Vanadium and chromium redox chemistryEdexcel International A Level Chemistry: Revision notes
Section 1
Vanadium oxidation states and colours
Vanadium shows four common oxidation states in aqueous solution, each with a different colour:
- +5: VO₂⁺, yellow
- +4: VO²⁺, blue
- +3: V³⁺, green
- +2: V²⁺, violet
The colour differs because the number of d electrons and the ligands change the d-orbital splitting.
Reducing yellow VO₂⁺ stepwise gives the sequence yellow, blue, green, violet.
Learn the sequence yellow, blue, green, violet for +5, +4, +3, +2.
Section 2
Redox interconversion of vanadium using E° values
Standard electrode potentials:
- VO₂⁺ + 2H⁺ + e⁻ → VO²⁺ + H₂O, E° = +1.00 V
- VO²⁺ + 2H⁺ + e⁻ → V³⁺ + H₂O, E° = +0.34 V
- V³⁺ + e⁻ → V²⁺, E° = −0.26 V
A reducing agent can reduce a species if its E° is more negative than that of the species being reduced, so that E°cell = E°(reduced species) − E°(reducing agent) is positive.
Zinc (Zn²⁺/Zn, −0.76 V) reduces every step, so excess zinc in acid takes VO₂⁺ all the way to V²⁺: Zn + 2VO₂⁺ + 4H⁺ → Zn²⁺ + 2VO²⁺ + 2H₂O is the first step.
Fe²⁺ (+0.77 V) can reduce VO₂⁺ (+1.00 V) to VO²⁺ only. Sn²⁺ (+0.15 V) reduces VO₂⁺ and VO²⁺ and stops at V³⁺.
A positive E°cell shows the reaction is feasible, not that it is fast or that it will happen if the reagent is the wrong way round.
Section 3
Reduction of dichromate(VI) with zinc
Dichromate(VI) is orange, Cr₂O₇²⁻ with chromium +6.
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O, E° = +1.33 V
Cr³⁺ + e⁻ → Cr²⁺, E° = −0.41 V
Zinc (−0.76 V) is more negative than both, so excess zinc in acid reduces it in two steps:
orange Cr₂O₇²⁻ → green Cr³⁺ → blue Cr²⁺
Overall for the first stage: Cr₂O₇²⁻ + 14H⁺ + 3Zn → 2Cr³⁺ + 3Zn²⁺ + 7H₂O, E°cell = 1.33 + 0.76 = +2.09 V.
Section 4
Oxidation of chromium(III) to chromate(VI)
Cr³⁺ is oxidised to chromate(VI) by hydrogen peroxide in alkaline conditions:
2Cr³⁺ + 3H₂O₂ + 10OH⁻ → 2CrO₄²⁻ + 8H₂O
The solution changes from green to yellow. Chromium goes from +3 to +6, and oxygen in H₂O₂ goes from −1 to −2.
To obtain dichromate(VI), the yellow solution is then acidified and turns orange.
Remember the order: oxidise in alkali first to give yellow chromate(VI), then acidify to give orange dichromate(VI).
Section 5
The chromate(VI) and dichromate(VI) equilibrium
Cr₂O₇²⁻ + H₂O ⇌ 2CrO₄²⁻ + 2H⁺
orange ⇌ yellow
- Add alkali: OH⁻ removes H⁺, the equilibrium shifts right, the solution turns yellow.
- Add acid: [H⁺] increases, the equilibrium shifts left, the solution turns orange.
Chromium is +6 on both sides, so this is not a redox reaction. It is an acid–base type change in position of equilibrium.
Do not describe the dichromate to chromate change as oxidation or reduction. The oxidation number of chromium is +6 in both.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Vanadium and chromium redox chemistry
- A student adds excess zinc to a yellow solution of acidified vanadate(V) ions, VO₂⁺, and sees the colour change in stages. Standard electrode potentials: VO₂⁺/VO²⁺ +1.00 V; VO²⁺/V³⁺ +0.34 V; V³⁺/V²⁺ −0.26 V; Zn²⁺/Zn −0.76 V.Calculate the standard cell potential for the reduction of V³⁺ to V²⁺ by zinc and state whether the reaction is feasible.2 marks
- A chemist has a blue solution containing VO²⁺ ions and several possible reagents. Standard electrode potentials: VO₂⁺/VO²⁺ +1.00 V; VO²⁺/V³⁺ +0.34 V; V³⁺/V²⁺ −0.26 V; Cr₂O₇²⁻/Cr³⁺ (acidic) +1.33 V; Fe³⁺/Fe²⁺ +0.77 V; Sn⁴⁺/Sn²⁺ +0.15 V; Zn²⁺/Zn −0.76 V.Use E° values to explain why Fe²⁺ ions can reduce VO₂⁺ to VO²⁺ but cannot reduce VO²⁺ to V³⁺.2 marks
- Excess zinc is added to an acidified solution of potassium dichromate(VI), which is orange. Standard electrode potentials: Cr₂O₇²⁻/Cr³⁺ (acidic) +1.33 V; Cr³⁺/Cr²⁺ −0.41 V; Zn²⁺/Zn −0.76 V.Describe and explain the colour changes seen as the excess zinc reduces the dichromate(VI).3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).