Balance the following oxidation–reduction reactions that occur in acidic solution using the half-reaction method. a. $\mathrm{Cr}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{NO}(g)$ b. $\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Ce}^{4+}(a q) \rightarrow \mathrm{CO}_{2}(a q)+\mathrm{Ce}^{3+}(a q)$ c. $\mathrm{SO}_{3}^{2-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{SO}_{4}^{2-}(a q)+\mathrm{Mn}^{2+}(a q)$

Using the rules for oxidation numbers, we find: Cr: 0 N in NO3⁻: +5 Cr³⁺: +3 N in NO: +2

Cr is being oxidized, from 0 to +3. N is being reduced, from +5 to +2.

Oxidation half-reaction: Cr(s) ⟶ Cr³⁺(aq) Reduction half-reaction: NO3⁻(aq) ⟶ NO(g)

Oxidation half-reaction is already balanced in terms of atoms. Reduction half-reaction should be balanced for oxygen atoms: NO3⁻(aq) ⟶ NO(g) + 2H2O(l) Now balance the hydrogen atoms: NO3⁻(aq) + 4H⁺(aq) ⟶ NO(g) + 2H2O(l)

Oxidation half-reaction: Cr(s) ⟶ Cr³⁺(aq) + 3e⁻ (add 3e⁻ to the right side) Reduction half-reaction is already balanced in terms of charge.

The oxidation half-reaction has 3e⁻, and the reduction half-reaction has 1e⁻, so we'll multiply the reduction half-reaction by 3 to get equal electrons transferred. Oxidation half-reaction: Cr(s) ⟶ Cr³⁺(aq) + 3e⁻ Reduction half-reaction x 3: 3NO3⁻(aq) + 12H⁺(aq) ⟶ 3NO(g) + 6H2O(l) Now add the half-reactions: Cr(s) + 3NO3⁻(aq) + 12H⁺(aq) ⟶ Cr³⁺(aq) + 3NO(g) + 6H2O(l)

We find that the number of atoms (Cr, N, O, and H) and charges are balanced in the combined equation. b. $\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Ce}^{4+}(a q) \rightarrow \mathrm{CO}_{2}(a q)+\mathrm{Ce}^{3+}(a q)$ Repeat the steps from part a: 1. Assign initial oxidation states: C in CH3OH: -2 O in CH3OH: -2 C in CO2: +4 Ce4⁺: +4 Ce3⁺: +3 2. Identify elements undergoing oxidation and reduction: C is being oxidized, from -2 to +4. Ce is being reduced, from +4 to +3. 3. Write down the unbalanced half-reactions: Oxidation half-reaction: CH3OH(aq) ⟶ CO2(aq) Reduction half-reaction: Ce4⁺(aq) ⟶ Ce3⁺(aq) 4. Balance the atoms in each half-reaction: Oxidation half-reaction: CH3OH(aq) ⟶ CO2(aq) + H2O(l) (balance O) Reduction half-reaction: already balanced 5. Balance the charges in each half-reaction: Oxidation half-reaction: CH3OH(aq) + 4H⁺(aq) ⟶ CO2(aq) + H2O(l) + 6e⁻ (balance charges) Reduction half-reaction: Ce4⁺(aq) + e⁻ ⟶ Ce3⁺(aq) 6. Multiply and add the half-reactions: Oxidation half-reaction: CH3OH(aq) + 4H⁺(aq) ⟶ CO2(aq) + H2O(l) + 6e⁻ Reduction half-reaction x 6: 6Ce4⁺(aq) + 6e⁻ ⟶ 6Ce3⁺(aq) Now add the half-reactions: CH3OH(aq) + 4H⁺(aq) + 6Ce4⁺(aq) ⟶ CO2(aq) + H2O(l) + 6Ce3⁺(aq) 7. Confirm that atoms and charges are balanced. c. $\mathrm{SO}_{3}^{2-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{SO}_{4}^{2-}(a q)+\mathrm{Mn}^{2+}(a q)$ Repeat the steps from part a: 1. Assign initial oxidation states: S in SO3²⁻: +4 O in SO3²⁻: -2 S in SO4²⁻: +6 Mn in MnO4⁻: +7 Mn²⁺: +2 2. Identify elements undergoing oxidation and reduction: S is being oxidized, from +4 to +6. Mn is being reduced, from +7 to +2. 3. Write down the unbalanced half-reactions: Oxidation half-reaction: SO3²⁻(aq) ⟶ SO4²⁻(aq) Reduction half-reaction: MnO4⁻(aq) ⟶ Mn²⁺(aq) 4. Balance the atoms in each half-reaction: Oxidation half-reaction is balanced. Reduction half-reaction: MnO4⁻(aq) ⟶ Mn²⁺(aq) + 4H2O(l) (balance O) Add 8H⁺ on left side: MnO4⁻(aq) + 8H⁺(aq) ⟶ Mn²⁺(aq) + 4H2O(l) (balance H) 5. Balance the charges in each half-reaction: Oxidation half-reaction: SO3²⁻(aq) + 2H⁺(aq) ⟶ SO4²⁻(aq) + 2e⁻ (balance charges) Reduction half-reaction is balanced. 6. Multiply and add the half-reactions: Oxidation half-reaction x 5: 5SO3²⁻(aq) + 10H⁺(aq) ⟶ 5SO4²⁻(aq) + 10e⁻ Reduction half-reaction x 5: 5MnO4⁻(aq) + 40H⁺(aq) ⟶ 5Mn²⁺(aq) + 20H2O(l) Now add the half-reactions: 5SO3²⁻(aq) + 10H⁺(aq) + 5MnO4⁻(aq) + 40H⁺(aq)⟶ 5SO4²⁻(aq) + 10e⁻ + 5Mn²⁺(aq) + 20H2O(l) 7. Confirm that atoms and charges are balanced.