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 0.22 V.

Answer 1

Cu^2+ + 2e ==> Cu Eo = 0.337 v
H2 ==> 2H^+ 2e Eo = 0.00 v
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Cu^2+ + H2 ==> Cu(s) + 2H^+ EoCell = 0.337
Ecell = 0.22 v

Ecell = Eocell - (0.0592/2) log Q
Plug in 0.22 for Ecell and 0.337 for Eocell and solve for Q for the reaction.
Then Qrxn = [Cu(s)][(H^+)^2]/[(Cu^2+)][H2]
Cu(s) is 1; (H^+)^2 is 1; (H2) in the denominator is 1 and Cu^2+ is the unknown. Solve for that. The unknown concn Cu^2+ should be less than 1 M
I obtained about 1E-4 but that's just a close estimate.
Post your work if you get stuck.

Answer 2

The answer I get is 0.0 for Cu^2+. Is that correct

Answer 3

0,99