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Q.

 LIST - 1 LIST - 2
A)One faradayi)Reduction
B)Anodeii)96500 coulomb
C)Cathode electronsiii)6.24×1018
D)1 coulombiv)Oxidation
  vi)Z × 96,500
 ABCD
1)viviiiii
2)iiiviiii
3)iiiviv
4)viiiiii

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a

2

b

1

c

3

d

4

answer is B.

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Detailed Solution

A) One Faraday = charge of 1 mole of electrons = 96500c 

B) Anode: oxidation takes place

C) Cathode : reduction 

D) 1 coulomb = 6.24 X1018 electrons .

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