• Every oxidation – reduction reaction always involves the interaction between a reducing agent and an oxidizing agent which brings about two opposing but complementary processes known as reduction and oxidation
• Oxidation is the addition of oxygen to a substance or the removal of hydrogen from a substance
• An oxidizing agent is one which transfers oxygen to another substance or removes hydrogen from that substance. For example:
  • CuO_(s) + H_2(g) → Cu_(s) + H₂O_(l)
  • The process of converting H₂ to H₂O is oxidation
  • CuO which releases the oxygen to hydrogen is the oxidizing agent
• Reduction is the addition of hydrogen to a substance or the removal of oxygen from a substance
• A reducing agent is a substance which transfers hydrogen to another substance or removes oxygen from that substance. For example:
  • 3Fe(s) + 4H₂O_(g) → Fe₃O_4(s) + 4H_2(g)
  • The conversion of H₂O_(g) to H_2(g) is reduction
  • Fe removes oxygen from water and hence is the reducing agent

Electron – Transfer Concept

• According to the electron – transfer concept,:
  • Oxidation is loss of electrons
  • Reduction is gain of electrons
  • An oxidizing agent is an electron acceptor
  • A reducing agent is an electron donor
  • NOTE: Oil Rig (an acronym) can be used to recall the above
    • Oxidation is loss of electrons
    • Reduction is gain of electrons
• For example:
  • Na(s) + ½Cl_2(g) → Na⁺Cl^–_(s)
  • Na, like all metals, is the reducing agent because it donates/loses its outermost electron
  • Cl is the oxidizing agent because it accepts the electron
• In all redox processes, the total number of electrons donated is equal to the total number of electrons accepted

Oxidation Number Concept

• According to the oxidation number concept,
  • Oxidation is the increase in oxidation number
  • Reduction is the decrease in oxidation number
  • A reducing agent is one whose oxidation number increases
  • An oxidizing agent is one whose oxidation number decreases
• For example:
  • Na(s) + ½Cl_2(g) → Na⁺Cl^–_(s)
  • Na is the reducing agent because its oxidation number increases from 0 to +1.
  • Cl is the oxidizing agent because its oxidation number decreases from 0 to -1

Rules for Assigning Oxidation Numbers

1. The oxidation number (o.n.) of an element in its free state or uncombined state is 0. For example:
   1. O.n. of Na as Na = 0
   2. O.n. of O as O₂ = 0, etc.
2. The o.n. of hydrogen in its compounds is +1 except in metallic hydrides where it is -1. For example:
   1. O.n. of H in H₂O, H₂SO₄, H₃PO₄, etc. = +1
   2. O.n. of H in NaH or CaH₂ is -1
3. The o.n. of O in its compounds is -2 except in peroxides where it is -1. For example:
   1. O.n. of O in H₂O, H₂SO₄, etc. = -2
   2. O.n. of O in H₂O₂, Na₂O₂, etc. = -1
4. The o.n. of any group one metal in its compounds is +1 while the o.n. of any group element in its compounds is +2. For example:
   1. O.n. of Na in NaOCl, Na₂CO₃ = +1
   2. O.n. of Mg in MgSO₄, Mg(NO₃)₂ = +2
5. The o.n. of an element which exists as a simple ion is equal to the charge on the ion. for example:
   1. O.n. of Fe in Fe²⁺ = +2
   2. O.n. of Al in AL³⁺ = +3
6. The o.n. of fluorine in all its compounds is -1
7. The sum of the oxidation numbers of all the atoms in a neutral compound is equal to zero. For example:
   1. For H₂SO₄, 2(o.n. of H) + o.n. of H + 4(o.n. of O) = 0
8. The sum of the oxidation numbers of all atoms in a charged compound or radical is equal to the overall charge on the charged compound. For example:
   1. For SO₄^2-, o.n. of S + 4(o.n. of O) = -2

Examples

1. Calculate the oxidation number of chromium in K₂Cr₂O₇
   • Let x represent the o.n. of Cr
   • Recall:
     • K is a group 1 metal and takes an o.n. of +1 in its compounds
     • O normally takes an o.n. of -2 in its compounds
   • Inserting the oxidation numbers we have:
     • 2(+1) + 2(x) + 7(-2) = 0
     • +2 + 2x – 14 = 0
     • 2x = 12
     • ∴ x = +6
2. Calculate the oxidation number of manganese in MnO₄^–
   • The overall charge of the radical is -1
   • Let y represent the o.n. of Mn
   • Inserting the oxidation numbers we know:
     • y + 4(-2) = -1
     • y – 8 = -1
     • ∴ y = +7

Common Redox Agents and Test for Redox Agents

• Common oxidizing agents include:
  • Oxygen
  • Potassium tetraoxomanganate(vii) solution (KMnO₄)
  • Concentrated Trioxonitrate(v) acid (HNO₃), etc.
• Common reducing agents include:
  • Metals
  • Carbon
  • Hydrogen
  • Hydrogen sulfide
  • carbon dioxide
• Tests for Oxidizing Agents:
  • An oxidizing agent will react with:
    • HCl solution to liberate chlorine gas
    • Hydrogen sulfide to form a precipitate of sulfur
    • Potassium iodide solution to liberate iodine
• Tests for Reducing Agents:
  • A reducing agent will react with:
    • acidified K₂Cr₂O₇ solution and change its orange color to green
    • acidified KMnO₄ solution and turn its purple color colorless

Balancing Redox Equations

• Non-redox reactions which do not involve the transfer of electrons or changes in oxidation numbers are usually balanced by inspection.
• Neutralization and double decomposition reactions are examples of non-redox reactions
• Some simple redox processes, such as the combustion of hydrogen, the reaction between hydrogen and copper (ii) oxide, etc., may be balanced by inspection
• However, redox equations often involve more than two reactants and their equations are often balances using oxidation number concept or half-reaction method

Balancing Redox Equations by the Half-Reaction (Ion-electron) Method

• The main steps for balancing redox equations by the half-reaction method are as follows:
  • Step 1: Separate the reaction into an oxidation half-reaction and a reduction half-reaction
  • Step 2: Balance the atoms other than H and O in each half reaction
  • Step 3: To the side that lacks oxygen atoms, add 1 mole of water for every O atom that is lacking
  • Step 4: To the side that lacks hydrogen atoms, add 1 mole of H⁺ for every H atom that is lacking
  • Step 5: Balance the charge by adding electrons to the more positive side in each half-equation
    • NOTE: The number of electrons to be added is always equal to the difference between the more net positive charge and the less net positive charge in each half-reaction
  • Step 6: Using appropriate integers, make the number of electrons gained equal the number lost and then add the two half-reactions
  • Step 7: Cancel anything that is the same on both sides to get the desired balanced equation
  • Step 8: If the reaction takes place in a basic solution, add to both sides of the equation the same number of OH^– as there are H⁺
  • Step 9: Combine H⁺ and OH^– to form H₂O and collect all H₂O algebraically to one side of the balanced equation

Balancing Redox Equations by the Half-Reaction (Ion-electron) Method

1. Balance the equation MnO₄^– + C₂O₄^2- → Mn²⁺ + CO₂ if the reaction occurs in an acidic medium
   • Step 1:
     1. MnO₄^– → Mn²⁺
     2. C₂O₄^2- → CO₂
   • Step 2:
     Leave (i) as it is but multiply CO₂ in (ii) by 2
     1. MnO₄^– → Mn²⁺
     2. C₂O₄^2- → 2CO₂
   • Step 3:
     In (i) add 4H₂O to Mn²⁺ and leave (ii) as it is
     1. MnO₄^– → Mn²⁺ + 4H₂O
     2. C₂O₄^2- → 2CO₂
   • Step 4:
     Add 8H⁺ to MnO₄^– and leave (ii) as it is
     1. 8H⁺ + MnO₄^– → Mn²⁺ + 4H₂O
        Net charge on left side: +8 + (-1) = +7
        Net charge on right side: +2 + 0 = +2
     2. C₂O₄^2- → 2CO₂
        Net charge on left side: -2
        Net charge on right side: 0
   • Step 5:
     Add 5 electrons to the left side of equation (i) to balance the charge as the difference between +7 and +2 is 5
     Add 2 electrons to the right side of equation (ii) as the difference between 0 and -2 is 2
     1. 5e + 8H⁺ + MnO₄^– → Mn²⁺ + 4H₂O
     2. C₂O₄^2- → 2CO₂ + 2e
   • Step 6:
     To make the number of electrons gained equal to the number of electrons lost, multiply equation (i) by 2 and equation (ii) by 5 and the resulting equation
     1. 10e + 16H⁺ + 2MnO₄^– → 2Mn²⁺ + 8H₂O
     2. 5C₂O₄^2- → 10CO₂ + 10e

     The resulting equation: 10e + 5C₂O₄^2- + 16H⁺ + 2MnO₄^– → 2Mn²⁺ + 8H₂O + 10CO₂ + 10e

   • Step 7:
     Cancel 10e from both sides
     5C₂O₄^2- + 16H⁺ + 2MnO₄^– → 2Mn²⁺ + 8H₂O + 10CO₂
2. Balance the equation SO₃^2- + MnO₄^– → SO₄^2- + MnO₂ for a reaction which occurs in a basic medium.
   • Following steps 1 to 7 for acidic solutions gives:
     2H⁺ + 3SO₃^2- + 2MnO₄^– → 3SO₄^2- + 2MnO₂ + H₂O
   • Step 8:
     Add 2OH^– to each side of the equation:
     2OH^– + 2H⁺ + 3SO₃^2- + 2MnO₄^– → 3SO₄^2- + 2MnO₂ + H₂O + 2OH^–
   • Step 9:
     Combine H⁺ and OH^– to H₂O (in place of 2H⁺ + OH^–, write 2H₂O):
     2H₂O + 3SO₃^2- + 2MnO₄^– → 3SO₄^2- + 2MnO₂ + H₂O + 2OH^–
   • Step 10:
     Delete an equal number of what is common from both sides of the equation:
     H₂O + 3SO₃^2- + 2MnO₄^– → 3SO₄^2- + 2MnO₂ + 2OH^–