Question

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 1

Answer:

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

MN7+ → Mn2+

n = 7– 2 = 5

Required charge will = 5 × 96487 Coulombs

= 482435 Coulombs

= 4.82 × 105 Coulombs