Question

Write the step of balancing equation by Ion Electron method and balance the following reactions :
(i) Al + NO₃^– ➝ Al(OH)₄^– + NH₃ (Basic)
(ii) MnO₄^– + Br^– ➝ Mn²⁺ + Br₂ (Acidic)
(iii) Cr₂O₇^2- + Fe³⁺ ➝ Cr³⁺ + H₂O + Fe³⁺(Acidic)

Answer 1

(i) 

(a) The unbalanced equation is :

First half reaction
Al ➝ Al(OH)₄^– (Oxidation)
(a) To balance oxygen atoms, add 4H₂O to left hand side
Al + 4H₂O ➝  Al(OH)₄^–

(b) To balance hydrogen atoms, add H+ to right side
Al + 4 H₂O➝ Al(OH)₄^– + 4H⁺
Since, reaction is in basic medium, OH^– is added to left side
Al + 4H₂O ➝ 4Al(OH)₄^– + 4H₂O

(c) To balance change, electrons are added to right side
Al + 4H₂O + 4OH^– ➝ Al(OH)₄^– + 4H₂O + 3\(\overline{e}\) …(1)

(a) To balance oxygen atoms, add 3H₂O to right side
NO₃^– ➝ NH₃ + 3H₂O

(b) NO₃^– + 9H₂O ➝ NH₃ + 3H₂O + 9 OH^–

(c) To balance charge, add electron to left side
NO₃^–+ 9H₂O+ 8\(\overline{e}\) ➝ NH₃ + 3H₂O + 9OH^– …(2)
Multiply eq. (1) By 8 and equation 2 by 3 and add, we get equation.

(ii) MnO^-₄ + Br^– ➝ Mn²⁺ + Br₂ (Acidic medium)

To balance oxygen atoms, H₂O is added on right side.
MnO^-₄ ➝ Mn²⁺ + 4H₂O

(b) To balance hydrogen, H+ ions are added to left side
MnO^-₄ + 8H⁺ ➝ Mn²⁺ + 4H₂O

(c) To balance charge, electrons are added left side
MnO^-₄ + 8H⁺ + 5e^– ➝ Mn²⁺ + 4 H₂O …(1)

or 2Br^– ➝ Br₂
To balance charge, 2\(\overline{e}\) are added right hand side
2Br^– ➝ Br₂ + 2\(\overline{e}\) …(2)
Now adding the eq. (1) and (2)

This is balanced equation.

(iii) (a) Balancing Cr, CrO^2-₇ ➝ 2Cr³⁺

(b) to balance oxygen atoms, 7H₂O is added to the right side.
CrO^2-₇ ➝ Cr³⁺ + 7H₂O

(c) To balance H atom H⁺ are added to the left side.
CrO^2-₇ + 14H⁺ ➝ 2Cr³⁺ + 7H₂O

(d) To balance charge, electrons are added to the left
side. CrO^2-₇ + 14H⁺\(6 \overline{e}\) ➝ 2Cr³⁺+7H₂O …(1)

To balance charge, electrons are added to the right side.
Fe²⁺ ➝ Fe³⁺ + e^– …(2)
Multiply equation (1) by 1 and equation 2 by 6 and add, we get.

This is balanced equation.