Question

Write balanced equations for the following reactions in basic solution. (a) \(\mathrm{SO}_{2}(g)+\mathrm{I}_{2}(a q) \longrightarrow \mathrm{SO}_{3}(g)+\mathrm{I}^{-}(a q)\) (b) \(\mathrm{Zn}(s)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NH}_{3}(a q)+\mathrm{Zn}^{2+}(a q)\) (c) \(\mathrm{ClO}^{-}(a q)+\mathrm{CrO}_{2}^{-}(a q) \longrightarrow \mathrm{Cl}^{-}(a q)+\mathrm{CrO}_{4}^{2-}(a q)\) (d) \(\mathrm{K}(s)+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{H}_{2}(g)\)

Short answer

2 Cl^−(aq) + CrO4^2−(aq) + 14 OH^−(aq)

Step-by-step solution

Break into half-reactions and identify oxidation and reduction

Separate the given equation into oxidation and reduction half-reactions: Oxidation half-reaction: SO2(g) -> SO3(g) Reduction half-reaction: I2(aq) -> 2 I^−(aq)

Balance half-reactions under basic conditions

For the oxidation half-reaction: 1. Balance the oxygen atoms: SO2(g) + H2O(l) -> SO3(g) + 2 H^+(aq) 2. Balance the hydrogen atoms: SO2(g) + H2O(l) -> SO3(g) + 2 H^+(aq) + 2 e^− 3. Balance the charges under basic conditions by adding same number of OH^-: SO2(g) + H2O(l) -> SO3(g) + 2 H2O(l) + 2 e^− For the reduction half-reaction: 1. Balance the Iodine atoms: I2(aq) -> 2 I^−(aq) 2. Balance the charges: I2(aq) + 2 e^− -> 2 I^−(aq)

Combine balanced half-reactions and simplify

Combine the balanced half-reactions: SO2(g) + H2O(l) + I2(aq) + 2 e^− -> SO3(g) + 2 H2O(l) + 2 I^−(aq) + 2 e^− Cancel 2 electrons and 2 H2O(l) on both sides: SO2(g) + I2(aq) -> SO3(g) + 2 I^−(aq) (b) Zn(s) + NO3^−(aq) -> NH3(aq) + Zn^2+(aq)

Break into half-reactions and identify oxidation and reduction

Separate the given equation into oxidation and reduction half-reactions: Oxidation half-reaction: Zn(s) -> Zn^2+(aq) Reduction half-reaction: NO3^−(aq) -> NH3(aq)

Balance half-reactions under basic conditions

For the oxidation half-reaction: 1. Balance the zinc atoms: Zn(s) -> Zn^2+(aq) 2. Balance the charges: Zn(s) -> Zn^2+(aq) + 2 e^− For the reduction half-reaction: 1. Balance the nitrogen atoms: NO3^−(aq) -> NH3(aq) 2. Balance the hydrogen atoms: NO3^−(aq) + 6 H^+(aq) -> NH3(aq) + 3 H2O(l) 3. Balance the charges under basic conditions: NO3^−(aq) + 6 H^+(aq) + 8 OH^−(aq) -> NH3(aq) + 3 H2O(l) + 5 e^−

Find the least common multiple of the electrons

The least common multiple of 2 e^−(from oxidation) and 5 e^−(from reduction) is 10 e^−.

Balance electrons and combine half-reactions

Multiply the oxidation half-reaction by 5, and the reduction half-reaction by 2, then combine: 5 Zn(s) -> 5 Zn^2+(aq) + 10 e^− 2(NO3^−(aq) + 6 H^+(aq) + 8 OH^−(aq)) -> 2(NH3(aq) + 3 H2O(l) + 5 e^−) 5 Zn(s) + 2 NO3^−(aq) + 12 H^+(aq) + 16 OH^−(aq) -> 5 Zn^2+(aq) + 10 e^− + 2 NH3(aq) + 6 H2O(l) + 10 e^− Cancel 10 electrons and 6 H2O(l) on both sides, then simplify: 5 Zn(s) + 2 NO3^−(aq) + 12 H^+(aq) + 10 OH^−(aq) -> 5 Zn^2+(aq) + 2 NH3(aq) (c) ClO^−(aq) + CrO2^−(aq) -> Cl^−(aq) + CrO4^2−(aq)

Break into half-reactions and identify oxidation and reduction

Separate the given equation into oxidation and reduction half-reactions: Oxidation half-reaction: ClO^−(aq) -> Cl^−(aq) Reduction half-reaction: CrO2^−(aq) -> CrO4^2−(aq)

Balance half-reactions under basic conditions

For the oxidation half-reaction: 1. Balance the chlorine atoms: ClO^−(aq) -> Cl^−(aq) 2. Balance the oxygen atoms: ClO^−(aq) + 2 H^+(aq) + H2O(l) -> Cl^−(aq) + 4 OH^−(aq) 3. Balance the charges under basic conditions: ClO^- + H2O -> Cl^- + 2 OH^- For the reduction half-reaction: 1. Balance the chromium atoms: CrO2^−(aq) -> CrO4^2−(aq) 2. Balance the oxygen atoms: CrO2^−(aq) + 2 H2O(l) -> CrO4^2−(aq) + 4 H^+(aq) 3. Balance the charges under basic conditions: CrO2^−(aq) + 4 H2O(l) -> CrO4^2−(aq) + 6 OH^−(aq) + 2 e^−

Balance electrons and combine half-reactions

Multiply the oxidation half-reaction by 2 to balance the electrons, then combine: 2(ClO^−(aq) + H2O(l)) -> 2(Cl^−(aq) + 2 OH^−(aq)) + 2 e^− CrO2^−(aq) + 4 H2O(l) -> CrO4^2−(aq) + 6 OH^−(aq) + 2 e^− 2 ClO^−(aq) + 2 H2O(l) + CrO2^−(aq) + 4 H2O(l) -> 2 Cl^−(aq) + 8 OH^−(aq) + CrO4^2−(aq) + 6 OH^−(aq) Cancel 2 electrons, and simplify the equation: 2 ClO^−(aq) + CrO2^−(aq) + 6 H2O(l) ->