Half-Reaction (Acidic and Basic) - Chemistry

Question

Balance the following chemical reaction by the half reaction method under both acidic and basic conditions. What was oxidized and what was reduced?

IrO₃ + V(OH)₄⁺² ------> Ir(OH)₅⁺³ + V⁺²

 

Answer 1

IrO₃ ===> Ir(OH)₅³⁺ ... oxidation half reaction because Ir goes from +6 to +8 oxidation number

IrO₃ + 2H2O ===> Ir(OH)₅³⁺ ... balanced for O

IrO₃ + 2H₂O + H⁺ ===> Ir(OH)₅³⁺ ... balanced for H (in acidic conditions)

IrO₃ + 2H₂O + H⁺ ===> Ir(OH)₅³⁺ + 2e- ... balanced for charge and a correctly balanced half reaction

V(OH)₄²⁺ ===> V²⁺ ... reduction half reaction because V goes from +6 to +2

V(OH)₄²⁺ ===> V²⁺ + 4H₂O ... balanced for O

V(OH)₄²⁺ + 4H⁺ ===> V²⁺ + 4H₂O ... balanced for H (in acidic conditions)

V(OH)₄²⁺ + 4H⁺ + 4e- ===> V²⁺ + 4H₂O ... balanced for charge and a correctly balanced half reaction

Multiply the oxidation half reaction by 2 to equalize the electrons and add the two half reactions:

2 IrO₃ + 4H₂O + 2H⁺ ===> 2 Ir(OH)₅³⁺ + 4e-

V(OH)₄²⁺ + 4H⁺ + 4e- ===> V²⁺ + 4H₂O

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2 IrO₃ + 4H₂O + 2H⁺ + V(OH)₄²⁺ + 4H⁺ + 4e- ===> 2 Ir(OH)₅³⁺ + 4e- + V²⁺ + 4H₂O

Eliminating common items from each side of the equation, we end up with...

2 IrO₃ + 2H⁺ + V(OH)₄²⁺ + 4H⁺ ===> 2 Ir(OH)₅³⁺ + V²⁺ OVERALL BALANCED REACTION

You should try doing this in basic conditions. You can follow the same idea as above but every time you add H+, you need to add an equal number of OH- moieties to the opposite side. They will eventually end up as H2O. Give it a try.