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

Explanation :

Let us rewrite the given equations again.

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

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

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

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.

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.

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