
How can you verify Faraday's first law of electrolysis experimentally?
Answer
506.4k+ views
Hint: In order to this question, first we will state the explanation of Faraday’s first law of electrolysis and then we will show the relation between the mass, current and the time, by which we can verify the Faraday's first law of electrolysis experimentally.
Complete step-by-step solution:
Faraday’s – First Law of Electrolysis:- It states that the amount of chemical reaction that takes place at any electrode under the influence of electrical energy is proportionate to the amount of electricity that passes through the electrolyte during electrolysis.
Consider an electrolyte cell with a battery, rheostat, key, and ammeter linked in series. Cleaning, drying, measuring, and finally placing the cathode into the cell. A current, ${I_1}$ , is passed for $t$ seconds. Current is measured with an ammeter. The cathode is taken out, cleaned, dried, and reweighed. While obtaining the mass ${m_1}$ of the deposited material.
The cathode is reinserted into the cell, and a new current, ${I_2}$ , is passed at the same moment. The deposit mass ${m_2}$ is calculated. It has been noted that:
$\because \dfrac{{{m_1}}}{{{m_2}}} = \dfrac{{{I_1}}}{{{I_2}}}$
so mass is directly proportional to the current passed.
i.e.. $m\alpha I$ …….eq(i)
The experiment is performed with the same $I$ current but various periods for \[{t_1}\,and{\text{ }}{t_2}\] . It is noted that if the deposit masses are \[{m_3}{\text{ }}and{\text{ }}{m_4}\] , respectively,
$\because \dfrac{{{m_3}}}{{{m_4}}} = \dfrac{{{t_1}}}{{{t_2}}}$ , So mass is directly proportional to the period;
i.e.. $m\alpha t$ ………..eq(ii)
Now, from equation (i) and (ii):-
$\therefore m\alpha It$
But, as we know $It = q$
So, $m\alpha q$
Hence, the equation proves Faraday’s first law of electrolysis.
Note: The mass of the substance created by electrolysis is proportional to the amount of electricity consumed, according to Faraday's first law. The following questions will provide you more practise and understanding on how to employ Faraday's Laws in calculations.
Complete step-by-step solution:
Faraday’s – First Law of Electrolysis:- It states that the amount of chemical reaction that takes place at any electrode under the influence of electrical energy is proportionate to the amount of electricity that passes through the electrolyte during electrolysis.
Consider an electrolyte cell with a battery, rheostat, key, and ammeter linked in series. Cleaning, drying, measuring, and finally placing the cathode into the cell. A current, ${I_1}$ , is passed for $t$ seconds. Current is measured with an ammeter. The cathode is taken out, cleaned, dried, and reweighed. While obtaining the mass ${m_1}$ of the deposited material.
The cathode is reinserted into the cell, and a new current, ${I_2}$ , is passed at the same moment. The deposit mass ${m_2}$ is calculated. It has been noted that:
$\because \dfrac{{{m_1}}}{{{m_2}}} = \dfrac{{{I_1}}}{{{I_2}}}$
so mass is directly proportional to the current passed.
i.e.. $m\alpha I$ …….eq(i)
The experiment is performed with the same $I$ current but various periods for \[{t_1}\,and{\text{ }}{t_2}\] . It is noted that if the deposit masses are \[{m_3}{\text{ }}and{\text{ }}{m_4}\] , respectively,
$\because \dfrac{{{m_3}}}{{{m_4}}} = \dfrac{{{t_1}}}{{{t_2}}}$ , So mass is directly proportional to the period;
i.e.. $m\alpha t$ ………..eq(ii)
Now, from equation (i) and (ii):-
$\therefore m\alpha It$
But, as we know $It = q$
So, $m\alpha q$
Hence, the equation proves Faraday’s first law of electrolysis.
Note: The mass of the substance created by electrolysis is proportional to the amount of electricity consumed, according to Faraday's first law. The following questions will provide you more practise and understanding on how to employ Faraday's Laws in calculations.
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