Important compounds — KMnO₄, K₂Cr₂O₇ — preparation, oxidizing reactions: Worked examples

The scenario matrix

Every question about these two compounds falls into one of these cells. We will hit every single one.

Cell	What varies	Example that covers it
A. Medium = acidic	MnO₄⁻ falls all the way to Mn²⁺ (5 e⁻)	Ex 1
B. Medium = neutral	MnO₄⁻ stops at MnO₂ (3 e⁻)	Ex 2
C. Medium = strong alkali	MnO₄⁻ stops at MnO₄²⁻ (1 e⁻)	Ex 3
D. Dichromate (acidic only)	Cr₂O₇²⁻ → 2Cr³⁺ (6 e⁻)	Ex 4
E. Non-redox "degenerate" case	Cr stays +6, only condensation	Ex 5
F. Zero / limiting input	What happens as [H⁺] → 0, or when the wrong acid is used	Ex 6
G. Real-world word problem	quantitative titration, find a concentration	Ex 7
H. Exam-style twist	equivalent weight / n-factor changing with medium	Ex 8

New tools we will earn as we go: n-factor (defined in Ex 7 the moment we need it) and the mole-ratio bridge from a balanced equation.

Cell A — Acidic KMnO₄

Worked example Ex 1. Balance MnO₄⁻ oxidising oxalate (C₂O₄²⁻) in acid

Forecast: Guess the whole-number coefficients before reading. How many MnO₄⁻ react with how many C₂O₄²⁻?

Step 1 — Write the two half-reactions. Why this step? A redox is two stories: something loses electrons, something gains them. Splitting them lets us count electrons on each side separately. $Reduction: M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$ $Oxidation: C_{2} O_{4}^{2 -} \to 2 C O_{2} + 2 e^{-}$

Step 2 — Make electrons cancel. Reduction gives 5, oxidation takes 2. LCM = 10. Why this step? Electrons are not created or destroyed. The 10 electrons the two oxalates release must be exactly the 10 the two permanganates swallow — otherwise charge isn't conserved. Multiply reduction ×2, oxidation ×5.

Step 3 — Add and cancel electrons. $2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O$

Verify: Charge left $= 2 (- 1) + 5 (- 2) + 16 (+ 1) = + 4$ . Charge right $= 2 (+ 2) + 0 + 0 = + 4$ . ✓ Atoms: Mn 2=2, C 10=10, O $8 + 20 = 28 = 20 + 8$ ✓, H $16 = 16$ ✓.

Cell B — Neutral / faintly alkaline KMnO₄

Worked example Ex 2. MnO₄⁻ oxidising iodide (I⁻) in

neutral solution In neutral medium there are almost no free H⁺, so MnO₄⁻ can only fall to MnO₂ (Mn +4), grabbing 3 electrons, not 5. Product I⁻ → I₂.

Forecast: Which is bigger here — the number of MnO₄⁻ or of I⁻? (Trap: it's not the same ratio as the acidic case.)

Step 1 — Neutral reduction half-reaction. Why this step? With no H⁺ to spend, oxygen is balanced using H₂O and OH⁻ instead. Mn: +7 → +4 = 3 e⁻. $M n O_{4}^{-} + 2 H_{2} O + 3 e^{-} \to M n O_{2} + 4 O H^{-}$ Check it: left charge $- 1 - 3 = - 4$ , right charge $- 4$ ✓; O $4 + 2 = 6 = 2 + 4$ ✓; H $4 = 4$ ✓.

Step 2 — Oxidation. $2 I^{-} \to I_{2} + 2 e^{-}$ .

Step 3 — Electrons: 3 and 2 → LCM 6. Reduction ×2, oxidation ×3. $2 M n O_{4}^{-} + 6 I^{-} + 4 H_{2} O \to 2 M n O_{2} + 3 I_{2} + 8 O H^{-}$ (2×2 H₂O = 4 H₂O; 2×4 OH⁻ = 8 OH⁻.)

Verify: Charge left $2 (- 1) + 6 (- 1) = - 8$ ; right $8 (- 1) = - 8$ ✓. Mn 2=2, I 6=6, O $8 + 4 = 12 = 4 + 8$ ✓, H $8 = 8$ ✓.

Cell C — Strong alkali KMnO₄

Worked example Ex 3. MnO₄⁻ in strongly alkaline solution grabs only

1 electron In concentrated OH⁻, MnO₄⁻ (+7) falls only to the green manganate MnO₄²⁻ (+6): a 1-electron drop. Take a generic 1-electron reducing agent, say sulphite oxidised to sulphate.

Forecast: How many MnO₄⁻ per one SO₃²⁻?

Step 1 — Reduction (1 e⁻): $M n O_{4}^{-} + e^{-} \to M n O_{4}^{2 -}$ . Why this step? Only the last, most loosely-held electron can be given back; there is no acid to help the ion break its Mn–O bonds further, so it stops at +6.

Step 2 — Oxidation of sulphite in base (2 e⁻): $S O_{3}^{2 -} + 2 O H^{-} \to S O_{4}^{2 -} + H_{2} O + 2 e^{-}$

Step 3 — Electrons 1 and 2 → LCM 2. Reduction ×2. $2 M n O_{4}^{-} + S O_{3}^{2 -} + 2 O H^{-} \to 2 M n O_{4}^{2 -} + S O_{4}^{2 -} + H_{2} O$

Verify: Charge left $2 (- 1) + (- 2) + 2 (- 1) = - 6$ ; right $2 (- 2) + (- 2) = - 6$ ✓. Mn 2=2, S 1=1, O $8 + 3 + 2 = 13 = 8 + 4 + 1$ ✓, H $2 = 2$ ✓.

Figure — Important compounds — KMnO₄, K₂Cr₂O₇ — preparation, oxidizing reactions

The staircase above shows all three cells A/B/C at once: how far MnO₄⁻ falls depends purely on how much H⁺ the medium hands it.

Cell D — Dichromate (acidic only)

Cell E — The degenerate (non-redox) case

Worked example Ex 5. Add acid to yellow chromate — what forms, and is it redox?

$2 C r O_{4}^{2 -} + 2 H^{+} ⇌ C r_{2} O_{7}^{2 -} + H_{2} O$ Forecast: Does Cr's oxidation state change? Guess before checking.

Step 1 — Assign oxidation states to Cr on both sides. Why this step? "Redox" is defined as a change in oxidation state. If nothing changes, it isn't redox, no matter how dramatic the colour flip. In $C r O_{4}^{2 -}$ : let Cr $= x$ . Oxygen is $- 2$ each: $x + 4 (- 2) = - 2 \Rightarrow x = + 6$ . In $C r_{2} O_{7}^{2 -}$ : $2 x + 7 (- 2) = - 2 \Rightarrow 2 x = 12 \Rightarrow x = + 6$ .

Step 2 — Compare. $+ 6 = + 6$ . No electrons moved. This is an acid–base condensation (two CrO₄ units glued by losing one H₂O), governed by Le Chatelier's Principle: add H⁺ → shift right (orange); add OH⁻ → shift left (yellow).

Verify (mass & charge): Cr $2 = 2$ ; O $8 = 7 + 1$ ; H $2 = 2$ ; charge left $2 (- 2) + 2 (+ 1) = - 2$ , right $- 2 + 0 = - 2$ ✓. Conclusion: not redox — the classic Trap 2.

Cell F — Zero / limiting input

Worked example Ex 6. Two limiting scenarios that break the naive rules

F(i): [H⁺] → 0 for KMnO₄. As you remove acid, the acidic path ( $M n O_{4}^{-} + 8 H^{+} + 5 e^{-}$ ) needs eight H⁺ per ion — when H⁺ runs out this path starves and cannot proceed. The system is forced onto the low-H⁺ neutral path (3 e⁻ → MnO₂). So the "5-electron" answer is a limit that only holds while acid is plentiful; at the H⁺ → 0 limit you must switch to Ex 2's equation. Colour signature: no colourless Mn²⁺, instead a brown MnO₂ precipitate.

Forecast: If someone titrates KMnO₄ using HCl as the acid, will the volume of KMnO₄ used be too high or too low?

F(ii): the wrong-acid degenerate case. With HCl, chloride is oxidised alongside your intended reductant: $2 M n O_{4}^{-} + 16 H^{+} + 10 C l^{-} \to 2 M n^{2 +} + 5 C l_{2} + 8 H_{2} O$ Why this matters: extra MnO₄⁻ is consumed making Cl₂, so you pour in more KMnO₄ than the true reaction needs → measured value is too high. Fix: dilute H₂SO₄ (sulphate isn't oxidised). See Standard Electrode Potential & E° values: $E^{\circ} (M n O_{4}^{-}) = + 1.51 V > E^{\circ} (C l_{2} / C l^{-}) = + 1.36 V$ , so MnO₄⁻ can oxidise Cl⁻.

Verify (F ii balancing): Charge left $2 (- 1) + 16 (+ 1) + 10 (- 1) = + 4$ ; right $2 (+ 2) = + 4$ ✓. Mn 2=2, Cl 10=10, O $8 = 8$ , H $16 = 16$ ✓.

Cell G — Real-world word problem (quantitative titration)

Worked example Ex 7. Find the concentration of an FeSO₄ solution

Statement: $25.0 mL$ of FeSO₄ is titrated in dilute H₂SO₄ against $0.0200 M$ KMnO₄; the end-point (first permanent pink) needs $20.0 mL$ of KMnO₄. Find [Fe²⁺].

Forecast: Fe²⁺ gives 1 e⁻, MnO₄⁻ takes 5. So you'll need roughly 5× as many moles of Fe²⁺ as of MnO₄⁻ — expect [Fe²⁺] noticeably larger than 0.02 M.

Step 1 — Moles of KMnO₄ used. Why this step? Moles = molarity × volume(L); this is our only fully-known quantity, so start here. $n (M n O_{4}^{-}) = 0.0200 \times 0.0200 L = 4.00 \times 1 0^{- 4} mol .$

Step 2 — Mole bridge from the balanced equation $M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$ . Why this step? The coefficients ARE the electron-conserving ratio (1 : 5). This is exactly n-factor bookkeeping: $1 \times 5 = 5 \times 1$ . $n (F e^{2 +}) = 5 \times 4.00 \times 1 0^{- 4} = 2.00 \times 1 0^{- 3} mol .$

Step 3 — Concentration. Divide by the Fe²⁺ volume in litres: $[F e^{2 +}] = \frac{2.00 \times 1 0 ^{- 3}}{0.0250} = 0.0800 M .$

Verify: Equivalents balance: $n (M n O_{4}^{-}) \times 5 = 4 \times 1 0^{- 4} \times 5 = 2 \times 1 0^{- 3}$ ; $n (F e^{2 +}) \times 1 = 2 \times 1 0^{- 3}$ ✓. And $0.08 > 0.02$ , matching the forecast. Units: $mol / L = M$ ✓.

Cell H — Exam twist (n-factor changes with medium)

Worked example Ex 8. The equivalent weight of KMnO₄ is

not a single number Statement: Molar mass of KMnO₄ = $158 g mol^{- 1}$ . Find its equivalent weight in (a) acidic, (b) neutral, (c) strongly alkaline medium.

Forecast: In which medium is the equivalent weight largest? (Hint: larger n-factor → smaller equivalent weight.)

Step 1 — Acidic, n = 5. $158/5 = 31.6 g eq^{- 1}$ . Step 2 — Neutral, n = 3. $158/3 = 52.67 g eq^{- 1}$ . Step 3 — Strong alkali, n = 1. $158/1 = 158 g eq^{- 1}$ . Why these steps? Fewer electrons transferred per formula unit → each electron "costs" more grams → bigger equivalent weight. The alkali case (1 e⁻) is the biggest, confirming the forecast.

Verify: $31.6 \times 5 = 158$ , $52.67 \times 3 = 158$ , $158 \times 1 = 158$ — all recover the molar mass ✓. Ordering $31.6 < 52.67 < 158$ tracks $5 > 3 > 1$ ✓.

Recall Self-test (cover the answers)

Every acidic reaction of MnO₄⁻ transfers how many electrons? ::: 5 (Mn +7 → +2) In neutral medium MnO₄⁻ becomes what, taking how many e⁻? ::: MnO₂ (Mn +4), 3 e⁻ In strong alkali MnO₄⁻ becomes what, taking how many e⁻? ::: green MnO₄²⁻ (+6), 1 e⁻ Dichromate transfers how many e⁻, in which medium? ::: 6 e⁻, acidic only Is chromate→dichromate a redox reaction? ::: No — Cr stays +6, acid–base condensation Why is the measured value too high if HCl is used with KMnO₄? ::: MnO₄⁻ also oxidises Cl⁻ to Cl₂, so extra KMnO₄ is consumed Equivalent weight formula ::: molar mass ÷ n-factor

Connections

Parent topic (Hinglish)
Balancing Redox Reactions (ion-electron method) — the half-reaction machinery used in every example
Volumetric Analysis / Titrations — the n-factor / mole-bridge tools of Ex 7–8
Standard Electrode Potential & E° values — why MnO₄⁻ can oxidise Cl⁻ (Ex 6)
Oxidation States of Transition Metals — the +7/+6/+4/+3 ladders
Le Chatelier's Principle — the equilibrium shift in Ex 5
Colour & d-d Transitions — the purple/green/brown colour signatures used as end-points