, from what I've seen in the solution manual.

Alice Chang 2H
Posts: 101
Joined: Fri Aug 30, 2019 12:18 am

Re: 6M.1: reading the cell diagram

I had a TON of trouble with this one too!

I eventually gave up and assumed the textbook made a mistake and flipped the entire cell diagram around. Or, we were supposed to flip the cell diagram ourselves (?) since we were asked for E(M2+/M) rather than E(M/M2+).

705121606
Posts: 68
Joined: Wed Sep 18, 2019 12:17 am

Re: 6M.1: reading the cell diagram

I also had trouble with this one, I read on another chemistry community question that when the ecell is negative we have to flip the cell around

Edmund Zhi 2B
Posts: 118
Joined: Sat Jul 20, 2019 12:16 am

Re: 6M.1: reading the cell diagram

705121606 wrote:I also had trouble with this one, I read on another chemistry community question that when the ecell is negative we have to flip the cell around

Update: I looked at the solutions manual and they did consider the cell to be flipped. Guess the textbook is wrong

Rodrigo2J
Posts: 102
Joined: Sat Jul 20, 2019 12:16 am

Re: 6M.1: reading the cell diagram

I got the same answer you did, I think the solutions had an error.

Rida Ismail 2E
Posts: 139
Joined: Sat Sep 07, 2019 12:16 am

Re: 6M.1: reading the cell diagram

I got the same answer as you. I am going to try to go to an office hour to ask this question, but my guess is that the book is wrong.