A cell is set up with copper and lead electrodes in contact withCuSO4(aq) and Pb

Question

A cell is set up with copper and lead electrodes in contact withCuSO4(aq) and Pb(NO3)2(aq), respectively, at 25°C.
          The standardreduction potentials are:
          Pb2+ +2e– Pb E° = –0.13 V
          Cu2+ +2e– Cu E° = +0.34 V
If sulfuric acid is added to the Pb(NO3)2 solution, forming aprecipitate of PbSO4, the cell potential:
A) increases B) decreases
C) is unchanged D) can't tell what will happen

Can you give an explanation for your answer as well.

Explanation / Answer

Reaction is spontaneous when E > 0 Pb    => Pb2+ + 2e- Cu2+ + 2e => Cu E = 0.34 - (-0.13) = 0.47 V   (we flip sign of Pbreaction because now lead is oxidized) E = E0 - RT/nF *ln(Pb2+/Cu2+) decreasing the Pb2+ concentration increases the cell potential fromthe above relation http://en.wikipedia.org/wiki/Nernst_equation

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