Balancing Redox Reactions (Oxidation Number Method) | Class 11 Chemistry

Balancing Redox Reactions

Oxidation Number Method | Class 11 Chemistry

1. The Oxidation Number Method

This method uses the change in the oxidation number of the oxidizing and reducing agents to balance the equation. It is based on the principle that the Total Increase in Oxidation Number must equal the Total Decrease in Oxidation Number.

2. Steps to Balance (General)

Write the skeletal equation for all reactants and products.
Assign Oxidation Numbers (ON) to all atoms and identify atoms undergoing change.
Calculate the change in ON per molecule and cross-multiply the formulae to equate the increase and decrease.
Balance all atoms other than Oxygen and Hydrogen.
Balance Oxygen by adding $H_2O$ molecules to the side deficient in oxygen.
Balance Hydrogen:
- Acidic Medium: Add $H^+$ ions to the side deficient in hydrogen.
- Basic Medium: Add $H_2O$ molecules to the side deficient in H, and add an equal number of $OH^-$ ions to the opposite side. (Or balance as acidic first, then add $OH^-$ to both sides to neutralize $H^+$).

3. Worked Example (Acidic Medium)

Reaction: Permanganate ion oxidizes Ferrous ion to Ferric ion in acidic medium.

Skeletal: $MnO_4^- + Fe^{2+} \rightarrow Mn^{2+} + Fe^{3+}$

Step 1: Identify Oxidation Numbers

Mn in $MnO_4^-$: $x + 4(-2) = -1 \Rightarrow x = +7$.
Mn in $Mn^{2+}$: $+2$.
Fe in $Fe^{2+}$: $+2$.
Fe in $Fe^{3+}$: $+3$.

Step 2: Calculate Change

Mn: $+7 \rightarrow +2$ (Decrease by 5).
Fe: $+2 \rightarrow +3$ (Increase by 1).

Step 3: Equalize Change (Cross Multiply)

Multiply Fe species by 5 and Mn species by 1.

$$ 1 MnO_4^- + 5 Fe^{2+} \rightarrow 1 Mn^{2+} + 5 Fe^{3+} $$

Step 4: Balance Oxygen

LHS has 4 Oxygen atoms. Add $4H_2O$ to RHS.

$$ MnO_4^- + 5 Fe^{2+} \rightarrow Mn^{2+} + 5 Fe^{3+} + 4H_2O $$

Step 5: Balance Hydrogen (Acidic)

RHS has 8 Hydrogen atoms. Add $8H^+$ to LHS.

                    Final Equation:

                    $$ MnO_4^- + 5 Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5 Fe^{3+} + 4H_2O $$

Check Charge Balance:

LHS: $-1 + 5(+2) + 8(+1) = -1 + 10 + 8 = +17$
RHS: $+2 + 5(+3) + 0 = +2 + 15 = +17$ (Balanced!)

4. Shortcut for Basic Medium

If the reaction is in a basic medium:

Balance the equation as if it were in Acidic Medium first (using $H^+$).
Add $OH^-$ ions to both sides equal to the number of $H^+$ ions.
Combine $H^+$ and $OH^-$ on the same side to form $H_2O$.
Cancel out excess $H_2O$ molecules from both sides.

Practice Quiz

Test your balancing skills.

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