Cu^2+(aq) + 2e ==> Cu(s) Eo = +0.34
Fe(s) ==> Fe^2+(aq) + 2e Eo = +0.44
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Cu^2+(aq) + Fe(s) =>Cu(s) + Fe^2+(aq)
Eo = 0.34 + 0.44 = 0.78v
Ecell = Eocell + (0.0592/n)*logQ
You know Eocell from above, Ecell from the problem, n = 2,
Q = (Fe^2+)(Cu)/(Fe)(Cu^2+) and all of this makes Q the only unknown.
In Q,[Fe^2+] is the unknown, (Cu) and (Fe)(solids) = 1 by definition, (Cu^2+) is given in the problem. Solve for Fe^2+.
A voltaic cell is based on the following two half reactions:
Fe2+(aq) + 2 e¯ ---> Fe(s); Eo = -0.44 V
Cu2+(aq) + 2 e¯ ---> Cu(s); Eo = +0.34 V
When [Cu2+] = 0.0500 M, Ecell = + 0.80 V.
What is the [Fe2+] for these conditions?
1 answer
1 answer