Redox chemistry of vanadium and chromiumEdexcel A-Level Chemistry: Revision notes

Vanadium shows four oxidation states in aqueous compounds, each a different colour:

• +5: $\mathrm{VO_2^+}$, yellow
• +4: $\mathrm{VO^{2+}}$, blue
• +3: $\mathrm{V^{3+}}$, green
• +2: $\mathrm{V^{2+}}$, violet

The colour changes show how far a reduction has gone, e.g. yellow → blue → green → violet when $\mathrm{VO_2^+}$ is reduced by zinc in acid.

Use the electrode potentials:

• $\mathrm{VO_2^+ + 2H^+ + e^- \rightleftharpoons VO^{2+} + H_2O}$, $E^\ominus = +1.00$ V
• $\mathrm{VO^{2+} + 2H^+ + e^- \rightleftharpoons V^{3+} + H_2O}$, $E^\ominus = +0.34$ V
• $\mathrm{V^{3+} + e^- \rightleftharpoons V^{2+}}$, $E^\ominus = -0.26$ V

A reducing agent can reduce a species if its own $E^\ominus$ is less positive (so $E^\ominus_{cell} > 0$).

• Zinc ($-0.76$ V) is below all three, so excess Zn/acid reduces $+5$ all the way to $+2$: e.g. $\mathrm{V^{3+} \rightarrow V^{2+}}$: $-0.26 - (-0.76) = +0.50$ V.
• Iron(II) ($+0.77$ V) reduces only $\mathrm{VO_2^+ \rightarrow VO^{2+}}$ ($1.00 - 0.77 = +0.23$ V); it cannot reduce $\mathrm{VO^{2+}}$ ($0.34 - 0.77 = -0.43$ V).
• Tin(II) ($+0.15$ V) reduces down to V³⁺ but not V²⁺.

Overall equation, zinc: $\mathrm{2VO_2^+ + 4H^+ + Zn \rightarrow 2VO^{2+} + 2H_2O + Zn^{2+}}$.

Orange dichromate(VI), $\mathrm{Cr_2O_7^{2-}}$ (Cr is $+6$), is reduced by zinc in acid:

$\mathrm{Cr_2O_7^{2-} + 14H^+ + 3Zn \rightarrow 2Cr^{3+} + 7H_2O + 3Zn^{2+}}$ (orange → green)

With excess zinc the Cr³⁺ is reduced further to blue $\mathrm{Cr^{2+}}$ (Cr is $+2$):

$\mathrm{2Cr^{3+} + Zn \rightarrow 2Cr^{2+} + Zn^{2+}}$

This is feasible because $E^\ominus$(Zn²⁺/Zn) $= -0.76$ V is more negative than $E^\ominus$(Cr³⁺/Cr²⁺) $= -0.41$ V: $E^\ominus_{cell} = -0.41 - (-0.76) = +0.35$ V. Dichromate(VI) with $E^\ominus = +1.33$ V is a strong oxidising agent.

$\mathrm{Cr^{3+}}$ (green) is oxidised by hydrogen peroxide in alkaline conditions. Add excess NaOH(aq), then $\mathrm{H_2O_2}$ and warm. Yellow chromate(VI), $\mathrm{CrO_4^{2-}}$, forms:

$\mathrm{2Cr^{3+} + 3H_2O_2 + 10OH^- \rightarrow 2CrO_4^{2-} + 8H_2O}$

In alkali, $E^\ominus$(HO₂⁻/OH⁻) $= +0.87$ V and $E^\ominus$(CrO₄²⁻/Cr(OH)₃) $= -0.13$ V, so $E^\ominus_{cell} = +1.00$ V: feasible. Acidifying the yellow chromate(VI) then gives orange dichromate(VI).

$\mathrm{2CrO_4^{2-} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} + H_2O}$ (yellow ⇌ orange)

• Add acid: $[\mathrm{H^+}]$ increases, equilibrium shifts right, solution turns orange.
• Add alkali: OH⁻ removes H⁺, equilibrium shifts left, solution turns yellow.

This is not redox: chromium is $+6$ in both ions, so the interconversion is a pH-dependent equilibrium. It can be reversed repeatedly.