
Faraday’s law of electrolysis fails when
A) Temperature is increased.
B) Inert electrodes are used.
C) A mixture of electrolytes is used.
D) In none of these cases.
Answer
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Hint: We can use faraday’s law of electrolysis to check whether the given conditions make the faraday’s law of electrolysis fail. Comparing the equation of the first law of electrolysis may give us a solution for this.
Complete step by step answer:
To check whether the faraday’s law of electrolysis fails under the given situation first we have to define the faraday’s law of electrolysis.The faraday’s first law of electrolysis states that the amount of any substances deposited or liberated at any of the electrodes is directly proportional to the quantity of electricity passed through the electrolyte.
\[W \propto Q\]
Where W is the amount of substance deposited or liberated in grams
And Q is the quantity of electricity in coulombs.
W = ZQ
Z is the constant of proportionality and is known as the electrochemical equivalent of the substance.
We know that,
Q = It
Where I is the current in amperes
And ‘t’ is the time in seconds.
Thus W = ZIt
From this, we understand that the faraday’s first law of electrolysis does not depend on temperature, inert electrodes or the mixture of electrolytes used.
Now we can check the Faraday’s second law of electrolysis
It states that when the same quantity of electricity is passed through solutions of different electrolytes connected in series, the amount of different substances deposited at the electrodes is directly proportional to their equivalent weights.
i.e., $\dfrac{{{W_1}}}{{{W_2}}} = \dfrac{{{E_1}}}{{{E_2}}}$
Where ${W_1}$ and ${W_2}$ are the weights of electrolyte 1 and electrolyte 2 deposited respectively and ${E_1}$ and ${E_2}$ are their respective equivalent weights.
From this also we can see that it does not depend on any of the temperatures, inert electrodes are used or a mixture of electrolytes is used.
Hence option (d) is correct.
Note: The faraday’s law of electrolysis mainly depends only on the amount of electricity passed and the time it is passed. Since the amount of any substance deposited or liberated depends of quantity of electricity Is passed
Complete step by step answer:
To check whether the faraday’s law of electrolysis fails under the given situation first we have to define the faraday’s law of electrolysis.The faraday’s first law of electrolysis states that the amount of any substances deposited or liberated at any of the electrodes is directly proportional to the quantity of electricity passed through the electrolyte.
\[W \propto Q\]
Where W is the amount of substance deposited or liberated in grams
And Q is the quantity of electricity in coulombs.
W = ZQ
Z is the constant of proportionality and is known as the electrochemical equivalent of the substance.
We know that,
Q = It
Where I is the current in amperes
And ‘t’ is the time in seconds.
Thus W = ZIt
From this, we understand that the faraday’s first law of electrolysis does not depend on temperature, inert electrodes or the mixture of electrolytes used.
Now we can check the Faraday’s second law of electrolysis
It states that when the same quantity of electricity is passed through solutions of different electrolytes connected in series, the amount of different substances deposited at the electrodes is directly proportional to their equivalent weights.
i.e., $\dfrac{{{W_1}}}{{{W_2}}} = \dfrac{{{E_1}}}{{{E_2}}}$
Where ${W_1}$ and ${W_2}$ are the weights of electrolyte 1 and electrolyte 2 deposited respectively and ${E_1}$ and ${E_2}$ are their respective equivalent weights.
From this also we can see that it does not depend on any of the temperatures, inert electrodes are used or a mixture of electrolytes is used.
Hence option (d) is correct.
Note: The faraday’s law of electrolysis mainly depends only on the amount of electricity passed and the time it is passed. Since the amount of any substance deposited or liberated depends of quantity of electricity Is passed
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