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 11.7 mL of a 0.0037 M solution of VO2+(aq), how many grams of Zn metal would be required to completely reduce the vanadium

Answer 1

The overall equation for the combined reactions become:
2VO₂⁺²₍aq₎ + 8H⁺₍aq₎ + 3Zn⁺²⁽s⁾ ⇒ 2V⁺²₍aq₎ + 3Zn⁺²₍aq₎ + 4H₂O₍l₎
The volume of solution is:
11.7/1000 = 0.0117 Litres
The moles of VO₂⁺² are:
0.0117 × 0.0037 = 4.3 × 10⁻⁵ mol
As per the equation, 2 moles of VO₂⁺² need 3 moles of 3Zn⁺²
Therefore, moles of Zn⁺² needed are:
6.5 × 10⁻⁵ mol
One mole Zn metal produces one mol of ions. So we need
6.5 × 10⁻⁵ mol of Zn metal
Mass required = moles × Molecular weight
Mass = 6.5 × 10⁻⁵ × 65
Mass = 0.0042 grams