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Faraday's Second Law

An aqueous so...

Question

An aqueous solution of NaCl on elctrolysis gives $H_{2} (g)$ , $C l_{2} (g)$ , and $N a O H$ according to the reaction
$2 C l^{⊖} (a q) + 2 H_{2} O \to 2 ⊖ O H (a q) + H_{2} (g) + C l_{2} (g)$
A Direct current (DC) of 25 A with a current efficiency of 62 % is passed through 20 L of NaCl solution (20% by weight). The molarity (M) of the solution with respect to hydroxide ion will be X. The value of X is ( asssume no loss due to evaporation):
(write your answer to nearest integer)

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Solution

(a) Following reaction takes place at anode and cathode
At anode :
$2 C l^{⊖} \to C l_{2} + 2 e^{-}$
At cathode : $2 H_{2} O + 2 e^{-} \to 2 ⊖ O H + H_{2}$
(b) Weight of $C l_{2} = 10^{3} g$
Z for $C l_{2} = \frac{35.5}{96500}$ $(∵ \frac{E q u i v a l e n t w e i g h t}{96500} = Z)$
Current efficiency = 62 % = 0.62
Current passed (I) = 25 5 0.62 A
According to first law of Faraday,
$M = Z \times I t$
$10^{3} = \frac{35.5}{96500} \times 25 \times 0.62 \times t$
T = 175374.83 S
Or t = 48.71 h
(c) Equivalent of $⊖ O H$ formed = Equivalent

of $C l_{2}$ formed = $\frac{10^{3}}{35.5} = 28.17$
Equivalent weight and mlecular weight of $⊖ O H$ ions are equal.
So, moles of $⊖ O H$ formed = 28.17
$∴ [⊖ O H] = \frac{M o l e s}{V o l u m e (i n l i t e r)} = \frac{28.17}{20} = 1.408 M$

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