Redox Titrations
Quantitative Analysis using Oxidation-Reduction | Class 11
1. Principle of Redox Titrations
Redox titrations involve a reaction between an Oxidizing Agent (Oxidant) and a Reducing Agent (Reductant). The equivalence point is reached when the number of gram equivalents of the oxidant equals that of the reductant.
Law of Equivalence:
$$ N_1 V_1 (\text{Oxidant}) = N_2 V_2 (\text{Reductant}) $$
where $N$ (Normality) = Molarity $\times$ n-factor.
2. Permanganometry (Using $KMnO_4$)
Potassium Permanganate ($KMnO_4$) is a powerful oxidizing agent. It acts as a Self-Indicator. The end point is marked by the appearance of a permanent Pink color.
n-factor Variations:
| Medium | Reaction | Change in ON | n-factor |
|---|---|---|---|
| Acidic | $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ | $+7 \to +2$ | 5 |
| Neutral / Weakly Alkaline | $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$ | $+7 \to +4$ | 3 |
| Strongly Alkaline | $MnO_4^- + e^- \rightarrow MnO_4^{2-}$ (Manganate) | $+7 \to +6$ | 1 |
Note: Dilute $H_2SO_4$ is used for acidic medium. $HCl$ is avoided because $KMnO_4$ oxidizes $Cl^-$ to $Cl_2$.
3. Dichrometry (Using $K_2Cr_2O_7$)
Potassium Dichromate ($K_2Cr_2O_7$) is an oxidizing agent used in acidic medium. It is not a self-indicator.
- n-factor: $$ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O $$ Change per Cr is $+6 \to +3 = 3$. Total change for 2 Cr atoms = 6.
- Indicator: Requires an external or internal indicator like Diphenylamine (turns blue/violet) or $K_3[Fe(CN)_6]$ (external).
4. Indicators in Redox Titrations
- Self Indicator: The titrant itself changes color (e.g., $KMnO_4$: Purple $\to$ Colorless).
- Internal Indicator: A chemical added to the solution that changes color at specific redox potential (e.g., Diphenylamine, Ferroin).
- Starch: Specific for Iodine titrations (Deep Blue with $I_2$).
5. Common Estimation Examples
- Estimation of $Fe^{2+}$: Using $KMnO_4$ or $K_2Cr_2O_7$. $Fe^{2+}$ oxidizes to $Fe^{3+}$ ($n=1$).
- Estimation of Oxalic Acid / Oxalate: Using $KMnO_4$.
$C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$ ($n=2$).
Note: This reaction is slow at room temp and requires heating to 60°C.
Practice Quiz
Test your quantitative skills.
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