Question

The quantity of electricity required to liberate112cc of hydrogen at STP from acidified water is
A) ${\text{965 C}}$
B) $9650{\text{ }}C$
C) ${\text{96500 C}}$
D) $4825{\text{ }}C$

Answer 1

If an electrolytic solution consists of more than two ions and the electrolysis is done, it is observed that all the ions are not discharged at the electrodes simultaneously but certain ions are liberated at the electrodes in preference to others. This is explained by preferential discharge theory. It states that if more than one type of ions are attracted towards a particular electrode, then the one discharged is the ion which requires least energy.
The potential at which the ion is discharged or deposited on the appropriate electrode is termed the discharge or deposition potential. The discharge potential of ${H^ + }$ ions is. lower than $N{a^ + }$ ions when platinum or most of the other metals are used as cathodes.

Complete step by step solution:
$2{H^ + } + 2{e^ - } \to {H_2}$
Therefore, 2 faradays of electricity is required for the production of one mole of ${H_2}$ gas.
$2 \times 96500\,cc\,$ of electricity is required for the production of $22.4\,l\,of\,{H_2}\,gas$
$1000\,cc = 1\,l$
$112\,cc = 0.112\,l$
For ${\text{22}}{\text{.4 }}\,l$ we require ${{2 \times\text 96500}}\,C$ of electricity
So, for $0.112\,l$ of hydrogen gas we required the electricity = $\dfrac{{193000 \times 0.112}}{{22.4}}$
$\therefore$ We require 965 C of electricity
Hence, the quantity of charge required = $965\,C$

**Hence, option “A” is correct

Note: **
The solution .of sodium chloride besides $Na{\text{ }} + {\text{ }}and{\text{ }}C{l^ - }$ ions possesses ${H^ + }{\text{ }}and{\text{ }}O{H^ - }$ ions due to ionisation of water. However, the number is small as water is a weak electrolyte. When potential difference is established across the two electrodes$,{\text{ }}N{a^ + }{\text{ }}and{\text{ }}{H^ + }$ ions move" towards cathode and $C{l^{ - {\text{ }}}}and{\text{ }}O{H^ - }{\text{ }}ions$ move towards anode. At cathode ${H^ + }$ ions are· discharged in preference to $N{a^ + }$ ions as the discharge potential of ${H^ + }$ ions is lower than $N{a^ + }$ ions. Similarly at anode, $C{l^ - }$ ions are discharged in preference to $O{H^ - }$ ions.