What reactions take place at the cathode and the anode when each of the following is electrolyzed? (Assume standard conditions.) a. \(1.0 M \mathrm{NiBr}_{2}\) solution b. \(1.0 \mathrm{M} \mathrm{AlF}_{3}\) solution c. \(1.0 \mathrm{M} \mathrm{MnI}_{2}\) solution

For this exercise, we must rely on reduction potentials of different species for figuring out half-reactions at the electrodes. For metals, reduction half-reaction can be written as: \( M^{n+} + ne^- \rightarrow M \). For halogen elements, the oxidation half-reactions are: \( 2X^- \rightarrow X_2 + 2e^- \) where X is a halogen (F, Cl, Br, I). Now let's analyze each case.

Given a \(\mathrm{NiBr}_{2}\) solution, we will have \(\mathrm{Ni}^{2+}\) and \(\mathrm{Br}^-\) ions in the solution. The cathode reaction will be the reduction of \(\mathrm{Ni}^{2+}\) and the anode reaction will be the oxidation of \(\mathrm{Br}^-\). Cathode (reduction) reaction: \[ \mathrm{Ni}^{2+} + 2e^- \rightarrow \mathrm{Ni} \] Anode (oxidation) reaction: \[ 2\mathrm{Br}^- \rightarrow \mathrm{Br}_{2} + 2e^- \]

Given a \(\mathrm{AlF}_{3}\) solution, we will have \(\mathrm{Al}^{3+}\) and \(\mathrm{F}^-\) ions in the solution. The cathode reaction will be the reduction of \(\mathrm{Al}^{3+}\) and the anode reaction will be the oxidation of \(\mathrm{F}^-\). Cathode (reduction) reaction: \[ \mathrm{Al}^{3+} + 3e^- \rightarrow \mathrm{Al} \] Anode (oxidation) reaction: \[ 2\mathrm{F}^- \rightarrow \mathrm{F}_{2} + 2e^- \] However, the oxidation of \(\mathrm{F^-}\) to \(\mathrm{F}_{2}\) has a very high standard potential. Since the electrolysis takes place under standard conditions, it's more likely that the water will undergo oxidation instead of the fluoride ions. Anode (oxidation) reaction (water instead of \(\mathrm{F}^-\)): \[ 2\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{O}_{2} + 4\mathrm{H}^+ + 4e^- \]

Given a \(\mathrm{MnI}_{2}\) solution, we will have \(\mathrm{Mn}^{2+}\) and \(\mathrm{I}^-\) ions in the solution. The cathode reaction will be the reduction of \(\mathrm{Mn}^{2+}\) and the anode reaction will be the oxidation of \(\mathrm{I}^-\) Cathode (reduction) reaction: \[ \mathrm{Mn}^{2+} + 2e^- \rightarrow \mathrm{Mn} \] Anode (oxidation) reaction: \[ 2\mathrm{I}^- \rightarrow \mathrm{I}_{2} + 2e^- \] In summary, during the electrolysis of each given solution, there will be a reduction half-reaction involving metal cations at the cathode and an oxidation half-reaction involving halogen anions or water at the anode.