Finding n-factor in Redox
Valency Factor Calculation for Redox Reactions
1. What is n-factor?
Definition: In redox reactions, the n-factor (valency factor) is the total number of moles of electrons lost or gained per mole of the reactant.
It links Molar Mass ($M$) and Equivalent Mass ($E$):
$$ E = \frac{M}{\text{n-factor}} $$
2. Simple Oxidation/Reduction (Single Element Changes)
When only one element in a molecule undergoes oxidation or reduction:
$$ n = |\text{Change in O.S.}| \times \text{No. of atoms in one molecule} $$
Common Oxidizing Agents (Oxidants)
| Species | Reaction | Change | n-factor |
|---|---|---|---|
| $KMnO_4$ (Acidic) | $Mn^{+7} \to Mn^{+2}$ | $|2 - 7| = 5$ | 5 |
| $KMnO_4$ (Neutral) | $Mn^{+7} \to Mn^{+4}O_2$ | $|4 - 7| = 3$ | 3 |
| $KMnO_4$ (Basic) | $Mn^{+7} \to Mn^{+6}O_4^{2-}$ | $|6 - 7| = 1$ | 1 |
| $K_2Cr_2O_7$ | $Cr_2^{+6} \to 2Cr^{+3}$ | $|3 - 6| \times 2$ | 6 |
Common Reducing Agents (Reductants)
- Oxalic Acid ($H_2C_2O_4$): $C_2^{+3} \to 2C^{+4}O_2$. Change per C is 1. Two C atoms $\Rightarrow n=2$.
- Mohr's Salt ($Fe^{2+}$): $Fe^{+2} \to Fe^{+3}$. Change is 1 $\Rightarrow n=1$.
- Sodium Thiosulphate ($Na_2S_2O_3$): Reacts with $I_2$ to form $S_4O_6^{2-}$. S goes from +2 to +2.5. Change is 0.5 per S. Two S atoms $\Rightarrow n = 0.5 \times 2 = 1$.
3. Disproportionation Reactions
A reaction where the same substance undergoes both oxidation and reduction. (e.g., $H_2O_2 \to H_2O + O_2$).
Logic: The substance acts as two parts: one part oxidizes ($n_{ox}$), one part reduces ($n_{red}$). The overall n-factor is derived from the fact that Equivalent Mass is the sum of equivalents for both roles.
$$ \frac{1}{n_{net}} = \frac{1}{n_{ox}} + \frac{1}{n_{red}} \quad \Rightarrow \quad n_{net} = \frac{n_{ox} \times n_{red}}{n_{ox} + n_{red}} $$
Example: Chlorine in Hot Alkali
Reaction: $3Cl_2 + 6OH^- \rightarrow 5Cl^- + ClO_3^- + 3H_2O$
- Reduction: $Cl_2 (0) \to 2Cl^- (-1)$. Change per atom = 1. Per molecule = 2. So, $n_{red} = 2$.
- Oxidation: $Cl_2 (0) \to 2ClO_3^- (+5)$. Change per atom = 5. Per molecule = 10. So, $n_{ox} = 10$.
$$ n_{factor} = \frac{10 \times 2}{10 + 2} = \frac{20}{12} = \frac{5}{3} = 1.67 $$
Equivalent Mass of $Cl_2 = M / (5/3) = 3M/5$.
Practice Quiz
Test your n-factor calculation skills.
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