Question

Balance the following reaction using Ion electron method :
(i) MnO₄^– + SO₃^2- → Mn²⁺ + SO₄^2-
(ii) Cr₂O₇^2- + H⁺ + I^– → Cr³⁺+ H₂O +I₂ (Acidic medium)
(iii) Cl₂ + OH^– → Cl^– + ClO₃^–
(iv) N₂H₄ + ClO₃^– → NO + Cl^– (Basic medium)

Answer 1

(i) MnO₄^– + SO₃^2- → Mn²⁺ + SO₄^2- (Acidic medium)

(a) To balance oxygen atoms, add H₂O to right side.
MnO₄^– ➝ Mn²⁺ + 4H₂O

(b) To balance hydrogen atoms, add H⁺ ions to left side.
MnO₄^– + 8H⁺ ➝ Mn²⁺ + 4H₂O

(c) To balance charge, add electrons to left side.
MnO₄^– + 8H⁺ + 5e^– ➝ Mn²⁺ + 4H₂O    …(1)

(a) To balance oxygen atom, add H₂O to left side.
SO₃^2- + H₂O ➝ SO₄^2-

(b) To balance hydrogen atom, add H+ ions to right side.
SO₃^2- + H₂O ➝ SO₄^2- + 2H⁺

(c) To balance charge, add electron to right side.
SO₃^2- + H₂O ➝ SO₄^2- + 2H⁺ + 2e^– …(2)
Multiply equation 1 by 2 and equation 2 by 5, and add, we get.

This is balanced equation.

(ii) Cr₂O₇^2- + H⁺ + I^– ➝ Cr³⁺ + H₂O + I₂

Balancing Cr, Cr₂O₇^2-  ➝ 2Cr³⁺

(a) To balance oxygen atoms, add H₂O to right side.
Cr₂O₇^2- ➝ 2Cr³⁺ + 7H₂O

(b) To balance hydrogen atoms, add H⁺ ions to left side.
Cr₂O₇^2- + 14H⁺ ➝ 2Cr³⁺ + 7H₂O

(c) To balance change, add electrons to left side.
Cr₂O₇^2- + 14H⁺ + 6e^–➝ 2Cr³⁺ +7H₂O   …(1)

(a) Balancing I, 2I^– ➝ I₂

(b) To balance charge, electron is added to right side.
2I^– ➝ I₂ + 2e^– …(2)
Multiply equation (1) by 1 and equation 2 by 3 and add, we get

This is balanced equation.

(iii) Cl₂ + OH^– ➝ Cl^– + ClO₃^–

Balancing Cl, Cl₂ ➝ 2ClO₃

(a) To balancing oxygen atoms, H₂O as added to left side.
Cl₂ + 6H₂O ➝ 2ClO₃^–

(b) To balance hydrogen atom, H⁺ are added to right side.
Cl₂ + 6H₂O ➝ 2CIO₃^– + 12H⁺
∴ Reaction is in basic medium OH^– are added to left side
∴ Cl₂+ 6H₂O + 12OH^– ➝ 2ClO₃^– + 12H₂O

(c) To balance charge, electrons are added to right side.
Cl₂ + 6H₂O + 12OH^– ➝ 2CO₃^– + 12H₂O + 10e^– …(1)

Balancing Cl, Cl₂ ➝ 2C1^–

(a) To balance charge, electrons are added to left side.
Cl₂ + 2e^– ➝ 2Cl^– …(2)
Multiply eq (1) by 1 and eq. (2) by 2 and add. we get

This is balanced equation.

(iv) N₂H₄ + ClO₃^– ➝ NO + Cl^– (Basic medium)

Balancing N, N₂H₄ ➝ 2NO

(a) To balance oxygen atoms, add H₂O to left side.
N₂H₄ + 2H₂O ➝ 2NO

(b) To balance hydrogen atoms, add H⁺ ions to right side.
N₂H₄ + 2H₂O ➝ 2NO + 8 H⁺
∴ Reaction is in basic medium, add OH^– to left side
N₂H₄ +2H₂O + 80H⁺ ➝ 2NO + 8H₂O + 8e^– …(1)

(a) To balance O-atoms, add H₂O to right side.
ClO₃^– ➝ Cl^– + 3H₂O

(b) To balance hydrogen atoms, add H⁺ ions to left side.
ClO₃^– + 6H₃⁺ ➝ Cl^– + 3H₂O
∴ Reaction is in basic medium , add OH^– to right side
ClO₃^– + 6H₃O ➝ Cl^– + 3H₂O + 6OH^–

(c) To balance charge, add electrons to left side.
ClO₃^– + 6H₂O + 6e ➝ Cl^– + 3H₂O + 6OH^– …(2)
Multiply equation 1 by 3 equation 2 by 4 and add, we get.

This is balanced equation.