Types of Redox Reaction

1. Combination reactions: A + B → C

For such a reaction to be redox reaction, either A or B or both should be in elemental form.

Ex: \(\overset{0}{\mathop{C}}\,\left( s \right)+\overset{0}{\mathop{{{O}_{2}}}}\,\left( g \right)\xrightarrow{\Delta }\overset{+4}{\mathop{C}}\,\overset{-2}{\mathop{{{O}_{2}}}}\,\left( g \right) \)

2 C⁺²O⁻²₂ (g) + O0₂ (g) → 2 C⁺⁴ O⁻²₂ (g)

Fe0 (s) + S0 (s) → Fe⁺²S⁻² (s)

2. Decomposition reactions: Breakdown of a compound to form 2 or more products

For a decomposition to be redox reaction, one or more of the products have to be in elemental form

Ex: \(2\overset{+1}{\mathop{{{H}_{2}}}}\,\overset{-2}{\mathop{0}}\,\left( l \right)\xrightarrow{\Delta }2\overset{0}{\mathop{{{H}_{2}}}}\,\left( g \right)+\overset{0}{\mathop{{{O}_{2}}\left( g \right)}}\, \)

3. Displacement Reactions: An ion (or atom) in a compound is replaced by an ion (or atom) of another element.

X + YZ → XZ + Y

a. Metal displacement: A metal in a compound is displaced by another metal in the uncombined state.

Ex: Cu⁺² S⁺⁴ O⁻²₄ + Zn0 (s) →Cu0 (S) + Zn⁺² S⁺⁴ O⁻²₄ (aq)

Note:  ZnSO₄ (aq) + Cu (s) → Zn (s) + Cu SO₄ (aq)
the above reaction does not take place as Zn is more electropositive than Cu.

Only a more electropositive element can displace a less electropositive element from its compound i.e., the displacing metal should be a better reducing agent than the metal that is displaced

b. Non – metal displacement: Includes

  • Hydrogen displacement: All alkali metals displace Hydrogen from cold water
    2K (s) + 2H₂O (l) →2KOH (aq) + H₂ (g)
    Alkaline earth metals Ca, Sr and Ba displace hydrogen from cold water
    Ca (s) + 2H₂O (l) → Ca (OH)₂ (aq) + H₂ (g)
    Less active metals like Mg & Fe react with steam to product dihydrogen gas.
    \(2Fe\left( s \right)+3{{H}_{2}}O\left( l \right)\xrightarrow{\Delta }F{{e}_{2}}{{O}_{3}}\left( s \right)+3{{H}_{2}}\left( g \right)\)
    Many metals displace hydrogen from acids
    Zn (s) + 2HCl (aq) → ZnCl₂ (aq) + H₂ (g)
    Zn (s) + H₂SO₄ (aq) → ZnSO₄ + H₂ (g)
  • Displacement of oxygen (rare):
    2H₂O (l) + 2F₂ (g) → 4HF (aq) + O₂ (g)
  • Displacement of halogen: Oxidizing power:
    F > Cl > Br > I
    Thus, Fluorine can replace Cl, Br & I
    Chlorine can replace Br & I
    Cl₂ (g) + 2KBr (aq) → 2KCl (aq) + Br₂ (l)
    Bromine can replace Iodine
    Br₂ (l) + 2KCl (aq) → 2KBr (aq) + I₂ (s)

4. Dispropartionation reactions:

  • An element undergoes oxidation and reduction simultaneously.
  • The reacting element undergoing reduction and oxidation simultaneously exists in the intermediate oxidation state so that both increase and decrease in oxidation is possible.
    Ex: Cu⁺¹₂ O (s) + H₂ SO₄ (aq) → Cu0 (s) + Cu⁺²SO₄ (aq) + H₂O (l)
    Here Cu is in +1 oxidation state in Cu₂O. After reacting with H₂SO₄ it forms Cu metal which has 0.
    Oxidation state and CuSO₄ where, its oxidation state is +2.
    So Cu undergoes oxidation to form CuSO₄ and reduction to form Cu metal.