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J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry 4th edition

Problem 18.97

What is Ecell of the following voltaic cell?

Cu(s)|Cu2+(aq, 0.10 M)||Ag2CrO4(saturated aq)|Ag(s)




Answer:

Assume 25 oC = 298 K

Anode (oxidation) reaction: Cu(s)Cu2+(aq) + 2 e    Eoox = –0.340 V

Cathode (reduction) reaction: Ag+(aq) + eAg(s)    Eored = +0.800 V

Net: 2 Ag+(aq) + Cu(s)Cu2+(aq) + 2 Ag(s)    Eocell = –0.340 + 0.800 = +0.460 V

The silver ion concentration in the cathode cell is found by doing a solubility problem:
Ag2CrO4(s)
2 Ag+(aq)
+
CrO42–(aq)
Ksp = [Ag+]e2[CrO42–]e = 1.1×10–12
Initial
0
0
Change
+ 2x
+ x
Equilibrium
2x
x

1.1×10–12 = [2x]2[x] = 4x3

x = 6.5×10–5

[Ag+] = 2x = 2(6.5×10–5) = 1.3×10–4 M

Now use the Nernst Equation to find Ecell

Ecell = +0.260 V