d- and f-Block Elements · Lesson 7 of 12
Chromate and Dichromate Chemistry
“Follow chromite oxidation, pH-dependent chromate equilibria and six-electron dichromate reactions.”
• Explain preparation of potassium dichromate from chromite with balanced equations. • Relate chromate and dichromate structures to their formulas. • Predict the effect of pH without confusing acid–base conversion with redox. • Construct and use acidic dichromate oxidation equations.
From an insoluble ore to a soluble oxoanion
Chromite, FeCr₂O₄, contains Cr(III). Producing chromate requires oxidation to Cr(VI), not merely dissolving the ore. Heating finely divided ore with sodium carbonate in an oxidising atmosphere supplies both oxygen and an alkaline medium. Chromium enters soluble yellow sodium chromate; iron forms a less soluble oxide that can be separated from the aqueous extract. This sequence illustrates chemical separation by deliberately changing oxidation state and solubility.
The balanced roasting equation is 4FeCr₂O₄+8Na₂CO₃+7O₂→8Na₂CrO₄+2Fe₂O₃+8CO₂. Oxygen is required: it accepts electrons while Cr(III) becomes Cr(VI) and Fe(II) becomes Fe(III). The carbonate provides sodium and the alkaline reaction environment; it is not the oxidising agent. Carbon remains +4 in both carbonate and carbon dioxide.
Problem
How many electrons are transferred in the equation with four FeCr₂O₄?
- 1.Eight Cr atoms each change +3 to +6, losing 8×3=24 electrons. Four Fe atoms each change +2 to +3, losing four more.
- 2.Total oxidation releases 28 electrons. Seven O₂ molecules contain fourteen O atoms; each changes 0 to −2 and accepts two electrons, a total of 28.
- 3.Check atoms: Fe 4, Cr 8, Na 16 and C 8 on each side; oxygen totals 16+24+14=54 on the left and 32+6+16=54 on the right. Electron and mass balance agree.
Filter the extract to separate the insoluble material, then acidify the yellow chromate solution. A molecular representation is 2Na₂CrO₄+H₂SO₄⇌Na₂Cr₂O₇+Na₂SO₄+H₂O. Sodium dichromate can crystallise as its dihydrate. Potassium dichromate is less soluble under suitable crystallisation conditions, so adding KCl allows its orange crystals to separate: Na₂Cr₂O₇+2KCl→K₂Cr₂O₇+2NaCl. The last equation represents salt conversion and crystallisation, not oxidation.
pH shifts a linked set of equilibria
Chromate is yellow and dichromate orange. A useful net equilibrium is 2CrO₄²⁻+2H⁺⇌Cr₂O₇²⁻+H₂O. Adding acid favours the side consuming H⁺; adding base removes H⁺ and favours chromate. In alkaline form the conversion is Cr₂O₇²⁻+2OH⁻⇌2CrO₄²⁻+H₂O. Real solutions also contain protonated chromate species, so the net equation is a compact description whose speciation depends on concentration as well as pH.
Problem
A dichromate solution turns yellow when sufficient base is added. Has chromium been reduced?
- 1.In dichromate, 2x+7(−2)=−2, so each Cr is +6. In chromate, x+4(−2)=−2, so Cr is also +6.
- 2.No chromium oxidation state changes and no electrons occur in the net conversion. It is an acid–base/speciation shift, despite the visible colour change.
- 3.A redox change to Cr(III), in contrast, requires three electrons per Cr and produces a chemically different lower oxidation state.
Four oxygen neighbours, then a shared corner
CrO₄²⁻ has a tetrahedral arrangement of oxygen atoms around chromium. Cr₂O₇²⁻ consists of two CrO₄ tetrahedra sharing one oxygen corner: four plus four oxygen positions minus the shared one gives seven distinct oxygen atoms. The bridge is Cr–O–Cr, with a representative angle near 126°. It is not a direct Cr–Cr bond and not an edge-sharing pair. Resonance and metal–oxygen bonding delocalise electron density, so a single line-bond drawing does not identify one permanently localised charge on each oxygen.
Acidic dichromate accepts six electrons
The reduction half-reaction is Cr₂O₇²⁻+14H⁺+6e⁻→2Cr³⁺+7H₂O. Balance Cr first, then oxygen with water, hydrogen with H⁺ and charge with electrons. Two chromium atoms each gain three electrons. The standard reduction potential in the stated acidic convention is about +1.33 V, so dichromate oxidises several suitable reductants. The final Cr(III) colour depends on its ligand environment; colour alone is not a stoichiometric measurement.
Problem
Construct the acidic dichromate reaction with Fe²⁺.
- 1.Fe²⁺→Fe³⁺+e⁻ supplies one electron per iron. Multiply this half-reaction by six to match dichromate’s six-electron demand.
- 2.Add the two halves and cancel electrons: Cr₂O₇²⁻+14H⁺+6Fe²⁺→2Cr³⁺+6Fe³⁺+7H₂O.
- 3.Charge left is −2+14+12=+24. Charge right is +6+18=+24; Cr,Fe,O,H counts also match.
- 4.The 1:6 mole ratio follows from electron exchange, not from the number of oxygen atoms or potassium ions.
| Reductant half-reaction | Full acidic dichromate reaction |
|---|---|
| 2I⁻→I₂+2e⁻ | Cr₂O₇²⁻+14H⁺+6I⁻→2Cr³⁺+3I₂+7H₂O |
| H₂S→S+2H⁺+2e⁻ | Cr₂O₇²⁻+8H⁺+3H₂S→2Cr³⁺+3S+7H₂O |
| Sn²⁺→Sn⁴⁺+2e⁻ | Cr₂O₇²⁻+14H⁺+3Sn²⁺→2Cr³⁺+3Sn⁴⁺+7H₂O |
| Fe²⁺→Fe³⁺+e⁻ | Cr₂O₇²⁻+14H⁺+6Fe²⁺→2Cr³⁺+6Fe³⁺+7H₂O |
For H₂S, the oxidation half generates H⁺, so adding it to the reduction half cancels some acid. The net requirement is eight H⁺ rather than fourteen. A frequent mistake is to retain fourteen and thereby violate hydrogen and charge balance. Sulfur formation gives a precipitate; iodine formation produces dissolved iodine or iodine-containing species depending on excess iodide.
Problem
How much Fe²⁺ reacts with 20.00 mL of 0.0200 mol L⁻¹ acidic dichromate?
- 1.Convert volume to litres: 20.00 mL=0.02000 L. Dichromate amount=cV=0.0200×0.02000=4.00×10⁻⁴ mol.
- 2.Multiply by six because each dichromate accepts six electrons and each Fe²⁺ supplies one: Fe²⁺ amount=2.40×10⁻³ mol.
- 3.If that Fe²⁺ occupied 25.00 mL, its concentration would be 0.00240/0.02500=0.0960 mol L⁻¹. Potassium spectator ions do not alter the ratio.
Potassium dichromate can serve as a primary standard because suitably pure material can be weighed reliably to prepare a known solution. Sodium dichromate’s higher solubility is useful in other oxidising applications. Chromium(VI) compounds have industrial and synthetic uses, but the lesson’s equations describe the chemical transformations rather than an unsupervised preparation protocol.
Quick Quiz
Quiz
What does increasing pH favour in a chromate/dichromate mixture?
Explanation: Base shifts the net equilibrium toward CrO₄²⁻.
How many electrons does one dichromate ion accept on reduction to Cr³⁺?
Explanation: Two Cr(VI) atoms each gain three electrons.
What is shared by the two tetrahedra in dichromate?
Explanation: Sharing one of eight oxygen positions leaves seven distinct O atoms.
How many Fe²⁺ ions react per dichromate ion?
Explanation: Each Fe²⁺ supplies one electron and dichromate needs six.
Which species oxidises chromite during alkaline roasting with air?
Explanation: Oxygen accepts the electrons lost by Cr and Fe.
Practice Problems
- Write the three chemical stages from chromite to K₂Cr₂O₇.
- Derive dichromate oxidation of H₂S to sulfur.
- Derive dichromate oxidation of iodide.
- How many moles of Sn²⁺ react with 0.00150 mol dichromate in acid?
- Why can yellow–orange interconversion occur without any change in d count?
- Why does KCl addition separate K₂Cr₂O₇ rather than oxidise chromium?
First roast: 4FeCr₂O₄+8Na₂CO₃+7O₂→8Na₂CrO₄+2Fe₂O₃+8CO₂. Then acidify: 2CrO₄²⁻+2H⁺⇌Cr₂O₇²⁻+H₂O. Finally suitable KCl addition and crystallisation give Na₂Cr₂O₇+2KCl→K₂Cr₂O₇+2NaCl. The first is redox, the second speciation, and the third salt separation.
Key Takeaways
• Chromite preparation requires oxidation of Cr(III) to Cr(VI) in an alkaline roast. • Acid favours dichromate and base favours chromate in the linked equilibrium. • Chromate and dichromate both contain Cr(VI); their interconversion is not redox. • Dichromate contains two corner-sharing tetrahedra. • Acidic dichromate accepts six electrons to produce two Cr³⁺ ions. • Electron accounting determines the ratios to iodide, sulfide, tin(II) and iron(II). • Full equations must cancel H⁺ produced by reductant half-reactions.