Question

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?

Answer 1

Answer : 0.0392 grams of Zn metal would be required to completely reduced the vanadium.

Explanation :

Let us rewrite the given equations again.

`2VO_(2)^(+) (aq)+ 4H^(+)(aq) + Zn (s)\rightarrow 2VO^(2+)(aq)+Zn^(2+)+2H2O(l)`
`2VO^(2+)(aq)+ 4H^(+)(aq) + Zn (s)\rightarrow 2V^(3+) (aq)+Zn^(2+)+2H2O(l)`

`2V^(3+) (aq)+ Zn (s)\rightarrow 2V^(2+)(aq)+Zn^(2+)(aq)`

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

`2VO_(2)^(+) (aq)+ 8H^(+) (aq) + 3Zn (s)\rightarrow 2V^(2+)(aq)+3Zn^(2+)(aq)+4H_(2)O(l)`

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

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

`12.1 mL * (1L)/(1000mL)* (0.033mol)/(L)=0.0003993mol NO_(2)^(+)`

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.

`0.0003993mol NO_(2)^(+)* (3mol Zn)/(2molNO_(2)^(+))=0.00059895mol Zn`

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.

`0.00059895mol Zn* (65.38gZn)/(1molZn)=0.0392gZn`

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