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Question: During electrolysis of acidified water, \({O_2}\) gas is formed at the anode. To produce \({O_2}\) g...

During electrolysis of acidified water, O2{O_2} gas is formed at the anode. To produce O2{O_2} gas at the anode at the rate of 0.224 ml0.224{\text{ ml}} per second at STPSTP , current passed is:

Explanation

Solution

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:
H2O→H++12O2+2e−{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 O2{O_2}:
Weightrq.  weight=I×t96500−−−(1)\dfrac{{Weight}}{{rq.\;weight}} = \dfrac{{I \times t}}{{96500}} - - - \left( 1 \right)
Any metal or any molecule.
We know that,
22400ml→32g22400ml \to 32g of O2{O_2} ,
[by STPSTP condition]
So,
We have to tell this
If 0.22ml→?0.22ml \to ? (x)\left( x \right)
Then how much O2{O_2} is deposited.
So, by unitary method.
That is x=0.224×3222400x = \dfrac{{0.224 \times 32}}{{22400}}
∴\therefore Weight of O2{O_2} deposited =0.00032g = 0.00032g
So substituting values in e.q-−−(1) - - \left( 1 \right)
(0.00032)weight/n−factor=I×1sec⁡96500\dfrac{{\left( {0.00032} \right)}}{{weight/n - factor}} = \dfrac{{I \times 1\sec }}{{96500}}
5.6×103ml=96500C5.6 \times {10^3}ml = 96500C
So, eq. weight of O2=32{O_2} = 32 (molecular weight)
And n−n - factor of O2=4{O_2} = 4
0.0323214=I×1  sec⁡96500\dfrac{{0.032}}{{3214}} = \dfrac{{I \times 1\;\sec }}{{96500}}
By showing the above.
Therefore, current; I=3.86AI = 3.86A
Alternative method:
As we know
IF =96500C=  1eq  of  O2  =22.44=5.6L = 96500C = \;1eq\;of\;{O_{2\; = }}\dfrac{{22.4}}{4} = 5.6L

5.6×103ml=96500C5.6 \times {10^3}ml = 96500C
O2=96500×0.2245.6×103ml.{O_2} = \dfrac{{96500 \times 0.224}}{{5.6 \times {{10}^3}ml.}}
O2=  3.86A{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.