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Redox Reactions - Concept of Oxidation and Reduction

Grade 11ICSEChemistry

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

🔑Concepts

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Classical Concept: Oxidation is the addition of oxygen or an electronegative element, or the removal of hydrogen or an electropositive element. Reduction is the addition of hydrogen or an electropositive element, or the removal of oxygen or an electronegative element.

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Electronic Concept: Oxidation is defined as the loss of one or more electrons by an atom or ion (de-electronation). Reduction is the gain of one or more electrons by an atom or ion (electronation).

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Oxidizing Agent (Oxidant): A substance that supplies oxygen, removes hydrogen, or accepts electrons. In the process, the oxidant itself gets reduced.

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Reducing Agent (Reductant): A substance that supplies hydrogen, removes oxygen, or donates electrons. In the process, the reductant itself gets oxidized.

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Oxidation Number (O.N.): It represents the charge which an atom appears to have when all other atoms are removed from it as ions. Oxidation involves an increase in O.N.O.N., while reduction involves a decrease in O.NO.N.

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Disproportionation Reaction: A specific type of redox reaction where the same element in a species is simultaneously oxidized and reduced, such as the decomposition of H2O2H_2O_2: 2H2O2ightarrow2H2O+O22H_2O_2 ightarrow 2H_2O + O_2.

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Stock Notation: Representing the oxidation state of a metal in a compound using Roman numerals in parentheses, e.g., FeCl2FeCl_2 as Iron(II) chloride and FeCl3FeCl_3 as Iron(III) chloride.

📐Formulae

Sum of Oxidation Numbers (Neutral Molecule)=0\text{Sum of Oxidation Numbers (Neutral Molecule)} = 0

Sum of Oxidation Numbers (Polyatomic Ion)=Charge of Ion\text{Sum of Oxidation Numbers (Polyatomic Ion)} = \text{Charge of Ion}

Oxidation: M→Mn++ne−\text{Oxidation: } M \rightarrow M^{n+} + ne^-

Reduction: X+ne−→Xn−\text{Reduction: } X + ne^- \rightarrow X^{n-}

Oxidation Number of Oxygen: −2 (general), −1 (peroxides), −0.5 (superoxides), +2 (in OF2)\text{Oxidation Number of Oxygen: } -2 \text{ (general), } -1 \text{ (peroxides), } -0.5 \text{ (superoxides), } +2 \text{ (in } OF_2)

💡Examples

Problem 1:

Determine the oxidation number of CrCr in the dichromate ion Cr2O72−Cr_2O_7^{2-}.

Solution:

Let the oxidation number of CrCr be xx. Since the sum of oxidation numbers must equal the charge of the ion (−2-2): 2(x)+7(−2)=−22(x) + 7(-2) = -2 2x−14=−22x - 14 = -2 2x=122x = 12 x=+6x = +6

Explanation:

The oxidation number of oxygen is taken as −2-2. Setting up the algebraic sum equal to the net charge of the ion allows us to solve for the unknown oxidation state of Chromium.

Problem 2:

Identify the oxidant and reductant in the following reaction: CuO+H2→Cu+H2OCuO + H_2 \rightarrow Cu + H_2O

Solution:

CuOCuO is the oxidant (oxidizing agent) and H2H_2 is the reductant (reducing agent).

Explanation:

In this reaction, CuCu changes from +2+2 in CuOCuO to 00 in metallic CuCu (decrease in O.N. = Reduction). HH changes from 00 in H2H_2 to +1+1 in H2OH_2O (increase in O.N. = Oxidation). Since CuOCuO causes the oxidation of H2H_2, it is the oxidant; since H2H_2 causes the reduction of CuOCuO, it is the reductant.

Problem 3:

Calculate the oxidation number of SS in Na2S4O6Na_2S_4O_6 (Sodium tetrathionate).

Solution:

Let the oxidation number of SS be xx. 2(+1)+4(x)+6(−2)=02(+1) + 4(x) + 6(-2) = 0 2+4x−12=02 + 4x - 12 = 0 4x=104x = 10 x=+2.5x = +2.5

Explanation:

In Na2S4O6Na_2S_4O_6, the average oxidation number of Sulfur is +2.5+2.5. However, structurally, two Sulfur atoms have an O.N. of 00 and two have an O.N. of +5+5.

Concept of Oxidation and Reduction Revision - Class 11 Chemistry ICSE