The reduction potentials of

Asked by Ujjwal Kumar | 22nd Sep, 2010, 05:07: PM

Expert Answer:

Oxidation is taking place at Cu electrode and reduction is at Ag.
Cu + 2Ag+......>Cu2+ + 2Ag
here, n=2
and, Eo=.80-.34=.46V
and by using the formula:
cell = Eo - (.0592/2) log10 (Cu2+/(Ag+)2 )
0=.46V-.0592/2 log(.1/(Ag+)2)
Ag+=5*10-9 M

Answered by  | 22nd Sep, 2010, 08:14: PM

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