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)

Show Answer

Answer:

Correct Answer: 39. ($1.4085 \mathrm{M}$)

Solution:

  1. 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 $\mathrm{HO}^{-}=\frac{28.17}{20}=1.4085 \mathrm{M}$



NCERT Chapter Video Solution

Dual Pane