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 11.8 mL of a 0.0031 M solution of VO2+(aq), how many grams of Zn metal would be required to completely reduce the vanadium? WebAssign will check your answer for the correct number of significant figures.

Answer 1

mass Zn metal required = 0.0036 g

Explanation

The overall equation is : 2 VO₂⁺ (aq) + 3 Zn (s) + 8 H⁺ (aq) 2 V²⁺ + 3 Zn²⁺ (aq) + 4 H₂O (l)

concentration VO₂⁺ = 0.0031 M

volume VO₂⁺ solution = 11.8 mL = 0.0118 L

moles VO₂⁺ = (concentration VO₂⁺) * (volume VO₂⁺ solution)

moles VO₂⁺ = (0.0031 M) * (0.0118 L)

moles VO₂⁺ = 3.658 x 10^-5 mol

moles Zn required = (moles VO₂⁺) * (3 moles Zn / 2 moles VO₂⁺)

moles Zn required = (3.658 x 10^-5 mol) * (3/2)

moles Zn required = 5.487 x 10^-5 mol

mass Zn required = (moles Zn required) * (molar mass Zn)

mass Zn required = (5.487 x 10^-5 mol) * (65.38 g/mol)

mass Zn required = 0.00359 g