Potassium dichromate (K₂Cr₂O₇): preparation
It is made from the ore chromite, FeCr₂O₄, in three steps.
- Fuse with soda in air: chromite + sodium carbonate + oxygen give yellow sodium chromate.
4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂ - Add sulphuric acid: chromate becomes orange dichromate.
2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O - Add KCl: potassium dichromate is less soluble, so it crystallises out.
Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl
K₂Cr₂O₇ forms orange-red crystals. Chromium is in the +6 state.
Chromate and dichromate change with pH
In water, chromate and dichromate turn into each other. It depends on pH (how acidic the solution is).
- Add acid → orange dichromate: 2CrO₄²⁻ + 2H⁺ → Cr₂O₇²⁻ + H₂O
- Add base → yellow chromate: Cr₂O₇²⁻ + 2OH⁻ → 2CrO₄²⁻ + H₂O
Chromium stays +6 in both. Only the shape changes, so this is not a redox reaction.
Structure of chromate and dichromate ions
Chromate, CrO₄²⁻: tetrahedral. One Cr in the middle, four O at the corners.
Dichromate, Cr₂O₇²⁻: two tetrahedra joined at one corner. The shared oxygen makes a bent Cr–O–Cr bridge (angle about 126°). It has six ends (terminal) Cr–O bonds, all the same length, and two longer bridge bonds.
Story steps 1 and 2 show this joining in 3D.
Dichromate as an oxidising agent
In acid, dichromate is a strong oxidising agent (it takes electrons from others):
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
Orange turns green. Each Cr drops +6 → +3, so two Cr take 6 electrons.
Examples (ionic):
- Iodide → iodine: Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 3I₂ + 7H₂O
- Iron(II) → iron(III): Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
- H₂S → sulphur: Cr₂O₇²⁻ + 8H⁺ + 3H₂S → 2Cr³⁺ + 3S + 7H₂O
- Tin(II) → tin(IV): Cr₂O₇²⁻ + 14H⁺ + 3Sn²⁺ → 2Cr³⁺ + 3Sn⁴⁺ + 7H₂O
Uses: primary standard in volumetric analysis (it is pure and does not absorb water), leather tanning, chrome plating, and in organic chemistry to oxidise alcohols.
Potassium permanganate (KMnO₄): preparation
It is made from pyrolusite, MnO₂.
- Fuse with KOH in air (or with KNO₃): green potassium manganate forms.
2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O - Make manganate change in neutral/acid solution (disproportionation, one Mn goes up and two go down):
3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O
Industrial method: manganate is oxidised electrolytically in alkaline solution: MnO₄²⁻ → MnO₄⁻ + e⁻ (green → purple).
In the lab, Mn²⁺ can also be oxidised to purple MnO₄⁻ by peroxodisulphate.
Properties and structure of KMnO₄
KMnO₄ is a dark purple, almost black, crystalline solid. It dissolves in water to give a deep purple solution. Mn is +7 (d⁰).
On heating to about 513 K it breaks down and gives oxygen: 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂
Both manganate (MnO₄²⁻, green) and permanganate (MnO₄⁻, purple) are tetrahedral. Manganate has one unpaired electron (paramagnetic); permanganate has none (diamagnetic).
Its colour does not come from d–d jumps (Mn⁷⁺ has no d-electrons). It comes from charge transfer: an electron jumps from O to Mn when light falls on it.
Permanganate as an oxidising agent in different media
How many electrons MnO₄⁻ takes depends on the medium:
- Acidic: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (5 e⁻, purple → colourless)
- Neutral or weakly alkaline: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻ (3 e⁻, brown solid)
- Strongly alkaline: MnO₄⁻ + e⁻ → MnO₄²⁻ (1 e⁻, green)
Acidic medium examples
- Fe²⁺ → Fe³⁺: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
- Oxalate → CO₂ (needs warming to about 333 K): 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O
- Iodide → iodine: 2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O
- H₂S → S, SO₂ (sulphite) → sulphate, nitrite → nitrate
Neutral / weakly alkaline examples
- Iodide → iodate: 2MnO₄⁻ + H₂O + I⁻ → 2MnO₂ + 2OH⁻ + IO₃⁻
- Thiosulphate → sulphate: 8MnO₄⁻ + 3S₂O₃²⁻ + H₂O → 8MnO₂ + 6SO₄²⁻ + 2OH⁻
- Mn²⁺ → MnO₂ (with a little zinc sulphate or ZnO)
Why is HCl not used to acidify KMnO₄? Permanganate would oxidise Cl⁻ to chlorine gas, wasting KMnO₄. Dilute H₂SO₄ is used instead.
Self indicator: in a titration the purple disappears as long as the reducing agent is left. One extra drop gives a lasting pale pink, which shows the end point. No other indicator is needed.
Uses: in volumetric analysis, as a disinfectant and antiseptic, for bleaching wool, cotton and silk, for decolourising oils, and in organic chemistry as an oxidant.
Try it: colour change you can see
1. Put one tiny crystal of KMnO₄ (from a chemist, "lal dawai") in a glass of water. Watch the purple spread without stirring.
2. Add a few drops of lemon juice and a pinch of sugar or a little hydrogen peroxide (ask an adult). The purple slowly fades: KMnO₄ is being reduced.
3. In the 3D free-play step, predict first: at pH 3, yellow or orange? In strong alkali, how many electrons does MnO₄⁻ take? Then move the slider and check.
Safety: KMnO₄ stains skin and is harmful to swallow. Never handle dichromate at home: chromium(VI) is toxic.
Key formulas and definitions
- 2CrO₄²⁻ (yellow) + 2H⁺ ⇌ Cr₂O₇²⁻ (orange) + H₂O
- Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
- MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (acidic)
- MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻ (neutral/weakly alkaline)
- MnO₄⁻ + e⁻ → MnO₄²⁻ (strongly alkaline)
- 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
- 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O
Worked examples
1. Find the oxidation state of Cr in K₂Cr₂O₇.
K is +1, O is −2. Let Cr = x. 2(+1) + 2x + 7(−2) = 0 → 2 + 2x − 14 = 0 → 2x = 12 → x = +6.
2. Find the oxidation state of Mn in KMnO₄ and in K₂MnO₄.
KMnO₄: +1 + x − 8 = 0 → x = +7. K₂MnO₄: +2 + x − 8 = 0 → x = +6.
3. Is the change CrO₄²⁻ → Cr₂O₇²⁻ a redox reaction?
Cr in CrO₄²⁻: x − 8 = −2 → +6. Cr in Cr₂O₇²⁻: 2x − 14 = −2 → +6. The oxidation state does not change, so it is not redox; it is an acid–base (condensation) change.
4. How many moles of Fe²⁺ are oxidised by 1 mol of KMnO₄ in acid?
Each MnO₄⁻ takes 5 e⁻ (Mn +7 → +2). Each Fe²⁺ gives 1 e⁻ (→ Fe³⁺). Electrons must match: 5 × 1 = 1 × n → n = 5 mol Fe²⁺.
5. How many moles of Fe²⁺ does 1 mol of K₂Cr₂O₇ oxidise in acid?
Each Cr goes +6 → +3 = 3 e⁻; two Cr → 6 e⁻ per dichromate. Each Fe²⁺ gives 1 e⁻. So 1 mol dichromate oxidises 6 mol Fe²⁺.
6. 20.0 mL of 0.02 M KMnO₄ exactly reacts with a FeSO₄ solution in acid. How many moles of Fe²⁺ were present?
Step 1: moles MnO₄⁻ = 0.02 × 20.0/1000 = 4.0 × 10⁻⁴ mol. Step 2: ratio MnO₄⁻ : Fe²⁺ = 1 : 5. Step 3: moles Fe²⁺ = 5 × 4.0 × 10⁻⁴ = 2.0 × 10⁻³ mol.
7. Balance: MnO₄⁻ + C₂O₄²⁻ → Mn²⁺ + CO₂ in acid.
Step 1: Mn +7 → +2 gains 5 e⁻. Each C₂O₄²⁻ → 2CO₂ loses 2 e⁻. Step 2: LCM of 5 and 2 is 10 → 2MnO₄⁻ and 5C₂O₄²⁻. Step 3: charges: left 2(−1) + 5(−2) = −12, right 2(+2) = +4, so add 16H⁺ on the left. Step 4: balance O with water: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O.
Common mistakes
- Calling chromate ⇌ dichromate a redox reaction. Cr is +6 in both; only pH changes it.
- Using HCl to acidify KMnO₄. It oxidises Cl⁻ to Cl₂, so the titre is wrong. Use dilute H₂SO₄.
- Writing 5 electrons for KMnO₄ in every medium. It is 5 in acid, 3 in neutral/weakly alkaline and 1 in strongly alkaline.
- Saying KMnO₄ is purple because of d–d transitions. Mn(+7) is d⁰; the colour is from charge transfer (O → Mn).