Balance the following oxidation–reduction reactions that occur in acidic solution using the half-reaction method. a. $\mathrm{Cu}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{NO}(g)$ b. $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cl}^{-}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{Cl}_{2}(g)$ c. $\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)$ d. $\mathrm{Mn}^{2+}(a q)+\mathrm{NaBiO}_{3}(s) \rightarrow \mathrm{Bi}^{3+}(a q)+\mathrm{MnO}_{4}-(a q)$ e. $\mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{Zn}(s) \rightarrow \mathrm{AsH}_{3}(g)+\mathrm{Zn}^{2+}(a q)$

Step 1: Separate the redox reaction into half-reactions Oxidation half-reaction: Cu → Cu²⁺ + 2e⁻ Reduction half-reaction: NO₃⁻ + e⁻ → NO Step 2: Balance each half-reaction Oxidation half-reaction is already balanced. For the reduction half-reaction, add 2H₂O to balance oxygen and 6H⁺ to balance hydrogen: NO₃⁻ + 2H₂O + 6H⁺ + 3e⁻ → NO + 6H₂O Step 3: Equalize the number of electrons in both half-reactions by multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 2: 3Cu → 3Cu²⁺ + 6e⁻ 2(NO₃⁻ + 2H₂O + 6H⁺ + 3e⁻) → 2(NO + 6H₂O) Step 4: Add the half-reactions back together and cancel electrons: 3Cu + 2NO₃⁻ + 4H₂O + 12H⁺ → 3Cu²⁺ + 2NO + 12H₂O Step 5: Simplify the reaction, if possible: 3Cu(s) + 2NO₃⁻(aq) + 4H₂O(l) + 12H⁺(aq) → 3Cu²⁺(aq) + 2NO(g) + 12H₂O(l) Removing 4H₂O from both sides: 3Cu(s) + 2NO₃⁻(aq) + 12H⁺(aq) → 3Cu²⁺(aq) + 2NO(g) + 8H₂O(l) The balanced reaction is: 3Cu(s) + 2NO₃⁻(aq) + 12H⁺(aq) → 3Cu²⁺(aq) + 2NO(g) + 8H₂O(l)

Step 1: Separate the redox reaction into half-reactions Oxidation half-reaction: Cl⁻ → Cl₂ + 2e⁻ Reduction half-reaction: Cr₂O₇²⁻ → 2Cr³⁺ Step 2: Balance each half-reaction Oxidation half-reaction is already balanced. For the reduction half-reaction, add 14H⁺ to balance the change and 7H₂O to balance oxygen: Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O Step 3: Equalize the number of electrons in both half-reactions by multiplying the oxidation half-reaction by 7 and the reduction half-reaction by 2: 7(Cl⁻ → Cl₂ + 2e⁻) 2(Cr₂O₇²⁻ + 14H⁺) → 2(2Cr³⁺ + 7H₂O) Step 4: Add the half-reactions back together and cancel electrons: 7Cl⁻ + 2Cr₂O₇²⁻ + 28H⁺ → 7Cl₂ + 4Cr³⁺ + 14H₂O The balanced reaction is: 7Cl⁻(aq) + 2Cr₂O₇²⁻(aq) + 28H⁺(aq) → 7Cl₂(g) + 4Cr³⁺(aq) + 14H₂O(l)

Step 1: Separate the redox reaction into half-reactions Oxidation half-reaction: Pb → Pb²⁺ + 2e⁻ Reduction half-reaction: PbO₂ + 4H⁺ + 2e⁻ → Pb²⁺ + 2H₂O Step 2: Balance each half-reaction Oxidation half-reaction is already balanced. Add one SO₄²⁻ to the reduction half-reaction: PbO₂ + 4H⁺ + SO₄²⁻ → PbSO₄ + 2H₂O Step 3: Equalize the number of electrons in both half-reactions by multiplying the oxidation half-reaction by 1 (no need to multiply): Step 4: Add half-reactions back together, and cancel electrons: Pb + PbO₂ + 4H⁺ + SO₄²⁻ → Pb²⁺ + PbSO₄ + 2H₂O Step 5: Simplify the reaction, if possible: Pb(s) + PbO₂(s) + H₂SO₄(aq) → PbSO₄(s) + 2H₂O(l) The balanced reaction is: Pb(s) + PbO₂(s) + H₂SO₄(aq) → PbSO₄(s) + 2H₂O(l)

Step 1: Separate the redox reaction into half-reactions Oxidation half-reaction: Mn²⁺ → MnO₄⁻ Reduction half-reaction: BiO₃⁻ → Bi³⁺ Step 2: Balance each half-reaction For the oxidation half-reaction, add 4H₂O to balance oxygen and 8H⁺ to balance hydrogen, and then subtract 5e⁻ to balance the charges: Mn²⁺ + 4H₂O → MnO₄⁻ + 8H⁺ + 5e⁻ For the reduction half-reaction, add 2H₂O to balance oxygen and 6H⁺ to balance hydrogen, and then add 3e⁻ to balance the charges: BiO₃⁻ + 6H⁺ + 3e⁻ → Bi³⁺ + 3H₂O Step 3: Equalize the number of electrons in both half-reactions by multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 5: 3(Mn²⁺ + 4H₂O → MnO₄⁻ + 8H⁺ + 5e⁻) 5(BiO₃⁻ + 6H⁺ + 3e⁻ → Bi³⁺ + 3H₂O) Step 4: Add the half-reactions back together and cancel electrons: 3Mn²⁺ + 12H₂O + 15BiO₃⁻ + 30H⁺ → 3MnO₄⁻ + 24H⁺ + 5Bi³⁺ + 15H₂O Step 5: Simplify the reaction: 3Mn²⁺(aq) + 12H₂O(l) + 15NaBiO₃(s) + 6H⁺(aq) → 3MnO₄⁻(aq) + 5Bi³⁺(aq) + 8H₂O(l) + 15Na⁺(aq) The balanced reaction is: 3Mn²⁺(aq) + 15NaBiO₃(s) + 6H⁺(aq) → 3MnO₄⁻(aq) + 5Bi³⁺(aq) + 8H₂O(l) + 15Na⁺(aq)

Step 1: Separate the redox reaction into half-reactions Oxidation half-reaction: Zn → Zn²⁺ + 2e⁻ Reduction half-reaction: H₃AsO₄ → AsH₃ Step 2: Balance each half-reaction Oxidation half-reaction is already balanced. For the reduction half-reaction, add 3H⁺ + 3e⁻ to balance hydrogen and charges: H₃AsO₄ + 3H⁺ + 3e⁻ → AsH₃ + 3H₂O Step 3: Equalize the number of electrons in both half-reactions by multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 2: 3(Zn → Zn²⁺ + 2e⁻) 2(H₃AsO₄ + 3H⁺ + 3e⁻ → AsH₃ + 3H₂O) Step 4: Add the half-reactions back together and cancel electrons: 3Zn + 2H₃AsO₄ + 6H⁺ → 3Zn²⁺ + 2AsH₃ + 6H₂O The balanced reaction is: 3Zn(s) + 2H₃AsO₄(aq) + 6H⁺(aq) → 3Zn²⁺(aq) + 2AsH₃(g) + 6H₂O(l)