Question

A Cu/Cu2+ concentration cell has a voltage of 0.23 V at 25 ?C. The concentration of Cu2+ in one of the half-cells is 1.5×10^?3 M .

Part A What is the concentration of Cu2+ in the other half-cell? (Assume the concentration in the unknown cell to be the lower of the two concentrations.)

Answer 1

A concentration cell consists of the same half reaction in both the anode and cathode; a voltage is produced by having different ion concentrations (molarities) in the two cells, denoted by M1 and M2.

Anode (oxidation): Cu(s) ==> Cu2+(M1) + 2e-
Cathode (reduction): Cu2+(M2) + 2e- ==> Cu(s)
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Net reaction: Cu2+(M2) ==> Cu2+(M1) . . .Eo = 0.00 V

Set up the Nernst equation for the cell reaction.

E cell = Eo - 0.059/n log Q
0.23 = 0.00 - 0.059/2 log ([M2] / [M1])
0.23 = -0.0295 log (M2/0.0015)
M2 = 2.40 x 10^-11 M

Thus, at 0.23 V the concentration of Cu2+ in the other half cell would be 2.40 x 10^-11 M