Question:

What is the stoichiometric coefficient of SO$_2$ in the following balanced reaction? \[ {MnO}_4^-(aq) + {SO}_2(g) \rightarrow {Mn}^{2+}(aq) + {HSO}_4^-(aq) \] (in acidic solution)

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Balancing redox reactions involves writing the oxidation and reduction half reactions, balancing the electrons, and then combining the half reactions.
Updated On: Mar 24, 2025
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The Correct Option is A

Solution and Explanation

Step 1: {Oxidation Half Reaction}
Write the oxidation half reaction: \[ {SO}_2(g) + 2 {H}_2{O}(l) \rightarrow {HSO}_4^-(aq) + 3 {H}^+(aq) + 2e^- \] Step 2: {Reduction Half Reaction}
Write the reduction half reaction: \[ {MnO}_4^-(aq) + 8 {H}^+(aq) + 5e^- \rightarrow {Mn}^{2+}(aq) + 4 {H}_2{O}(l) \] Step 3: {Balance the Electrons}
To balance the electrons, multiply the oxidation half reaction by 5 and the reduction half reaction by 2: \[ 5 {SO}_2(g) + 10 {H}_2{O}(l) \rightarrow 5 {HSO}_4^-(aq) + 15 {H}^+(aq) + 10e^- \] \[ 2 {MnO}_4^-(aq) + 16 {H}^+(aq) + 10e^- \rightarrow 2 {Mn}^{2+}(aq) + 8 {H}_2{O}(l) \] Step 4: {Combine and Simplify}
Combine the two half reactions: \[ 2 {MnO}_4^-(aq) + 5 {SO}_2(g) + 16 {H}^+(aq) \rightarrow 2 {Mn}^{2+}(aq) + 5 {HSO}_4^-(aq) + 8 {H}_2{O}(l) \] Thus, the stoichiometric coefficient of SO$_2$ is 5.
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