Question

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

Short answer

During the electrolysis of the given molten compounds: a. For molten NiBr₂, cathode reaction: \(Ni^{2+}(l) + 2e^- \rightarrow Ni(s)\), and anode reaction: \(2Br^-(l) \rightarrow Br_2(g) + 2e^-\). b. For molten AlF₃, cathode reaction: \(Al^{3+}(l) + 3e^- \rightarrow Al(s)\), and anode reaction: \(6F^-(l) \rightarrow 3F_2(g) + 6e^-\). c. For molten MnI₂, cathode reaction: \(Mn^{2+}(l) + 2e^- \rightarrow Mn(s)\), and anode reaction: \(2I^-(l) \rightarrow I_2(g) + 2e^-\).

Step-by-step solution

a. Reactions in electrolysis of molten NiBr₂

First, let's identify the cations and anions in the compound. NiBr₂ contains the cations Ni²⁺ and anions Br⁻. During electrolysis: - Reduction takes place at the cathode. - Oxidation takes place at the anode. Reduction half-reaction at the cathode: \[Ni^{2+}(l) + 2e^- \rightarrow Ni(s)\] Oxidation half-reaction at the anode: \[2Br^-(l) \rightarrow Br_2(g) + 2e^-\]

b. Reactions in electrolysis of molten AlF₃

First, let's identify the cations and anions in the compound. AlF₃ contains the cations Al³⁺ and anions F⁻. Reduction half-reaction at the cathode: \[Al^{3+}(l) + 3e^- \rightarrow Al(s)\] Oxidation half-reaction at the anode: \[6F^-(l) \rightarrow 3F_2(g) + 6e^-\]

c. Reactions in electrolysis of molten MnI₂

First, let's identify the cations and anions in the compound. MnI₂ contains the cations Mn²⁺ and anions I⁻. Reduction half-reaction at the cathode: \[Mn^{2+}(l) + 2e^- \rightarrow Mn(s)\] Oxidation half-reaction at the anode: \[2I^-(l) \rightarrow I_2(g) + 2e^-\] In summary, during the electrolysis of the given molten compounds, the respective metal cations (Ni²⁺, Al³⁺, and Mn²⁺) get reduced at the cathode, and the respective halogen anions (Br⁻, F⁻, and I⁻) get oxidized at the anode.