Potassium Dichromate: K2Cr2O7

K₂Cr₂O₇

Uses:
1. In leather industry
2. Oxidizing agent for the preparation of azo compounds in organic chemistry
Preparation:

This can be prepared from Chromite ore (FeCr₂O₄) in three steps.

Step1: fusion of chromite ore (FeCr₂O₄) with sodium or potassium carbonate in free access of air(O₂). This results Yellow solution of sodium chromate(Na₂CrO₄).

4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂

Step2:Acidification of the resulting yellow chromate solution with H₂SO₄ gives orange crystals of sodium dichromate dihydrate(Na₂Cr₂O₇.2H₂O)
              2Na₂CrO₄ + 2H₊ → Na₂Cr₂O₇ + 2Na₊ + H₂O
Step3:Potassium dichromate is less soluble than sodium dichromate. Therefore, treating of sodium dichromate(Na₂Cr₂O₇) with KCl gives orange crystals of K₂Cr₂O₇.
                       Na₂Cr₂O₄ + 2KCl → K₂Cr₂O₇ + 2NaCl
Interconversion between chromate(CrO₄²⁻) and dichromate(Cr₂O₇²⁻)
As the pH changes Both yellow chromate (CrO₄²⁻) and orange dichromate (Cr₂O₇²⁻) interconvert each other. (Remember: BOY: Basic Medium ⟹ Orange to Yellow). In acidic medium yellow chromate changes to orange dichromate and in basic medium orange dichromate changes to yellow.
               2CrO₄²⁻.(Yellow) + 2H⁺ → Cr₂O₇²⁻ (Orange)+ H₂O
             Cr₂O₇²⁻(Orange) + 2OH⁻ → 2CrO₄²⁻(Yellow) + H₂O
Structure of Chromate and Dichromate
Structure of Chromate:
In chromate Cr(VI) is tetrahedrally surrounded four oxygen atoms.

Structure of Dichromate:
In dichromate ion, two tetrahedral CrO₄ units share one oxygen atom with Cr-O-Cr bond angle 126^o.

Chemical properties
Sodium and potassium chromates are strong oxidizing agents. Sodium dichromate used as oxidizing agent in organic chemistry and K₂Cr₂O₇ used as primary standard in volumetric analysis.
Oxidizing properties of K₂Cr₂O₇ .
In acidic medium, Cr₂O₇²⁻ reduces to Cr³⁺ and it oxidizes the following:
Iodides to Iodine: I⁻ → I₂
Ferrous to Ferric: Fe²⁺ → Fe³⁺
Tin(II) to Tin(IV): Sn²⁺ → Sn⁴⁺
Sulfide to Sulfur: H₂S → S
Reduction of Cr₂O₇²⁻ in acidic medium: 14H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O (E^o = 1.33V). Addition of this half reaction to the above four half reactions results the following balanced redox reactions.
1. 14H⁺ + Cr₂O₇²⁻ + 6I⁻ → 2Cr³⁺ + 3I₂ + 7H₂O
2. 14H⁺ + Cr₂O₇²⁻ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
3. 14H⁺ + Cr₂O₇²⁻ + 3Sn²⁺ → 2Cr³⁺ + 3Sn⁴⁺ + 7H₂O
4. 8H⁺ + Cr₂O₇²⁻ + 3H₂S→ 2Cr³⁺ + 3S + 7H₂O