Question

Given the equation

5VO2++Mn2+→5VO2++MnO4−

how many H2O molecules should be added to the right side of the equation to balance the oxygen atoms

Answer 1

Given equation:

5VO₂⁺ + Mn²⁺ = 5VO²⁺ + MnO₄^-

Here, two half reactions are involved:

1. Reduction half reaction (RHR)

5VO₂⁺ = 5VO²⁺

5VO₂⁺ = 5VO²⁺ + 5H₂O (balancing oxygen atoms)

5VO₂⁺ + 10H⁺ = 5VO²⁺ + 5H₂O (balancing hydrogen atoms)

Balanced RHR: 5VO₂⁺ + 10H⁺ + 5e^- = 5VO²⁺ + 5H₂O (balancing charge)

2. Oxidation half reaction (OHR)

Mn²⁺ = MnO₄^-

Mn²⁺ + 4H₂O = MnO₄^- (balancing oxygen atoms)​

Mn²⁺ + 4H₂O = MnO₄^- + 8H⁺ (balancing hydrogen atoms)​

Balanced OHR: Mn²⁺ + 4H₂O = MnO₄^- + 8H⁺ + 5e^- (balancing charge)​

By adding both the balanced RHR and OHR, we will get the complete balanced equation as follows.

5VO₂⁺ + Mn²⁺ + 2H⁺ = 5VO²⁺ + MnO₄^- + H₂O

Therefore, one H₂O molecule should be added to the right side of the equation to balance the oxygen atoms