Balancing Redox Reactions: Oxidation Number Method
Welcome to Lecture 10 of the CHEMCA Bridge Course! Redox balancing goes far beyond classical inspections because both atomic count and electronic charges must be balanced. In this session, Abhishek Sengar Sir teaches a highly structured Oxidation Number Method to balance reactions in both Acidic ($H^+$) and Basic ($OH^-$) media.
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In-Depth Lecture Notes & Summary
Why is Redox Balancing Unique?
In simple chemical reactions, we balance elements by checking their atom counts on both sides. In redox reactions, this is insufficient because:
1. Mass must be conserved: Number of atoms of each element on the left side must equal those on the right side.
2. Charge must be conserved: The net sum of charges on the reactant side must equal the net sum of charges on the product side.
3. Water ($H_2O$), $H^+$, or $OH^-$ addition: Based on whether the reaction happens in an acidic or basic medium, you may need to add hydrogen and oxygen-carrying ions that aren't originally written in the skeletal equation.
The 5 Steps of the Oxidation Number Method
Abhishek Sengar Sir outlines 5 standard, algorithmic steps to balance any redox system:
Find Oxidation Numbers: Write down the individual oxidation numbers of all atoms in the reaction to identify elements undergoing changes.
Balance Active Atoms: Identify oxidation/reduction, then balance the atom count of ONLY those elements whose oxidation numbers are changing.
Equate Oxidation Number Changes: Calculate the total increase/decrease in oxidation number. Cross-multiply with suitable integers to equalize the gain and loss of electrons.
Balance Net Charges: Add $H^+$ ions in an acidic medium or $OH^-$ ions in a basic medium to balance net ionic charge on both sides.
Balance Oxygen & Hydrogen: Add $H_2O$ molecules to the appropriate deficient side to balance oxygen atoms (hydrogen will balance automatically).
Solved Example: Acidic Medium
Let's review the balancing process for: $$\text{MnO}_4^- + \text{Fe}^{2+} \to \text{Mn}^{2+} + \text{Fe}^{3+} \quad (\text{Acidic Medium})$$
• Oxidation States: $Mn$ changes $+7 \to +2$ (decrease $= 5$ per atom). $Fe$ changes $+2 \to +3$ (increase $= 1$ per atom).
• Equating Changes: Multiply $Fe$ species by $5$ to balance the electronic transfer: $$\text{MnO}_4^- + 5\text{Fe}^{2+} \to \text{Mn}^{2+} + 5\text{Fe}^{3+}$$
• Charge Balance (Acidic $\implies$ add $H^+$):
Left side charge: $-1 + 5(+2) = +9$
Right side charge: $+2 + 5(+3) = +17$
Add $8\text{H}^+$ to the left side to balance charges at $+17$:
$$\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \to \text{Mn}^{2+} + 5\text{Fe}^{3+}$$
• Oxygen Balance (add $H_2O$): Left side has 4 oxygen atoms; right side has none. Add $4\text{H}_2\text{O}$ to the right: $$\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \to \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$$
Solved Example: Basic Medium
Let's review the balancing process for basic conditions: $$\text{MnO}_4^- + \text{I}^- \to \text{MnO}_2 + \text{IO}_3^- \quad (\text{Basic Medium})$$
• Oxidation States: $Mn$ changes $+7 \to +4$ (decrease $= 3$). $I$ changes $-1 \to +5$ (increase $= 6$).
• Equating Changes: Multiply $Mn$ species by $2$ to balance the electron change: $$2\text{MnO}_4^- + \text{I}^- \to 2\text{MnO}_2 + \text{IO}_3^-$$
• Charge Balance (Basic $\implies$ add $OH^-$):
Left side charge: $2(-1) + (-1) = -3$
Right side charge: $-1$
Add $2\text{OH}^-$ to the right side to balance charges at $-3$:
$$2\text{MnO}_4^- + \text{I}^- \to 2\text{MnO}_2 + \text{IO}_3^- + 2\text{OH}^-$$
• Oxygen Balance (add $H_2O$):
Left side oxygens: $8$. Right side oxygens: $4 + 3 + 2 = 9$.
Add $1\text{H}_2\text{O}$ to the left side:
$$2\text{MnO}_4^- + \text{I}^- + \text{H}_2\text{O} \to 2\text{MnO}_2 + \text{IO}_3^- + 2\text{OH}^-$$
Redox Stepper Solver
Choose any of the 5 redox reactions discussed in class. Click through the 5 steps to see how each coefficient and atomic balance is constructed.
Lecture 10 Concept Test
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Redox Balancing Block?
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