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Redox Reactions - Concept of Oxidation and Reduction

Grade 11CBSEChemistry

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

🔑Concepts

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Redox reactions involve the simultaneous occurrence of oxidation and reduction processes.

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Classical Concept: Oxidation is the addition of oxygen/electronegative elements or the removal of hydrogen/electropositive elements. Reduction is the addition of hydrogen/electropositive elements or the removal of oxygen/electronegative elements.

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Electronic Concept: Oxidation is defined as the loss of electrons (e−e^-) by an atom, ion, or molecule. Reduction is defined as the gain of electrons (e−e^-) by an atom, ion, or molecule (OIL RIG: Oxidation Is Loss, Reduction Is Gain).

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Oxidation Number (O.N.): It is the formal charge that an atom appears to have in a substance when electrons are counted according to specific rules. Oxidation involves an increase in the oxidation number, while reduction involves a decrease in the oxidation number.

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Oxidizing Agent (Oxidant): A substance that supplies oxygen, removes hydrogen, or accepts electrons. It undergoes reduction itself.

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Reducing Agent (Reductant): A substance that supplies hydrogen, removes oxygen, or loses electrons. It undergoes oxidation itself.

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Fractional Oxidation State: In some compounds like Fe3O4Fe_3O_4 or Mn3O4Mn_3O_4, the oxidation state is calculated as an average, though the individual atoms may have different integral oxidation states.

📐Formulae

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

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

∑(Oxidation numbers of all atoms in a neutral molecule)=0\sum (\text{Oxidation numbers of all atoms in a neutral molecule}) = 0

∑(Oxidation numbers of all atoms in a polyatomic ion)=Charge of the ion\sum (\text{Oxidation numbers of all atoms in a polyatomic ion}) = \text{Charge of the ion}

Oxidation Number of alkali metals (Group 1)=+1\text{Oxidation Number of alkali metals (Group 1)} = +1

Oxidation Number of alkaline earth metals (Group 2)=+2\text{Oxidation Number of alkaline earth metals (Group 2)} = +2

Oxidation Number of Fluorine=−1 (always)\text{Oxidation Number of Fluorine} = -1 \text{ (always)}

💡Examples

Problem 1:

Calculate the oxidation number of CrCr in K2Cr2O7K_2Cr_2O_7.

Solution:

Let the oxidation state of CrCr be xx. The oxidation state of KK is +1+1 and OO is −2-2. 2(+1)+2(x)+7(−2)=02(+1) + 2(x) + 7(-2) = 0 2+2x−14=02 + 2x - 14 = 0 2x=122x = 12 x=+6x = +6

Explanation:

In a neutral compound, the sum of the oxidation states of all atoms must be zero. Solving the algebraic equation gives the oxidation state of Chromium as +6+6.

Problem 2:

Identify the oxidizing and reducing agents in the reaction: H2S(g)+Cl2(g)→2HCl(g)+S(s)H_2S(g) + Cl_2(g) \rightarrow 2HCl(g) + S(s)

Solution:

  1. Oxidation state of SS changes from −2-2 in H2SH_2S to 00 in SS (Increase, so H2SH_2S is oxidized).
  2. Oxidation state of ClCl changes from 00 in Cl2Cl_2 to −1-1 in HClHCl (Decrease, so Cl2Cl_2 is reduced).

Explanation:

H2SH_2S is the reducing agent because it loses electrons (increases oxidation state of SS). Cl2Cl_2 is the oxidizing agent because it gains electrons (decreases oxidation state of ClCl).

Problem 3:

Calculate the oxidation number of MnMn in the permanganate ion MnO4−MnO_4^-.

Solution:

Let the oxidation state of MnMn be xx. The oxidation state of OO is −2-2. x+4(−2)=−1x + 4(-2) = -1 x−8=−1x - 8 = -1 x=+7x = +7

Explanation:

For polyatomic ions, the sum of the oxidation states must equal the net charge of the ion, which is −1-1 in this case.

Concept of Oxidation and Reduction Revision - Class 11 Chemistry CBSE