Calculate \(E^{\circ}\) for the following cells: (a) \(\mathrm{Pb}\left|\mathrm{PbSO}_{4} \| \mathrm{Pb}^{2+}\right| \mathrm{Pb}\) (b) \(\mathrm{Pt}\left|\mathrm{Cl}_{2}\right| \mathrm{ClO}_{3}^{-} \| \mathrm{O}_{2}\left|\mathrm{H}_{2} \mathrm{O}\right| \mathrm{Pt}\) (c) \(\mathrm{Pt}\left|\mathrm{OH}^{-}\right| \mathrm{O}_{2} \| \mathrm{ClO}_{3}^{-}, \mathrm{Cl}^{-} \mid \mathrm{Pt} \quad\) (basic medium)

For each cell, we need to identify which reactions are taking place at the cathode (reduction) and the anode (oxidation). (a) Pb|PbSO4|Pb2+|Pb: Here, the cathode and anode are both made of lead, so the half-reactions are: Cathode (reduction): Pb2+ + 2e- → Pb Anode (oxidation): Pb → Pb2+ + 2e- (b) Pt|Cl2|ClO3-|O2|H2O|Pt: Here, we have three distinct half-reactions taking place. Cathode (reduction): Cl2 + 2e → 2Cl- Anode (oxidation): 2ClO3- → Cl2 + 2O2 + 2e- (c) Pt|OH-|O2|ClO3-, Cl-|Pt (basic medium): The half reactions taking place in basic medium include combining OH- ions. Cathode (reduction): O2 + 2H2O + 4e- → 4OH- Anode (oxidation): 2ClO3- + 12OH- → 6H2O + 2Cl- + 10e-

Using the Nernst equation, we can calculate the E° potential. The equation is given by: E° = E°cathode - E°anode (a) For cell (a), the half-reactions have the same E°, which cancels out, resulting in: E° = E°Pb2+/Pb - E°Pb2+/Pb = 0 (b) For cell (b), use the E° values for Cl2/Cl- and ClO3-/O2 half reactions: E° = E°Cl2/Cl- - E°ClO3-/O2 Unfortunately, we do not have enough information provided in this exercise to directly find the values for these half-reactions. They would usually be found in a table of standard electrode potentials. (c) For cell (c), use the E° values for OH-/O2 and ClO3-/Cl- half reactions: E° = E°OH-/O2 - E°ClO3-/Cl- Similar to cell (b), we don't have enough information to directly find the value for these half-reactions without a table of standard electrode potentials.