(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, CrO2-7 ➝ 2Cr3+
(b) to balance oxygen atoms, 7H2O is added to the right side.
CrO2-7 ➝ Cr3+ + 7H2O
(c) To balance H atom H+ are added to the left side.
CrO2-7 + 14H+ ➝ 2Cr3+ + 7H2O
(d) To balance charge, electrons are added to the left
side. CrO2-7 + 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.