Chemistry | MHT CET 2020 | Discussion Page

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

$6.642\times 10^{25}$

$1.505\times 10^{25}$

$6.022\times 10^{25}$

$2.4088\times 10^{25}$

Option D

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)

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

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