Calculate E cell for the following reaction at 25deg celcius

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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!
 

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Charles Link
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:welcome: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. :smile: ## \\ ## 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.
 
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