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Standard XII

Chemistry

Faraday's First Law

An aqueous so...

Question

An aqueous solution of sodium chloride on electroysis 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 of $25$ ampere with a current efficiency $62$ % is passed through $20 L$ of $N a C l$ solution $(20$ % by mass). Write down the reactions taking place at the anode and cathode. How long will it take to produce $1 k g$ of $C l_{2}$ ? Assume no loss due to evaporation.

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Solution

Reactions at anode and cathode are:
$2 C l^{-} \to C l_{2} + 2 e^{-}$ (at anode)
$2 H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}$ (at cathode)
$1 k g$ of $C l_{2} = \frac{1000}{71.0} = 14.08 m o l e$
Charge to produce one mole of $C l_{2} = 2 \times 96500 c o u l o m b$
Charge to produce $14.08$ mole of $C l_{2} = 2 \times 96500 \times 14.08 c o u l o m b$
Effective current $= \frac{62}{100} \times 25.0 = 15.5 a m p e r e$
$T i m e = \frac{C h a r g e}{C u r r e n t} = \frac{2 \times 96500 \times 14.08}{15.5}$
$- 175318.7 s e c o n d = 48.699 h o u r$
$O H^{-}$ ions produced $= 2 \times m o l e s o f C l_{2}$
$= 3 \times 14.08 = 28.16 m o l e$
$M o l a r i t y = \frac{M o l e}{V o l u m e} = \frac{28.16}{20} = 1.408 M$ .

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