Short Notes: Redox Reactions

7.1 Basic Concepts

Redox reaction is a chemical reaction in which there is a transfer of electrons between species. In a redox process one species loses electrons and another gains electrons.

Oxidation is the process of loss of electrons or an increase in oxidation number.

Reduction is the process of gain of electrons or a decrease in oxidation number.

Oxidising agent is the species that causes oxidation of another species and itself gets reduced.

Reducing agent is the species that causes reduction of another species and itself gets oxidised.

Disproportionation is a special redox reaction in which a single substance is both oxidised and reduced.

7.2 Oxidation Number Rules

• Oxidation number (O.N.) of an atom in a compound is the hypothetical charge it would have if all bonds were ionic.
• Free element has O.N. = 0.
• Monatomic ion has O.N. equal to the ion charge.
• Hydrogen generally has O.N. = +1; in metal hydrides (for example NaH) H has O.N. = -1.
• Oxygen generally has O.N. = -2; in peroxides (for example H2O2) O has O.N. = -1; in OF2 oxygen has O.N. = +2.
• Alkali metals (group 1) have O.N. = +1 in their compounds; alkaline earth metals (group 2) have O.N. = +2.
• Halogens generally have O.N. = -1 unless combined with oxygen or with a more electronegative halogen.
• The sum of oxidation numbers in a neutral molecule equals 0; in a polyatomic ion equals the ion charge.

Examples:

• O.N. of O in H2O is -2 and of H is +1.
• O.N. of Mn in KMnO4: K = +1, O = -2 (×4) so Mn = +7.
• O.N. of S in SO4^2-: O = -2 (×4) so S = +6.

7.3 Balancing Redox Reactions

Redox equations must be balanced for both mass and charge. Two common approaches are the oxidation-reduction (half-reaction) method and the oxidation number method. The half-reaction method is systematic and works well in acidic and basic media.

7.3.1 Half-Reaction Method (Acidic Medium)

1. Write separate oxidation and reduction half-reactions.
2. Balance all atoms except hydrogen and oxygen.
3. Balance oxygen atoms by adding H2O molecules.
4. Balance hydrogen atoms by adding H+ ions.
5. Balance the charge of each half-reaction by adding electrons (e-).
6. Multiply the half-reactions by appropriate factors so that the number of electrons lost equals the number gained.
7. Add the half-reactions and cancel species that appear on both sides (including electrons).
8. Check that atoms and charge are balanced; if necessary, simplify coefficients to smallest whole numbers.

Worked example (acidic medium): Balance the reaction between permanganate ion and ferrous ion:

Unbalanced equation:

$\; \mathrm{MnO_4^- + Fe^{2+} \rightarrow Mn^{2+} + Fe^{3+}}$

Write the oxidation and reduction half-reactions.

$\; \mathrm{MnO_4^- \rightarrow Mn^{2+}}$

$\; \mathrm{Fe^{2+} \rightarrow Fe^{3+}}$

Balance atoms other than O and H in each half-reaction.

$\; \mathrm{MnO_4^- \rightarrow Mn^{2+}}$

Balance oxygen by adding water to the side lacking oxygen.

$\; \mathrm{MnO_4^- \rightarrow Mn^{2+} + 4H_2O}$

Balance hydrogen by adding H+.

$\; \mathrm{8H^+ + MnO_4^- \rightarrow Mn^{2+} + 4H_2O}$

Balance charge by adding electrons.

$\; \mathrm{8H^+ + MnO_4^- + 5e^- \rightarrow Mn^{2+} + 4H_2O}$

Now balance the iron half-reaction by charge.

$\; \mathrm{Fe^{2+} \rightarrow Fe^{3+} + e^-}$

Equalise electrons by multiplying the iron half-reaction by 5.

$\; \mathrm{5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^-}$

Add the two half-reactions and cancel electrons and any other species that appear on both sides.

$\; \mathrm{8H^+ + MnO_4^- + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}}$

Verify atoms and charge balance. The equation is balanced for mass and charge.

7.3.2 For Basic Medium

• First balance the equation as if in acidic medium following the half-reaction method.
• For every H+ added, add an equal number of OH- to both sides to neutralise H+ and form H2O.
• Combine H+ and OH- to make H2O and cancel water molecules on both sides where possible.
• Finally, simplify coefficients to smallest whole numbers and verify mass and charge balance.

Worked example (basic medium): Balance the reaction:

Unbalanced equation:

$\; \mathrm{Cl_2 + OH^- \rightarrow Cl^- + ClO^-}$

Write the half-reactions.

$\; \mathrm{Cl_2 \rightarrow Cl^-}$

$\; \mathrm{Cl_2 \rightarrow ClO^-}$

Balance chlorine atoms in each half-reaction and then oxygen and hydrogen where needed.

$\; \mathrm{Cl_2 + 2e^- \rightarrow 2Cl^-}$

$\; \mathrm{Cl_2 + 2OH^- \rightarrow 2ClO^- + 2H^+ + 2e^-}$

Convert the acid-medium second half into basic medium by adding OH- to neutralise H+.

For each H+ add one OH- to both sides.

$\; \mathrm{Cl_2 + 2OH^- + 2OH^- \rightarrow 2ClO^- + 2H_2O + 2e^-}$

Simplify water and hydroxide where possible.

$\; \mathrm{Cl_2 + 2OH^- \rightarrow Cl^- + ClO^- + H_2O}$

Balance electrons and combine half-reactions. Verify final equation for mass and charge balance.

Additional Points, Tips and Common Mistakes

• Always assign oxidation numbers first to identify which atoms are oxidised and reduced.
• When balancing in basic medium, do not add OH- until the equation is balanced for acidic medium; then convert H+ to H2O by adding OH-.
• Check both mass and charge at the end. Charges must balance on both sides for ionic equations.
• For redox in organic reactions, follow the change in oxidation number of carbon atoms to identify oxidation and reduction.
• Use smallest whole-number coefficients by dividing by common factors at the end.

Summary

Understanding oxidation numbers and the half-reaction method is essential for balancing redox reactions systematically. Memorise the key oxidation number rules, practise assigning oxidation states, and use the stepwise half-reaction approach for acidic and basic media. Verify final equations for both mass and charge balance.