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

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Chemistry

Bullet

27 March, 03:28

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

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Leland Bowen · 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