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Redox Reactions - Oxidation Number

Grade 11CBSEChemistry

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

🔑Concepts

Oxidation Number (O.N.) is defined as the formal charge that an atom would carry in a molecule or ion if the shared electrons in a covalent bond were assigned completely to the more electronegative element.

Rules for assigning O.N.: (1) In free or elementary state (e.g., H2H_2, O2O_2, P4P_4, S8S_8, NaNa), the O.N. of each atom is 00. (2) For monatomic ions, the O.N. is equal to the charge of the ion (e.g., Na+Na^+ is +1+1, Mg2+Mg^{2+} is +2+2, ClCl^- is 1-1).

Oxygen Rule: In most compounds, the O.N. of oxygen is 2-2. Exceptions include peroxides (e.g., H2O2H_2O_2, Na2O2Na_2O_2) where it is 1-1, superoxides (e.g., KO2KO_2) where it is 12-\frac{1}{2}, and in OF2OF_2 where it is +2+2.

Hydrogen Rule: The O.N. of hydrogen is +1+1 when bonded to non-metals. However, in metal hydrides (e.g., LiHLiH, CaH2CaH_2), it is 1-1.

Halogen Rule: Fluorine, being the most electronegative element, always has an O.N. of 1-1. Other halogens (ClCl, BrBr, II) are typically 1-1 unless combined with oxygen or fluorine.

In a neutral molecule, the algebraic sum of the oxidation numbers of all constituent atoms must be 00. In a polyatomic ion, the sum must equal the net charge of the ion.

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

Stock Notation: The oxidation state of a metal in a compound is represented by a Roman numeral in parentheses after the symbol of the metal (e.g., FeOFeO is Iron(II) oxide, Fe2O3Fe_2O_3 is Iron(III) oxide).

📐Formulae

(ni×O.N.i)=0 (For neutral compounds)\sum (n_i \times O.N._i) = 0 \text{ (For neutral compounds)}

(ni×O.N.i)=Charge on the ion (For polyatomic ions)\sum (n_i \times O.N._i) = \text{Charge on the ion} \text{ (For polyatomic ions)}

Oxidation: ΔO.N.>0\text{Oxidation: } \Delta O.N. > 0

Reduction: ΔO.N.<0\text{Reduction: } \Delta O.N. < 0

💡Examples

Problem 1:

Calculate the oxidation number of SS in H2SO4H_2SO_4.

Solution:

2(+1)+x+4(2)=0    2+x8=0    x=+62(+1) + x + 4(-2) = 0 \implies 2 + x - 8 = 0 \implies x = +6

Explanation:

The O.N. of HH is +1+1 and OO is 2-2. Let the O.N. of SS be xx. Since the molecule is neutral, the sum of oxidation numbers is zero.

Problem 2:

Calculate the oxidation number of CrCr in Cr2O72Cr_2O_7^{2-}.

Solution:

2(x)+7(2)=2    2x14=2    2x=12    x=+62(x) + 7(-2) = -2 \implies 2x - 14 = -2 \implies 2x = 12 \implies x = +6

Explanation:

Let the O.N. of CrCr be xx. The sum of oxidation numbers must equal the charge of the ion, which is 2-2.

Problem 3:

Identify the oxidant and reductant in the reaction: Zn(s)+2H+(aq)Zn2+(aq)+H2(g)Zn(s) + 2H^+(aq) \rightarrow Zn^{2+}(aq) + H_2(g).

Solution:

Reductant: ZnZn, Oxidant: H+H^+

Explanation:

The O.N. of ZnZn increases from 00 to +2+2 (Oxidation), so ZnZn is the reducing agent (reductant). The O.N. of HH decreases from +1+1 to 00 (Reduction), so H+H^+ is the oxidizing agent (oxidant).

Problem 4:

Calculate the oxidation number of MnMn in KMnO4KMnO_4.

Solution:

1(+1)+x+4(2)=0    1+x8=0    x=+71(+1) + x + 4(-2) = 0 \implies 1 + x - 8 = 0 \implies x = +7

Explanation:

Potassium (KK) is an alkali metal with O.N. +1+1, and Oxygen (OO) is 2-2. The sum of the oxidation numbers in the neutral compound KMnO4KMnO_4 is 00.

Oxidation Number - Revision Notes & Key Formulas | CBSE Class 11 Chemistry