Balancing Redox Equations: MDCAT Chemistry notes
Balancing Redox Equations for MDCAT: oxidation number method and ion-electron (half-reaction) method, with fully worked acidic-medium examples.
Two methods
A redox equation is balanced when both the atoms and the charges are balanced, and the electrons lost equal the electrons gained. The FSc course uses two methods, and MDCAT asks which steps belong to which method.
| Oxidation number method | Ion-electron (half-reaction) method |
|---|---|
| Assign oxidation numbers to all atoms | Split the reaction into oxidation and reduction half-reactions |
| Identify the atoms whose oxidation number changes | Balance atoms other than O and H in each half |
| Find the increase and decrease per atom | Balance O by adding $\mathrm{H_2O}$ |
| Multiply so that total increase equals total decrease | Balance H by adding $\mathrm{H^+}$ (acidic medium) |
| Balance the remaining atoms by inspection (usually H and O last) | Balance charge by adding electrons, then equalize electrons and add the halves |
Splitting into half-reactions is not a step of the oxidation number method.
Worked example 1: oxidation number method
Balance: $\mathrm{Cu + HNO_3 \to Cu(NO_3)_2 + NO + H_2O}$
- Changes: Cu goes from 0 to +2 (increase 2). N goes from +5 in $\mathrm{HNO_3}$ to +2 in NO (decrease 3).
- Equalize: $2 \times 3 = 3 \times 2$. Take 3 Cu and 2 NO.
- $\mathrm{3Cu \to 3Cu(NO_3)_2}$ needs 6 nitrate N. Adding the 2 N that become NO gives 8 $\mathrm{HNO_3}$.
- 8 H gives 4 $\mathrm{H_2O}$. Check O: 24 on the left; 18 + 2 + 4 = 24 on the right.
Worked example 2: ion-electron method (acidic)
Balance: $\mathrm{MnO_4^- + Fe^{2+} \to Mn^{2+} + Fe^{3+}}$ in acid.
Reduction half: $\mathrm{MnO_4^- \to Mn^{2+}}$. Add 4 $\mathrm{H_2O}$ on the right for oxygen, then 8 $\mathrm{H^+}$ on the left for hydrogen. The charge on the left is +7 and on the right +2, so add 5 electrons on the left:
$$\mathrm{MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O}$$Oxidation half: $\mathrm{Fe^{2+} \to Fe^{3+} + e^-}$. Multiply by 5.
$$\mathrm{MnO_4^- + 5Fe^{2+} + 8H^+ \to Mn^{2+} + 5Fe^{3+} + 4H_2O}$$Charge check: $-1 + 10 + 8 = 17$ on the left and $2 + 15 = 17$ on the right.
Worked example 3: dichromate half-reaction
$\mathrm{Cr_2O_7^{2-} \to 2Cr^{3+}}$: add 7 $\mathrm{H_2O}$ on the right, 14 $\mathrm{H^+}$ on the left, then 6 electrons (Cr changes from +6 to +3, so 3 electrons for each of the two Cr atoms):
$$\mathrm{Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O}$$Basic medium
Balance first as if the solution were acidic. Then add as many $\mathrm{OH^-}$ ions to both sides as there are $\mathrm{H^+}$. Combine $\mathrm{H^+ + OH^- \to H_2O}$ and cancel any water that appears on both sides.
Common MDCAT traps
- Oxygen is balanced with $\mathrm{H_2O}$, not with $\mathrm{O_2}$ or $\mathrm{OH}$ (in acidic medium).
- Half-reactions belong only to the ion-electron method.
- Electrons in the final equation must cancel completely. If any are left over, the multipliers are wrong.
- Check charge as well as atoms. An equation can be atom-balanced but charge-unbalanced.
Quick revision
- Electrons lost = electrons gained in every balanced redox equation.
- Permanganate in acid takes 5 electrons ($+7 \to +2$).
- Dichromate in acid takes 6 electrons ($2 \times (+6 \to +3)$).
- Acidic medium: $\mathrm{H_2O}$ for O, $\mathrm{H^+}$ for H, electrons for charge.