The balanced redox reactions for the sequential reduction of vanadium are given

Question

The balanced redox reactions for the sequential reduction of vanadium are given below. reduction from + 5 to + 4 2 VO_2^+(aq) + 4H^+ (aq) + zn(s) rightarrow 2 VO^2+ (aq) + zn^2+ 9aq) + 2 H_2O(l) reduction from + 4t o + 3: 2 VO^2+ (aq) + Zn(s) + 4 H^+ (aq) rightarrow 2 V^3 + (aq) + Zn^2 + (aq) + 2H_2O(l) reduction from + 3 to +2: 2 V^3+ (aq) + Zn(s) rightarrow 2 V^2+(aq) + Zn^2 + (aq) If you had 12.0 mL of a 0.0040 M solution of VO_2^+(aq), how many grams of Zn metal would be required to completely reduce the vanadium?

Explanation / Answer

A:- The overall equation is as follows,

2 VO2+(aq) + 3 Zn(s) + 8 H+(aq) -----------> 2 V2+(aq) + 3 Zn2+(aq) + 4 H2O(l)

So 2 moles VO2+(aq) requires 3 moles of Zn(s) , the number of moles VO2+ is calculated as,

Number moles = molarity * volume (in L)

                      = 0.0040 M * 0.012 L

                      = 4.8 * 10^-5 moles

Number of moles Zn = (2 * 4.8 *10^-5) / 3

                             = 3.2 * 10^-5 moles

Mass of Zn required = 3.2 * 10^-5 * atomic mass of Zn

                             = 3.2 *10^-5 * 65.38

      Mass of Zn required = 0.00209 grams

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