Chemistry · Ch 6 — Redox Reactions
Ion Electron Method (Half Reaction Method)
Ion Electron Method (Half Reaction Method)
The ion-electron (half-reaction) method balances a redox equation in seven steps by keeping the oxidation process and the reduction process completely separate until the very end. Step I: write the unbalanced equation, assign oxidation numbers to every atom in the reactants and products, and split the equation into two half equations -- the one whose element's oxidation number rises is the oxidation half, the one whose element's oxidation number falls is the reduction half. Step II: within each half equation on its own, balance every atom except O and H, then balance oxygen by adding H2O to whichever side of that half equation has fewer O atoms. Step III: balance hydrogen in each half equation by adding H⁺ ions to whichever side has fewer H atoms. Step IV: balance the charge on each half equation separately by adding the appropriate number of electrons -- to the PRODUCT side of the oxidation half equation (since oxidation releases electrons as a product) and to the REACTANT side of the reduction half equation (since reduction consumes electrons as a reactant). Step V: multiply each half equation by whatever whole-number factor is needed so that both half equations carry the identical number of electrons, then add the two half equations together and cancel the electrons (and any other species, such as H2O or H⁺, that appear on both sides) to leave one combined, balanced equation. Step VI: if the reaction is in basic medium, add OH⁻ ions -- equal in number to whatever H⁺ ions remain -- to both sides of the combined equation, and combine any H⁺ and OH⁻ that end up on the same side into H2O. Step VII: check that the final equation bala …
Worked out. Worked example balancing Mn²⁺(aq) + ClO3⁻(aq) → MnO2(s) + ClO2⁻(aq) by the half-reaction method. Splitting into half reactions from the oxidation-number changes (Mn: +2 to +4, oxidation; Cl: +5 to +3, reduction): oxidation half Mn²⁺(aq) → MnO2(s); reduction half ClO3⁻(aq) → ClO2⁻(aq). Balancing O with H2O: oxidation becomes Mn²⁺(aq) + 2H2O(l) → MnO2(s); reduction becomes ClO3⁻(aq) → ClO2⁻(aq) + H2O(l). Balancing H with H⁺: oxidation becomes Mn²⁺(aq) + 2H2O(l) → MnO2(s) + 4H⁺(aq); reduction becomes ClO3⁻(aq) + 2H⁺(aq) → ClO2⁻(aq) + H2O(l). Balancing charge with electrons: oxidation half Mn²⁺(aq) + 2H2O(l) → MnO2(s) + 4H⁺(aq) + 2e⁻; reduction half ClO3⁻(aq) + 2H⁺(aq) + e⁻ → ClO2⁻(aq) + H2O(l). Multiplying the reduction half by 2 to match 2 electrons and adding the two halves, the H2O, H⁺ and electrons that appear on both sides ca …
Worked out. Worked example balancing H2O2(aq) + ClO4⁻(aq) → ClO2⁻(aq) + O2(g) by the half-reaction method. Splitting into half reactions (O in H2O2 at -1 rising to 0 in O2, oxidation; Cl in ClO4⁻ at +7 falling to +3 in ClO2⁻, reduction): oxidation half H2O2(aq) → O2(g); reduction half ClO4⁻(aq) → ClO2⁻(aq). Balancing O with H2O: reduction becomes ClO4⁻(aq) → ClO2⁻(aq) + 2H2O(l) (oxidation needs none, both sides already have 2 O). Balancing H with H⁺: oxidation becomes H2O2(aq) → O2(g) + 2H⁺(aq); reduction becomes ClO4⁻(aq) + 4H⁺(aq) → ClO2⁻(aq) + 2H2O(l). Balancing charge with electrons: oxidation half H2O2(aq) → O2(g) + 2H⁺(aq) + 2e⁻; reduction half ClO4⁻(aq) + 4H⁺(aq) + 4e⁻ → ClO2⁻(aq) + 2H2O(l). Multiplying the oxidation half by 2 to match 4 electrons and adding, then cancelling common H⁺, H2O and electrons on both sides, g …
An in-text activity table giving four unbalanced half equations for students to classify, with the first given as a worked example: Cl⁻(aq) → Cl2(g) is classified 'oxidation' (Cl rises from -1 to 0). The remaining three, to be classified the same way: OCl⁻(aq) → Cl⁻(aq) -- Cl falls from +1 to -1, so this is reduction; Fe(OH)2 → Fe(OH)3 -- Fe rises from +2 to +3, so this is oxidation …