Balancing Redox Reactions

Balancing Equations balanced atoms. Redox equations in water need more: the electrons lost must equal the electrons gained, and the charges must balance, with $\mathrm{H_2O}$, $\mathrm{H^+}$ or $\mathrm{OH^-}$ often joining in. The half-reaction method, built on Oxidation and Reduction, handles all of it.

Key idea

Half-reaction method (acidic solution): (1) Split into oxidation and reduction half-reactions. (2) Balance atoms other than O and H. (3) Balance O with $\mathrm{H_2O}$. (4) Balance H with $\mathrm{H^+}$. (5) Balance charge with electrons. (6) Multiply the halves so the electrons match, add them, and cancel anything on both sides. In basic solution, finish by adding enough $\mathrm{OH^-}$ to both sides to turn each $\mathrm{H^+}$ into $\mathrm{H_2O}$.

Why it works

Each half-reaction shows one partner's change on its own. Water supplies or absorbs oxygen atoms, and $\mathrm{H^+}$ supplies hydrogen, because in acidic water both are plentiful. Electrons are added to the more positive side until the charges match; their number is the change in oxidation number. Since electrons can't be left over, the halves must be scaled to lose and gain the same number. In base, $\mathrm{H^+}$ is scarce, so we neutralize it: $\mathrm{H^+} + \mathrm{OH^-} \to \mathrm{H_2O}$.

Balance each half with $\mathrm{H_2O}$, $\mathrm{H^+}$ and electrons; the two electron counts must match before the halves are added.

Worked example: Permanganate and iron(II), in acid

Balance $\mathrm{MnO_4^-} + \mathrm{Fe^{2+}} \to \mathrm{Mn^{2+}} + \mathrm{Fe^{3+}}$ in acidic solution.

Oxidation: $\mathrm{Fe^{2+}} \to \mathrm{Fe^{3+}} + e^-$.
Reduction: $\mathrm{MnO_4^-} \to \mathrm{Mn^{2+}}$. Add $4\,\mathrm{H_2O}$ on the right for the O, then $8\,\mathrm{H^+}$ on the left for the H: $\mathrm{MnO_4^-} + 8\,\mathrm{H^+} \to \mathrm{Mn^{2+}} + 4\,\mathrm{H_2O}$.
Charge: left $-1 + 8 = +7$, right $+2$. Add $5e^-$ on the left: $\mathrm{MnO_4^-} + 8\,\mathrm{H^+} + 5e^- \to \mathrm{Mn^{2+}} + 4\,\mathrm{H_2O}$.
Multiply the iron half by 5 and add.

Answer: $$\mathrm{MnO_4^-} + 5\,\mathrm{Fe^{2+}} + 8\,\mathrm{H^+} \to \mathrm{Mn^{2+}} + 5\,\mathrm{Fe^{3+}} + 4\,\mathrm{H_2O}$$

Worked example: Dichromate and iodide, in acid

Balance $\mathrm{Cr_2O_7^{2-}} + \mathrm{I^-} \to \mathrm{Cr^{3+}} + \mathrm{I_2}$ in acidic solution.

Oxidation: $2\,\mathrm{I^-} \to \mathrm{I_2} + 2e^-$.
Reduction: $\mathrm{Cr_2O_7^{2-}} \to 2\,\mathrm{Cr^{3+}}$; add $7\,\mathrm{H_2O}$ right, $14\,\mathrm{H^+}$ left; charge $-2 + 14 = +12$ against $+6$: add $6e^-$ left.
Multiply the iodide half by 3: $6e^-$ each way. Add.

Answer: $$\mathrm{Cr_2O_7^{2-}} + 6\,\mathrm{I^-} + 14\,\mathrm{H^+} \to 2\,\mathrm{Cr^{3+}} + 3\,\mathrm{I_2} + 7\,\mathrm{H_2O}$$

Worked example: In basic solution

Balance $\mathrm{MnO_4^-} + \mathrm{I^-} \to \mathrm{MnO_2} + \mathrm{I_2}$ in basic solution.

Reduction as if in acid: $\mathrm{MnO_4^-} + 4\,\mathrm{H^+} + 3e^- \to \mathrm{MnO_2} + 2\,\mathrm{H_2O}$. Oxidation: $2\,\mathrm{I^-} \to \mathrm{I_2} + 2e^-$.
Match electrons (6): $$2\,\mathrm{MnO_4^-} + 8\,\mathrm{H^+} + 6\,\mathrm{I^-} \to 2\,\mathrm{MnO_2} + 4\,\mathrm{H_2O} + 3\,\mathrm{I_2}$$
Add $8\,\mathrm{OH^-}$ to both sides: the left's $8\,\mathrm{H^+}$ become $8\,\mathrm{H_2O}$; cancel 4 waters.

Answer: $$2\,\mathrm{MnO_4^-} + 6\,\mathrm{I^-} + 4\,\mathrm{H_2O} \to 2\,\mathrm{MnO_2} + 3\,\mathrm{I_2} + 8\,\mathrm{OH^-}$$

Common mistake

Balancing the atoms but not the charge, or adding $\mathrm{H^+}$ in a basic solution and never converting it to $\mathrm{OH^-}$. "$\mathrm{MnO_4^-} + \mathrm{Fe^{2+}} \to \mathrm{Mn^{2+}} + \mathrm{Fe^{3+}}$" with $\mathrm{H^+}$ and water added looks balanced in atoms but has 5 times too few iron ions. Check both: count each kind of atom, then add up the charges on each side; they must agree. In base, no $\mathrm{H^+}$ may remain.

Balancing Redox Reactions

Practice

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