Chemistry worksheet showing half-reactions and balanced redox equations for various elements and compounds.

Problem Analysis:

The task involves writing and balancing half-reactions for oxidation and reduction processes, as well as completing and balancing reactions using the half-reaction method. Let's solve each part step by step.

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Problem 15: Write half-reactions for the oxidation and reduction process

#### (a) Fe²⁺ + MnO₄⁻ → Fe³⁺ + Mn²⁺
1. Identify oxidation and reduction:
- Oxidation: Fe²⁺ → Fe³⁺ (Fe is oxidized)
- Reduction: MnO₄⁻ → Mn²⁺ (Mn is reduced)

2. Write half-reactions:
- Oxidation half-reaction:
\[
\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-
\]
- Reduction half-reaction:
\[
\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}
\]

#### (b) I₂ + S²⁻ → I⁻ + S
1. Identify oxidation and reduction:
- Oxidation: S²⁻ → S (S is oxidized)
- Reduction: I₂ → I⁻ (I is reduced)

2. Write half-reactions:
- Oxidation half-reaction:
\[
\text{S}^{2-} \rightarrow \text{S} + 2e^-
\]
- Reduction half-reaction:
\[
\text{I}_2 + 2e^- \rightarrow 2\text{I}^-
\]

#### (c) S²⁻ + NO₃⁻ → S + NO
1. Identify oxidation and reduction:
- Oxidation: S²⁻ → S (S is oxidized)
- Reduction: NO₃⁻ → NO (N is reduced)

2. Write half-reactions:
- Oxidation half-reaction:
\[
\text{S}^{2-} \rightarrow \text{S} + 2e^-
\]
- Reduction half-reaction:
\[
\text{NO}_3^- + 4\text{H}^+ + 3e^- \rightarrow \text{NO} + 2\text{H}_2\text{O}
\]

#### (d) NH₃ + O₂ → N₂ + H₂O
1. Identify oxidation and reduction:
- Oxidation: NH₃ → N₂ (N is oxidized)
- Reduction: O₂ → H₂O (O is reduced)

2. Write half-reactions:
- Oxidation half-reaction:
\[
2\text{NH}_3 \rightarrow \text{N}_2 + 6e^- + 6\text{H}^+
\]
- Reduction half-reaction:
\[
\text{O}_2 + 4\text{H}^+ + 4e^- \rightarrow 2\text{H}_2\text{O}
\]

---

Problem 16: Complete and balance each reaction using the half-reaction method

#### (a) Fe²⁺ + MnO₄⁻ → Fe³⁺ + Mn²⁺
1. Half-reactions:
- Oxidation:
\[
\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-
\]
- Reduction:
\[
\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}
\]

2. Balance electrons:
- Multiply the oxidation half-reaction by 5:
\[
5\text{Fe}^{2+} \rightarrow 5\text{Fe}^{3+} + 5e^-
\]
- The reduction half-reaction already has 5 electrons.

3. Combine half-reactions:
\[
5\text{Fe}^{2+} + \text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5e^-
\]
Cancel out electrons:
\[
5\text{Fe}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O}
\]

#### (b) S²⁻ + NO₃⁻ → S + NO
1. Half-reactions:
- Oxidation:
\[
\text{S}^{2-} \rightarrow \text{S} + 2e^-
\]
- Reduction:
\[
\text{NO}_3^- + 4\text{H}^+ + 3e^- \rightarrow \text{NO} + 2\text{H}_2\text{O}
\]

2. Balance electrons:
- Multiply the oxidation half-reaction by 3:
\[
3\text{S}^{2-} \rightarrow 3\text{S} + 6e^-
\]
- Multiply the reduction half-reaction by 2:
\[
2\text{NO}_3^- + 8\text{H}^+ + 6e^- \rightarrow 2\text{NO} + 4\text{H}_2\text{O}
\]

3. Combine half-reactions:
\[
3\text{S}^{2-} + 2\text{NO}_3^- + 8\text{H}^+ + 6e^- \rightarrow 3\text{S} + 2\text{NO} + 4\text{H}_2\text{O} + 6e^-
\]
Cancel out electrons:
\[
3\text{S}^{2-} + 2\text{NO}_3^- + 8\text{H}^+ \rightarrow 3\text{S} + 2\text{NO} + 4\text{H}_2\text{O}
\]

#### (c) NH₃ + O₂ → N₂ + H₂O
1. Half-reactions:
- Oxidation:
\[
2\text{NH}_3 \rightarrow \text{N}_2 + 6e^- + 6\text{H}^+
\]
- Reduction:
\[
\text{O}_2 + 4\text{H}^+ + 4e^- \rightarrow 2\text{H}_2\text{O}
\]

2. Balance electrons:
- Multiply the oxidation half-reaction by 2:
\[
4\text{NH}_3 \rightarrow 2\text{N}_2 + 12e^- + 12\text{H}^+
\]
- Multiply the reduction half-reaction by 3:
\[
3\text{O}_2 + 12\text{H}^+ + 12e^- \rightarrow 6\text{H}_2\text{O}
\]

3. Combine half-reactions:
\[
4\text{NH}_3 + 3\text{O}_2 + 12\text{H}^+ + 12e^- \rightarrow 2\text{N}_2 + 6\text{H}_2\text{O} + 12e^-
\]
Cancel out electrons:
\[
4\text{NH}_3 + 3\text{O}_2 \rightarrow 2\text{N}_2 + 6\text{H}_2\text{O}
\]

#### (d) Mn²⁺ + IO₃⁻ → I₂ + MnO₂
1. Half-reactions:
- Oxidation:
\[
\text{Mn}^{2+} \rightarrow \text{MnO}_2 + 2e^- + 2\text{H}_2\text{O}
\]
- Reduction:
\[
2\text{IO}_3^- + 12\text{H}^+ + 10e^- \rightarrow \text{I}_2 + 6\text{H}_2\text{O}
\]

2. Balance electrons:
- Multiply the oxidation half-reaction by 5:
\[
5\text{Mn}^{2+} \rightarrow 5\text{MnO}_2 + 10e^- + 10\text{H}_2\text{O}
\]
- The reduction half-reaction already has 10 electrons.

3. Combine half-reactions:
\[
5\text{Mn}^{2+} + 2\text{IO}_3^- + 12\text{H}^+ + 10e^- \rightarrow 5\text{MnO}_2 + \text{I}_2 + 6\text{H}_2\text{O} + 10e^- + 10\text{H}_2\text{O}
\]
Cancel out electrons and combine water molecules:
\[
5\text{Mn}^{2+} + 2\text{IO}_3^- + 12\text{H}^+ \rightarrow 5\text{MnO}_2 + \text{I}_2 + 16\text{H}_2\text{O}
\]

#### (e) K + Na₂S₂O₃ → Na₂S₄O₆ + KI
1. Half-reactions:
- Oxidation:
\[
\text{Na}_2\text{S}_2\text{O}_3 + \text{H}_2\text{O} \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2\text{e}^- + 2\text{OH}^-
\]
- Reduction:
\[
\text{K} + \text{H}_2\text{O} + \text{e}^- \rightarrow \text{KI} + 2\text{H}^+
\]

2. Balance electrons:
- Multiply the reduction half-reaction by 2:
\[
2\text{K} + 2\text{H}_2\text{O} + 2\text{e}^- \rightarrow 2\text{KI} + 4\text{H}^+
\]

3. Combine half-reactions:
\[
\text{Na}_2\text{S}_2\text{O}_3 + \text{H}_2\text{O} + 2\text{K} + 2\text{H}_2\text{O} + 2\text{e}^- \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2\text{e}^- + 2\text{OH}^- + 2\text{KI} + 4\text{H}^+
\]
Cancel out electrons and adjust for neutralization:
\[
\text{Na}_2\text{S}_2\text{O}_3 + 2\text{K} + 2\text{H}_2\text{O} \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2\text{KI}
\]

---

Final Answers:

15.
(a)
- Oxidation: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$
- Reduction: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$

(b)
- Oxidation: $\text{S}^{2-} \rightarrow \text{S} + 2e^-$
- Reduction: $\text{I}_2 + 2e^- \rightarrow 2\text{I}^-$

(c)
- Oxidation: $\text{S}^{2-} \rightarrow \text{S} + 2e^-$
- Reduction: $\text{NO}_3^- + 4\text{H}^+ + 3e^- \rightarrow \text{NO} + 2\text{H}_2\text{O}$

(d)
- Oxidation: $2\text{NH}_3 \rightarrow \text{N}_2 + 6e^- + 6\text{H}^+$
- Reduction: $\text{O}_2 + 4\text{H}^+ + 4e^- \rightarrow 2\text{H}_2\text{O}$

16.
(a) $5\text{Fe}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O}$
(b) $3\text{S}^{2-} + 2\text{NO}_3^- + 8\text{H}^+ \rightarrow 3\text{S} + 2\text{NO} + 4\text{H}_2\text{O}$
(c) $4\text{NH}_3 + 3\text{O}_2 \rightarrow 2\text{N}_2 + 6\text{H}_2\text{O}$
(d) $5\text{Mn}^{2+} + 2\text{IO}_3^- + 12\text{H}^+ \rightarrow 5\text{MnO}_2 + \text{I}_2 + 16\text{H}_2\text{O}$
(e) $\text{Na}_2\text{S}_2\text{O}_3 + 2\text{K} + 2\text{H}_2\text{O} \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2\text{KI}$

Boxed Final Answer:
\[
\boxed{
\begin{aligned}
&\text{(a)} \ 5\text{Fe}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O} \\
&\text{(b)} \ 3\text{S}^{2-} + 2\text{NO}_3^- + 8\text{H}^+ \rightarrow 3\text{S} + 2\text{NO} + 4\text{H}_2\text{O} \\
&\text{(c)} \ 4\text{NH}_3 + 3\text{O}_2 \rightarrow 2\text{N}_2 + 6\text{H}_2\text{O} \\
&\text{(d)} \ 5\text{Mn}^{2+} + 2\text{IO}_3^- + 12\text{H}^+ \rightarrow 5\text{MnO}_2 + \text{I}_2 + 16\text{H}_2\text{O} \\
&\text{(e)} \ \text{Na}_2\text{S}_2\text{O}_3 + 2\text{K} + 2\text{H}_2\text{O} \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2\text{KI}
\end{aligned}
}
\]