Write equations for the half-reactions that occur at the anode and cathode for the electrolysis of each of the following aqueous solutions. NaBr, PbI2, Na2SO4

For NaBr (aq) hydrogen gas is formed at the cathode. Hydrogen in water is reduced to hydrogen gas. For NaBr (aq), bromine, Br2 (l), will be produced instead of oxygen gas at the anode.

For sodium sulfate hydrogen gas is formed at the cathode. Hydrogen in water is reduced to hydrogen gas. Oxygen gas will be produced at the anode.

Should someone suggest that sodium metal is formed at the cathode, rest assured that that can't happen. Sodium metal reacts with water to make Na+.

Lead (II) iodide is insoluble in water. Therefore, not much will happen since there is no electrolyte.