Question

Balance the following reaction using Ion Electron method,
(i) NO^-₃ + H⁺ + I^–  ➝ NO+ H₂O + I₂
(ii) CrO₄^2- + SO₃^2- + OH^– ➝ CrO₂^2- + SO^2-₄
(iii) MnO^-₄ + Fe₃O₄ + OH^– ➝ Fe₂O₃ + MnO₂
(iv) P₄ + OH^– ➝ PH₃ + HPO^-₂

Answer 1

(i)  NO^-₃ + H⁺ + I^– ➝ NO + H₂O + I₂

(a) To balance oxygen atoms add H₂O to right side.
NO^-₃ ➝ NO + 2H₂O

(b) To balance hydrogen atoms add H⁺ ions to left side.
NO^-₃ + 4H⁺ ➝ NO + 2H₂O

(c) To balance charge, add electrons to left side.
NO^-₃ + 4H⁺ + 3e^– ➝ NO + 2H₂O     …. (1)

Balancing I + 2I^– ➝ I₂

(a) To balancing charge, electrons are added to right side.
2I^– ➝ I₂ + 2e^–
Multiply eq (1) by 2 and eq. (2) by 3 and add, we get

This is balanced equation.

(ii)  CrO^2-₄ + SO₃^2- + OH^– ➝ CrO^2-₂ + SO₄^2-

(a) To balance oxygen atoms add H₂O to right side.
CrO^2-₄ ➝ CrO^2-₂ + 2H₂O

(b) To balance hydrogen atoms add H⁺ ions to left side.
CrO^2-₄ + 4H⁺ ➝ CrO^2-₂ + 2H₂O

(c) To balance charge, electrons are added to right side.
SO^2-₃ + 2OH^–  ➝ SO^2-₄ + 2H₂O + 2e^– … (2)
Multiply eq 1 by 1 and eq. 2 by 2 and add, we get

This is balanced equation.

(iii) MnO^-₄ + Fe₃O₄ + OH^– ➝ Fe₂O₃ + MnO₂

(a) To balance oxygen atoms add H₂O to right side.
MnO^-₄ ➝ MnO₂ + 2H₂O
(b) To balance hydrogen atoms add H⁺ ions to left side.
MnO^-₄ + 4H⁺ ➝ MnO₂ + 2H₂O
∴ Reaction is in basic medium, OH^–  ions are added to right side
MnO^-₄ + 4H₂O ➝ MnO₂ + 2H₂O + 4OH^–
(c) To balance charge, electrons are added to right side.
MnO^-₄ + 4H₂O + 3e^– ➝  MnO₂ + 2H₂O + 4OH^– … (1)

Balancing Fe, 2Fe₃O₄ ➝ 3Fe₂O₃

(a) To balance oxygen atoms, add H₂O to left side.
2Fe₃O₄ + H₂O ➝ 3Fe₂O₃

(b) To balance hydrogen atoms, add H⁺ ions to right side.
2Fe₃O₄ + H₂O  ➝ 3Fe₂O₃ + 2H₂O
∴ Reaction is in basic medium, so add OH^– ions to left side.
2Fe₃O₄ + H₂O + 20H^–  ➝ 3Fe₂O₃ + 2H₂O

(c) To balance charge, add electrons to right side,
2Fe₃O₄ + H₂O + 2OH^–  ➝ 3Fe₂O₃ + 2H₂O + 2e^– … (2)
Multiply eq. (1) by 2 and eq. (2) by 3 and add, we get

This is balanced equation.

(iv) P₄ + OH^–  ➝ PH₃ + HPO₂^–

Balancing P, P₄ ➝ 4PH₃
(a) To balance H-atoms, add H⁺ ions to left side.
P₄ + 12H⁺ ➝  4PH₃
∴ Reaction is in basic medium, add OH” ions to right side.
P₄ + 12H₂O ➝ 4PH₃ + 12OH^–

(b) To balance charge, electrons add to left side.
P₄ + 12H₂O + 12e^– ➝ 4PH₃ + 12OH^– …( 1)

(a) To balance hydrogen atoms, add H₂O to left side.
P₄ + 8H₂O ➝ 4HPO₂^–

(b) To balance hydrogen atom, add H⁺ ions to right side.
P₄ + 8H₂O ➝ 4HPO₂^– + 12H⁺
∴ Reaction is in basic medium, add OH^– to left side
P₄ + 8H₂O + 12OH^– ➝ 4 HPO₂^– +12 H₂O

(c) To balance charge, electrons are added to right side.
P₄ + 8H₂O + 12OH^– ➝ 4HPO₂^– + 12H₂O + 8e^– …(2)
Multiply eq. (1) by 2 and eq (2) by 3, and add we get.

This is balanced equation.