Mastering n-Factor: Disproportionation & Complex Redox | ChemCa

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Advanced n-Factor Calculation

Unlock the secrets of calculating the valency factor for Disproportionation reactions and compounds where multiple elements undergo redox changes.

Redox Equivalence

Case 1: Multiple Elements undergoing Change

In some compounds, more than one element undergoes oxidation or reduction simultaneously. If both elements are oxidized (or both reduced), the n-factor is the sum of the individual electron changes per molecule.

Example A: Ferrous Oxalate ($FeC_2O_4$)

$FeC_2O_4$ is oxidized by $KMnO_4$ in acidic medium to form $Fe^{3+}$ and $CO_2$.

$$ Fe^{+2} \longrightarrow Fe^{+3} + 1e^- \quad (\Delta n_1 = 1) $$ $$ C_2^{+3}O_4^{2-} \longrightarrow 2C^{+4}O_2 + 2e^- \quad (\Delta n_2 = 2 \times (4-3) = 2) $$ $$ \text{Total n-factor} = \Delta n_1 + \Delta n_2 = 1 + 2 = \mathbf{3} $$

Example B: Arsenic Sulfide ($As_2S_3$)

$As_2S_3$ is oxidized to Arsenate ($AsO_4^{3-}$) and Sulfate ($SO_4^{2-}$).

$$ As_2^{+3} \longrightarrow 2As^{+5} \quad (\text{Change} = 2 \times |5-3| = 4e^-) $$ $$ S_3^{-2} \longrightarrow 3S^{+6} \quad (\text{Change} = 3 \times |6-(-2)| = 3 \times 8 = 24e^-) $$ $$ \text{Total n-factor} = 4 + 24 = \mathbf{28} $$

Case 2: Disproportionation Reactions

In a disproportionation reaction, the same element in a compound is both oxidized and reduced. To find the n-factor of the reactant molecule, we cannot simply add the n-factors.

Formula

$$ n_{factor} = \frac{n_{ox} \times n_{red}}{n_{ox} + n_{red}} $$

Where $n_{ox}$ is electron loss in oxidation part and $n_{red}$ is electron gain in reduction part.

Solved Example: Chlorine in Basic Medium

Reaction: $Cl_2 + OH^- \longrightarrow Cl^- + ClO_3^- + H_2O$

Oxidation Half

$$ Cl_2^0 \longrightarrow 2Cl^{+5}O_3^- $$ $$ \Delta e^- = 2 \times 5 = 10 \implies n_{ox} = 10 $$

Reduction Half

$$ Cl_2^0 \longrightarrow 2Cl^{-1} $$ $$ \Delta e^- = 2 \times 1 = 2 \implies n_{red} = 2 $$

Calculation:

$$ n_{factor}(Cl_2) = \frac{10 \times 2}{10 + 2} = \frac{20}{12} = \frac{5}{3} \approx 1.67 $$

Balancing via n-Factor Method

The n-factor method (or Oxidation Number method) is the fastest way to balance redox reactions. The core principle is the Law of Equivalence: Total electrons lost must equal total electrons gained.

Identify n-factors for the Oxidizing Agent ($n_A$) and Reducing Agent ($n_B$).
Cross-multiply: Assign $n_B$ as the coefficient for Reactant A, and $n_A$ as the coefficient for Reactant B.
Balance the main atoms (other than O and H).
Balance Oxygen by adding $H_2O$.
Balance Hydrogen by adding $H^+$ (acidic) or $OH^-$ (basic).

Master Example: Dichromate + Ferrous Oxalate

Cr₂O₇²⁻ + FeC₂O₄ + H⁺ ⟶ Cr³⁺ + Fe³⁺ + CO₂ + H₂O

1. Find n-factors

$$ \text{Oxidant: } Cr_2^{+6} \to 2Cr^{+3} $$ $$ n_1 = 2 \times (6-3) = \mathbf{6} $$

$$ \text{Reductant: } FeC_2O_4 \to Fe^{+3} + 2CO_2 $$ $$ n_2 = (1) + 2(1) = \mathbf{3} $$

2. Cross Multiply Ratio

Ratio of n-factors = $6 : 3$ or $2 : 1$.

Cross multiply coefficients:

Coefficient of $Cr_2O_7^{2-}$ gets 1
Coefficient of $FeC_2O_4$ gets 2

3. Final Balance

1 Cr₂O₇²⁻ + 2 FeC₂O₄ + __ H⁺ ⟶ 2 Cr³⁺ + 2 Fe³⁺ + 4 CO₂ + __ H₂O

(Balance O: Left=7+8=15, Right=8. Add 7 H₂O. Then balance H with 14 H⁺)

Cr₂O₇²⁻ + 2 FeC₂O₄ + 14 H⁺ ⟶ 2 Cr³⁺ + 2 Fe³⁺ + 4 CO₂ + 7 H₂O

Summary Cheat Sheet

Reaction Type	Example	n-Factor Formula
Simple Redox	MnO₄⁻ → Mn²⁺	\|Ox State Change\| × No. of Atoms
Multiple Oxidized	FeS₂ → Fe³⁺ + SO₂	Sum of individual n-factors
Disproportionation	H₂O₂ → H₂O + O₂	(n₁ × n₂) / (n₁ + n₂)

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