Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: brentr on December 09, 2008, 12:47:56 AM
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A Cu/Cu^2+ concentration cell has a voltage of 0.23V at 25C. The concentration of Cu^2+ in one of the half cells is 1.6x10^-3M. What is the concentration of Cu^2+ in the other half-cell?
This is what I have worked out, but the answer does not seem to be correct...
Oxidation half cell :
Cu ---- > Cu+2 + 2e
Reduction half cell :
Cu+2 + 2e ------> Cu
Net cell Reaction:
Cu + Cu+2 ------> Cu+2 + Cu
Ecell = E0cell - 0.059 / 2 log [ Cu+2 ] / [ Cu+2 ]
0.23 V = + 0.340 V - 0.059 / 2 log [ 1.6 x 10-3 M ] / [ Cu+2 ]
[ Cu+2 ] = 3.048 x 10-7 M
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In concentration cell E0 doesn't matter, it should cancel out.