Balancing redox half-equations template

Step-by-step ion–electron (half-equation) method in acidic solution for GCSE, A-level and NCERT Class 11: MnO₄⁻/Fe²⁺ (5 e⁻) and Cr₂O₇²⁻/I⁻ (6 e⁻) worked through each step, with charge checks, oxidation-state changes and colour changes.

What the template shows

• Balance atoms, then O with H₂O, H with H⁺ and charge with electrons; equalise the electrons and add
• Manganate(VII) oxidises iron(II) in acid
• Dichromate(VI) oxidises iodide in acid
• charge: −1 + 8 − 5 = +2 on both sides · Mn +7 → +2
• Other half
• × 5 so that both half-equations transfer 5 e⁻
• Purple MnO₄⁻ turns colourless: self-indicating titration (acidify with dilute H₂SO₄, not HCl).
• charge: −2 + 14 − 6 = +6 on both sides · each Cr +6 → +3
• × 3 so that both half-equations transfer 6 e⁻
• Orange Cr₂O₇²⁻ turns green Cr³⁺; the iodine released colours the solution brown.
• OIL RIG: oxidation is loss of electrons, reduction is gain. Electrons appear on the right of an oxidation half-equation and on the left of a reduction.
• In acid, balance O with H₂O and H with H⁺; in alkaline solution, then add the same number of OH⁻ to both sides to turn H⁺ into H₂O.
• The electrons gained must equal the electrons lost: multiply the half-equations so the electrons cancel. Check that atoms and total charge balance in the final equation.
• The change in oxidation state equals the electrons per atom: Mn +7 → +2 takes 5 e⁻; each Cr +6 → +3 takes 3 e⁻, 6 e⁻ for Cr₂O₇²⁻.
• Every part of this figure is editable · sources on the template card

Drawn with 5 shapes – every one editable in MolDraw.

Sources

• Housecroft & Sharpe, Inorganic Chemistry, 5th ed., Pearson 2018, ch. 8 Textbook
• NCERT, Chemistry Part II, Class XI, ch. 7 Redox Reactions Textbook