Ion Electron Method for Alkaline Redox Reaction

Learn how to balance a redox reaction in alkaline medium using the ion electron method with chromium and iodate ions. This concept helps solve JEE...

❓ Question

Balance the ionic equation in alkaline medium:

C r (O H)_{3} + I O_{3}^{-} \to I^{-} + C r O_{4}^{2 -}

🖼 Question Image

This is a redox reaction in alkaline medium.

👉 Chromium gets oxidised
👉 Iodate gets reduced
👉 Use oxidation number method or half-reaction method carefully.

For chromium:

Cr(OH)_3

Cr = +3

In product:

CrO_4^{2-}

Cr = +6

So chromium undergoes oxidation.

For iodine:

IO_3^-

I = +5

In product:

I^-

I = -1

So iodine undergoes reduction.

Chromium:

Cr^{+3} \rightarrow Cr^{+6}

Loss of:

3e^-

Iodine:

I^{+5} \rightarrow I^{-1}

Gain of:

6e^-

LCM of electrons:

6

So multiply chromium reaction by 2.

2Cr(OH)_3 + IO_3^- \rightarrow I^- + 2CrO_4^{2-}

Balance oxygen and hydrogen using:

H_2O \text{ and } OH^-

Final balanced equation:

2Cr(OH)_3 + IO_3^- + 4OH^- \rightarrow I^- + 2CrO_4^{2-} + 5H_2O

❌ Acidic medium method use kar dena

❌ Electron balance bhool jaana

❌ $OH^-$ and $H_2O$ galat side add kar dena

Remember:

\text{Alkaline medium} \Rightarrow OH^- \text{ used for balancing}

\boxed{ 2Cr(OH)_3 + IO_3^- + 4OH^- \rightarrow I^- + 2CrO_4^{2-} + 5H_2O }