Balance the following redox reaction under acidic aqueous conditions using the smallest whole-number coefficients possible. What is the coefficient of HSO3-? MnO4-(aq) + HSO3-(aq)... Balance the following redox reaction under acidic aqueous conditions using the smallest whole-number coefficients possible. What is the coefficient of HSO3-? MnO4-(aq) + HSO3-(aq) → Mn2+(aq) + SO3^(2-)(aq) Read more

Steps to Solve

1. Identify Oxidation States Determine the oxidation states of each element in the reactants and products:

• In $MnO_4^-$, Mn is in the +7 oxidation state.
• In $HSO_3^-$, S is in the +4 oxidation state.
• In $Mn^{2+}$, Mn is in the +2 oxidation state.
• In $SO_3^{2-}$, S is still in the +4 oxidation state.

2. Determine Oxidation and Reduction Identify what is being oxidized and what is being reduced:

• Mn is reduced from +7 to +2 (gaining electrons).
• S is not oxidized or reduced—it changes oxidation state but remains the same in terms of total electrons.

3. Write Half-Reactions Write the half-reactions for oxidation and reduction:

• Reduction half-reaction for Mn: $$ MnO_4^{-} + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O $$
• Oxidation half-reaction for HSO3- (acting as a reducing agent): $$ 2HSO_3^{-} \rightarrow 2SO_3^{2-} + 2H^+ + 2e^- $$

4. Balance Electrons Make sure the number of electrons lost equals the number of electrons gained:

• Multiply the oxidation half-reaction by 5 to balance electrons: $$ 5(2HSO_3^{-} \rightarrow 2SO_3^{2-} + 2H^+ + 2e^-) $$ This leads to: $$ 10HSO_3^{-} \rightarrow 10SO_3^{2-} + 10H^+ + 10e^- $$

5. Combine Half-Reactions Add the balanced half-reactions together: $$ MnO_4^{-} + 10HSO_3^{-} + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O + 10SO_3^{2-} + 10H^+ + 10e^- $$

6. Simplify the Reaction Combine and simplify to get the final balanced reaction: $$ MnO_4^{-} + 5HSO_3^{-} + 4H^+ \rightarrow Mn^{2+} + 5SO_3^{2-} + 2H_2O $$