Chemistry · Ch 6 — Redox Reactions
The Oxidation Number Method
The Oxidation Number Method
The oxidation number method balances a redox equation in five steps. Step I: write the unbalanced (skeletal) equation and balance every atom EXCEPT hydrogen and oxygen first; then identify which atoms undergo a change in oxidation number, and by how much per atom. Step II: for the oxidised species, note the increase in oxidation number per atom (and hence the total increase, once the number of atoms is accounted for); for the reduced species, note the decrease per atom (and the total decrease). Find the smallest whole-number multiplying factors for each species that make the TOTAL increase equal the TOTAL decrease, and insert these as coefficients in the equation. Step III: balance oxygen atoms by adding H2O molecules to whichever side has fewer O atoms (one H2O supplies exactly one O atom), then balance hydrogen atoms by adding H⁺ ions to whichever side now has fewer H atoms. Step IV: if the reaction takes place in basic (not acidic) medium, add OH⁻ ions -- in a number equal to however many H⁺ ions were added in Step III -- to BOTH sides of the equation, and wherever an H⁺ and an OH⁻ now sit on the same side, combine that pair into one H2O molecule; this s …
Worked out. Worked example balancing the reaction of potassium permanganate with ferrous sulfate by the oxidation number method. Step 1, skeletal equation: MnO4⁻(aq) + Fe²⁺(aq) → Mn²⁺(aq) + Fe³⁺(aq). Step 2: Fe goes from +2 to +3 (increase of 1 per atom) while Mn goes from +7 to +2 (decrease of 5 per atom); to equalise the total increase and decrease, 5 atoms of Fe²⁺ are needed for every 1 Mn: MnO4⁻(aq) + 5Fe²⁺(aq) → Mn²⁺(aq) + 5Fe³⁺(aq). Step 3: balance O by adding 4H2O to the right (MnO4⁻ has 4 O, the right side has none), then balance H (now 8 H on the right, from the 4H2O) by adding 8H⁺ to the left. Step 4: the medium is acidic, so this step is skipped. Final balanced equation: MnO4⁻(aq) + 5Fe²⁺(aq) + 8H⁺(aq) → Mn²⁺(aq) + 5Fe³⁺(aq) + …
Worked out. Worked example balancing CuO + NH3 → Cu + N2 + H2O by the oxidation number method. Step I: balancing N first gives CuO + 2NH3 → Cu + N2 + H2O. Step II: assigning oxidation numbers, Cu goes from +2 (in CuO) to 0 (decrease of 2 per atom) while N goes from -3 (in NH3) to 0 (in N2, an increase of 3 per atom, and since 2 N atoms are already balanced the total increase per N2 formed is 6); equalising the total increase (which needs 3 x Cu at a decrease of 2 each = 6) and total decrease, 3 atoms of Cu and 2 atoms of N (i.e. one N2) are needed: 3CuO + 2NH3 → 3Cu + N2 + H2O. Step III: balance O by adding 2 more H2O to the right side (3 O on the left needs 3 H2O, so the coefficient of H2O becomes 3): 3CuO + 2NH3 → 3Cu + N2 + 3H2O. Step IV: charges are already balanced (all species neutral). Final b …