All subjects · Chemistry · WCH15 (Unit 5)

Redox chemistry of vanadium and chromium

Vanadium and chromium each show several coloured oxidation states. Electrode potentials explain which reagents move them from one state to another.

Vanadium (spec 17.18, 17.19)

Oxidation state	Ion	Colour
+5	VO2+	yellow
+4	VO²+	blue
+3	V³+	green
+2	V²+	violet

Reducing vanadium

Zinc and acid reduce vanadium(V) step by step: yellow -> (green, a yellow and blue mixture) -> blue -> green -> violet. E° values:

VO2+ + 2H+ + e- -> VO²+ + H2O, +1.00 V
VO²+ + 2H+ + e- -> V³+ + H2O, +0.34 V
V³+ + e- -> V²+, -0.26 V
Zn2+ + 2e- -> Zn, -0.76 V

Zinc is a strong enough reducing agent for every step (V3+ to V2+: -0.26 - (-0.76) = +0.50 V). A weaker reducing agent such as tin (Sn2+/Sn, -0.14 V) reduces vanadium only as far as V3+, because -0.26 - (-0.14) is negative.

Chromium (spec 17.20)

Reduction: zinc and acid reduce orange Cr2O7²- to green Cr3+, then to blue Cr2+. Cr2+ is easily oxidised by air, so air is kept out (the hydrogen produced helps).
Oxidation: add excess NaOH to Cr3+ to make green [Cr(OH)6]3-, then warm with hydrogen peroxide, giving yellow chromate(VI): 2[Cr(OH)6]3- + 3H2O2 -> 2CrO4²- + 2OH- + 8H2O. Acidify to give orange dichromate(VI).

Chromate and dichromate (spec 17.21)

2CrO4²-(aq) + 2H+(aq) ⇌ Cr2O7²-(aq) + H2O(l)

yellow chromate(VI) in alkali; orange dichromate(VI) in acid. Adding acid shifts it right, adding alkali shifts it left. It is not a redox reaction: chromium is +6 in both.

Worked example

Use E° values to show that zinc can reduce chromium(III) to chromium(II). Cr3+ + e- -> Cr2+, -0.41 V; Zn2+ + 2e- -> Zn, -0.76 V.

Cr3+ is reduced and Zn is oxidised.
E°cell = -0.41 - (-0.76) = +0.35 V

Answer: E°cell is positive, so the reduction is feasible: green to blue

Key idea: Vanadium: +5 yellow, +4 blue, +3 green, +2 violet; zinc/acid reduces all the way. Chromium: Zn/acid gives Cr3+ (green) then Cr2+ (blue); H2O2 in alkali oxidises Cr3+ to CrO4²- (yellow); acid turns chromate into orange dichromate (not redox).

Check yourself

What colour is vanadium in the +4 oxidation state?
- Yellow
- Green
- Violet
- Blue
Show answer
Blue. VO²+ is blue.
What is the oxidation number of vanadium in VO2+?
Show answer
+5. x + 2(-2) = +1, so x = +5.
Calculate E°cell for zinc reducing V3+ to V2+. (V3+/V2+ -0.26 V; Zn2+/Zn -0.76 V)
Show answer
+0.50 V. E°cell = -0.26 - (-0.76) = +0.50 V, so the reduction is feasible.
Tin (Sn2+/Sn, E° = -0.14 V) is added to acidified vanadium(V). What is the final colour? (VO2+/VO²+ +1.00 V; VO²+/V3+ +0.34 V; V3+/V2+ -0.26 V)
- Green, because tin reduces vanadium only as far as V3+
- Violet, because tin reduces vanadium to V2+
- Yellow, because no reaction occurs
- Blue, because tin reduces vanadium only to VO²+
Show answer
Green, because tin reduces vanadium only as far as V3+. Steps to V3+ give positive E°cell (+1.14 V and +0.48 V); V3+ to V2+ gives -0.12 V.
What colour is the final solution when dichromate(VI) is reduced fully by zinc and acid in the absence of air?
- Green
- Orange
- Blue
- Yellow
Show answer
Blue. Cr2O7²- (orange) -> Cr3+ (green) -> Cr2+ (blue).
Acid is added to a yellow solution of potassium chromate(VI). What is seen and why?
- It turns orange, because the equilibrium shifts to form dichromate(VI)
- It turns green, because chromium is reduced
- It turns blue, because chromium(II) forms
- No change, because chromium is already +6
Show answer
It turns orange, because the equilibrium shifts to form dichromate(VI). 2CrO4²- + 2H+ ⇌ Cr2O7²- + H2O shifts right.
Which reagents oxidise chromium(III) to chromium(VI)?
- Zinc and hydrochloric acid
- Dilute sulfuric acid
- Hydrogen peroxide in alkaline solution, warmed
- Aqueous ammonia only
Show answer
Hydrogen peroxide in alkaline solution, warmed. H2O2 in excess NaOH turns green [Cr(OH)6]3- into yellow CrO4²-.

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Last updated 7 November 2026. Spotted a mistake? Tell us.