How Much Charge is Required For the Following Reductions (i) 1 Exercise Chapter 3 Electrochemistry Chemistry Class

Question 12:How much charge is required for the following reductions:
(i) 1 mol of Al³⁺ to Al.
(ii) 1 mol of Cu²⁺ to Cu.
(iii) 1 mol of MnO^4– to Mn²⁺.

Answer:(i)
Formula required charge n × F
n = difference of charge on ions   
F is constant and equal to 96487 Coulombs
Here n = 3
Hence required charge = 3 × 96487 Coulombs
                                  = 289461 Coulombs
                                  = 2.89  ×10 ^–5 Coulombs
(ii)

n = 2
plug the value in formula we get
Required charge= 2 × 96487 Coulombs
= 192974 Coulombs
= 1.93 × 10⁵ Coulombs
(iii)
Charge on Mn in MnO^4–
Charge on Oxygen is – 2
Mn + 4O     = – 1
Mn +4(–2)   = – 1
Mn              = +7
So our reaction is
MN⁷⁺  → Mn²⁺
n = 7– 2  = 5  
Required charge will  = 5 × 96487 Coulombs
= 482435 Coulombs
= 4.82 × 10⁵ Coulombs