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Standard XII

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 NaOH according to the reaction
$2 C l_{(a q)}^{-} + 2 H_{2} O ⇌ 2 O H_{(a q)}^{-} + H_{2 (g)} + C l_{2 (g)}$
A direct current of 25 amperes with a current efficiency of 62% is passed through 20 liters of NaCl solution (20% by weight). Write down the reactions taking place at the anode and the cathode. How long will it take to produce 1 kg of $C l_{2}$ ? in hrs. (write the value to the nearest integer).

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Solution

$2 N a C l ⇌ 2 N a^{+} + 2 C l^{-}$
$2 H_{2} O ⇌ 2 H^{+} + 2 O H^{-}$
At anode:
$2 C l^{-} \to C l_{2} + 2 e^{-}$
At cathode :
$2 H^{+} + 2 e^{-} \to H_{2}$
Net reaction (8 5 5:)
$2 N a C l + 2 H_{2} O ⇌ 2 O H^{-} + H_{2} + C l_{2} + 2 N a^{+}$
$1000 = \frac{35.5}{96500} \times 25.62 \times t$
$t = 175374.83 s e c = 48.74 h r$ .
moles of $O H^{-} = \frac{1000}{71} \times 2$
volume $= 20 L$
$[O H^{-}] = \frac{2000}{71 \times 20} = 1.408 M$

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2NaCl⇌2Na++2Cl−2H2O⇌2H++2OH−At anode:2Cl−→Cl2+2e−At cathode :2H++2e−→H2Net reaction (8 5 5:)2NaCl+2H2O⇌2OH−+H2+Cl2+2Na+1000=35.596500×25.62×tt=175374.83sec=48.74hr.moles of OH−=100071×2volume =20L[OH−]=200071×20=1.408M