Answer
1. $ \mathrm{CuSO _{4}}$ $(\mathrm{aq})$ yields $\mathrm{O}_{2}(\mathrm{g})$ at the Pt anode and $\mathrm{Cu}(\mathrm{s})$ at the Pt cathode.
2. $\mathrm{NaI(aq)}$ yields $\mathrm{H}_{2}(\mathrm{g})$ at the cathode and $\mathrm{I _{2}}$ at the anode.
3. $\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}), \mathrm{NaOH}(\mathrm{aq}),$ and $\mathrm{KNO}_{3}(\mathrm{aq})$ all yield $\mathrm{H}_{2}(\mathrm{g})$ at the cathode and $\mathrm{O}_{2}(\mathrm{g})$ at the anode.
Work Step by Step
See answer.
