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Redox Reactions - Oxidation Number Method

Grade 11ICSEChemistry

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

🔑Concepts

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Oxidation is defined as an increase in the oxidation number of an element, while Reduction is defined as a decrease in the oxidation number.

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Oxidation Number (O.N.) is the formal charge an atom would bear if all bonds to it were considered 100%100\% ionic.

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An Oxidizing Agent (Oxidant) is a substance that undergoes a decrease in O.N., while a Reducing Agent (Reductant) is a substance that undergoes an increase in O.N.

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In a balanced redox reaction, the total increase in oxidation number must be equal to the total decrease in oxidation number.

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Rules for O.N.: Elements in their free state (e.g., H2,O2,NaH_2, O_2, Na) have an O.N. of 00. Fluorine is always −1-1. Oxygen is usually −2-2 (except in peroxides where it is −1-1, and in OF2OF_2 where it is +2+2). Hydrogen is usually +1+1 (except in metal hydrides where it is −1-1).

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In acidic medium, oxygen atoms are balanced by adding H2OH_2O to the side deficient in oxygen, and hydrogen atoms are balanced by adding H+H^+ ions.

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In basic medium, oxygen and hydrogen are balanced by adding H2OH_2O and OH−OH^- ions respectively, or by neutralizing H+H^+ with an equal number of OH−OH^- ions on both sides.

📐Formulae

Total Increase in O.N.=Total Decrease in O.N.\text{Total Increase in O.N.} = \text{Total Decrease in O.N.}

∑(Oxidation Number of all atoms)=Net Charge of species\sum (\text{Oxidation Number of all atoms}) = \text{Net Charge of species}

Change in O.N. per molecule=∣Change per atom∣×Number of atoms in formula\text{Change in O.N. per molecule} = |\text{Change per atom}| \times \text{Number of atoms in formula}

n1×ΔO.N.1=n2×ΔO.N.2n_1 \times \Delta O.N._1 = n_2 \times \Delta O.N._2

💡Examples

Problem 1:

Balance the following equation by the Oxidation Number Method (Acidic Medium): K2Cr2O7+FeSO4+H2SO4→Cr2(SO4)3+Fe2(SO4)3+K2SO4+H2OK_2Cr_2O_7 + FeSO_4 + H_2SO_4 \rightarrow Cr_2(SO_4)_3 + Fe_2(SO_4)_3 + K_2SO_4 + H_2O

Solution:

K2Cr2O7+6FeSO4+7H2SO4→Cr2(SO4)3+3Fe2(SO4)3+K2SO4+7H2OK_2Cr_2O_7 + 6FeSO_4 + 7H_2SO_4 \rightarrow Cr_2(SO_4)_3 + 3Fe_2(SO_4)_3 + K_2SO_4 + 7H_2O

Explanation:

  1. Assign O.N.: CrCr in K2Cr2O7K_2Cr_2O_7 is +6+6, FeFe in FeSO4FeSO_4 is +2+2. In products, CrCr is +3+3 and FeFe is +3+3. 2. Change in O.N.: CrCr decreases by 33 (per atom), total decrease for Cr2=3×2=6Cr_2 = 3 \times 2 = 6. FeFe increases by 11 (per atom). 3. Equate: Multiply FeSO4FeSO_4 by 66 to match the decrease of 66. 4. Balance other atoms: KK and SO42−SO_4^{2-} are balanced next. 5. Balance OO and HH: Add 7H2O7H_2O to the right to balance OO from K2Cr2O7K_2Cr_2O_7, then 7H2SO47H_2SO_4 provides the necessary HH.

Problem 2:

Balance the redox reaction in basic medium: MnO4−+I−→MnO2+I2MnO_4^- + I^- \rightarrow MnO_2 + I_2

Solution:

2MnO4−+6I−+4H2O→2MnO2+3I2+8OH−2MnO_4^- + 6I^- + 4H_2O \rightarrow 2MnO_2 + 3I_2 + 8OH^-

Explanation:

  1. Assign O.N.: MnMn changes from +7+7 to +4+4 (decrease of 33). II changes from −1-1 to 00 (increase of 11). 2. To equalize, multiply MnMn species by 22 (total decrease 66) and II species by 66 (total increase 66). 3. Skeleton: 2MnO4−+6I−→2MnO2+3I22MnO_4^- + 6I^- \rightarrow 2MnO_2 + 3I_2. 4. Balance charge: Left side is −8-8, right is 00. Add 8OH−8OH^- to the right. 5. Balance HH and OO by adding 4H2O4H_2O to the left.
Oxidation Number Method - Revision Notes & Key Formulas | ICSE Class 11 Chemistry