Discussion Forum



1)

How many electrons are involved in the reaction, when 0.40 F of electricity is passed through an electrolytic solution ?


A) $6.642\times 10^{25}$

B) $1.505\times 10^{25}$

C) $6.022\times 10^{25}$

D) $2.4088\times 10^{25}$

Answer:

Option D

Explanation:

Given,

 Quantity of charge  (Q) passed= 0.40 F

                                             =0.40 x96500 C

                                       =38600 C

 we know that q= ne-

 where, n= number of electrons

 $e^{-}$  = charge on electron  (1.6020 x 10-19 C)

 $\therefore$      $n=\frac{Q}{e^{-}}=\frac{38600 C}{1.6020\times 10^{-19}C}=2.4088\times 10^{23}$