Chemistry · General chemistry I · Worked example

Balance a redox reaction with half-reactions

Balance the reaction of permanganate with iron(II) in acidic solution: MnO₄⁻(aq) + Fe²⁺(aq) → Mn²⁺(aq) + Fe³⁺(aq).

\begin{array}{l} {MnO}_{4}^{-} + {Fe}^{2 +} \\ \to {Mn}^{2 +} + {Fe}^{3 +} \end{array}

Split into half-reactions

Manganese goes from +7 in MnO₄⁻ to +2, so it is reduced. Iron goes from +2 to +3, so it is oxidized. Write one half-reaction for each.

{MnO}_{4}^{-} \to {Mn}^{2 +}, {Fe}^{2 +} \to {Fe}^{3 +}

\begin{array}{c} {MnO}_{4}^{-} \to {Mn}^{2 +} \\ {Fe}^{2 +} \to {Fe}^{3 +} \end{array}

Balance oxygen with water

MnO₄⁻ has four oxygen atoms, so add 4H₂O to the product side.

{MnO}_{4}^{-} \to {Mn}^{2 +} + 4 H_{2} O

Balance hydrogen with H⁺

The four water molecules hold eight hydrogen atoms. In acidic solution, supply them as 8H⁺ on the reactant side.

{MnO}_{4}^{-} + 8 H^{+} \to {Mn}^{2 +} + 4 H_{2} O

\begin{array}{c} {MnO}_{4}^{-} + 8 H^{+} \\ \to {Mn}^{2 +} + 4 H_{2} O \end{array}

Balance charge with electrons

The left side has charge −1 + 8 = +7 and the right side +2. Five electrons on the left make both sides +2. The iron half-reaction releases one electron.

{MnO}_{4}^{-} + 8 H^{+} + 5 e^{-} \to {Mn}^{2 +} + 4 H_{2} O

\begin{array}{c} {MnO}_{4}^{-} + 8 H^{+} + 5 e^{-} \\ \to {Mn}^{2 +} + 4 H_{2} O \end{array}

Make the electrons cancel

Multiply the iron half-reaction by 5 so it releases the five electrons the manganese half-reaction gains, then add the two. The electrons cancel.

{MnO}_{4}^{-} (aq) + 8 H^{+} (aq) + 5 {Fe}^{2 +} (aq) \to {Mn}^{2 +} (aq) + 5 {Fe}^{3 +} (aq) + 4 H_{2} O (l)

\begin{array}{c} {MnO}_{4}^{-} (aq) + 8 H^{+} (aq) \\ + 5 {Fe}^{2 +} (aq) \to {Mn}^{2 +} (aq) \\ + 5 {Fe}^{3 +} (aq) + 4 H_{2} O (l) \end{array}

Check atoms and charge

Manganese, iron, oxygen and hydrogen each match. The charge is −1 + 8 + 10 = +17 on the left and +2 + 15 = +17 on the right.

Atom and charge count
Count	Reactant side	Product side
Mn	1	1
Fe	5	5
O	4	4
H	8	8
Charge	+17	+17

Result

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

Your turn

Balance the half-reaction Cr₂O₇²⁻ → Cr³⁺ in acidic solution.

Show the answer and explanation

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.

Balance Cr with 2Cr³⁺, oxygen with 7H₂O and hydrogen with 14H⁺. The charge is −2 + 14 = +12 on the left and +6 on the right, so six electrons go on the left.

{Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 {Cr}^{3 +} + 7 H_{2} O

\begin{array}{c} {Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \\ \to 2 {Cr}^{3 +} + 7 H_{2} O \end{array}

Keep exploring

Open the half-reaction in Reaction balance & redox and switch the model to a basic half-reaction: OH⁻ and H₂O now balance the oxygen and hydrogen.

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