Q.Balance the following equations by the oxidation number method.

Balancing Redox Equations by Oxidation Number Method

Concept: In a redox reaction, the total increase in oxidation number (oxidation) must equal the total decrease in oxidation number (reduction). Balance electrons transferred, then balance atoms and charge.

(i) $\ce{Fe^{2+} + H^+ + Cr2O7^{2-} -> Cr^{3+} + Fe^{3+} + H2O}$

Oxidation: $\ce{Fe^{2+} -> Fe^{3+}}$ (ON: +2 → +3, increase of 1)

Reduction: $\ce{Cr2O7^{2-} -> Cr^{3+}}$ (ON of Cr: +6 → +3, decrease of 3 per Cr, total 6 for 2 Cr)

To balance electrons: 6 Fe²⁺ oxidised, 1 Cr₂O₇²⁻ reduced.

(ii) $\ce{I2 + NO3^- -> NO2 + IO3^-}$ (acidic medium)

Oxidation: $\ce{I2 -> IO3^-}$ (ON: 0 → +5, increase of 5 per I, total 10 for I₂)

Reduction: $\ce{NO3^- -> NO2}$ (ON of N: +5 → +4, decrease of 1)

Balance electrons: 1 I₂ oxidised, 10 NO₃⁻ reduced.

(iii) $\ce{I2 + S2O3^{2-} -> I^- + S4O6^{2-}}$

Reduction: $\ce{I2 -> I^-}$ (ON: 0 → −1, decrease of 1 per I, total 2 for I₂)

Oxidation: $\ce{S2O3^{2-} -> S4O6^{2-}}$ (ON of S: +2 → +2.5, increase of 0.5 per S, total 1 per S₂O₃²⁻)

Balance electrons: 1 I₂ reduced, 2 S₂O₃²⁻ oxidised.

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Redox reactions involve the transfer of electrons. The oxidation-number method tracks this transfer by identifying which atoms are oxidised (lose electrons, oxidation number increases) and which are reduced (gain electrons, oxidation number decreases). The core principle: the total increase in oxidation number must equal the total decrease, because electrons lost by one species are gained by another.

The strategy is systematic:

1. Assign oxidation numbers to every atom.
2. Identify what is oxidised and what is reduced.
3. Calculate the change per atom.
4. Multiply by coefficients so that total increase = total decrease.
5. Balance the skeleton equation with those coefficients.
6. Add H⁺ (in acid) or OH⁻ (in base) and H₂O to balance oxygen and hydrogen.
7. Verify atom and charge balance.

(i) $\ce{Fe^{2+} + H^+ + Cr2O7^{2-} -> Cr^{3+} + Fe^{3+} + H2O}$

1. Assign oxidation numbers.

   • $\ce{Fe^{2+}}$: Fe is $+2$.
   • $\ce{Cr2O7^{2-}}$: Each Cr is $+6$ (since $2x + 7(-2) = -2 \Rightarrow x = +6$).
   • $\ce{Cr^{3+}}$: Cr is $+3$.
   • $\ce{Fe^{3+}}$: Fe is $+3$.

2. Identify changes.

   • Fe: $+2 \to +3$, increase of $1$ per Fe atom (oxidation).
   • Cr: $+6 \to +3$, decrease of $3$ per Cr atom (reduction). Since there are 2 Cr in dichromate, total decrease per dichromate ion is $2 \times 3 = 6$.

3. Equalise electron transfer.

   • To balance, we need 6 Fe atoms oxidised (total increase $6 \times 1 = 6$) for every 1 dichromate reduced (total decrease $6$).

4. Write the skeleton with these coefficients:

5. Balance oxygen with water.
   • Left: 7 O in dichromate. Right: need 7 O, so add $7 \ce{H2O}$ on the right.

6. Balance hydrogen with H⁺.
   • Right: $7 \times 2 = 14$ H. Add $14 \ce{H^+}$ on the left.

7. Check charge balance.
   • Left: $6(+2) + 14(+1) + (-2) = 12 + 14 - 2 = 24$.
   • Right: $2(+3) + 6(+3) = 6 + 18 = 24$. ✓

Balanced equation (i):

(ii) $\ce{I2 + NO3^- -> NO2 + IO3^-}$

1. Assign oxidation numbers.

   • $\ce{I2}$: I is $0$.
   • $\ce{IO3^-}$: I is $+5$ (since $x + 3(-2) = -1 \Rightarrow x = +5$).
   • $\ce{NO3^-}$: N is $+5$.
   • $\ce{NO2}$: N is $+4$.

2. Identify changes.

   • I: $0 \to +5$, increase of $5$ per I atom. Since $\ce{I2}$ has 2 atoms, total increase per $\ce{I2}$ is $2 \times 5 = 10$ (oxidation).
   • N: $+5 \to +4$, decrease of $1$ per N atom (reduction).

3. Equalise electron transfer.

   • Need 10 N atoms reduced (total decrease $10 \times 1 = 10$) for every 1 $\ce{I2}$ oxidised (total increase $10$).

4. Write the skeleton:

5. Balance oxygen with water.
   • Left: $10 \times 3 = 30$ O. Right: $10 \times 2 + 2 \times 3 = 20 + 6 = 26$ O. Deficit of $4$ O on the right, so add $4 \ce{H2O}$ on the right.

6. Balance hydrogen with H⁺.
   • Right: $4 \times 2 = 8$ H. Add $8 \ce{H^+}$ on the left.

7. Check charge balance.
   • Left: $10(-1) + 8(+1) = -10 + 8 = -2$.
   • Right: $2(-1) = -2$. ✓

Balanced equation (ii):

(iii) $\ce{I2 + S2O3^{2-} -> I^- + S4O6^{2-}}$

1. Assign oxidation numbers.

   • $\ce{I2}$: I is $0$.
   • $\ce{I^-}$: I is $-1$.
   • $\ce{S2O3^{2-}}$ (thiosulphate): Average S oxidation state is $+2$ (since $2x + 3(-2) = -2 \Rightarrow x = +2$).
   • $\ce{S4O6^{2-}}$ (tetrathionate): Average S oxidation state is $+2.5$ (since $4x + 6(-2) = -2 \Rightarrow x = +2.5$).

2. Identify changes.

   • I: $0 \to -1$, decrease of $1$ per I atom. Since $\ce{I2}$ has 2 atoms, total decrease per $\ce{I2}$ is $2 \times 1 = 2$ (reduction).
   • S: $+2 \to +2.5$, increase of $0.5$ per S atom. Since $\ce{S2O3^{2-}}$ has 2 S atoms, total increase per thiosulphate is $2 \times 0.5 = 1$. Since $\ce{S4O6^{2-}}$ is formed from 2 thiosulphate ions (4 S atoms), total increase is $4 \times 0.5 = 2$ per tetrathionate formed, or equivalently $1$ per thiosulphate consumed (oxidation).

3. Equalise electron transfer.

   • Need 2 thiosulphate oxidised (total increase $2 \times 1 = 2$) for every 1 $\ce{I2}$ reduced (total decrease $2$).

4. Write the skeleton:

5. Check atom balance.

   • I: $2 = 2$. ✓
   • S: $2 \times 2 = 4 = 4$. ✓
   • O: $2 \times 3 = 6 = 6$. ✓

6. Check charge balance. …