Preparation and Properties of K2Cr2O7 | chemca

chemca.in

Home Articles Quizzes About

Inorganic Chemistry

Potassium Dichromate ($K_2Cr_2O_7$)

Preparation from Chromite Ore, Properties, and Oxidizing Action.

By chemca Team • Updated Jan 2026

Potassium Dichromate is a very important orange-colored crystalline solid used extensively as an oxidizing agent in volumetric analysis and organic synthesis. It is prepared industrially from Chromite ore ($FeCr_2O_4$).

1. Preparation from Chromite Ore

Step 1: Conversion to Sodium Chromate

The finely powdered Chromite ore is fused with Sodium Carbonate ($Na_2CO_3$) in excess of air (Oxygen) in a reverberatory furnace.

$$ 4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \xrightarrow{\Delta} \underset{\text{Sodium Chromate (Yellow)}}{8Na_2CrO_4} + 2Fe_2O_3 + 8CO_2 $$

Step 2: Conversion to Sodium Dichromate

The yellow solution of sodium chromate is filtered and acidified with concentrated Sulphuric acid to give orange Sodium Dichromate.

$$ 2Na_2CrO_4 + H_2SO_4 \rightarrow \underset{\text{Sodium Dichromate (Orange)}}{Na_2Cr_2O_7} + Na_2SO_4 + H_2O $$

Step 3: Conversion to Potassium Dichromate

Sodium dichromate is highly soluble (deliquescent). It is treated with Potassium Chloride ($KCl$). Being less soluble, $K_2Cr_2O_7$ separates out as orange crystals.

$$ Na_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7 \downarrow + 2NaCl $$

2. Physical & Chemical Properties

Action of Heat: On strong heating, it decomposes to evolve Oxygen.

$$ 4K_2Cr_2O_7 \xrightarrow{\Delta} 4K_2CrO_4 + 2Cr_2O_3 + 3O_2 $$

Effect of pH (Chromate-Dichromate Equilibrium):

Chromates ($CrO_4^{2-}$) and Dichromates ($Cr_2O_7^{2-}$) are interconvertible depending on pH.

• In Acidic Medium (pH < 7): Chromate (Yellow) changes to Dichromate (Orange).
• In Basic Medium (pH > 7): Dichromate (Orange) changes to Chromate (Yellow).

$$ \underset{\text{Yellow}}{2CrO_4^{2-}} + 2H^+ \rightleftharpoons \underset{\text{Orange}}{Cr_2O_7^{2-}} + H_2O $$

Structure: The chromate ion is tetrahedral. The dichromate ion consists of two tetrahedra sharing one oxygen atom (bond angle $126^\circ$).

3. Oxidizing Properties (Acidic Medium)

Strong Oxidizing Agent

In acidic medium ($H_2SO_4$), $K_2Cr_2O_7$ accepts electrons and gets reduced to Chromium(III) ion (Green).

General Reaction: $$ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O $$

Equivalent Weight = Mol. Wt / 6

Important Redox Reactions:

1. Iodide to Iodine:

$$ Cr_2O_7^{2-} + 14H^+ + 6I^- \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O $$

2. Ferrous to Ferric:

$$ Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O $$

3. Hydrogen Sulphide to Sulphur:

$$ Cr_2O_7^{2-} + 8H^+ + 3H_2S \rightarrow 2Cr^{3+} + 3S \downarrow \text{ (Colloidal)} + 7H_2O $$

4. Sulphite to Sulphate:

$$ Cr_2O_7^{2-} + 8H^+ + 3SO_3^{2-} \rightarrow 2Cr^{3+} + 3SO_4^{2-} + 4H_2O $$

4. Chromyl Chloride Test

Test for Chloride Ions

When solid Potassium Dichromate is heated with a solid chloride salt (e.g., $NaCl$) and concentrated $H_2SO_4$, reddish-brown vapors of Chromyl Chloride ($CrO_2Cl_2$) are evolved.

$$ K_2Cr_2O_7 + 4NaCl + 6H_2SO_4 \xrightarrow{\Delta} 2KHSO_4 + 4NaHSO_4 + \underset{\text{Red Vapours}}{2CrO_2Cl_2} + 3H_2O $$

Confirmatory Step: The red vapors dissolve in NaOH to form a yellow solution of Sodium Chromate.

$$ CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 \text{ (Yellow)} + 2NaCl + 2H_2O $$

Knowledge Check

Test your understanding of Potassium Dichromate

© 2026 chemca.in. All rights reserved.

Optimized for learning Inorganic Chemistry.

Quiz Revision Notes Downloads Class Notes Video Lectures