Topic 17 · page 10 of 20

Chromate(VI) and dichromate(VI)

Yellow chromate(VI) and orange dichromate(VI) are linked by an acid-base equilibrium. Alkali moves it to yellow and acid moves it to orange. Chromium is +6 in both, so it is not redox.

9 min read17.21

Read first: Chromium: reduction and oxidation

The equilibrium 17.21

Dichromate(VI) and chromate(VI) are in equilibrium in water:

\ce{Cr2O7^2-(aq) + H2O(l) <=> 2CrO4^2-(aq) + 2H+(aq)}

Ion	Colour
Cr₂O₇²⁻, dichromate(VI)	orange
CrO₄²⁻, chromate(VI)	yellow

The hydrogen ions are on the right. The position of the equilibrium therefore depends on the acidity of the solution. Apply Le Chatelier's principle.

Add alkali. The OH⁻ ions remove H⁺ ions by neutralisation. The equilibrium shifts right to replace them, and the solution turns yellow.
Add acid. The concentration of H⁺ rises. The equilibrium shifts left to remove the extra, and the solution turns orange.

\ce{Cr2O7^2-(aq) + 2OH-(aq) -> 2CrO4^2-(aq) + H2O(l)}

The change is reversible, and you can swap the colour back and forth by adding alkali and acid in turn.

Worked exampleIs it a redox reaction?

In Cr₂O₇²⁻: 2x + 7(−2) = −2, so 2x = +12 and x = +6.
In CrO₄²⁻: x + 4(−2) = −2, so x = +6.
The oxidation number of chromium does not change, and neither does that of oxygen or hydrogen. This is not redox. It is an acid-base equilibrium: H⁺ ions are lost or gained.

The equilibrium constant, from Topic 13, leaves water out because it is the solvent:

K_\text{c} = \frac{[\ce{CrO4^2-}]^2[\ce{H+}]^2}{[\ce{Cr2O7^2-}]}

A sudden rise in [H⁺] makes the fraction larger than K_c, so the system responds by turning chromate into dichromate until the fraction equals K_c again.

The whole chromium story 17.21

A chain of chromium species with their colours. Cr2+, blue, goes by oxidation in air to Cr3+, green; zinc and acid take Cr2O7 2- orange to Cr3+ and on to Cr2+. From Cr3+, excess alkali gives [Cr(OH)6]3-, dark green; hydrogen peroxide gives CrO4 2- yellow; acid and alkali interconvert CrO4 2- and Cr2O7 2-. — The chromium species and what links them. Only the steps between chromate(VI) and dichromate(VI) are not redox.

Change	Reagent	Type of reaction
Cr₂O₇²⁻ → Cr³⁺ → Cr²⁺	zinc and acid	reduction
Cr³⁺ → [Cr(OH)₆]³⁻	excess NaOH	deprotonation and amphoteric behaviour
[Cr(OH)₆]³⁻ → CrO₄²⁻	H₂O₂, alkaline	oxidation
CrO₄²⁻ ⇌ Cr₂O₇²⁻	acid or alkali	acid-base equilibrium

Check yourself

State the colour of chromate(VI) and of dichromate(VI) ions.
A yellow solution of potassium chromate(VI) turns orange when dilute sulfuric acid is added. Explain.
Orange potassium dichromate(VI) solution is made alkaline with sodium hydroxide. State the colour change and explain it.
Show by oxidation numbers that the change in question 2 is not a redox reaction.
Write the expression for K_c for the dichromate(VI) / chromate(VI) equilibrium.

Answers cover these until you have tried the questions

Chromate(VI), CrO₄²⁻: yellow. Dichromate(VI), Cr₂O₇²⁻: orange.
Adding acid raises [H⁺]. The equilibrium Cr₂O₇²⁻ + H₂O ⇌ 2CrO₄²⁻ + 2H⁺ shifts to the left to remove the extra H⁺, so Cr₂O₇²⁻ forms. The solution turns orange.
Orange to yellow. The OH⁻ ions remove H⁺ ions, so the equilibrium shifts to the right, forming chromate(VI).
Cr is +6 in CrO₄²⁻ (x − 8 = −2) and +6 in Cr₂O₇²⁻ (2x − 14 = −2). There is no change in oxidation number, so the change is not redox: it is an acid-base equilibrium.
K_c = $\dfrac{[\ce{CrO4^2-}]^2[\ce{H+}]^2}{[\ce{Cr2O7^2-}]}$ (water is left out because it is the solvent).