ChemistryISC Class 12 · d- and f-Block Elements

Potassium Dichromate and Potassium Permanganate: ISC Class 12

Updated 15 October 2026 · Fact-checked

Potassium dichromate is made from chromite ore via sodium chromate and is a strong oxidant in acid, being reduced from orange Cr(+6) to green Cr³⁺. Potassium permanganate is made from pyrolusite via green manganate; it is reduced to Mn²⁺ in acid, MnO₂ in neutral or alkaline solution, and acts as its own indicator.

Acid or alkali only swaps yellow chromate and orange dichromate (Cr stays +6); a reductant in acid changes Cr +6 to green Cr +3.

Illustration drawn for QuestionBuddy.

Understand Potassium Dichromate and Potassium Permanganate

Both compounds contain a metal in its group oxidation state (+6 for Cr, +7 for Mn), which is why both are powerful oxidising agents and why their ions are coloured although the metal has no d electrons: the colour comes from charge transfer, not d-d transitions.

Potassium dichromate, K₂Cr₂O₇: preparation. Chromite ore, FeCr₂O₄, is fused with sodium carbonate in free access of air, giving yellow sodium chromate: $4 F e C r_{2} O_{4} + 8 N a_{2} C O_{3} + 7 O_{2} \to 8 N a_{2} C r O_{4} + 2 F e_{2} O_{3} + 8 C O_{2}$ . The filtered yellow solution is acidified with sulphuric acid to give orange sodium dichromate: $2 N a_{2} C r O_{4} + H_{2} S O_{4} \to N a_{2} C r_{2} O_{7} + N a_{2} S O_{4} + H_{2} O$ . Sodium dichromate is more soluble, so potassium chloride is added and the less soluble potassium salt crystallises: $N a_{2} C r_{2} O_{7} + 2 K C l \to K_{2} C r_{2} O_{7} + 2 N a C l$ .

Structure and the effect of pH. The chromate ion $C r O_{4}^{2 -}$ is tetrahedral. The dichromate ion $C r_{2} O_{7}^{2 -}$ is two tetrahedra sharing one corner oxygen, with a Cr–O–Cr angle of 126°. The two ions interconvert with pH, and Cr stays +6 in both, so this is not a redox change: acid turns yellow chromate orange, $2 C r O_{4}^{2 -} + 2 H^{+} \to C r_{2} O_{7}^{2 -} + H_{2} O$ ; alkali turns orange dichromate yellow, $C r_{2} O_{7}^{2 -} + 2 O H^{-} \to 2 C r O_{4}^{2 -} + H_{2} O$ .

Oxidising action of dichromate. In acid: $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$ ( $E^{\circ} = 1.33$ V); the colour changes from orange to green. It oxidises Fe²⁺ to Fe³⁺, I⁻ to I₂, H₂S to S, Sn²⁺ to Sn⁴⁺ and SO₂ to sulphate. Molecular equations: $K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} + 6 F e S O_{4} \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 3 F e_{2} (S O_{4})_{3} + 7 H_{2} O$ ; $K_{2} C r_{2} O_{7} + 4 H_{2} S O_{4} + 3 H_{2} S \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 3 S + 7 H_{2} O$ ; $K_{2} C r_{2} O_{7} + H_{2} S O_{4} + 3 S O_{2} \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + H_{2} O$ . K₂Cr₂O₇ is a primary standard in volumetric analysis (pure, stable, not hygroscopic); titrations of Fe²⁺ with it need an internal indicator such as diphenylamine.

Potassium permanganate, KMnO₄: preparation. Pyrolusite (MnO₂) is fused with KOH and an oxidising agent (air or KNO₃) to give dark green potassium manganate: $2 M n O_{2} + 4 K O H + O_{2} \to 2 K_{2} M n O_{4} + 2 H_{2} O$ . In neutral or acidic solution manganate disproportionates: $3 M n O_{4}^{2 -} + 4 H^{+} \to 2 M n O_{4}^{-} + M n O_{2} + 2 H_{2} O$ . Commercially the manganate is oxidised electrolytically in alkaline solution to permanganate. On heating at 513 K, $2 K M n O_{4} \to K_{2} M n O_{4} + M n O_{2} + O_{2}$ .

Structure and magnetism. Both $M n O_{4}^{-}$ and $M n O_{4}^{2 -}$ are tetrahedral, with π bonding between oxygen p orbitals and manganese d orbitals. Manganate (Mn +6, one unpaired electron) is green and paramagnetic; permanganate (Mn +7, no unpaired electron) is deep purple and diamagnetic.

Oxidising action of permanganate depends on the medium.
(i) Acidic: $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$ (5 electrons; purple to colourless). With Mohr's salt: $M n O_{4}^{-} + 8 H^{+} + 5 F e^{2 +} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$ , or $2 K M n O_{4} + 8 H_{2} S O_{4} + 10 F e S O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 5 F e_{2} (S O_{4})_{3} + 8 H_{2} O$ . With oxalic acid at about 333 K, CO₂ bubbles off: $2 K M n O_{4} + 3 H_{2} S O_{4} + 5 H_{2} C_{2} O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 8 H_{2} O + 10 C O_{2}$ . With KI, iodine is liberated: $2 K M n O_{4} + 8 H_{2} S O_{4} + 10 K I \to 6 K_{2} S O_{4} + 2 M n S O_{4} + 5 I_{2} + 8 H_{2} O$ . It also oxidises nitrite to nitrate and sulphite to sulphate.
(ii) Neutral or faintly alkaline: $M n O_{4}^{-} + 2 H_{2} O + 3 e^{-} \to M n O_{2} + 4 O H^{-}$ (3 electrons; brown MnO₂ precipitates). With KI in alkaline solution, iodide becomes iodate: $2 K M n O_{4} + K I + H_{2} O \to 2 M n O_{2} + 2 K O H + K I O_{3}$ . With H₂S in neutral solution, sulphur separates: $2 K M n O_{4} + 3 H_{2} S \to 2 K O H + 2 M n O_{2} + 2 H_{2} O + 3 S$ .

Titrations. KMnO₄ acts as its own indicator: in acidic titrations the first excess drop gives a permanent pale pink. Dilute H₂SO₄ is used; HCl is avoided because permanganate oxidises it to Cl₂, and HNO₃ is avoided because it is itself an oxidant. Oxalate titrations are warmed because the reaction starts slowly; once some Mn²⁺ forms, it speeds up the reaction.

Uses. K₂Cr₂O₇: leather tanning, oxidant for preparing azo compounds, primary standard. KMnO₄: analytical reagent, oxidant in organic chemistry, bleaching wool, cotton and silk, decolourising oils.

Potassium Dichromate and Potassium Permanganate: formulas and key facts

Dichromate in acid: $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$; $E^{\circ} = 1.33$ V; orange to green; 6 electrons per ion.
Permanganate in acid: $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$; Purple to colourless; equivalent mass $= M /5$ .
Permanganate in neutral or alkaline solution: $M n O_{4}^{-} + 2 H_{2} O + 3 e^{-} \to M n O_{2} + 4 O H^{-}$; Brown MnO₂; equivalent mass $= M /3$ .
Chromate-dichromate equilibrium: $2 C r O_{4}^{2 -} + 2 H^{+} ⇌ C r_{2} O_{7}^{2 -} + H_{2} O$; Acid gives orange dichromate; alkali gives yellow chromate. Cr is +6 in both.
Titration stoichiometry: $n_{M n O_{4}^{-}} : n_{F e^{2 +}} = 1 : 5, n_{C r_{2} O_{7}^{2 -}} : n_{F e^{2 +}} = 1 : 6$; From electrons accepted per oxidant ion (acid medium).

How to solve Potassium Dichromate and Potassium Permanganate questions

ISC asks preparation equations, 'complete and balance' reactions in a stated medium, reasons (colour, magnetism, choice of acid) and titration calculations. Use half-equations to build every equation.

1Note the medium. Acid: Mn goes to +2 (5 e⁻) and Cr to +3 (3 e⁻ per Cr). Neutral or alkaline: Mn goes to +4 as MnO₂ (3 e⁻).
2Write the reducing agent's half-equation (Fe²⁺ → Fe³⁺ + e⁻; H₂C₂O₄ → 2CO₂ + 2H⁺ + 2e⁻; H₂S → S + 2H⁺ + 2e⁻; I⁻ → ½I₂ + e⁻).
3Multiply the two half-equations so the electrons cancel, add, and cancel H⁺ and H₂O that appear on both sides.
4For a molecular equation, add K⁺ and SO₄²⁻ as spectators (K₂SO₄, MnSO₄, Cr₂(SO₄)₃) and count H₂SO₄ last.
5Check atoms and charge on both sides before you finish.
6For titrations, convert volume × molarity to moles, use the mole ratio from the ionic equation, then convert to mass or molarity.

Potassium Dichromate and Potassium Permanganate in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Permanganate titration finds the strength of an antiseptic

Suppose a pharmacy lab checks a bottle of hydrogen peroxide antiseptic. A 10.0 mL portion of a diluted sample is acidified with dilute H₂SO₄ and titrated with 0.020 M KMnO₄; 24.0 mL are needed for a permanent pale pink colour. The reaction is $2 M n O_{4}^{-} + 6 H^{+} + 5 H_{2} O_{2} \to 2 M n^{2 +} + 5 O_{2} + 8 H_{2} O$ .

Find the molarity of H₂O₂ in the diluted sample, and explain why no separate indicator is needed.

Try it first, then show the solutionSolution

Moles of $M n O_{4}^{-}$ $= 0.020 \times 0.0240 = 4.8 \times 1 0^{- 4}$ mol.
From the equation, 2 mol $M n O_{4}^{-}$ react with 5 mol H₂O₂: moles of H₂O₂ $= \frac{5}{2} \times 4.8 \times 1 0^{- 4} = 1.2 \times 1 0^{- 3}$ mol.
Molarity $= \frac{1.2 \times 1 0 ^{- 3}}{0.0100} = 0.12$ M.
Purple permanganate becomes colourless Mn²⁺ while H₂O₂ remains; the first excess drop stays pink, marking the end point.

Answer: 0.12 M H₂O₂ in the diluted sample. KMnO₄ is its own indicator.

Memory hook: Purple in, colourless out, until the reducer runs out: then pink stays.

Tricks to remember Potassium Dichromate and Potassium Permanganate

Short memory aids. Say each one aloud once, then use it on a question the same day.

Five, three, one for manganese
Permanganate takes 5 electrons in acid (to Mn²⁺), 3 in neutral or weakly alkaline solution (to MnO₂), and 1 in strongly alkaline solution (to MnO₄²⁻).
Use it when: Products and equivalent masses in each medium.
Chromite goes yellow, then orange
Ore → yellow Na₂CrO₄ (fusion with Na₂CO₃ in air) → orange Na₂Cr₂O₇ (acid) → orange K₂Cr₂O₇ (KCl). Yellow-orange-orange, Na-Na-K.
Use it when: Three-step preparation equations.
Orange to green means Cr went to +3
Any time acidified dichromate turns green, Cr(+6) has become Cr³⁺ and something has been oxidised.
Use it when: Observation-based questions.

Potassium Dichromate and Potassium Permanganate: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Oxidation state change in neutral medium · 1 mark · MCQBoard level

In neutral solution, KMnO₄ oxidises H₂S to sulphur. The manganese-containing product, and the oxidation state of Mn in it, are:
(a) Mn²⁺, +2
(b) MnO₂, +4
(c) K₂MnO₄, +6
(d) Mn₂O₃, +3

Try it first, then show the solutionSolution

In neutral solution permanganate is reduced to MnO₂.
Mn is +7 in $M n O_{4}^{-}$ and +4 in MnO₂.

Answer: (b) MnO₂, in which Mn is +4 (reduced from +7).

Example 2: Dichromate and sulphur dioxide · 2 marks · Short answerBoard level

Write the balanced molecular equation for the reaction of acidified potassium dichromate with sulphur dioxide, and state the colour change observed.

Try it first, then show the solutionSolution

SO₂ is oxidised to sulphate; Cr(+6) is reduced to Cr³⁺ (½ mark).
$K_{2} C r_{2} O_{7} + H_{2} S O_{4} + 3 S O_{2} \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + H_{2} O$ (1 mark).
Orange solution turns green (½ mark).

Answer: $K_{2} C r_{2} O_{7} + H_{2} S O_{4} + 3 S O_{2} \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + H_{2} O$ ; orange to green.

Example 3: From pyrolusite to permanganate · 3 marks · Short answerHarder board level

(a) Write equations for the preparation of potassium permanganate from pyrolusite, by fusion followed by disproportionation in acid solution. (b) Manganate is green and paramagnetic; permanganate is purple and diamagnetic. Explain the difference in magnetism.

Try it first, then show the solutionSolution

(a) $2 M n O_{2} + 4 K O H + O_{2} \to 2 K_{2} M n O_{4} + 2 H_{2} O$ (1 mark).
$3 M n O_{4}^{2 -} + 4 H^{+} \to 2 M n O_{4}^{-} + M n O_{2} + 2 H_{2} O$ (1 mark).
(b) In $M n O_{4}^{2 -}$ , Mn is +6 ( $3 d^{1}$ ): one unpaired electron, so paramagnetic. In $M n O_{4}^{-}$ , Mn is +7 ( $3 d^{0}$ ): no unpaired electron, so diamagnetic (1 mark).

Answer: (a) Fusion of MnO₂ with KOH in air gives K₂MnO₄, which disproportionates in acid to KMnO₄ and MnO₂. (b) Mn(+6) has one unpaired electron; Mn(+7) has none.

Example 4: Volume of permanganate for Mohr's salt · 3 marks · Short answerHarder board level

0.632 g of KMnO₄ ( $M = 158$ g mol⁻¹) is dissolved in water to make 250 mL of solution. What volume of this solution will react exactly with 25.0 mL of 0.100 M Fe²⁺ (Mohr's salt) solution in acid?

Try it first, then show the solutionSolution

Molarity of KMnO₄ $= \frac{0.632/158}{0.250} = 0.0160$ M (1 mark).
$M n O_{4}^{-} + 8 H^{+} + 5 F e^{2 +} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$ : moles of Fe²⁺ $= 2.50 \times 1 0^{- 3}$ , so moles of $M n O_{4}^{-}$ $= 5.00 \times 1 0^{- 4}$ (1 mark).
Volume $= \frac{5.00 \times 1 0 ^{- 4}}{0.0160} = 0.03125$ L $= 31.3$ mL (1 mark).

Answer: About 31.3 mL.

Example 5: Identify the manganese compounds · 5 marks · Long answerHOTS / competency

A black mineral W is fused with KOH in air to give a dark green compound X. X dissolved in water and acidified gives a purple solution of Y and a brown precipitate of W. Crystals of Y, heated to 513 K, give X, W and a gas Z. When acidified Y is warmed with oxalic acid, the purple colour disappears and gas bubbles are seen.
(a) Identify W, X, Y and Z. [2]
(b) Write balanced equations for the formation of X and for the heating of Y. [2]
(c) Write the balanced molecular equation for the reaction of Y with oxalic acid in dilute H₂SO₄ and name the gas. [1]

Try it first, then show the solutionSolution

(a) W = MnO₂ (pyrolusite), X = K₂MnO₄ (potassium manganate), Y = KMnO₄, Z = O₂ (2 marks).
(b) $2 M n O_{2} + 4 K O H + O_{2} \to 2 K_{2} M n O_{4} + 2 H_{2} O$ (1 mark).
$2 K M n O_{4} \to K_{2} M n O_{4} + M n O_{2} + O_{2}$ (1 mark).
(c) $2 K M n O_{4} + 3 H_{2} S O_{4} + 5 H_{2} C_{2} O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 8 H_{2} O + 10 C O_{2}$ ; the gas is CO₂ (1 mark).

Answer: (a) W MnO₂, X K₂MnO₄, Y KMnO₄, Z O₂. (b) As shown. (c) The gas is carbon dioxide.

Where students lose marks in Potassium Dichromate and Potassium Permanganate

The slips examiners see most often, and what to write instead.

Mistake: Unbalanced KMnO₄ and oxalic acid equation
ISC examiners reported correct products but an unbalanced equation for KMnO₄ with oxalic acid (2025).
Wrong working
$K M n O_{4} + H_{2} S O_{4} + H_{2} C_{2} O_{4} \to K_{2} S O_{4} + M n S O_{4} + H_{2} O + C O_{2}$
Right working
$2 K M n O_{4} + 3 H_{2} S O_{4} + 5 H_{2} C_{2} O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 8 H_{2} O + 10 C O_{2}$
Do this: Balance electrons first: 2 Mn take 10 e⁻, 5 oxalic acid give 10 e⁻; then fix K, S, H and O.
Usually costs: 1 mark of 2.
Mistake: Wrong products for dichromate and H₂S
Examiners noted wrong product formulae for K₂Cr₂O₇ with H₂S (2025).
Wrong working
$K_{2} C r_{2} O_{7} + H_{2} S \to K_{2} S + C r_{2} O_{3} + H_{2} O$
Right working
$K_{2} C r_{2} O_{7} + 4 H_{2} S O_{4} + 3 H_{2} S \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 3 S + 7 H_{2} O$
Do this: In acid, Cr ends as Cr³⁺ (Cr₂(SO₄)₃) and sulphide is oxidised to yellow sulphur.
Usually costs: 1 to 2 marks.
Mistake: Not stating the change in oxidation state
Examiners noted that candidates explaining why pink KMnO₄ decolourises with Fe²⁺ did not mention Mn going from +7 to +2 (2025).
Do this: Say: purple $M n O_{4}^{-}$ (Mn +7) is reduced to almost colourless Mn²⁺ (Mn +2).
Usually costs: 1 mark.
Mistake: Naming the wrong gas
In a 2024 MCQ many candidates chose SO₂ for the gas evolved when KMnO₄ is heated with acidified oxalic acid.
Do this: Oxalic acid is oxidised to CO₂; sulphuric acid only supplies H⁺.
Usually costs: 1 mark.

Exam tips

Learn the three dichromate preparation equations and the two permanganate ones exactly; they are asked as straight recall.
In 'complete and balance' questions, always use the medium stated: acid gives Mn²⁺, neutral gives MnO₂.
Use molecular equations when the question gives formulae like KMnO₄ + H₂SO₄ + ...; use ionic equations when ions are given.
Mention the colour change in observation questions: purple to colourless, orange to green, yellow to orange.
Remember that chromate ⇌ dichromate is not a redox reaction; Cr stays +6.

Practice questions on Potassium Dichromate and Potassium Permanganate

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The shape of the permanganate ion is:
(a) square planar
(b) octahedral
(c) tetrahedral
(d) trigonal planar
Hint
Four oxygen atoms around Mn.
Answer
(c) tetrahedral.
Q2 · 2 marks · Short answerBoard level
Write the ionic equation for the oxidation of oxalate ions by acidified permanganate. Why is the solution warmed to about 333 K?
Hint
Oxalate becomes carbon dioxide.
Answer
$2 M n O_{4}^{-} + 16 H^{+} + 5 C_{2} O_{4}^{2 -} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O$ . The reaction is slow at room temperature; warming speeds it up (Mn²⁺ formed later also speeds it up).
Q3 · 2 marks · Short answerHarder board level
Write the balanced ionic equation for the reaction of acidified dichromate with tin(II) ions, and give the change in oxidation state of chromium and of tin.
Hint
Each Sn²⁺ gives 2 electrons; each dichromate takes 6.
Answer
$C r_{2} O_{7}^{2 -} + 14 H^{+} + 3 S n^{2 +} \to 2 C r^{3 +} + 3 S n^{4 +} + 7 H_{2} O$ . Cr: +6 to +3; Sn: +2 to +4.
Q4 · 3 marks · Short answerHarder board level
Write balanced ionic equations for the oxidation, by acidified KMnO₄, of (a) nitrite ions and (b) sulphite ions. In each case state the change in oxidation state of the element oxidised.
Hint
Both need 6 H⁺ for 2 permanganate ions.
Answer
(a) $2 M n O_{4}^{-} + 6 H^{+} + 5 N O_{2}^{-} \to 2 M n^{2 +} + 5 N O_{3}^{-} + 3 H_{2} O$ ; N +3 to +5. (b) $2 M n O_{4}^{-} + 6 H^{+} + 5 S O_{3}^{2 -} \to 2 M n^{2 +} + 5 S O_{4}^{2 -} + 3 H_{2} O$ ; S +4 to +6.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Permanganate titrations are not carried out in the presence of hydrochloric acid.
Reason (R): Acidified KMnO₄ oxidises HCl to chlorine, so extra titrant is used up.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Which acid is itself oxidised?
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
0.245 g of pure K₂Cr₂O₇ ( $M = 294$ g mol⁻¹) is used up exactly in oxidising Fe²⁺ ions in an acidified solution. What mass of iron(II) was present? Why is K₂Cr₂O₇ preferred to KMnO₄ as a primary standard? (Fe = 56)
Hint
1 mol dichromate oxidises 6 mol Fe²⁺.
Answer
Moles of $C r_{2} O_{7}^{2 -}$ $= \frac{0.245}{294} = 8.33 \times 1 0^{- 4}$ ; Fe²⁺ $= 6 \times 8.33 \times 1 0^{- 4} = 5.00 \times 1 0^{- 3}$ mol $= 0.280$ g. K₂Cr₂O₇ can be obtained very pure, is stable and not hygroscopic, so a standard solution can be made by weighing; KMnO₄ solutions slowly decompose and must be standardised.
Q7 · 3 marks · Short answerHOTS / competency
Find the equivalent mass of KMnO₄ ( $M = 158$ ) when it acts as an oxidant in (a) acidic, (b) neutral and (c) strongly alkaline solution. Give the manganese product in each case.
Hint
Equivalent mass = molar mass ÷ electrons gained.
Answer
(a) Mn²⁺, 5 e⁻: $\frac{158}{5} = 31.6$ . (b) MnO₂, 3 e⁻: $\frac{158}{3} = 52.7$ . (c) $M n O_{4}^{2 -}$ , 1 e⁻: 158.
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student titrates warm, acidified oxalic acid with KMnO₄ solution. The first few drops of purple titrant take a long time to lose their colour, but later drops decolourise almost instantly. No indicator is added.
(i) Why are the later drops decolourised faster? [1]
(ii) How does the student know the end point has been reached? [1]
(iii) Why must dilute H₂SO₄, not dilute HNO₃, be used to acidify the solution? [1]
Hint
Think about a product of the reaction.
Answer
(i) Mn²⁺ ions formed in the reaction speed up the oxidation (the reaction catalyses itself). (ii) KMnO₄ is its own indicator: the first drop in excess gives a permanent pale pink colour. (iii) HNO₃ is itself an oxidising agent and would oxidise some of the oxalic acid, giving a wrong titre; H₂SO₄ is not oxidised and does not oxidise.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Ore of chromium used to make K₂Cr₂O₇?Show answer
Chromite, FeCr₂O₄.
2.Cr–O–Cr bond angle in the dichromate ion?Show answer
126°.
3.Product of KMnO₄ in neutral solution?Show answer
MnO₂ (brown).
4.Why is manganate green and paramagnetic?Show answer
Mn(+6) has one unpaired d electron.
5.What does KMnO₄ give on heating at 513 K?Show answer
K₂MnO₄, MnO₂ and O₂.
6.What happens to iodide with KMnO₄ in alkaline solution?Show answer
It is oxidised to iodate: $2 K M n O_{4} + K I + H_{2} O \to 2 M n O_{2} + 2 K O H + K I O_{3}$ .

Potassium Dichromate and Potassium Permanganate: frequently asked questions

How is potassium dichromate prepared from chromite ore?

Chromite is fused with sodium carbonate in air to give sodium chromate; acidifying gives sodium dichromate; adding KCl crystallises the less soluble potassium dichromate.

Why is KMnO₄ called a self-indicator?

Its purple colour disappears as long as reducing agent remains; the first excess drop gives a lasting pink colour, so no separate indicator is needed.

What is the product of KMnO₄ in acidic, neutral and alkaline media?

Acidic: Mn²⁺. Neutral or faintly alkaline: MnO₂. Strongly alkaline: manganate, MnO_4^{2-}.

Is the conversion of chromate to dichromate a redox reaction?

No. Chromium is +6 in both ions; only the pH changes the equilibrium.

Why are permanganate and dichromate ions coloured though the metal has no d electrons?

The colour comes from charge transfer between oxygen and the metal, not from d-d transitions.