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tamaranim1 [39]
3 years ago
9

Balance each of the following redox reactions occurring in acidic aqueous solution.

Chemistry
1 answer:
dmitriy555 [2]3 years ago
4 0

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>

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