
Faraday constant F, Avogadro number N and electronic charge e are related with each other by:
A. $F = \dfrac{N}{e}$
B. $F = \dfrac{e}{N}$
C. $F = N\,e$
D. $F = N\,{e^2}$
Answer
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Hint: In this question, we need to establish a relation between the faraday’s constant, Avogadro’s number and the electronic charge. For this, we will use the basic definition of the Faraday’s constant.
Complete step by step answer:
The Faraday constant F represents the amount of electric charge carried by the one mole of electrons. The symbol F used here is Faraday constant, this symbol has been used due to Michael Faraday’s.Mole is equivalent to the $6.02 \times {10^{23}}$ which is also known as the Avogadro number and is denoted by N.Charge on an electron is given as $1.6 \times {10^{ - 19}}{\text{ coulomb}}$ and is denoted by $e$ so, the charge on N numbers of electrons can be given by unitary method i.e., $Ne$.Now, following the basic definition of the Faraday’s constant which is defined as the amount of electric charge carried by the one mole of electrons, we can say that, $F = Ne$
Hence, option (C) is correct.
Note:Faraday’s applied this reaction occurring at the both electrode in an electrochemical cell separately i.e. to the information of both oxidation and reduction produced, and he applied equally well to galvanic (spontaneous) reaction and to the electrolytic (non-spontaneous or driven) reactions.
Complete step by step answer:
The Faraday constant F represents the amount of electric charge carried by the one mole of electrons. The symbol F used here is Faraday constant, this symbol has been used due to Michael Faraday’s.Mole is equivalent to the $6.02 \times {10^{23}}$ which is also known as the Avogadro number and is denoted by N.Charge on an electron is given as $1.6 \times {10^{ - 19}}{\text{ coulomb}}$ and is denoted by $e$ so, the charge on N numbers of electrons can be given by unitary method i.e., $Ne$.Now, following the basic definition of the Faraday’s constant which is defined as the amount of electric charge carried by the one mole of electrons, we can say that, $F = Ne$
Hence, option (C) is correct.
Note:Faraday’s applied this reaction occurring at the both electrode in an electrochemical cell separately i.e. to the information of both oxidation and reduction produced, and he applied equally well to galvanic (spontaneous) reaction and to the electrolytic (non-spontaneous or driven) reactions.
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