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Oxidation - Reduction (Redox) Reactions

  • 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) + H2(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) → Fe3O4(s) + 4H2(g)
    • The conversion of H2O(g) to H2(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 iloss of electrons
      • Reduction igain of electrons
  • For example:
    • Na(s) + ½Cl2(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) + ½Cl2(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 SO42-, 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 + C2O42- → Mn2+ + CO2 if the reaction occurs in an acidic medium
    • Step 1:
      1. MnO4 → Mn2+
      2. C2O42- → CO2
    • Step 2:
      Leave (i) as it is but multiply CO2 in (ii) by 2

      1. MnO4 → Mn2+
      2. C2O42- → 2CO2
    • Step 3:
      In (i) add 4H2O to Mn2+ and leave (ii) as it is

      1. MnO4 → Mn2+ + 4H2O
      2. C2O42- → 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. C2O42- → 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. C2O42- → 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. 5C2O42- → 10CO2 + 10e

      The resulting equation: 10e + 5C2O42- + 16H+ + 2MnO4 → 2Mn2+ + 8H2O + 10CO2 + 10e

    • Step 7:
      Cancel 10e from both sides
      5C2O42- + 16H+ + 2MnO4 → 2Mn2+ + 8H2O + 10CO2
  2. Balance the equation SO32- + MnO4 → SO42- + MnO2 for a reaction which occurs in a basic medium.
    • Following steps 1 to 7 for acidic solutions gives:
      2H+ + 3SO32- + 2MnO4 → 3SO42- + 2MnO2 + H2O
    • Step 8:
      Add 2OH to each side of the equation:
      2OH + 2H+ + 3SO32- + 2MnO4 → 3SO42- + 2MnO2 + H2O + 2OH
    • Step 9:
      Combine H+ and OH to H2O (in place of 2H+ + OH, write 2H2O):
      2H2O + 3SO32- + 2MnO4 → 3SO42- + 2MnO2 + H2O + 2OH
    • Step 10:
      Delete an equal number of what is common from both sides of the equation:
      H2O + 3SO32- + 2MnO4 → 3SO42- + 2MnO2 + 2OH
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