## 6M.1

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

KeiannaPineda1B
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### 6M.1

6M.1 A student was given a standard Cu(s)uCu21(aq) half-cell and another half-cell containing an unknown metal M in 1.00 m M(NO3)2(aq) and formed the cell M(s)uM1(aq)uuCu21(aq)uCu(s). The cell potential was found to be 20.689 V. What is the value of E8(M21/M)?

Could someone please explain step by step on how to get to the answer?

Posts: 74
Joined: Fri Sep 28, 2018 12:29 am

### Re: 6M.1

I got 1.04 V but the solution is -0.349 and I do not know why. Here's my attempted answer:

Oxidation (Anode): M-->M^+ + e^-
Reduction (Cathode): Cu2+ +2e- ------> Cu
Ecell= E(cathode)-E(Anode)= -0.689V
-0.689V= 0.34V- E(Anode)
E(Anode)=1.04V

If anyone knows why this is wrong can you say why?? Thank you!

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