A voltaic cell consists of a standard hydrogen electrode in one half-cell and a Cu/Cu2+ half-cell. Calculate [Cu2+] when E cell is 0.160 V.
H2(g) -------------> 2H^+ (aq) + 2e^- E0 = 0v
Cu^2+ (aq) + 2e^- -----------> Cu(s) E0 = 0.34v
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Cu^2+ (aq) + H2(g) -----------> 2H^+ (aq) + Cu(s) E0cell = 0.34v
n = 2
Ecell = 0.16v
Ecell = E0cell - 0.0592/n log[H^+]^2/[Cu^2+]PH2
0.16 = 0.34 - 0.0592/2 log1/[Cu^2+]*1
0.16 = 0.34 - 0.0296log1/[Cu^2+]
0.16 -0.34 = -0.0296log1/[Cu^2+]
-0.18 = -0.0296log1/[Cu^2+]
log1/[Cu^2+] = 0.18/0.0296
log1/[Cu^2+] = 6.08
1/[Cu^2+] = 10^6.08
1/[Cu^2+] = 1.2*10^6
[Cu^2+] = 8.34*10^-7 M >>>>answer
A voltaic cell consists of a standard hydrogen electrode in one half-cell and a Cu/Cu2+ half-cell....