In this scenario both half reactions are shown as a reduction reaction (gaining electrons). The reaction with the highest reduction potential (positive E0 value) is the most likely to occur.
In this case Ag+ + e- => Ag(s) is the reduction reaction that will occur in the cell as it has the highest reduction potential (+0.8V), and this will occur at the cathode. For the redox reaction to occur, we must invert the other reduction reaction to make it an oxidation reaction. In doing so, we also reverse the charge. This will occur at the anode.
i.e. Pb(s) => Pb2+ + 2e- (E0 = +0.13V)
If we consolidated both half reactions we'd have to balance the products and reactants, but this does not change the electrical potentials.
The overall electrical potential E0(cell) = E0(cathode) - E(anode)
i.e. E0(cell) = 0.8 - 0.13
i.e. E0(cell) = 0.67 V
