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Question: What is the stoichiometric coefficient of SO$_2$ in the following balance reaction? MnO$_4^-$ (aq) ...

What is the stoichiometric coefficient of SO2_2 in the following balance reaction?

MnO4−_4^- (aq) + SO2_2(g) ⟶\longrightarrow Mn2+^{2+}(aq) + HSO4−_4^-(aq)

(in acidic solution)

A

5

B

4

C

3

D

2

Answer

The stoichiometric coefficient of SO2_2 in the balanced reaction is 5.

Explanation

Solution

To balance the given redox reaction in acidic solution, we will use the ion-electron method. The unbalanced reaction is:

MnO4−_4^- (aq) + SO2_2(g) ⟶\longrightarrow Mn2+^{2+}(aq) + HSO4−_4^-(aq)

Step 1: Identify the oxidation and reduction half-reactions.

Manganese is reduced from +7 in MnO4−_4^- to +2 in Mn2+^{2+}.
Sulfur is oxidized from +4 in SO2_2 to +6 in HSO4−_4^-.

Reduction half-reaction: MnO4−_4^- ⟶\longrightarrow Mn2+^{2+}
Oxidation half-reaction: SO2_2 ⟶\longrightarrow HSO4−_4^-

Step 2: Balance atoms other than O and H.

Mn and S atoms are already balanced in both half-reactions.

Step 3: Balance O atoms by adding H2_2O.

Reduction: MnO4−_4^- ⟶\longrightarrow Mn2+^{2+} + 4H2_2O (4 O on left, 0 on right   ⟹  \implies add 4H2_2O to right)
Oxidation: SO2_2 + 2H2_2O ⟶\longrightarrow HSO4−_4^- (2 O on left, 4 O on right   ⟹  \implies add 2H2_2O to left)

Step 4: Balance H atoms by adding H+^+ (since the solution is acidic).

Reduction: MnO4−_4^- + 8H+^+ ⟶\longrightarrow Mn2+^{2+} + 4H2_2O (0 H on left, 8 H on right   ⟹  \implies add 8H+^+ to left)
Oxidation: SO2_2 + 2H2_2O ⟶\longrightarrow HSO4−_4^- + 3H+^+ (4 H on left, 1 H on right   ⟹  \implies add 3H+^+ to right)

Step 5: Balance charge by adding electrons (e−^-).

Reduction: MnO4−_4^- + 8H+^+ + 5e−^- ⟶\longrightarrow Mn2+^{2+} + 4H2_2O (Left charge: -1 + 8 = +7; Right charge: +2. Add 5e−^- to left)
Oxidation: SO2_2 + 2H2_2O ⟶\longrightarrow HSO4−_4^- + 3H+^+ + 2e−^- (Left charge: 0; Right charge: -1 + 3 = +2. Add 2e−^- to right)

Step 6: Multiply the half-reactions by appropriate integers to equalize the number of electrons transferred.

Multiply the reduction half-reaction by 2 (to gain 10e−^-).
Multiply the oxidation half-reaction by 5 (to lose 10e−^-).

2 ×\times (MnO4−_4^- + 8H+^+ + 5e−^- ⟶\longrightarrow Mn2+^{2+} + 4H2_2O)
  ⟹  \implies 2MnO4−_4^- + 16H+^+ + 10e−^- ⟶\longrightarrow 2Mn2+^{2+} + 8H2_2O

5 ×\times (SO2_2 + 2H2_2O ⟶\longrightarrow HSO4−_4^- + 3H+^+ + 2e−^-)
  ⟹  \implies 5SO2_2 + 10H2_2O ⟶\longrightarrow 5HSO4−_4^- + 15H+^+ + 10e−^-

Step 7: Add the balanced half-reactions and cancel common species on both sides.

(2MnO4−_4^- + 16H+^+ + 10e−^- ⟶\longrightarrow 2Mn2+^{2+} + 8H2_2O)

  • (5SO2_2 + 10H2_2O ⟶\longrightarrow 5HSO4−_4^- + 15H+^+ + 10e−^-)

2MnO4−_4^- + 16H+^+ + 10e−^- + 5SO2_2 + 10H2_2O ⟶\longrightarrow 2Mn2+^{2+} + 8H2_2O + 5HSO4−_4^- + 15H+^+ + 10e−^-

Cancel 10e−^- from both sides.
Cancel 15H+^+ from 16H+^+ on the left, leaving 1H+^+ on the left.
Cancel 8H2_2O from 10H2_2O on the left, leaving 2H2_2O on the left.

The balanced equation is:

2MnO4−_4^- (aq) + 5SO2_2(g) + H+^+ (aq) + 2H2_2O (l) ⟶\longrightarrow 2Mn2+^{2+}(aq) + 5HSO4−_4^-(aq)

Therefore, the stoichiometric coefficient of SO2_2 in the balanced reaction is 5.