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

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)