Question

During electrolysis of acidified water, ${O_2}$ gas is formed at the anode. To produce ${O_2}$ gas at the anode at the rate of $0.224{\text{ ml}}$ per second at $STP$ , current passed is:

Answer 1

Hint:
As we know that electrolysis is very important as a stage in the separation of elements from naturally occurring sources like we can use ones using an electrolytic cell. And in this question we have to tell how much current passed.

Complete step by step answer:
Firstly,
Electrolysis is done of acidified water like electrolysis of --- is done at anode.
So following is the chemical reaction at anode:
${H_2}O \to {H^ + } + \dfrac{1}{2}{O_2} + 2{e^ - }$
As we have to pass out the current, so by faraday's law-
As we know that, when some quantity of electricity is passed through solutions of different electrolytes connected in series, the mass of substances produced at the electrode is directly proportional to the equivalent weights.
So talking about equivalent of ${O_2}$:
$\dfrac{{Weight}}{{rq.\;weight}} = \dfrac{{I \times t}}{{96500}} - - - \left( 1 \right)$
Any metal or any molecule.
We know that,
$22400ml \to 32g$ of ${O_2}$ ,
[by $STP$ condition]
So,
We have to tell this
If $0.22ml \to ?$ $\left( x \right)$
Then how much ${O_2}$ is deposited.
So, by unitary method.
That is $x = \dfrac{{0.224 \times 32}}{{22400}}$
$\therefore $ Weight of ${O_2}$ deposited $ = 0.00032g$
So substituting values in e.q-$ - - \left( 1 \right)$
$\dfrac{{\left( {0.00032} \right)}}{{weight/n - factor}} = \dfrac{{I \times 1\sec }}{{96500}}$
$5.6 \times {10^3}ml = 96500C$
So, eq. weight of ${O_2} = 32$ (molecular weight)
And $n - $ factor of ${O_2} = 4$
$\dfrac{{0.032}}{{3214}} = \dfrac{{I \times 1\;\sec }}{{96500}}$
By showing the above.
Therefore, current; $I = 3.86A$
Alternative method:
As we know
IF $ = 96500C = \;1eq\;of\;{O_{2\; = }}\dfrac{{22.4}}{4} = 5.6L$

$5.6 \times {10^3}ml = 96500C$
${O_2} = \dfrac{{96500 \times 0.224}}{{5.6 \times {{10}^3}ml.}}$
${O_2} = \;3.86A$
Hence, the correct option is D.

Note: When an electric current is passed through acidified water, it decomposes to five hydrogen as well as oxygen gas.
Moreover, hydrogen gas is obtained at the cathode, and the oxygen gas is obtained at the anode. Furthermore, electrolysis is used in metallurgical processes, like extraction and purification of metals from ores.