Oxidation and reduction are always found to go hand in hand during a redox reaction. Whenever an element or compound is oxidized, another element or another compound must be reduced.

- An oxidant is reduced and simultaneously the reductant is oxidized. It follows from our above discussion that we should be able to balance oxidation-reduction reactions on the basis of

__Ion-Electron Method.____Oxidation Number Method.__

### Ion-Electron Method:

- Ascertain the Reactants and Products, and their chemical formula's.
- Set up the partial equations representing the reduction of the oxidant, and the oxidation of the reductant.
- If the reaction occurs in acid medium use a requisite number of
**H⁺**for balancing the number of atoms involved in the partial equation. for alkaline medium use**OH⁻**ions. - Balancing the charge in the partial equations by adding a suitable number of electrons. These electrons indicate the electrons involves in the oxidation or the reduction half reactions.
- Multiply each partial equation by a suitable factor so each of the two partial equations involves the same number of electrons.
- Add the partial equations and cancel out species which appear on both sides of the equations.

**In aqueous acid medium potassium permanganate oxidizes a ferrous ion to the ferric state:**

- In this reaction, permanganate ions are the oxidant and ferrous ion the reductant. The left-hand side of the ultimate equation will carry,

**MnO₄⁻**(

**KMnO₄**),

**H⁺**(

**H₂SO₄**) and

**Fe²⁺**(

**FeSO₄**).

- The Right-hand side will have as products,

**Mn²⁺ (MnSO₄), H₂O and Fe³⁺ [Fe₂(SO₄)₃]**. The partial equation representing the reduction of the oxidant

**MnO₄⁻**will involve,

- MnO₄⁻ → Mn²⁺

- Since the reaction occurs in an acid medium we utilize eight H⁺ ions to balance the four oxygen atoms in MnO₄⁻.

- Thus, MnO₄⁻+8H⁺ → Mn²⁺+4H₂O

- Since the above partial equation is still unbalanced from the viewpoint of charge, the equation is balanced by bringing in five electrons:

- MnO₄⁻ + 8H⁺ + 5e ⇆ Mn²⁺ + 4H₂O

- If we now apply the above steps to obtain the partial equation representing the oxidation of reductant we have the following:

- Fe⁺² ⇆ Fe⁺³ + e

- This equation on multiplication by

**5**and then adding to the partial equation of the oxidant gives the final balanced equation:

- MnO₄⁻ + 8H⁺ + 5e ⇆ Mn²⁺ + 4H₂O
5×( Fe⁺² ⇆ Fe⁺³ + e)

- MnO₄⁻ + 8H⁺ + 5Fe⁺² → Mn²⁺ + 5Fe⁺³ + 4H₂O

**The Oxidation of potassium iodide by potassium dichromate in dilute acid medium.**

- In this reaction, dichromate is reduced to (

**+6**) state to (

**+3**) state and iodide ion is oxidized to elementary iodine. Taking the dichromate side, balancing the atoms and charges provides the partial equation:

- Cr₂O₇⁻² + 14H⁺ + 6e ⇆ 2Cr⁺³ + 7H₂O
(Reduction of Oxidant)

__Considering the case of iodide ion we get:__

- 2I⁻ ⇆ I₂ + 2e
(Oxidation of Reductant)

- Multiplying the above equation by

**3**and adding these two partial equations we have the following final equation:

Cr₂O₇⁻² + 14H⁺ + 6e ⇆ 2Cr⁺³ + 7H₂O |

3 (2I⁻ ⇆ I₂ + 2e) |

Cr₂O₇⁻²+14H⁺+6I⁻→2Cr⁺³+3I₂+7H₂O |

**The reaction of permanganate ion with Sodium Stannite (Na₂SnO₂) in alkaline solution in which manganese dioxide (MnO₂) and Sodium Stannate(Na₂SnO₃) are formed.**

- MnO₄⁻ → MnO₂

- The Partial equation representing the reduction of oxidant is, Since the medium is alkaline we put the requisite number of OH⁻ ions to effect atom balance as:

- MnO₄⁻ + 2H₂O → MnO₂ + 4OH⁻

- To balance the charge on both sides three electrons are added on the left-hand side:

- MnO₄⁻ + 2H₂O + 3e ⇆ MnO₂ + 4OH⁻

- For the reductant, the balance partial equation is:

- SnO₂⁻² + OH⁻ ⇆ SnO₃⁻² + H₂O + 2e

- To balance the number of electrons involved in the two partial equations the first equation is multiplied by 2 and the second by 3 and on the adding, we have the final form of the ionic equation:

- 2MnO₄⁻ + 4H₂O + 6e ⇆ 2MnO₂ + 8OH⁻
3SnO₂⁻² + 3OH⁻ ⇆ 3SnO₃⁻² + 3H₂O + 6e

- 2MnO₄⁻+3SnO₂⁻²+H₂O→2MnO₂+3SnO₃⁻²+2OH⁻

**Oxidation of Mn⁺² to MnO₄⁻ by PbO₂ or NaBiO₃ in Acid Medium: **

Balancing Oxidation-Reduction Reactions by oxidation number method |

**Oxidation Number Method:**

- Assuming that the oxidation number of oxygen remains unchanged manganese in MnO₄⁻ has an oxidation number +7. Calculation of the decrease and increase in oxidation number gives:
- Putting the right factors the decreases and increases in oxidation numbers are balanced, giving,

**MnO₄⁻ + 5Fe⁺² → Mn⁺² + 5Fe⁺³** - Since the reaction medium is acidic and the changes are not equal on the two sides of the above expression, H⁺ ions are added and the requisite number of H₂O written:

**MnO₄⁻+8H⁺+5Fe⁺²→Mn⁺²+ 5Fe⁺³+4H₂O**

Balancing Oxidation-Reduction Reactions by oxidation number method |

**The Oxidation of iodide ion by dichromate ion in acid medium: **

- The reaction represented as,
Cr₂O₇⁻² + I⁻ → 2Cr⁺³ + I₂

- The oxidation number of Cr decreases by 2 and the oxidation number of Sn increases by 1. Equalizing the increase and decrease in oxidation number,

- Cr₂O₇⁻² + 6I⁻ → 2Cr⁺³ + 3I₂

- The medium is acidic and because the ionic charges are not equal on the two sides, fourteen

**H⁺**ions are added to the left-hand side and requisite number of H₂O added to the left-hand side.

- Cr₂O₇⁻² + 6I⁻ + 8H⁺ → 2Cr⁺³ + 3I₂ +7H₂O

**Oxidation of sodium stannite to stannate in alkaline medium:**

- This reaction represented as,
MnO₄⁻ + SnO₂⁻² → MnO₂ + SnO₃⁻²

- The oxidation number of Mn decreases by 3 and the oxidation number of Sn increases by 2. Equalizing the increase and decrease in oxidation number,

- 2MnO₄⁻ + 3SnO₂⁻² → 2MnO₂ + 3SnO₃⁻²

- Because the medium is alkaline and because the ionic charges are not equal on the two sides, two OH⁻ ions are added to the right-hand side.

- 2MnO₄⁻ + 3SnO₂⁻² → 2MnO₂ + 3SnO₃⁻² + 2OH⁻

- Since there are two hydrogen atoms on the right and none on the left, H₂O is added to make atom balance. The final equation is:

- 2MnO₄⁻ + 3SnO₂⁻² + H₂O → 2MnO₂ + 3SnO₃⁻² + 2OH⁻

- Problem 1:

- Express by ion electron method the reduction of permanganate to manganous stat by hydrogen peroxide in acid medium.

- Answer:

- This Reaction occurs in acid medium and the partial equation is:
MnO₄⁻ + 8H⁺ + 5e ⇆ Mn⁺² + 4H₂O
(Reduction of Oxidant)

- In acid medium H₂O₂ will give O₂ and the partial equation being:
H₂O₂ ⇆ 2H⁺ + O₂ +2e
(Oxidation of Reductant)

- The first equation is multiplying by 2 and the second is 5 to have electron balanced. We have,

2MnO₄⁻ + 16H⁺ + 10e ⇆ 2Mn⁺² + 8H₂O 5H₂O₂ ⇆ 10H⁺ + 5O₂ +10e |

2MnO₄⁻+5H₂O₂+6H⁺⇆2Mn⁺²+8H₂O+5O₂ |

- Problem 2:

- Express by ion - electron method the reduction of nitrate ion to ammonia by aluminum in aqueous NaOH.

- Answer:

- In alkaline medium, we use H₂O and OH⁻ ion according to convenience. The partial equation with charge balance is:

- NO₃⁻ + 6H₂O + 8e ⇆ NH₃ + 9OH⁻

- Metallic aluminum will go over to aluminate ion, the partial equation being:

- Al + 4OH⁻ ⇆ AlO₂⁻ + 2H₂O +3e

- Multiplying by right factors for electron balance we have the balanced equation:

- 3NO₃⁻ + 18H₂O + 24e ⇆ 3NH₃ + 27OH⁻
8Al + 32OH⁻ ⇆ 8AlO₂⁻ + 16H₂O +24e
3NO₃⁻+8Al+2H₂O+5OH⁻→3NH₃+8AlO₂⁻

- Problem 3:

- Use the Oxidation Number method to balance the reaction of iodide ion and iodate ion in an acid medium to liberate iodine.

- Answer:

- The reaction represented as,
I⁻ + IO₃⁻ → I₂

- Representation of the above equation with an oxidation number of iodine,
I⁻(-1) + IO₃⁻(+5) → I₂(0)

- Since I⁻ (-1) oxidation number of iodide increases by 1 and IO₃⁻(+5) oxidation number decreases by 5. Putting the right factors the decrease and increase in oxidation number balanced:

- 5I⁻ + IO₃⁻ → 3I₂

- Since the medium is acidic and the charges are unequal on two sides of the above expression we put Six H⁺ and write three H₂O to balanced the equation:

- I⁻ + IO₃⁻ + 6H⁺ → I₂ + 3H₂O

- Problem 4:

- Use the oxidation number method to balance the reaction between Sulphurus acid and dichromate in acidic medium.

- Answer:

- The reaction represented as,
SO₃⁻² + Cr₂O₇⁻² → SO₄⁻² + 2Cr⁺³

- Oxidation of two Cr decreases by 2 × (+3) = 6 and the oxidation number of S increases by 2. Equalizing the above equation increase and decrees in oxidation number,

- 3SO₃⁻² + Cr₂O₇⁻² → 3SO₄⁻² + 2Cr⁺³

- Because of the medium is acidic and the ionic charges are not equal on the two sides, eight H⁺ ions are added to the Left-hand side.

- 3SO₃⁻² + Cr₂O₇⁻² + 8H⁺ → 3SO₄⁻² + 2Cr⁺³

- Since there are eight hydrogen ion on the left and none on the right, four H₂O added to make atom balance. Thus the final equation is:

- 3SO₃⁻²+Cr₂O₇⁻²+8H⁺→3SO₄⁻²+2Cr⁺³+4H₂O