Redox titration — permanganometry and dichromatometry procedures

Question

Compare permanganometry and dichromatometry titrations. Write the balanced equations for KMnO $_4$ titration with FeSO $_4$ in acidic medium. Why is KMnO $_4$ called a self-indicator?

(JEE Main + CBSE 11 pattern)

Solution — Step by Step

KMnO $_4$ is a strong oxidising agent. In acidic medium (H $_2$ SO $_4$ ):

\text{MnO}_4^- + 8\text{H}^+ + 5e^- \to \text{Mn}^{2+} + 4\text{H}_2\text{O}

The equivalence factor (n-factor) of KMnO $_4$ in acidic medium = 5.

For titration with FeSO $_4$ (Mohr’s salt):

\text{Fe}^{2+} \to \text{Fe}^{3+} + e^- \quad (\text{n-factor} = 1)

Balanced equation:

2\text{KMnO}_4 + 10\text{FeSO}_4 + 8\text{H}_2\text{SO}_4 \to 2\text{MnSO}_4 + 5\text{Fe}_2(\text{SO}_4)_3 + \text{K}_2\text{SO}_4 + 8\text{H}_2\text{O}

KMnO $_4$ is deep purple. When it reacts with the reducing agent, it is converted to colourless Mn $^{2+}$ . As long as the reducing agent is present, every drop of KMnO $_4$ is decolourised.

At the equivalence point, the very next drop of KMnO $_4$ has no reducing agent left to react with — it remains purple and imparts a permanent pink colour to the solution. This colour change marks the endpoint. No external indicator needed — hence the name “self-indicator.”

Feature	Permanganometry	Dichromatometry
Titrant	KMnO $_4$	K $_2$ Cr $_2$ O $_7$
Colour	Purple → colourless	Orange → green
Self-indicator?	Yes (pink endpoint)	No (needs diphenylamine indicator)
n-factor (acidic)	5	6
Stability	Unstable, must be standardised	Very stable, can be a primary standard
Medium	Always acidic (H $_2$ SO $_4$ )	Acidic (H $_2$ SO $_4$ or HCl)

Dichromate half-reaction: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \to 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$

flowchart TD
    A["Redox Titration"] --> B["Permanganometry (KMnO₄)"]
    A --> C["Dichromatometry (K₂Cr₂O₇)"]
    B --> D["n-factor = 5 in acidic medium"]
    B --> E["Self-indicator: purple → pink endpoint"]
    C --> F["n-factor = 6 in acidic medium"]
    C --> G["Needs indicator: diphenylamine"]
    B --> H["Not a primary standard"]
    C --> I["Primary standard (very stable)"]

Why This Works

Redox titrations work on the principle that at the equivalence point, moles of oxidant $\times$ n-factor = moles of reductant $\times$ n-factor. For KMnO $_4$ vs FeSO $_4$ :

n_1 \times 5 = n_2 \times 1

So 1 mole of KMnO $_4$ reacts with 5 moles of FeSO $_4$ . This stoichiometry comes directly from balancing the electrons transferred.

KMnO $_4$ is used in acidic medium (not alkaline or neutral) for quantitative titrations because in acidic medium, the change is from Mn $^{7+}$ to Mn $^{2+}$ (gain of 5 electrons per Mn), giving a clean, well-defined reaction.

Alternative Method — Equivalent Weight Approach

Using equivalent weights: $\text{Eq. wt.} = \frac{\text{Molar mass}}{\text{n-factor}}$

KMnO $_4$ : Eq. wt. $= 158/5 = 31.6$

FeSO $_4$ : Eq. wt. $= 152/1 = 152$

At equivalence: milliequivalents of KMnO $_4$ = milliequivalents of FeSO $_4$

$N_1 V_1 = N_2 V_2$

For JEE, always check the medium before assigning the n-factor of KMnO $_4$ . In acidic: n = 5 (Mn $^{7+}$ to Mn $^{2+}$ ). In neutral/slightly alkaline: n = 3 (Mn $^{7+}$ to Mn $^{4+}$ as MnO $_2$ ). In strongly alkaline: n = 1 (Mn $^{7+}$ to Mn $^{6+}$ as manganate). The default in problems is acidic unless stated otherwise.

Common Mistake

Students use HCl as the acid in permanganometry. HCl reacts with KMnO $_4$ itself (HCl is oxidised to Cl $_2$ ), consuming extra KMnO $_4$ and giving wrong results. Always use H $_2$ SO $_4$ with KMnO $_4$ . In contrast, dichromatometry CAN use HCl because K $_2$ Cr $_2$ O $_7$ does not oxidise HCl under normal conditions.