A2 Chemistry 9701: The Vanadium Oxidation-State Colour Series and Zinc Reduction

Ammonium vanadate(V), $\text{NH}_4\text{VO}_3$ , dissolves in dilute sulfuric acid to give a yellow solution containing the $\text{VO}_2^+$ ion. Adding excess zinc powder to this acidified solution reduces vanadium successively through a series of distinctly coloured oxidation states, ending at the pale violet $\text{V}^{2+}(aq)$ ion.

Standard electrode potentials for the relevant half-reactions are:

$\text{VO}_2^+(aq) + 2\text{H}^+(aq) + e^- \rightleftharpoons \text{VO}^{2+}(aq) + \text{H}_2\text{O}(l) \qquad E^{\ominus} = +1.00\ \text{V}$ $\text{VO}^{2+}(aq) + 2\text{H}^+(aq) + e^- \rightleftharpoons \text{V}^{3+}(aq) + \text{H}_2\text{O}(l) \qquad E^{\ominus} = +0.34\ \text{V}$ $\text{V}^{3+}(aq) + e^- \rightleftharpoons \text{V}^{2+}(aq) \qquad E^{\ominus} = -0.26\ \text{V}$ $\text{Zn}^{2+}(aq) + 2e^- \rightleftharpoons \text{Zn}(s) \qquad E^{\ominus} = -0.76\ \text{V}$

(a) State the oxidation state of vanadium in each of $\text{VO}_2^+$ , $\text{VO}^{2+}$ , $\text{V}^{3+}$ and $\text{V}^{2+}$ . [2]

(b) State the colour of each of these four vanadium species, in the order they would be observed as the zinc reduction proceeds from $\text{VO}_2^+$ to $\text{V}^{2+}$ . [2]

(c) By combining the first half-equation above with the reverse of the zinc half-equation, construct the overall balanced ionic equation for the reduction of $\text{VO}_2^+(aq)$ to $\text{VO}^{2+}(aq)$ by zinc, and calculate $E^{\ominus}_{cell}$ for this reaction. [3]

(d) Explain, using the standard electrode potentials given, why zinc is able to reduce vanadium all the way down to $\text{V}^{2+}(aq)$ , whereas a weaker reducing agent (with a standard electrode potential between $-0.26\ \text{V}$ and $+0.34\ \text{V}$ ) might succeed in producing $\text{VO}^{2+}(aq)$ or $\text{V}^{3+}(aq)$ but fail to reduce $\text{V}^{3+}(aq)$ further to $\text{V}^{2+}(aq)$ . [2]

Part (a): Oxidation states

In $\text{VO}_2^+$ : oxygen is $-2$ each, so $2\times(-2)=-4$ ; for an overall charge of $+1$ , $\text{V}=+1-(-4)=+5$ .

In $\text{VO}^{2+}$ : oxygen contributes $-2$ ; for an overall charge of $+2$ , $\text{V}=+2-(-2)=+4$ .

$\text{V}^{3+}$ and $\text{V}^{2+}$ have no oxygen attached, so vanadium’s oxidation state equals the ion’s charge directly: $+3$ and $+2$ respectively.

$\text{VO}_2^+:\ +5 \qquad \text{VO}^{2+}:\ +4 \qquad \text{V}^{3+}:\ +3 \qquad \text{V}^{2+}:\ +2$

Part (b): Colour sequence

As zinc reduces vanadium stepwise from $+5$ to $+2$ , the classic colour sequence observed is:

$\underbrace{\text{VO}_2^+}_{\text{yellow}} \rightarrow \underbrace{\text{VO}^{2+}}_{\text{blue}} \rightarrow \underbrace{\text{V}^{3+}}_{\text{green}} \rightarrow \underbrace{\text{V}^{2+}}_{\text{violet}}$

Part (c): Ionic equation and $E^{\ominus}_{cell}$ for the first reduction step

The vanadium half-equation transfers 1 electron; the zinc half-equation transfers 2 electrons. Multiplying the vanadium equation by 2 so electrons cancel, and reversing the zinc equation (since zinc is oxidised):

$2\text{VO}_2^+(aq) + 4\text{H}^+(aq) + 2e^- \rightarrow 2\text{VO}^{2+}(aq) + 2\text{H}_2\text{O}(l)$ $\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-$

Adding and cancelling the $2e^-$ : $2\text{VO}_2^+(aq) + 4\text{H}^+(aq) + \text{Zn}(s) \rightarrow 2\text{VO}^{2+}(aq) + 2\text{H}_2\text{O}(l) + \text{Zn}^{2+}(aq)$

(Charge check: LHS $2(+1)+4(+1)+0=+6$ ; RHS $2(+2)+0+(+2)=+6$ (balanced. Atom check: V: 2=2; O: LHS $2\times2=4$ , RHS $2(1)+2(1)=4$ ; H: LHS 4, RHS $2\times2=4$ ; Zn: 1=1) all balanced.)

$E^{\ominus}_{cell} = E^{\ominus}_{reduction} - E^{\ominus}_{oxidation} = 1.00 - (-0.76) = +1.76\ \text{V}$

Part (d): Why zinc drives the reduction all the way to V²⁺

Each vanadium reduction step is feasible only if it is combined with an oxidation whose own $E^{\ominus}$ is more negative than that step’s $E^{\ominus}$ (so that $E^{\ominus}_{cell}=E^{\ominus}_{reduction}-E^{\ominus}_{oxidation}$ is positive).

The three vanadium steps have $E^{\ominus}$ values $+1.00\ \text{V}$ , $+0.34\ \text{V}$ and $-0.26\ \text{V}$ . Each successive step is a weaker oxidising agent than the last, making it progressively harder to reduce. The final step, $\text{V}^{3+}/\text{V}^{2+}$ ( $-0.26\ \text{V}$ ), is therefore the most difficult.

$E^{\ominus}(\text{Zn}^{2+}/\text{Zn}) = -0.76\ \text{V}$

is more negative than all three vanadium potentials, including $-0.26\ \text{V}$ . So combining zinc’s oxidation with any of the three vanadium reductions gives a positive $E^{\ominus}_{cell}$ , and zinc can drive the sequence all the way to $\text{V}^{2+}$ .

A weaker reducing agent with a standard electrode potential between $-0.26\ \text{V}$ and $+0.34\ \text{V}$ would still give a positive (feasible) $E^{\ominus}_{cell}$ for the first two, easier steps (down to $\text{VO}^{2+}$ or $\text{V}^{3+}$ ), since its potential is more negative than $+1.00\ \text{V}$ and $+0.34\ \text{V}$ . However, because its potential is less negative than $-0.26\ \text{V}$ , combining it with the final $\text{V}^{3+}/\text{V}^{2+}$ step would give a negative $E^{\ominus}_{cell}$ , not feasible, so reduction would stop at $\text{V}^{3+}$ .

Final answers

(a) $\text{VO}_2^+$ : $+5$ ; $\text{VO}^{2+}$ : $+4$ ; $\text{V}^{3+}$ : $+3$ ; $\text{V}^{2+}$ : $+2$ .
(b) Yellow $\rightarrow$ blue $\rightarrow$ green $\rightarrow$ violet.
(c) $2\text{VO}_2^+(aq)+4\text{H}^+(aq)+\text{Zn}(s)\rightarrow2\text{VO}^{2+}(aq)+2\text{H}_2\text{O}(l)+\text{Zn}^{2+}(aq)$ ; $E^{\ominus}_{cell}=+1.76\ \text{V}$ .
(d) Zinc’s $E^{\ominus}=-0.76\ \text{V}$ is more negative than all three vanadium potentials (including the hardest, $-0.26\ \text{V}$ ), giving positive $E^{\ominus}_{cell}$ throughout; a weaker reducing agent (between $-0.26\ \text{V}$ and $+0.34\ \text{V}$ ) would fail only the final, hardest step.