Balancing by the half-reaction method.

Step through the half-reaction method: electrons lost and gained are drawn as counters, multipliers equalise them, then H2O, H+ (or OH−) are added until an atom tally and a charge tally both read zero.

Balance MnO4⁻ + I⁻ in basic medium with no H⁺ left in the final equation; goal checks atom and charge tallies are both zero.

Leaving H⁺ in a basic-medium answer; electrons lost ≠ electrons gained.

• Oxidation number method: write the skeletal equation, find the atoms whose oxidation numbers change, and multiply so that the total increase equals the total decrease.
• Then balance the charge (with H⁺ in acid or OH⁻ in base), then hydrogen with water, and finally check oxygen.
• Example: dichromate and sulphite in acid give Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O. Each Cr falls by 3 (6 for the pair) and each S rises by 2, so three sulphites are needed.
• Half-reaction method: split the reaction into oxidation and reduction halves, balance the atoms other than O and H, then O with H₂O and H with H⁺ (in acid). Add electrons to balance charge, scale the halves to equal electron counts and add them.
• Example: Fe²⁺ with dichromate in acid gives 6Fe²⁺ + Cr₂O₇²⁻ + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O, since each Fe²⁺ gives one electron and one Cr₂O₇²⁻ takes six.
• In basic medium, balance as if acidic and then add as many OH⁻ to both sides as there are H⁺, combining H⁺ + OH⁻ into water. Permanganate and iodide in base give 2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻.
• A balanced redox equation must balance atoms of every element and the total charge on both sides.

Teaching this idea? Put the simulation on your own page or LMS with this code.