Calculate E cell for the following reaction at 25deg celcius

[Biggins1]

Homework Statement

Calculate E cell for the following reaction at 25deg celcius

3 CU(s) + 2 MnO4-(aq) + 8 H+(aq) →2 MnO2(s) + 3 CU^2+(aq) + 4 H2O(l)

Given: [CU^2+] = 0.010M, [MnO4-] = 2.0M, [H+] = 1.0M

I’ve posted a picture of my work, if someone could just let me know if I answered the question correctly and if not guide me as to how to correctly complete the problem. Thank you!

 

[Charles Link]

You need to keep the homework template in place on homeworks please. Also welcome to PF.
I needed to review this material=I had it 40+ years ago in a chemistry sequence.
You didn't show all of the work.
For $MnO_4^-$, I looked it up in a handbook and got $MnO_4^-+4H^++3e^- \rightarrow MnO_2+2H_2O$ has $\mathcal{E}=+1.679$ $\\$ $Cu \rightarrow Cu^{++}+2e^-$ has $\mathcal{E}=-.340$. $\\$ I think the EMF for the whole process is just the sum of these, with each the whole reaction being the result of what is going on in each half-cell. I think you can then apply the Nernst equation to get the result when the concentrations are different from 1.0 M. $\\$ Your notes are rather difficult to read=please try to be a little neater and show more of the work. Also, I'm showing more than I normally would to a student because I am on a little bit of a learning curve here, having to relearn something that was even somewhat tricky 40+ years ago. $\\$ Edit: One thing puzzles me here=the final answer is supposed to be an EMF, but there are no electrons transferred in the complete reaction. I believe I added the half reactions together properly, and the numbers I got from the CRC handbook are correct. It looks to me like one half-reaction is taking place in each of two cells. Perhaps this last part is assumed, but they could have been more clear in the statement of the problem. $\\$ Additional editing: I did some review of the book University Chemistry by Mahan. And apparently the reaction proceeds in the forward direction, but the $H^+$ takes time to react with the $MnO_4^-$, and it takes time for the $Cu^{++}$ to enter the solution, (being originally in the form of a $Cu$ electrode), for the reaction to occur. $\\$ Perhaps @Chestermiller and/or @Borek can provide some helpful inputs. Edit: I think I have it figured out=(it took a little work, but I think I solved it correctly). $\\$ Editing: @Biggins1 Upon further study of your notes, your answer looks correct, and is in agreement with the answer that I got. $\\$ Additional note: The lowest line on your notes should read: $2MnO_4^-+8H^++6e^- \rightarrow 2MnO_2+4H_2O=(MnO_4^-+4H^++3e^- \rightarrow MnO_2+2H_2O )$ x $2$, in order to arrive at the $z=6$ in the Nernst equation. (You incorrectly have $3e^-$ in the same equation that has $8H^+$). Also, one more minor correction: you wrote $[Cu^{++}]^2$ instead of $[Cu^{++}]^3$, but you correctly wrote $(.01)^3$ when you computed it in the next line. $\\$ One additional suggestion would be to call the EMF's $\mathcal{E}_{cell}^o$ and $\mathcal{E}_{cell}$ to distinguish the two. Initially, it was very difficult to follow what you were trying to compute.