Problem 109 - Chemistry An Atoms First Approach

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 \(\mathrm{MnI}_{2}\)

Step-by-Step Solution

Answer

For each compound, the electrolysis reactions at the cathode and anode are: a. \(\mathrm{NiBr}_{2}\): Cathode: \(\mathrm{Ni^{2+}} + 2e^{-} \rightarrow \mathrm{Ni}\) Anode: \(2\: \mathrm{Br^{-}} \rightarrow \mathrm{Br_{2}} + 2e^{-}\) b. \(\mathrm{AlF}_{3}\): Cathode: \(\mathrm{Al^{3+}} + 3e^{-} \rightarrow \mathrm{Al}\) Anode: \(2\: \mathrm{F^{-}} \rightarrow \mathrm{F_{2}} + 2e^{-}\) c. \(\mathrm{MnI}_{2}\): Cathode: \(\mathrm{Mn^{2+}} + 2e^{-} \rightarrow \mathrm{Mn}\) Anode: \(2\: \mathrm{I^{-}} \rightarrow \mathrm{I_{2}} + 2e^{-}\)

Step 1: Understand the molten compounds

The molten compounds provided are: a. \(\mathrm{NiBr}_{2}\: (Nickel\,bromide)\) b. \(\mathrm{AlF}_{3}\: (Aluminum\,fluoride)\) c. \(\mathrm{MnI}_{2}\: (Manganese\,iodide)\) Next, identify the ions present in the compounds.

Step 2: Identify the ions present in each compound

The ions present in the compounds are: a. \(\mathrm{NiBr}_{2}\): \(\mathrm{Ni^{2+}}\) and \(\mathrm{Br^{-}}\) b. \(\mathrm{AlF}_{3}\): \(\mathrm{Al^{3+}}\) and \(\mathrm{F^{-}}\) c. \(\mathrm{MnI}_{2}\): \(\mathrm{Mn^{2+}}\) and \(\mathrm{I^{-}}\)

Step 3: Determine the cathode reaction

During electrolysis, metal cations are reduced at the cathode. Reduction occurs by gaining electrons. The cathode reactions for the molten compounds are as follows: a. \(\mathrm{NiBr}_{2}\): \(\mathrm{Ni^{2+}} + 2e^{-} \rightarrow \mathrm{Ni}\) b. \(\mathrm{AlF}_{3}\): \(\mathrm{Al^{3+}} + 3e^{-} \rightarrow \mathrm{Al}\) c. \(\mathrm{MnI}_{2}\): \(\mathrm{Mn^{2+}} + 2e^{-} \rightarrow \mathrm{Mn}\)

Step 4: Determine the anode reaction

During electrolysis, non-metal anions are oxidized at the anode. Oxidation occurs by losing electrons to form elemental substances. The anode reactions for the molten compounds are: a. \(\mathrm{NiBr}_{2}\): \(2\: \mathrm{Br^{-}} \rightarrow \mathrm{Br_{2}} + 2e^{-}\) b. \(\mathrm{AlF}_{3}\): \(2\: \mathrm{F^{-}} \rightarrow \mathrm{F_{2}} + 2e^{-}\) c. \(\mathrm{MnI}_{2}\): \(2\: \mathrm{I^{-}} \rightarrow \mathrm{I_{2}} + 2e^{-}\)

Step 5: Combine the cathode and anode reactions

For each compound, combine the cathode and anode reactions: a. \(\mathrm{NiBr}_{2}\) electrolysis reactions: Cathode: \(\mathrm{Ni^{2+}} + 2e^{-} \rightarrow \mathrm{Ni}\) Anode: \(2\: \mathrm{Br^{-}} \rightarrow \mathrm{Br_{2}} + 2e^{-}\) b. \(\mathrm{AlF}_{3}\) electrolysis reactions: Cathode: \(\mathrm{Al^{3+}} + 3e^{-} \rightarrow \mathrm{Al}\) Anode: \(2\: \mathrm{F^{-}} \rightarrow \mathrm{F_{2}} + 2e^{-}\) c. \(\mathrm{MnI}_{2}\) electrolysis reactions: Cathode: \(\mathrm{Mn^{2+}} + 2e^{-} \rightarrow \mathrm{Mn}\) Anode: \(2\: \mathrm{I^{-}} \rightarrow \mathrm{I_{2}} + 2e^{-}\)