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Redox Reactions - Balancing Redox Reactions

Grade 11ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Oxidation state (or oxidation number) is the formal charge an atom would carry if all bonds to it were completely ionic. In a redox reaction, balancing ensures that the total increase in oxidation number equals the total decrease.

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The Oxidation Number Method involves assigning oxidation numbers to all atoms, identifying which atoms change their oxidation state, and then multiplying the species by suitable coefficients to equalize the total increase and decrease in oxidation numbers.

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The Ion-Electron Method (Half-Reaction Method) involves splitting the overall reaction into two half-reactions: the oxidation half and the reduction half. These are balanced separately for atoms and charge, then recombined.

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In Acidic Medium balancing: Oxygen atoms are balanced by adding H2OH_2O molecules, and Hydrogen atoms are balanced by adding H+H^+ ions.

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In Basic Medium balancing: Balance the reaction as if in an acidic medium using H+H^+ ions, then add an equal number of OH−OH^- ions to both sides of the equation to neutralize H+H^+ into H2OH_2O.

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Conservation of Charge: The net charge on the reactant side must equal the net charge on the product side in the final balanced equation.

📐Formulae

∑(Oxidation Numbers)=Net Charge of the species\sum (\text{Oxidation Numbers}) = \text{Net Charge of the species}

Total electrons lost (Oxidation)=Total electrons gained (Reduction)\text{Total electrons lost (Oxidation)} = \text{Total electrons gained (Reduction)}

In Acidic Medium: O→H2O,H→H+\text{In Acidic Medium: } O \rightarrow H_2O, \quad H \rightarrow H^+

In Basic Medium: H++OH−→H2O\text{In Basic Medium: } H^+ + OH^- \rightarrow H_2O

💡Examples

Problem 1:

Balance the following redox reaction in acidic medium using the Ion-Electron method: MnO4−+Fe2+→Mn2++Fe3+MnO_4^- + Fe^{2+} \rightarrow Mn^{2+} + Fe^{3+}

Solution:

MnO4−+5Fe2++8H+→Mn2++5Fe3++4H2OMnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O

Explanation:

  1. Split into half-reactions: Oxidation: Fe2+→Fe3++e−Fe^{2+} \rightarrow Fe^{3+} + e^-; Reduction: MnO4−+5e−→Mn2+MnO_4^- + 5e^- \rightarrow Mn^{2+}. 2. Balance Oxygen in the reduction half by adding 4H2O4H_2O to the products. 3. Balance Hydrogen by adding 8H+8H^+ to the reactants. 4. Multiply the oxidation half by 5 to equalize electrons: 5Fe2+→5Fe3++5e−5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^-. 5. Add the half-reactions and cancel the electrons.

Problem 2:

Balance the following reaction in basic medium: P4→PH3+HPO2−P_4 \rightarrow PH_3 + HPO_2^-

Solution:

P4+2H2O+4OH−→2PH3+2HPO2−P_4 + 2H_2O + 4OH^- \rightarrow 2PH_3 + 2HPO_2^-

Explanation:

This is a disproportionation reaction where Phosphorus (P4P_4) is both oxidized and reduced. Oxidation half: P4→4HPO2−P_4 \rightarrow 4HPO_2^-. Reduction half: P4→4PH3P_4 \rightarrow 4PH_3. After balancing atoms and charges using H2OH_2O, H+H^+, and e−e^-, convert to basic medium by adding OH−OH^- to neutralize H+H^+, then simplify.

Balancing Redox Reactions - Revision Notes & Key Formulas | ICSE Class 11 Chemistry