The number of electrons delivered at the cathode during electrolysis by a current of `1` ampere in 60 seconds is (charger on electron `= 1.60 xx 10^(-19)C`)

Options

A. `6xx10^(23)`
B. `6xx10^(20)`
C. `3.75xx10^(20)`
D. `7.48xx10^(23)`

Answer

C

Explanation

(c ) From Faraday's firs law of electrolysis,
`(w)/( E)=(it)/(96500)`….(i)
Given, i=1A,t=60s
Putting these values in Eq. (i) , we get
`(w)/( E)=(1xx60)/(96500) or (w)/( E)=(6)/(9650)`
= Number of moles of electrons
`:.` Number of electrons `=(6)/(9650)xx6.022xx10^(23)=3.75xx10^(20)`