Balancing oxidation reduction reactions

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

  1. Ascertain the Reactants and Products, and their chemical formula's.
  2. Set up the partial equations representing the reduction of the oxidant, and the oxidation of the reductant. 
  3. 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.
  4. 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.
  5. Multiply each partial equation by a suitable factor so each of the two partial equations involves the same number of electrons.
  6. Add the partial equations and cancel out species which appear on both sides of the equations.
    An aqueous acid medium, potassium permanganate oxidizes ferrous state 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
    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.
    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⁻3
SnO₂⁻² + 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
Balancing oxidation reaction

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:
How Balancing Oxidation Reduction Reactions by oxidation number method?
Balancing oxidation-reduction reactions
    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
    Oxidation of iodide ion by dichromate ion in acid medium
The reaction represented as Cr₂O₇⁻² + I⁻ → 2Cr⁺³ + I₂
    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
    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
    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 an 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
    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
    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 the 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
    Use the oxidation number method to balance the reaction between Sulfurous 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 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

Balancing oxidation reduction reactions in Ion - electron method and oxidation number method with examples and related problems with solutions

[Chemical kinetics] [column1]

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