Question

HOI + NO = I2 + NO3-1 balance the following reaction in acid

Answer 1

       +1    +2       0      +5

    HOI + NO I₂ + NO₃^-1

Here the oxidation number of I decreases from +1 to 0 & the oxidation number of N increases from +2 to +5

So the reduction half reaction is HOI   I₂

   the oxidation half reaction is NO NO₃^-1

Balancing reduction half reaction. HOI   I₂

Balancing I atoms : 2 HOI   I₂   

Balancing O atoms ( by adding H₂O on deficient side) : 2 HOI   I₂    + 2H₂O

Balance H atoms ( by adding H+ on the deficient side ) : 2 HOI +2H⁺ I₂    + 2H₂O

Balance charge ( by adding electrons ) : 2 HOI +2H⁺ +2e^- I₂    + 2H₂O    ---(1)

Balancing oxidation half reaction. NO NO₃^-1

Balancing O atoms ( by adding H₂O on deficient side) : NO+2H₂O NO₃^-1

Balance H atoms ( by adding H+ on the deficient side ) :NO+2H₂O NO₃^-1 +4H⁺

Balance charge ( by adding electrons ) : NO+2H₂O NO₃^-1 +4H⁺ +3e-    ---(2)

The overall reactions is obtained by addind both half reactions so that electrons balance each other

(3xEqn(1)) + (2xEqn(2) ) gives

6 HOI +6H⁺ +6e^- +2NO+4H₂O 3 I₂    + 6H₂O + 2NO₃^-1 +8H⁺ +6e^-

6 HOI +2NO 3 I₂    + 2H₂O + 2NO₃^-1 +2H⁺  

This is the balanced equation.