Answer:
Part A
: Zn(s) + Sn²⁺(aq) → Zn²⁺(aq) + Sn(s).
Part B
: 3Mg(s) + 2Cr³⁺(aq) → 3Mg²⁺(aq) + 2Cr(s).
Part C: 3MnO₄⁻ + 24H⁺ + 5Al → 5Al³⁺ + 3Mn²⁺ + 12H₂O.
Explanation:
<em>Part A
: Zn(s) + Sn²⁺(aq) → Zn²⁺(aq) + Sn(s), Express your answer as a chemical equation. Identify all of the phases in your answer.
</em>
- It is balanced as written: Zn(s) + Sn²⁺(aq) → Zn²⁺(aq) + Sn(s).
The two half reactions are:
The oxidation reaction: Zn(s) → Zn²⁺(aq) + 2e.
The reduction reaction: Sn²⁺(aq) + 2e → Sn(s).
- To obtain the net redox reaction, we add the two-half reactions as the no. of electrons in the two-half reactions is equal.
So, the net chemical equation is:
<em>Zn(s) + Sn²⁺(aq) → Zn²⁺(aq) + Sn(s).</em>
<em>Part B: Mg(s) + Cr³⁺(aq) → Mg²⁺(aq) + Cr(s), Express your answer as a chemical equation. Identify all of the phases in your answer.
</em>
- To balance and write the net chemical equation, we should write the two-half reactions:
The two half reactions are:
The oxidation reaction: Mg(s) → Mg²⁺(aq) + 2e.
The reduction reaction: Cr³⁺(aq) + 3e → Cr(s).
- To obtain the net redox reaction, we multiply the oxidation reaction by 3 (3Mg(s) → 3Mg²⁺(aq) + 6e) and the reduction reaction by 2 (2Cr³⁺(aq) + 6e → 2Cr(s)) to equalize the no. of electrons in the two-half reactions.
So, the net redox reaction will be:
<em>3Mg(s) + 2Cr³⁺(aq) → 3Mg²⁺(aq) + 2Cr(s).</em>
<em>Part C
: MnO⁴⁻(aq) + Al(s) → Mn²⁺(aq) + Al³⁺(aq), Express your answer as a chemical equation. Identify all of the phases in your answer.</em>
- To balance and write the net chemical equation, we should write the two-half reactions:
The two half reactions are:
The oxidation reaction: Al → Al³⁺ + 3e.
The reduction reaction: MnO₄⁻ + 8H⁺ + 5e → Mn²⁺ + 4H₂O.
- To obtain the net redox reaction, we multiply the oxidation reaction by 5 (5Al → 5Al³⁺ + 15e) and the reduction reaction by 3 (3MnO₄⁻ + 24H⁺ + 15e → 3Mn²⁺ + 12H₂O) to equalize the no. of electrons in the two-half reactions.
So, the net redox reaction will be:
<em>3MnO₄⁻ + 24H⁺ + 5Al → 5Al³⁺ + 3Mn²⁺ + 12H₂O.</em>
