Notes › Physical Chemistry

Redox Reactions

1. Oxidation and Reduction

In electron terms, oxidation is loss of electrons and reduction is gain (OIL RIG). The species that gets reduced is the oxidising agent; the species that gets oxidised is the reducing agent.

2. Oxidation Number Rules

  • Elements in the free state: 0 (Na, $O_2$, $P_4$, $S_8$).
  • Fluorine is always −1. Oxygen is usually −2, but −1 in peroxides ($H_2O_2$), −½ in superoxides ($KO_2$) and +2 in $OF_2$.
  • Hydrogen is +1 with non-metals, −1 in metal hydrides (NaH).
  • Alkali metals +1, alkaline earth metals +2.
  • The sum of oxidation numbers equals the charge on the species.

⚠️ Watch the Structure

Average oxidation numbers can hide reality. In $CrO_5$ the oxidation number of Cr is +6, not +10, because four of the oxygens are in peroxide linkages. In $H_2SO_5$ (Caro's acid), S is +6 for the same reason. In $Fe_3O_4$ the average is $+8/3$: one $Fe^{2+}$ and two $Fe^{3+}$.

3. Types of Redox Reactions

  • Combination: $C + O_2 \to CO_2$
  • Decomposition: $2KClO_3 \to 2KCl + 3O_2$
  • Displacement: $Zn + CuSO_4 \to ZnSO_4 + Cu$ (a more reactive metal displaces a less reactive one).
  • Disproportionation: one element is both oxidised and reduced, e.g. $2H_2O_2 \to 2H_2O + O_2$ and $Cl_2 + 2OH^- \to Cl^- + ClO^- + H_2O$.

4. Balancing Redox Equations

Ion–electron (half-reaction) method, acidic medium

  • Split into oxidation and reduction half-reactions.
  • Balance atoms other than O and H.
  • Balance O with $H_2O$, then H with $H^+$.
  • Balance charge with electrons, equalise electrons, and add.
  • In basic medium, add $OH^-$ to both sides to neutralise each $H^+$.

Example: $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$ and $Fe^{2+} \to Fe^{3+} + e^-$ combine to give $MnO_4^- + 5Fe^{2+} + 8H^+ \to Mn^{2+} + 5Fe^{3+} + 4H_2O$.

5. Equivalent Weight and n-factor

Equivalent weight = molar mass ÷ n-factor. For redox species, the n-factor is the number of electrons gained or lost per formula unit.

  • $KMnO_4$: 5 in acidic ($\to Mn^{2+}$), 3 in neutral or weakly basic ($\to MnO_2$), 1 in strongly basic ($\to MnO_4^{2-}$).
  • $K_2Cr_2O_7$ in acidic medium: 6 ($2Cr^{6+} \to 2Cr^{3+}$).

🎯 Titration Shortcut

At the end point, milliequivalents of oxidant = milliequivalents of reductant: $N_1V_1 = N_2V_2$, or $n_1M_1V_1 = n_2M_2V_2$ using n-factors.

Practice questions

  1. The oxidation number of Cr in K₂Cr₂O₇ is:

    1. +3
    2. +6
    3. +7
    4. +12
    Answer

    B. 2(+1) + 2x + 7(-2) = 0 x = +6.

  2. The oxidation number of oxygen in OF₂ is:

    1. −2
    2. −1
    3. +1
    4. +2
    Answer

    D. Fluorine is always −1, so x + 2(-1) = 0 x = +2.

  3. Which reaction is a disproportionation?

    1. 2H₂O₂ → 2H₂O + O₂
    2. Zn + Cu²⁺ → Zn²⁺ + Cu
    3. C + O₂ → CO₂
    4. NaCl + AgNO₃ → AgCl + NaNO₃
    Answer

    A. In H₂O₂ oxygen is −1; it goes to −2 in water (reduced) and 0 in O₂ (oxidised).

  4. The n-factor of KMnO₄ in acidic medium is:

    1. 1
    2. 3
    3. 5
    4. 7
    Answer

    C. Mn goes from +7 in MnO₄⁻ to +2 in Mn²⁺: five electrons gained.

  5. How many moles of Fe²⁺ are oxidised by one mole of MnO₄⁻ in acidic medium?

    1. 1
    2. 2
    3. 5
    4. 6
    Answer

    C. MnO₄⁻ gains 5 electrons and each Fe²⁺ loses 1, so 5 mol Fe²⁺.

  6. In the reaction Zn + CuSO₄ → ZnSO₄ + Cu, the oxidising agent is:

    1. Zn
    2. Cu²⁺
    3. SO₄²⁻
    4. Cu
    Answer

    B. Cu²⁺ gains electrons (is reduced), so it is the oxidising agent.

← Ionic Equilibrium · Solid State →