
What is the amount of chlorine evolved when 2 ampere of current is passed for 30 minutes in an aqueous solution of NaCl?:
(A) $ 66g $
(B) $ 1.32g $
(C) $ 33g $
(D) $ 99g $
Answer
511.8k+ views
Hint :To calculate the amount evolved in aqueous solution can be calculated using Faraday's first law. As we have given the value of current passed through the solution and also we have time for how long the current is passed.
Formula for Faraday’s first law is-
$\Rightarrow m = Eit $
Complete Step By Step Answer:
We have, current (i) = 2 amp.
and $time = 30\;minutes = 30\; \times \;60 \;seconds$
According to Faraday's First Law-
$\Rightarrow m = Eit $
$\Rightarrow m = \dfrac{Eit}{96500} \to (1)$
$Here, E = \dfrac{\text{Atomic Mass}}{\text{Valency}}$
${\text{Atomic Mass of Chlorine}} = 35.5$
${\text{Valency of Chlorine}} = 1$
${\text{E}} = \dfrac{35.5}{1} = 35.5$
${\text{Atomic Mass of Chlorine}} = 35.5$
By putting all values in equation 1-
$m = \dfrac{35.5 \times 2 \times 30 \times 60}{96500}$
$m = 1.324g$
Amount of Chlorine evolved is 1.324g.
Hence Option (B) is the correct option.
Note :
While calculating the amount of elements using Faraday’s first law, first of all change the time into seconds. We are using 96500 here which is the value of 1 mole electron’s charge that is $[6.023 \times 10^{23}] \times [1.6 \times 10^{-19}] \; Coulombs$.
Formula for Faraday’s first law is-
$\Rightarrow m = Eit $
Complete Step By Step Answer:
We have, current (i) = 2 amp.
and $time = 30\;minutes = 30\; \times \;60 \;seconds$
According to Faraday's First Law-
$\Rightarrow m = Eit $
$\Rightarrow m = \dfrac{Eit}{96500} \to (1)$
$Here, E = \dfrac{\text{Atomic Mass}}{\text{Valency}}$
${\text{Atomic Mass of Chlorine}} = 35.5$
${\text{Valency of Chlorine}} = 1$
${\text{E}} = \dfrac{35.5}{1} = 35.5$
${\text{Atomic Mass of Chlorine}} = 35.5$
By putting all values in equation 1-
$m = \dfrac{35.5 \times 2 \times 30 \times 60}{96500}$
$m = 1.324g$
Amount of Chlorine evolved is 1.324g.
Hence Option (B) is the correct option.
Note :
While calculating the amount of elements using Faraday’s first law, first of all change the time into seconds. We are using 96500 here which is the value of 1 mole electron’s charge that is $[6.023 \times 10^{23}] \times [1.6 \times 10^{-19}] \; Coulombs$.
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