posted by Anonymous on .
A cell is set up with copper and lead electrodes in contact with CuSO4(aq) and Pb(NO3)2(aq), respectively, at 25˚C. The standard reduction potentials are:
Pb2+ + 2e- --> Pb E˚=-0.13 V
Cu2+ + 2e- --> Cu E˚=+.34 V
If sulfuric acid is added to the Pb(NO3)2 solution, forming a precipitate of PbSO4, the cell potential:increase, decrease or remain the same
E = Eo -0.06/2*log(1/(Pb^+2)
If Pb^+2 + SO4^= -->PbSO4.
Therefore, the addition of sulfate causes the PbSO4 to ppt thus decreasing the (Pb^+2). What does that do to the E value for Pb and how will that affect the cell voltage?