Balancing redox equations — ion-electron (half-reaction) method, oxidation-number method

WHY do ordinary balancing methods fail here?

For a reaction like $\mathrm{MnO_4^- + Fe^{2+} \to Mn^{2+} + Fe^{3+}}$ (acidic), trial-and-error fails because:

• Oxygen and hydrogen appear from the solvent (water, $\mathrm{H^+}$, $\mathrm{OH^-}$), not written in the skeleton.
• Charge must balance, not just atoms.

So we need a systematic method that tracks electrons explicitly.

Method 1 — Ion–Electron (Half-Reaction) Method

HOW (acidic medium), step by step:

1. Split the skeleton into an oxidation half and a reduction half.
2. Balance all atoms except O and H.
3. Balance O by adding $\mathrm{H_2O}$.
4. Balance H by adding $\mathrm{H^+}$.
5. Balance charge by adding $e^-$ to the more-positive side.
6. Multiply each half so electrons cancel; add and simplify.

Basic medium: do all of the above as if acidic, then add equal $\mathrm{OH^-}$ to both sides to neutralize every $\mathrm{H^+}$ (forming $\mathrm{H_2O}$), and cancel duplicate waters.

[!example] Worked Example 1 — $\mathrm{MnO_4^-}$ + $\mathrm{Fe^{2+}}$ (acidic)

Reduction half: $\mathrm{MnO_4^- \to Mn^{2+}}$

• Balance O with water: $\mathrm{MnO_4^- \to Mn^{2+} + 4H_2O}$ Why? Left has 4 O; add 4 waters right to supply them.
• Balance H with $\mathrm{H^+}$: $\mathrm{MnO_4^- + 8H^+ \to Mn^{2+} + 4H_2O}$ Why? Right now has 8 H (from $4\mathrm{H_2O}$); add 8 $\mathrm{H^+}$ left.
• Balance charge: Left $= -1+8 = +7$, Right $= +2$. Add $5e^-$ left. $\mathrm{MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O}$ Why? $+7 + (-5) = +2$. ✔ Mn goes $+7 \to +2$, gains 5 electrons — matches!

Oxidation half: $\mathrm{Fe^{2+} \to Fe^{3+} + e^-}$ Why? Charge $+2 \to +3$ means loss of 1 electron.

Equalize electrons: LCM of 5 and 1 is 5 → multiply Fe half by 5. $\mathrm{MnO_4^- + 8H^+ + 5Fe^{2+} \to Mn^{2+} + 4H_2O + 5Fe^{3+}}$

Check: charge left $= -1+8+10 = +17$; right $= +2+15 = +17$. ✔

[!example] Worked Example 2 — Basic medium: $\mathrm{MnO_4^- + I^- \to MnO_2 + I_2}$

Reduction: $\mathrm{MnO_4^- \to MnO_2}$

• O: $\mathrm{MnO_4^- \to MnO_2 + 2H_2O}$
• H: $\mathrm{MnO_4^- + 4H^+ \to MnO_2 + 2H_2O}$
• charge: left $-1+4=+3$, right $0$ → add $3e^-$ left: $\mathrm{MnO_4^- + 4H^+ + 3e^- \to MnO_2 + 2H_2O}$

Oxidation: $\mathrm{2I^- \to I_2 + 2e^-}$

Equalize (LCM 6): ×2 reduction, ×3 oxidation: $\mathrm{2MnO_4^- + 8H^+ + 6I^- \to 2MnO_2 + 4H_2O + 3I_2}$

Convert to basic: add $8\,\mathrm{OH^-}$ to both sides. Left $8\mathrm{H^+}+8\mathrm{OH^-}=8\mathrm{H_2O}$; cancel 4 waters on the right: $\mathrm{2MnO_4^- + 4H_2O + 6I^- \to 2MnO_2 + 8OH^- + 3I_2}$ Why add OH⁻? In base there is no free $\mathrm{H^+}$; neutralizing it into water gives a physically real equation.

Method 2 — Oxidation-Number Method

HOW:

1. Assign oxidation numbers to every atom.
2. Identify atoms that go up (oxidized) and down (reduced); compute Δ per atom.
3. Multiply species by factors so total increase = total decrease.
4. Balance remaining atoms (O with $\mathrm{H_2O}$, H with $\mathrm{H^+}$/$\mathrm{OH^-}$), then charge.

[!example] Worked Example 3 — $\mathrm{Cu + HNO_3 \to Cu(NO_3)_2 + NO_2 + H_2O}$ (acidic)

• Cu: $0 \to +2$, loses 2 electrons.
• N in some $\mathrm{HNO_3}$: $+5 \to +4$ (in $\mathrm{NO_2}$), gains 1 each. Why only some? The nitrate in $\mathrm{Cu(NO_3)_2}$ stays $+5$ (spectator); only the N that becomes $\mathrm{NO_2}$ is reduced.
• Balance electrons: 1 Cu (loses 2) needs 2 N reduced (gain 1 each).

$\mathrm{Cu + 4HNO_3 \to Cu(NO_3)_2 + 2NO_2 + 2H_2O}$ Check: N: left 4 = right $2+2$ ✔; H: 4 = 4 ✔; O: 12 = $6+4+2$ ✔.

Flashcards

Connections

• Oxidation states / oxidation number rules
• Oxidizing and reducing agents
• Electrochemical cells & standard electrode potentials
• Stoichiometry & mole ratios
• Acid–base neutralization (for the OH⁻ step)
• Conservation of mass and charge

Concept Map

Hinglish (regional understanding)