Electrochemistry 1 Question 40
39. An aqueous solution of $\mathrm{NaCl}$ on electrolysis gives $\mathrm{H} _{2}(g), \mathrm{Cl} _{2}(g)$ and $\mathrm{NaOH}$ according to the reaction.
$ 2 \mathrm{Cl}^{-}(a q)+2 \mathrm{H} _{2} \mathrm{O} \rightleftharpoons 2 \mathrm{OH}^{-}(a q)+\mathrm{H} _{2}(g)+\mathrm{Cl} _{2}(g) $
A direct current of $25 \mathrm{A}$ with a current efficiency of $62 %$ is passed through $20 \mathrm{L}$ of $\mathrm{NaCl}$ solution ( $20 %$ by weight). Write down the reactions taking place at the anode and cathode. How long will it take to produce $1 \mathrm{kg}$ of $\mathrm{Cl} _{2}$ ? What will be the molarity of the solution with respect to hydroxide ion? (Assume no loss due to evaporation)
(1992, 3M)
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Answer:
Correct Answer: 39. ($1.4085 \mathrm{M}$)
Solution:
- At anode
$ 2 \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl} _{2}+2 e^{-} $
At cathode $2 \mathrm{H} _{2} \mathrm{O}+2 e^{-} \longrightarrow \mathrm{H} _{2}+2 \mathrm{OH}^{-}$
$1 \mathrm{kg} \mathrm{Cl} _{2}=\frac{1000}{35.5}$ equivalent of $\mathrm{Cl} _{2}=28.17$ equivalent
$\Rightarrow$ Theoretical electricity requirement $=28.17 \mathrm{F}$
$\because \quad$ Efficiency is only $62 %$
$\therefore$ Electricity requirement (experimental)
$ \begin{array}{rlrl} = & \frac{28.17 \times 100}{62} \mathrm{F} & =45.44 \mathrm{F} \\ \Rightarrow & & 45.44 \times 96500 & =25 t \text { (in second) } \\ \Rightarrow & & t & =48.72 \mathrm{h} \end{array} $
Also, gram equivalent of $\mathrm{HO}^{-}$produced $=28.17$
$\Rightarrow$ Molarity of $\mathrm{HO}^{-}=\frac{28.17}{20}=1.4085 \mathrm{M}$
