How much charge is required for the following reductions

Q.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 MnO4^– to Mn²⁺.

Solution:

(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 MnO4^–

Charge on Oxygen is – 2

Mn + 4O     = – 1

Mn +4(–2)   = – 1

Mn              = +7

So our reaction is

MN ⁷⁺ à Mn ²⁺

n = 7– 2  = 5  

Required charge will  = 5 × 96487 Coulombs

                                = 482435 Coulombs

                                = 4.82 × 10⁵ Coulombs