Part 4: Vanadium and chromium redox chemistry

Use observed colours and balanced half-equations to track oxidation states, and calculate whether a reducing or oxidising agent can complete each stage.

Distinguish VO₂⁺ from VO²⁺

Vanadium in acidic solution can be reduced successively from +5 to +2 by zinc. The observed progression is yellow → blue → green → violet, although mixtures during conversion can give intermediate appearances. VO₂⁺ has x − 4 = +1, so vanadium is +5; VO²⁺ has x − 2 = +2, so vanadium is +4. The subscript and superscript are chemically different.

Zinc supplies electrons by oxidation to Zn²⁺. Acid supplies H⁺ to convert oxygen-containing vanadium ions into water as reduction proceeds. Protect the reduced solution from unnecessary air exposure because oxygen can reoxidise low oxidation states. Follow the teacher’s small-scale method and controls: soluble vanadium compounds are hazardous, and zinc/acid can generate flammable hydrogen.

A reducing agent may stop at an intermediate state

For Zn²⁺ + 2e⁻ ⇌ Zn, E° = −0.76 V. Pair each vanadium reduction with zinc oxidation: +1.00 − (−0.76) = +1.76 V; +0.34 − (−0.76) = +1.10 V; −0.26 − (−0.76) = +0.50 V. All three are positive, so reduction through to V²⁺ is thermodynamically feasible under the stated standard comparison.

For Sn²⁺ + 2e⁻ ⇌ Sn with E° = −0.14 V, the corresponding values are +1.14 V, +0.48 V and −0.12 V. Tin metal is therefore predicted to reduce V(V) through V(IV) to V(III), but not the final V(III) → V(II) step under standard conditions. State the final species and colour: V³⁺, green. A negative potential is a condition-dependent thermodynamic conclusion, not an assertion that no chemistry could ever occur.

To write an overall zinc equation, multiply vanadium half-equations to accept zinc’s two electrons. For the last step: 2V³⁺ + Zn → 2V²⁺ + Zn²⁺. For the whole +5 to +2 change, each vanadium accepts three electrons; two vanadium species need six, supplied by three zinc atoms.

Zinc reduces chromium(VI) to chromium(II), with air excluded

Acidified orange dichromate(VI) can be reduced first to chromium(III), commonly observed green, then to blue chromium(II) using zinc. Cr²⁺ is readily oxidised by air, so maintaining an oxygen-excluding atmosphere in an appropriate teacher-controlled apparatus is necessary to observe/preserve the reduced state. Hydrogen produced by zinc/acid can help exclude air in the established demonstration; never seal a gas-generating system or introduce flames.

Use the given reduction values: Cr₂O₇²⁻/Cr³⁺ = +1.33 V, Cr³⁺/Cr²⁺ = −0.41 V and Cr²⁺/Cr = −0.91 V; Zn²⁺/Zn = −0.76 V. The zinc-driven step emfs are +2.09 V, +0.35 V and −0.15 V. Thus reduction to Cr²⁺ is predicted, but reduction onward to chromium metal is not under standard conditions. Testing the last step prevents an unsupported claim of complete reduction to metal.

Iron(II), with E°(Fe³⁺/Fe²⁺) = +0.77 V, reduces dichromate to Cr³⁺ because 1.33 − 0.77 = +0.56 V. It does not reduce Cr³⁺ to Cr²⁺ because −0.41 − 0.77 = −1.18 V. The reductant matters as much as the initial chromium species. Chromium(VI) compounds require stringent laboratory controls and appropriate hazardous-waste collection.

Oxidise chromium(III) in alkaline conditions

Add sufficient NaOH to form the soluble green [Cr(OH)₆]³⁻ species, then hydrogen peroxide can oxidise chromium(III) to yellow chromate(VI). Hydrogen peroxide is reduced to OH⁻ in this alkaline reaction. Conditions change both the chromium species and the appropriate potential values; the acidic dichromate/Cr³⁺ couple is not the right isolated description of the alkaline starting solution.

For a numerical comparison, Pearson’s June 2025 Q6(f) supplies CrO₄²⁻ + 4H₂O + 3e⁻ ⇌ Cr(OH)₃(s) + 5OH⁻, E° = −0.13 V, and H₂O₂ + 2e⁻ ⇌ 2OH⁻, E° = +0.88 V. For oxidation of that solid hydroxide by peroxide, E°cell = 0.88 − (−0.13) = +1.01 V. The balanced reaction is 2Cr(OH)₃ + 3H₂O₂ + 4OH⁻ → 2CrO₄²⁻ + 8H₂O. The −0.13 V value belongs to the stated solid-hydroxide pair; do not silently assign it to the different dissolved [Cr(OH)₆]³⁻ pair.

To balance, reverse the chromium reduction, double it to release six electrons, and triple the peroxide reduction to accept six. Cancel electrons, water or hydroxide appearing on both sides. Chromium changes from +3 to +6: this is redox. Subsequent acidification produces orange dichromate without further changing the oxidation state. Follow the specified method before acidification, including management of any excess peroxide.

Chromate and dichromate interconvert without redox

Yellow chromate and orange dichromate are connected by 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O. Adding acid increases [H⁺] and moves the equilibrium towards dichromate. Adding alkali removes H⁺ by neutralisation and favours chromate. Chromium is +6 on both sides, so the colour change is not evidence of electron transfer.

Check the equation: two Cr, eight O and two H occur on each side, with total charge −2 on each side. Combining it with the alkaline peroxide oxidation explains the full route from Cr(III) to Cr(VI) dichromate: alkaline oxidation first, then acid–base equilibrium shift. These are two different processes with different reagents and purposes.

In an unfamiliar reaction map, label each arrow with reagent, conditions, observation and oxidation-number change. A single colour arrow can conceal ligand exchange, acid–base chemistry or redox; establish the species and the electron accounting before classifying it.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.

Sources