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14 August, 22:46

Balance the following redox reaction if it occurs in acidic solution. What are the coefficients in front of H + and Fe3 + in the balanced reaction? Fe2 + (aq) + MnO4 - (aq) → Fe3 + (aq) + Mn2 + (aq) Balance the following redox reaction if it occurs in acidic solution. What are the coefficients in front of H + and Fe3 + in the balanced reaction? Fe2 + (aq) + MnO4 - (aq) → Fe3 + (aq) + Mn2 + (aq) H + = 8, Fe3 + = 1 H + = 8, Fe3 + = 5 H + = 2, Fe3 + = 3 H + = 3, Fe3 + = 2 H + = 5, Fe3 + = 1

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  1. 14 August, 23:09
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    H⁺ = 8; Fe³⁺ = 5

    Explanation:

    In order to balance a redox reaction, we use the ion-electron method.

    Step 1: Identify both half-reactions.

    Reduction: MnO₄⁻ (aq) → Mn²⁺ (aq)

    Oxidation: Fe²⁺ (aq) → Fe³⁺ (aq)

    Step 2: Balance the masses using H⁺ and H₂O where appropiate.

    8 H⁺ (aq) + MnO₄⁻ (aq) → Mn²⁺ (aq) + 4 H₂O (l)

    Fe²⁺ (aq) → Fe³⁺ (aq)

    Step 3: Balance the charges adding electrons where appropiate.

    8 H⁺ (aq) + MnO₄⁻ (aq) + 5 e⁻ → Mn²⁺ (aq) + 4 H₂O (l)

    Fe²⁺ (aq) → Fe³⁺ (aq) + 1 e⁻

    Step 4: Multiply both half-reactions by numbers that assure that the number of electrons gained is equal to the number of electrons lost.

    1 * [8 H⁺ (aq) + MnO₄⁻ (aq) + 5 e⁻ → Mn²⁺ (aq) + 4 H₂O (l) ]

    5 * [Fe²⁺ (aq) → Fe³⁺ (aq) + 1 e⁻]

    Step 5: Add both half-reactions

    8 H⁺ (aq) + MnO₄⁻ (aq) + 5 e⁻ + 5 Fe²⁺ (aq) → Mn²⁺ (aq) + 4 H₂O (l) + 5 Fe³⁺ (aq) + 5 e⁻

    8 H⁺ (aq) + MnO₄⁻ (aq) + 5 Fe²⁺ (aq) → Mn²⁺ (aq) + 4 H₂O (l) + 5 Fe³⁺ (aq)
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