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Unit 6: Oxidation and Reduction

Chemistry - Class 11

This chapter delves into the fundamental concepts of oxidation and reduction, covering their definitions from traditional and electronic perspectives. It details the assignment of oxidation numbers, methods for balancing complex redox reactions, and the principles of electrolysis, including Faraday's quantitative laws.

No MCQ questions available for this chapter.

Unit 6: Oxidation and Reduction

1. General Concept of Oxidation and Reduction

The terms oxidation and reduction were initially used to describe reactions involving oxygen and hydrogen. Over time, their definitions have evolved to encompass a broader range of chemical processes, particularly those involving electron transfer and changes in oxidation states.

Oxidation

Oxidation can be defined in several ways:

  • Addition of Oxygen: A substance gains oxygen atoms.
    Example: 2Mg(s) + O2(g) → 2MgO(s) (Magnesium is oxidized)
  • Removal of Hydrogen: A substance loses hydrogen atoms.
    Example: H2S(g) + Cl2(g) → 2HCl(g) + S(s) (Hydrogen sulfide is oxidized)
  • Loss of Electrons: A species loses one or more electrons. This is the most fundamental definition in modern chemistry.
    Example: Na → Na+ + e- (Sodium is oxidized)
  • Increase in Oxidation Number: The oxidation number of an atom in a compound increases.
    Example: In Fe2+ → Fe3+ + e-, the oxidation number of iron increases from +2 to +3.

Reduction

Reduction is the opposite of oxidation and can be defined as:

  • Removal of Oxygen: A substance loses oxygen atoms.
    Example: CuO(s) + H2(g) → Cu(s) + H2O(l) (Copper oxide is reduced)
  • Addition of Hydrogen: A substance gains hydrogen atoms.
    Example: C2H4(g) + H2(g) → C2H6(g) (Ethene is reduced)
  • Gain of Electrons: A species gains one or more electrons.
    Example: Cl + e- → Cl- (Chlorine is reduced)
  • Decrease in Oxidation Number: The oxidation number of an atom in a compound decreases.
    Example: In Cl2 + 2e- → 2Cl-, the oxidation number of chlorine decreases from 0 to -1.

Redox Reactions

Redox reactions (short for Reduction-Oxidation reactions) are chemical reactions where both oxidation and reduction occur simultaneously. One substance loses electrons (gets oxidized) while another substance gains those electrons (gets reduced). It's impossible for one to happen without the other.

Example: When zinc metal reacts with copper(II) sulfate solution:

Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)

Here, Zinc (Zn) loses electrons and is oxidized (Zn → Zn2+ + 2e-), while Copper(II) ions (Cu2+) gain electrons and are reduced (Cu2+ + 2e- → Cu).

2. Electronic Concept of Oxidation and Reduction

The electronic concept provides a more fundamental understanding of redox reactions, focusing on the transfer of electrons.

Oxidation: Loss of Electrons

According to the electronic concept, oxidation is the process involving the loss of one or more electrons by an atom, ion, or molecule. The species that undergoes oxidation becomes more positive or less negative in charge.

Example:

  • Sodium atom losing an electron: Na → Na+ + e-
  • Iron(II) ion losing an electron: Fe2+ → Fe3+ + e-

Reduction: Gain of Electrons

Conversely, reduction is the process involving the gain of one or more electrons by an atom, ion, or molecule. The species that undergoes reduction becomes more negative or less positive in charge.

Example:

  • Chlorine atom gaining an electron: Cl + e- → Cl-
  • Copper(II) ion gaining electrons: Cu2+ + 2e- → Cu

Oxidizing Agent (Oxidant)

An oxidizing agent (or oxidant) is a substance that causes another substance to be oxidized. In doing so, the oxidizing agent itself gains electrons and gets reduced. It accepts electrons from the reducing agent.

Characteristics:

  • Gains electrons
  • Gets reduced
  • Causes oxidation
Examples: O2, Cl2, KMnO4, K2Cr2O7

Reducing Agent (Reductant)

A reducing agent (or reductant) is a substance that causes another substance to be reduced. In doing so, the reducing agent itself loses electrons and gets oxidized. It donates electrons to the oxidizing agent.

Characteristics:

  • Loses electrons
  • Gets oxidized
  • Causes reduction
Examples: H2, C, Na, Zn, Fe2+

Consider the reaction: Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)

  • Zn loses electrons (Zn → Zn2+ + 2e-), so Zn is oxidized and acts as the reducing agent.
  • Cu2+ gains electrons (Cu2+ + 2e- → Cu), so Cu2+ is reduced and acts as the oxidizing agent.

3. Oxidation Number and Rules for Assigning

The oxidation number (or oxidation state) is a hypothetical charge an atom would have if all its bonds were 100% ionic, meaning all electrons in a bond are assigned to the more electronegative atom. It is a useful tool for tracking electron transfer in redox reactions, even for covalent compounds.

Rules for Assigning Oxidation Numbers

  1. The oxidation number of an element in its free (uncombined) state is zero.
    Examples: Na (in Na metal), O2, Cl2, P4, S8 all have an oxidation number of 0.
  2. For a monoatomic ion, the oxidation number is equal to the charge on the ion.
    Examples: Na+ has +1, Cl- has -1, Mg2+ has +2, Al3+ has +3.
  3. Hydrogen (H) usually has an oxidation number of +1.
    Exception: In metal hydrides (e.g., NaH, CaH2), hydrogen is bonded to a less electronegative metal and has an oxidation number of -1.
  4. Oxygen (O) usually has an oxidation number of -2.
    Exceptions:
    • In peroxides (e.g., H2O2, Na2O2), oxygen has an oxidation number of -1.
    • In superoxides (e.g., KO2), oxygen has an oxidation number of -1/2.
    • When bonded to fluorine (e.g., OF2), oxygen has an oxidation number of +2 (since fluorine is more electronegative).
  5. Fluorine (F) always has an oxidation number of -1 in all its compounds because it is the most electronegative element.
  6. The sum of the oxidation numbers of all atoms in a neutral compound is zero.
    Example: In H2O, (2 * +1) + (-2) = 0.
  7. The sum of the oxidation numbers of all atoms in a polyatomic ion is equal to the charge on the ion.
    Example: In SO4^2-, S + (4 * -2) = -2. So, S = +6.
  8. Group 1 metals (Li, Na, K, etc.) always have an oxidation number of +1 in compounds.
    Group 2 metals (Be, Mg, Ca, etc.) always have an oxidation number of +2 in compounds.
  9. Other common elements with fixed oxidation numbers in most compounds:
    • Aluminium (Al) = +3
    • Silver (Ag) = +1
    • Zinc (Zn) = +2
    • Copper (Cu) = +2 (though Cu can also be +1)

Example Calculation: Find the oxidation number of Manganese (Mn) in KMnO4.

Let the oxidation number of Mn be x.

Using rules 2, 4, and 6:

  • K is a Group 1 metal, so its ON is +1.
  • O usually has an ON of -2.
  • The sum of ONs in a neutral compound is 0.

(+1) + (x) + (4 * -2) = 0

1 + x - 8 = 0

x - 7 = 0

x = +7

So, the oxidation number of Mn in KMnO4 is +7.

4. Balancing Redox Reactions

Balancing redox reactions is crucial for stoichiometric calculations. There are two primary methods:

Oxidation Number Method

This method focuses on balancing the total increase in oxidation number with the total decrease in oxidation number.

  1. Write the unbalanced skeletal equation.
  2. Assign oxidation numbers to all atoms and identify the atoms whose oxidation numbers change.
  3. Identify the species oxidized and reduced. Calculate the change in oxidation number per atom for each.
  4. Calculate the total change in oxidation number for the oxidized and reduced species by multiplying the change per atom by the number of atoms undergoing the change.
  5. Cross multiply to balance electron transfer. Make the total increase in oxidation number equal to the total decrease in oxidation number by multiplying the appropriate species by suitable coefficients.
  6. Balance remaining atoms by inspection.
    • Balance all atoms other than oxygen and hydrogen first.
    • Balance oxygen atoms by adding H2O molecules to the side deficient in oxygen.
    • Balance hydrogen atoms by adding H+ ions (for acidic medium) or H2O and OH- ions (for basic medium).

Example (Acidic Medium): Balance the reaction: K2Cr2O7 + FeSO4 + H2SO4 → K2SO4 + Cr2(SO4)3 + Fe2(SO4)3 + H2O

  1. Unbalanced equation: K2Cr2O7 + FeSO4 + H2SO4 → K2SO4 + Cr2(SO4)3 + Fe2(SO4)3 + H2O
  2. Assign ONs:
    • In Cr2O7^2-, Cr = +6
    • In FeSO4, Fe = +2
    • In Cr2(SO4)3, Cr = +3
    • In Fe2(SO4)3, Fe = +3
  3. Identify changes:
    • Cr: +6 → +3 (decrease of 3 per Cr atom)
    • Fe: +2 → +3 (increase of 1 per Fe atom)
  4. Total change:
    • For Cr2O7^2- (2 Cr atoms): 2 * (-3) = -6 (total decrease)
    • For Fe2+ (1 Fe atom): 1 * (+1) = +1 (total increase)
  5. Balance electron transfer: To balance the total change of -6 with +1, we need 6 Fe atoms.
    Multiply FeSO4 by 6 on the reactant side, and since Fe is in Fe2(SO4)3 on the product side, we need 3 Fe2(SO4)3 (3 * 2 = 6 Fe atoms).
    K2Cr2O7 + 6FeSO4 + H2SO4 → K2SO4 + Cr2(SO4)3 + 3Fe2(SO4)3 + H2O
  6. Balance remaining atoms:
    • K: 2 on both sides (balanced)
    • Cr: 2 on both sides (balanced)
    • Fe: 6 on both sides (balanced)
    • Balance SO4^2- ions (or S atoms) and then H and O.
      Product side SO4: 1 (from K2SO4) + 3 (from Cr2(SO4)3) + 3*3 (from 3Fe2(SO4)3) = 1 + 3 + 9 = 13 SO4.
      Reactant side SO4: 6 (from 6FeSO4). We need 13 - 6 = 7 more SO4. Add 7 H2SO4.
      K2Cr2O7 + 6FeSO4 + 7H2SO4 → K2SO4 + Cr2(SO4)3 + 3Fe2(SO4)3 + H2O
    • Balance H: Reactant side has 7 * 2 = 14 H atoms (from 7H2SO4). Add 7 H2O to the product side.
      K2Cr2O7 + 6FeSO4 + 7H2SO4 → K2SO4 + Cr2(SO4)3 + 3Fe2(SO4)3 + 7H2O
    • Check O atoms:
      Reactant O: 7 (from K2Cr2O7) + 6*4 (from FeSO4) + 7*4 (from H2SO4) = 7 + 24 + 28 = 59 O.
      Product O: 4 (from K2SO4) + 3*4 (from Cr2(SO4)3) + 3*3*4 (from 3Fe2(SO4)3) + 7 (from 7H2O) = 4 + 12 + 36 + 7 = 59 O.
      The equation is balanced.

Ion-Electron (Half Reaction) Method

This method separates the overall redox reaction into two half-reactions (oxidation and reduction), balances each independently, and then combines them.

  1. Write the ionic equation (if not already given) and identify the species undergoing oxidation and reduction.
  2. Split the reaction into two half-reactions: one for oxidation and one for reduction.
  3. Balance atoms other than O and H in each half-reaction.
  4. Balance oxygen atoms (O) by adding H2O molecules to the side deficient in oxygen.
  5. Balance hydrogen atoms (H):
    • In acidic medium: Add H+ ions to the side deficient in hydrogen.
    • In basic medium: Add H2O to the side deficient in hydrogen, and an equal number of OH- ions to the opposite side. (Alternatively, balance with H+ first, then add OH- to both sides to neutralize H+ to H2O).
  6. Balance the charge in each half-reaction by adding electrons (e-) to the more positive side (or less negative side).
  7. Equalize the number of electrons in both half-reactions by multiplying one or both half-reactions by appropriate integers.
  8. Add the two balanced half-reactions and cancel any common terms (electrons, H+, OH-, H2O) present on both sides.

Example (Acidic Medium): Balance MnO4- + Fe2+ → Mn2+ + Fe3+

  1. Ionic equation: MnO4- + Fe2+ → Mn2+ + Fe3+
  2. Half-reactions:
    • Oxidation: Fe2+ → Fe3+
    • Reduction: MnO4- → Mn2+
  3. Balance atoms (other than O, H): Fe and Mn are balanced.
  4. Balance O (add H2O):
    • Oxidation: Fe2+ → Fe3+ (no O)
    • Reduction: MnO4- → Mn2+ + 4H2O (4 O on left, so 4 H2O on right)
  5. Balance H (add H+ for acidic medium):
    • Oxidation: Fe2+ → Fe3+ (no H)
    • Reduction: 8H+ + MnO4- → Mn2+ + 4H2O (8 H on right, so 8 H+ on left)
  6. Balance charge (add e-):
    • Oxidation: Fe2+ → Fe3+ + e- (charge +2 on left, +3 on right; add 1e- to right)
    • Reduction: 8H+ + MnO4- + 5e- → Mn2+ + 4H2O (charge (+8 - 1) = +7 on left, +2 on right; add 5e- to left)
  7. Equalize electrons: Multiply oxidation half-reaction by 5.
    • Oxidation: 5Fe2+ → 5Fe3+ + 5e-
    • Reduction: 8H+ + MnO4- + 5e- → Mn2+ + 4H2O
  8. Add half-reactions and cancel electrons:
    5Fe2+ + 8H+ + MnO4- + 5e- → 5Fe3+ + 5e- + Mn2+ + 4H2O
    Final balanced equation: 5Fe2+ + 8H+ + MnO4- → 5Fe3+ + Mn2+ + 4H2O

Example (Basic Medium): Balance Cr(OH)3 + ClO- → CrO4^2- + Cl-

  1. Ionic equation: Cr(OH)3 + ClO- → CrO4^2- + Cl-
  2. Half-reactions:
    • Oxidation: Cr(OH)3 → CrO4^2-
    • Reduction: ClO- → Cl-
  3. Balance atoms (other than O, H): Cr and Cl are balanced.
  4. Balance O (add H2O):
    • Oxidation: Cr(OH)3 + H2O → CrO4^2- (3 O from OH + 1 O from H2O = 4 O on left; 4 O on right)
    • Reduction: ClO- → Cl- + H2O (1 O on left, so 1 H2O on right)
  5. Balance H (add H2O and OH- for basic medium):
    • Oxidation: Cr(OH)3 + H2O → CrO4^2- + 5H+ (Balance H temporarily with H+).
      Now, add 5 OH- to both sides to neutralize H+ to H2O:
      Cr(OH)3 + H2O + 5OH- → CrO4^2- + 5H2O
      Simplify H2O: Cr(OH)3 + 5OH- → CrO4^2- + 4H2O
    • Reduction: ClO- + 2H+ → Cl- + H2O (Balance H temporarily with H+).
      Now, add 2 OH- to both sides:
      ClO- + 2H2O → Cl- + H2O + 2OH-
      Simplify H2O: ClO- + H2O → Cl- + 2OH-
  6. Balance charge (add e-):
    • Oxidation: Cr(OH)3 + 5OH- → CrO4^2- + 4H2O + 3e- (charge -5 on left, -2 on right; add 3e- to right to get -5)
    • Reduction: ClO- + H2O + 2e- → Cl- + 2OH- (charge -1 on left, -3 on right; add 2e- to left to get -3)
  7. Equalize electrons: Multiply oxidation half-reaction by 2, and reduction half-reaction by 3.
    • Oxidation: 2Cr(OH)3 + 10OH- → 2CrO4^2- + 8H2O + 6e-
    • Reduction: 3ClO- + 3H2O + 6e- → 3Cl- + 6OH-
  8. Add half-reactions and cancel common terms:
    2Cr(OH)3 + 10OH- + 3ClO- + 3H2O + 6e- → 2CrO4^2- + 8H2O + 6e- + 3Cl- + 6OH-
    Cancel 6e- from both sides.
    Cancel 6OH- from 10OH- (leaves 4OH- on left).
    Cancel 3H2O from 8H2O (leaves 5H2O on right).
    Final balanced equation: 2Cr(OH)3 + 4OH- + 3ClO- → 2CrO4^2- + 5H2O + 3Cl-

5. Electrolysis

Electrolysis is a process that uses electrical energy to drive non-spontaneous chemical reactions. It involves the decomposition of an electrolyte by passing an electric current through it.

Qualitative Aspect of Electrolysis

An electrolytic cell is the apparatus used for electrolysis. It consists of:

  • Electrolyte: A substance (molten ionic compound or aqueous solution of an ionic compound) that conducts electricity due to the movement of free ions.
  • Anode: The positive electrode, where oxidation occurs (anions lose electrons).
  • Cathode: The negative electrode, where reduction occurs (cations gain electrons).

The products of electrolysis depend on several factors:

  • The nature of the electrolyte (molten vs. aqueous).
  • The concentration of the electrolyte.
  • The nature of the electrodes (inert like Pt or C, or active like Cu).
  • The position of ions in the electrochemical series (for aqueous solutions, competition between ions and water).

Examples of Electrolysis:

  • Electrolysis of Molten NaCl:
    • Electrolyte: Molten NaCl (contains Na+ and Cl- ions)
    • At Cathode (reduction): Na+(l) + e- → Na(l) (Sodium metal is produced)
    • At Anode (oxidation): 2Cl-(l) → Cl2(g) + 2e- (Chlorine gas is produced)
    • Overall reaction: 2NaCl(l) &xrightarrow{electrolysis} 2Na(l) + Cl2(g)
  • Electrolysis of Aqueous CuSO4 using Inert Electrodes (e.g., Platinum):
    • Electrolyte: Aqueous CuSO4 (contains Cu2+, SO4^2-, H+, OH- from water dissociation)
    • At Cathode (reduction): Competing reactions: Cu2+ + 2e- → Cu (E° = +0.34V) and 2H2O + 2e- → H2 + 2OH- (E° = -0.83V at pH 7). Since Cu2+ has a higher reduction potential, Copper metal is deposited.
    • At Anode (oxidation): Competing reactions: 2SO4^2- → S2O8^2- + 2e- (E° = -2.01V) and 2H2O → O2 + 4H+ + 4e- (E° = -1.23V). Water is oxidized more easily than sulfate ions. So, Oxygen gas is produced.
    • Overall reaction: 2CuSO4(aq) + 2H2O(l) &xrightarrow{electrolysis} 2Cu(s) + O2(g) + 2H2SO4(aq)
  • Electrolysis of Brine (Concentrated Aqueous NaCl) using Inert Electrodes:
    • Electrolyte: Concentrated aqueous NaCl (contains Na+, Cl-, H+, OH- from water)
    • At Cathode (reduction): Competing reactions: Na+ + e- → Na (E° = -2.71V) and 2H2O + 2e- → H2 + 2OH- (E° = -0.83V at pH 7). Water is reduced preferentially. So, Hydrogen gas is produced, and the solution becomes alkaline due to OH- formation.
    • At Anode (oxidation): Competing reactions: 2Cl- → Cl2 + 2e- (E° = -1.36V) and 2H2O → O2 + 4H
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