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Standard VII

Chemistry

Faraday's Second Law

An aqueous so...

Question

An aqueous solution of NaCl on electrolysis gives $H_{2} (g), C l_{2} (g)$ and $N a O H$ according to the reaction :
$2 C l^{-} (a q) + 2 H_{2} O (l) ⟶ 2 O H^{-} (a q) + H_{2} (g) + C l_{2} (g)$
A direct current of 25 amperes with a current efficiency of 62% is passed. Calculate the time required to produce 1 kg $C l_{2} (g)$ .

24.66 h r

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48.7 h r

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96.3 h r

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1.2 h r

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Solution

The correct option is B $48.7 h r$
Moles of chlorine in 1 kg are:
$Moles (n_{C l_{2}}) = \frac{Mass}{Molecular mass} n_{C l_{2}} = \frac{1000}{71} = 14.08 m o l$

Reaction at anode:
$2 C l^{-} (a q) + 2 e^{-} ⇌ C l_{2} (g)$
From the above reaction,
$Charge required to produce 1 mol of C l_{2} = 2 F$
$Charge required to produce 14.08 mol of C l_{2} = 2 F \times 14.08 = 28.16 F = Q$

$Current efficiency = \frac{Current actually used in electricity}{Current passed}$
$0.62 = \frac{Current actually used in electricity (i)}{25}$
$i = 0.25 \times 62 = 15.5 A$
$Time (t) = \frac{Q}{i} t = \frac{28.16 \times 96500}{15.5 \times 60 \times 60} h r \approx 48.7 h r$

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An aqueous solution of NaCl on electrolysis gives H2(g),Cl2(g) and NaOH according to the reaction : 2Cl−(aq)+2H2O(l)⟶2OH−(aq)+H2(g)+Cl2(g) A direct current of 25 amperes with a current efficiency of 62% is passed. Calculate the time required to produce 1 kg Cl2(g).