The balanced redox reactions for the sequential reduction of vanadium are given below.
reduction from +5 to +4:
2 VO2+(aq) + 4 H +(aq) + Zn(s) → 2 VO2+(aq) + Zn2+(aq) + 2 H2O(l)
reduction from +4 to +3:
2 VO2+(aq) + Zn(s) + 4 H +(aq) → 2 V3+(aq) + Zn2+(aq) + 2 H2O(l)
reduction from +3 to +2:
2 V3+(aq) + Zn(s) → 2 V2+(aq) + Zn2+(aq)
If you had 12.1 mL of a 0.0033 M solution of VO2+(aq), how many grams of Zn metal would be required to completely reduce the vanadium?

[tex]2VO_{2}^{+} (aq)+ 4H^{+}(aq) + Zn (s)\rightarrow 2VO^{2+}(aq)+Zn^{2+}+2H2O(l)[/tex][tex]2VO^{2+}(aq)+ 4H^{+}(aq) + Zn (s)\rightarrow 2V^{3+} (aq)+Zn^{2+}+2H2O(l)[/tex]

[tex]2V^{3+} (aq)+ Zn (s)\rightarrow 2V^{2+}(aq)+Zn^{2+}(aq)[/tex]

On adding above equations, we get the following combined equation.

[tex]2VO_{2}^{+} (aq)+ 8H^{+} (aq) + 3Zn (s)\rightarrow 2V^{2+}(aq)+3Zn^{2+}(aq)+4H_{2}O(l)[/tex]

We have 12.1 mL of 0.033 M solution of VO₂⁺.

Let us find the moles of VO₂⁺ from this information.

[tex]12.1 mL \times \frac{1L}{1000mL}\times \frac{0.033mol}{L}=0.0003993mol NO_{2}^{+}[/tex]

From the combined equation, we can see that the mole ratio of VO₂⁺ to Zn is 2:3.

Let us use this as a conversion factor to find the moles of Zn.

[tex]0.0003993mol NO_{2}^{+}\times \frac{3mol Zn}{2molNO_{2}^{+}}=0.00059895mol Zn[/tex]

Let us convert the moles of Zn to grams of Zn using molar mass of Zn.

Molar mass of Zn is 65.38 g/mol.

[tex]0.00059895mol Zn\times \frac{65.38gZn}{1molZn}=0.0392gZn[/tex]

We need 0.0392 grams of Zn metal to completely reduce vanadium.