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Concept 03 / 04Redox Reactions

Balancing Redox Equations: Oxidation Number and Half-Reaction Methods

Balancing redox equations by the oxidation number method and by the ion-electron (half-reaction) method in acidic and basic media, with worked examples for permanganate, dichromate, and disproportionation reactions, and checks for mass and charge balance.

Figure 1
A balanced redox equation must conserve atoms and charge, and the electrons lost by the reductant must equal the electrons gained by the oxidant. Oxidation number method. 1. Write the skeleton equation and assign ONs; identify the atoms whose ON changes. 2. Find the increase in ON per formula unit of reductant and the decrease per formula unit of oxidant. 3. Multiply the species so total increase == total decrease. 4. Balance other atoms except O and H; then balance O with H2O\mathrm{H_{2}O} and H with H+\mathrm{H^{+}} (acidic) or with H2O\mathrm{H_{2}O} and OH−\mathrm{OH^{-}} (basic); check charge. Example: MnO4−+Fe2+→Mn2++Fe3+\mathrm{MnO_{4}^{-} + Fe^{2+} \rightarrow Mn^{2+} + Fe^{3+}} (acidic). Mn falls by 55; Fe rises by 11; so 1 MnO4−1\,\mathrm{MnO_{4}^{-}} to 5 Fe2+5\,\mathrm{Fe^{2+}}. Balance O with 4H2O4\mathrm{H_{2}O} and H with 8H+8\mathrm{H^{+}}: MnO4−+5Fe2++8H+→Mn2++5Fe3++4H2O\mathrm{MnO_{4}^{-} + 5Fe^{2+} + 8H^{+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_{2}O}. Charge check: left −1+10+8=+17-1 + 10 + 8 = +17; right +2+15=+17+2 + 15 = +17. Half-reaction (ion-electron) method (see figure). Example: dichromate oxidising iron(II) in acid. Reduction: Cr2O72−→2Cr3+\mathrm{Cr_{2}O_{7}^{2-} \rightarrow 2Cr^{3+}}; add 7H2O7\mathrm{H_{2}O} on the right, 14H+14\mathrm{H^{+}} on the left, then 6e−6e^{-} on the left: Cr2O72−+14H++6e−→2Cr3++7H2O\mathrm{Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O}. Oxidation: Fe2+→Fe3++e−\mathrm{Fe^{2+} \rightarrow Fe^{3+} + e^{-}}, multiplied by 66. Sum: Cr2O72−+6Fe2++14H+→2Cr3++6Fe3++7H2O\mathrm{Cr_{2}O_{7}^{2-} + 6Fe^{2+} + 14H^{+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_{2}O}. Basic medium. Balance as if acidic, then add as many OH−\mathrm{OH^{-}} to both sides as there are H+\mathrm{H^{+}}; combine H++OH−\mathrm{H^{+} + OH^{-}} into water and cancel. Alternatively, balance O by adding 2OH−2\mathrm{OH^{-}} on the O-deficient side and H2O\mathrm{H_{2}O} on the other. Disproportionation. Write the same reactant in both half-reactions, balance each, and add. For Cl2→Cl−+ClO3−\mathrm{Cl_{2} \rightarrow Cl^{-} + ClO_{3}^{-}} in base: reduction Cl2+2e−→2Cl−\mathrm{Cl_{2} + 2e^{-} \rightarrow 2Cl^{-}} (times 55); oxidation Cl2+12OH−→2ClO3−+6H2O+10e−\mathrm{Cl_{2} + 12OH^{-} \rightarrow 2ClO_{3}^{-} + 6H_{2}O + 10e^{-}}. Adding and dividing by 22: 3Cl2+6OH−→5Cl−+ClO3−+3H2O\mathrm{3Cl_{2} + 6OH^{-} \rightarrow 5Cl^{-} + ClO_{3}^{-} + 3H_{2}O}. Key Takeaways: - Electrons lost == electrons gained; atoms and charge both balance. - Acidic: O with H2O\mathrm{H_{2}O}, H with H+\mathrm{H^{+}}. Basic: then neutralise H+\mathrm{H^{+}} with OH−\mathrm{OH^{-}}. - MnO4−\mathrm{MnO_{4}^{-}} gains 5e−5e^{-} in acid; Cr2O72−\mathrm{Cr_{2}O_{7}^{2-}} gains 6e−6e^{-}. - Pitfall: always check charge at the end; a mass-balanced equation can still be wrong. Brain-Teaser 1: Balance MnO4−+C2O42−→Mn2++CO2\mathrm{MnO_{4}^{-} + C_{2}O_{4}^{2-} \rightarrow Mn^{2+} + CO_{2}} in acidic solution. Solution: Reduction: MnO4−+8H++5e−→Mn2++4H2O\mathrm{MnO_{4}^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_{2}O} (×2\times 2). Oxidation: C2O42−→2CO2+2e−\mathrm{C_{2}O_{4}^{2-} \rightarrow 2CO_{2} + 2e^{-}} (×5\times 5). Sum: 2MnO4−+5C2O42−+16H+→2Mn2++10CO2+8H2O\mathrm{2MnO_{4}^{-} + 5C_{2}O_{4}^{2-} + 16H^{+} \rightarrow 2Mn^{2+} + 10CO_{2} + 8H_{2}O}. Charge: −2−10+16=+4-2 - 10 + 16 = +4 on both sides. Brain-Teaser 2: Balance MnO4−+I−→MnO2+IO3−\mathrm{MnO_{4}^{-} + I^{-} \rightarrow MnO_{2} + IO_{3}^{-}} in basic solution. Solution: Mn falls by 33 (+7+7 to +4+4); I rises by 66 (−1-1 to +5+5), so 2 MnO4−2\,\mathrm{MnO_{4}^{-}} per I−\mathrm{I^{-}}. Oxygen: 88 on the left, 4+3=74 + 3 = 7 on the right; add H2O\mathrm{H_{2}O} on the left and 2OH−2\mathrm{OH^{-}} on the right: 2MnO4−+I−+H2O→2MnO2+IO3−+2OH−\mathrm{2MnO_{4}^{-} + I^{-} + H_{2}O \rightarrow 2MnO_{2} + IO_{3}^{-} + 2OH^{-}}. Charge: −3-3 on each side; O: 99 each; H: 22 each. Brain-Teaser 3: How many moles of Fe2+\mathrm{Fe^{2+}} are oxidised by one mole of Cr2O72−\mathrm{Cr_{2}O_{7}^{2-}}, and by one mole of MnO4−\mathrm{MnO_{4}^{-}}, in acid? Solution: Dichromate accepts 6e−6e^{-} (two Cr, each +6+6 to +3+3), so it oxidises 6 mol6\,\mathrm{mol} Fe2+\mathrm{Fe^{2+}}. Permanganate accepts 5e−5e^{-}, oxidising 5 mol5\,\mathrm{mol}.

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Balancing Redox Equations: Oxidation Number and Half-Reaction Methods — DPP

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