Question

Permanganate (VII) ion, in basic solution oxidizes iodide ion 1- to produce molecular iodine (I₂) and manganese (IV) oxide (MnO₂). Write a balanced ionic equation to represent this redox reaction.

Answer 1

Step-1: Write pie skeletal equation for the given reaction.

MnO₄^-(aq) + I^-(aq) → MnO₂(s) + I₂(s)  ...(i)

Step-2: Write the O.N. of all the elements above their respective symbols.

(ON of Mn from +7t +4 i.e., 3 unit charge)   ...(ii)

I^- → I₂⁰  ....(iii)

(ON of Iodine from -1 to 0 i.e., Laui's charge)

Oxidation half equation: 

I^-(aq) → I₂(s)  ...(iv)

Reduction half equation: 

MnO₄^-(aq) → MnO₂(s)   ...(v)

Step-4: To balance oxidation half Eq. … (ii)

(a) Balance all atoms 2I^–(aq) → I₂(s) … (iv)

(b) Balance O.N. by adding electrons.

2I^–(aq) → I₂(s) + 2e^– …(v)

Charge on either side of Eq(v) is balanced. Thus, Eq. (v) represents the balanced oxidation half equation.

Step-5: Balance the reduction half equation (iii) Balance O.N. by adding electrons.

Step-6: To balance the electrons lost in Eq. (v) and gained in Eq. (viii) Equation (v) x 3 + Equation (viii) x 2 we have,

This represents are final balanced redox equation.