To determine the correctly balanced spontaneous reaction equation for the voltaic cell with the given cell notation [tex]\( \text{Al (s)} \left| \text{Al}^{3+}(aq) \| \text{Ni}^{2+}(aq) \right| \text{Ni (s)} \)[/tex], let's follow these steps:
1. Identify Oxidation and Reduction Half-Reactions:
- Oxidation Half-Reaction (Anode):
Since aluminum is being oxidized, the oxidation half-reaction is:
[tex]\[
\text{Al} (s) \rightarrow \text{Al}^{3+} (aq) + 3e^-
\][/tex]
- Reduction Half-Reaction (Cathode):
Since nickel is being reduced, the reduction half-reaction is:
[tex]\[
\text{Ni}^{2+} (aq) + 2e^- \rightarrow \text{Ni} (s)
\][/tex]
2. Balance Electrons in Both Half-Reactions:
- To balance electrons, the number of electrons lost in the oxidation reaction must equal the number of electrons gained in the reduction reaction. Here, we need a common multiple for the electrons:
- Multiplying the oxidation half-reaction by 2 (to have [tex]\(6e^-\)[/tex]):
[tex]\[
2\text{Al} (s) \rightarrow 2\text{Al}^{3+}(aq) + 6e^-
\][/tex]
- Multiplying the reduction half-reaction by 3 (to have [tex]\(6e^-\)[/tex]):
[tex]\[
3\text{Ni}^{2+} (aq) + 6e^- \rightarrow 3\text{Ni} (s)
\][/tex]
3. Add the Two Half-Reactions Together:
- Now, combine the two balanced half-reactions:
[tex]\[
2\text{Al} (s) \rightarrow 2\text{Al}^{3+}(aq) + 6e^-
\][/tex]
[tex]\[
3\text{Ni}^{2+} (aq) + 6e^- \rightarrow 3\text{Ni} (s)
\][/tex]
- The electrons will cancel out:
[tex]\[
3\text{Ni}^{2+}(aq) + 2\text{Al}(s) \rightarrow 2\text{Al}^{3+}(aq) + 3\text{Ni}(s)
\][/tex]
4. Conclusion:
- The correctly balanced spontaneous reaction equation is:
[tex]\[
3\text{Ni}^{2+}(aq) + 2\text{Al}(s) \rightarrow 2\text{Al}^{3+}(aq) + 3\text{Ni}(s)
\][/tex]
Therefore, the correct answer is:
[tex]\[ 3 \text{Ni}^{2+}(aq) + 2 \text{Al}(s) \rightarrow 2 \text{Al}^{3+}(aq) + 3 \text{Ni}(s) \][/tex]
