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)

• A. 5
• B. 4
• C. 3
• D. 2

Hint:

Balancing redox reactions involves writing the oxidation and reduction half reactions, balancing the electrons, and then combining the half reactions.

Answer 1

The Correct Option is A

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 Electrons
To balance 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 balanced 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.