Balance the following redox reaction under acidic aqueous conditions using the coefficients possible. What is the coefficient of HSO3-? MnO4-(aq) + HSO3-(aq) → Mn2+(aq) + SO43-(aq) Balance the following redox reaction under acidic aqueous conditions using the coefficients possible. What is the coefficient of HSO3-? MnO4-(aq) + HSO3-(aq) → Mn2+(aq) + SO43-(aq) Read more

Steps to Solve

1. Identify Oxidation and Reduction Reactions

Determine which species are being oxidized and reduced. Here, $MnO_4^-$ is reduced to $Mn^{2+}$, and $HSO_3^-$ is oxidized to $SO_4^{2-}$.

2. Write the Half-Reactions

Write the half-reactions for the oxidation and reduction processes:

• Reduction: $$ MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O $$

• Oxidation: $$ HSO_3^- + H_2O \rightarrow SO_4^{2-} + 2H^+ + 2e^- $$

3. Balance the Electrons

To balance the overall reaction, ensure that the number of electrons lost equals the number gained. Multiply the oxidation half-reaction by 5 (to match the 5 electrons from the reduction half-reaction):

$$ 5(HSO_3^- + H_2O \rightarrow SO_4^{2-} + 2H^+ + 2e^-) $$

This simplifies to: $$ 5HSO_3^- + 5H_2O \rightarrow 5SO_4^{2-} + 10H^+ + 10e^- $$

4. Combine Half-Reactions

Add the balanced half-reactions together:

$$ MnO_4^- + 8H^+ + 5e^- + 5HSO_3^- + 5H_2O \rightarrow Mn^{2+} + 4H_2O + 5SO_4^{2-} + 10H^+ + 10e^- $$

5. Simplify the Reaction

Cancel out the common species on both sides (e.g., $H_2O$ and $H^+$):

$$ MnO_4^- + 5HSO_3^- + H_2O \rightarrow Mn^{2+} + 5SO_4^{2-} + 2H^+ $$

6. Identify the Coefficient of HSO₃⁻

From the balanced equation, the coefficient of $HSO_3^-$ is 5.