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Difficulty: MediumBalancing Redox Equations and Half-Reactions
Consider the unbalanced redox reaction occurring in acidic solution:
a AsO33−(aq)+b MnO4−(aq)+c H+(aq)→d AsO43−(aq)+e Mn2+(aq)+f H2O(l)\text{a AsO}_3^{3-}(\text{aq}) + \text{b MnO}_4^-(\text{aq}) + \text{c H}^+(\text{aq}) \rightarrow \text{d AsO}_4^{3-}(\text{aq}) + \text{e Mn}^{2+}(\text{aq}) + \text{f H}_2\text{O}(\text{l})
When this ionic equation is balanced using the smallest possible whole-number coefficients, what is the value of the coefficient cc for H+\text{H}^+?

Answer: 6

Answer

The coefficient c for hydrogen ions (H⁺) in the balanced redox equation is 6.
Balancing the oxidation half-reaction shows that each arsenite ion produces 2 electrons and 2 H⁺ ions. The reduction half-reaction shows that each permanganate ion consumes 5 electrons and 8 H⁺ ions. Multiplying the oxidation half-reaction by 5 and the reduction half-reaction by 2 balances the total electron transfer at 10 electrons. Combining the equations gives 16 H⁺ on the left and 10 H⁺ on the right, which simplifies to 6 H⁺ on the reactant side.

Step-by-Step Solution

1
Balance the oxidation half-reaction (arsenite to arsenate)
AsO₃³⁻ + H₂O → AsO₄³⁻ + 2H⁺ + 2e⁻
Arsenic changes oxidation state from +3 to +5, releasing 2 electrons. Oxygen is balanced with H₂O and hydrogen with H⁺.
2
Balance the reduction half-reaction (permanganate to manganese(II))
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Manganese changes oxidation state from +7 to +2, consuming 5 electrons in acidic medium.
3
Equalize the electrons transferred in both half-reactions
5(AsO₃³⁻ + H₂O → AsO₄³⁻ + 2H⁺ + 2e⁻) and 2(MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O)
The total number of electrons gained and lost must equal 10 electrons.
4
Combine the half-reactions and cancel common terms
5 AsO₃³⁻ + 2 MnO₄⁻ + 6 H⁺ → 5 AsO₄³⁻ + 2 Mn²⁺ + 3 H₂O
Subtracting 10 H⁺ and 5 H₂O from both sides leaves 6 H⁺ on the reactant side.

Key Concept

Balancing Ion-Electron Redox Half-Reactions in Acidic Medium
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Topics

Balancing Redox Equations and Half-Reactions

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