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In producing chlorine through electrolysis 100 watt power at 125 V is being consumed. How much chlorine per minute is liberated ? E.C.E. of chlorine is $0.367 \times 10^{-6}$ kg/coulomb :-
A
17.6 mg
B
21.3 mg
C
24.3 mg
D
13.6 mg
Detailed Solution
By Faraday's first law of electrolysis, the mass liberated is $m = ZIt$, where $Z$ is the electrochemical equivalent.
Power consumed: $P = VI \Rightarrow I = \dfrac{P}{V} = \dfrac{100}{125} = 0.8$ A
Time: $t = 1$ minute $= 60$ s
$m = ZIt = (Z)\left(\dfrac{P}{V}\right)(t) = (0.367 \times 10^{-6})\left(\dfrac{100}{125}\right)(60)$
$m = 0.367 \times 10^{-6} \times 0.8 \times 60 = 0.367 \times 10^{-6} \times 48$
$m = 1.76 \times 10^{-5}$ kg
$m = 17.6$ mg
Power consumed: $P = VI \Rightarrow I = \dfrac{P}{V} = \dfrac{100}{125} = 0.8$ A
Time: $t = 1$ minute $= 60$ s
$m = ZIt = (Z)\left(\dfrac{P}{V}\right)(t) = (0.367 \times 10^{-6})\left(\dfrac{100}{125}\right)(60)$
$m = 0.367 \times 10^{-6} \times 0.8 \times 60 = 0.367 \times 10^{-6} \times 48$
$m = 1.76 \times 10^{-5}$ kg
$m = 17.6$ mg