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How much charge in Faraday is required for the reduction of 1 mole of ${\text{A}}{{\text{g}}^{\text{ + }}}$to Ag?(A) ${\text{19}}{\text{.29 x 1}}{{\text{0}}^{\text{4}}}$ C(B) 96487 C(C) ${\text{38}}{\text{.59 x 1}}{{\text{0}}^{\text{4}}}$ C(D) 4824 C

Last updated date: 11th Jun 2024
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Hint: Faraday is a unit of electricity and calculated from number of electrons and magnitude of charge. Reduction process is a process where electrons are gained.

Formula used: The charge, q can be found from the formula ${\text{q = n x F}}$.

Complete step by step answer: It is given that 1 mole of ${\text{A}}{{\text{g}}^{\text{ + }}}$to Ag is used for reduction.
To find his charge in faraday.
We know that charge, q can be calculated from the formula ${\text{q = n x F}}$
Where n is the number of electrons and F is faraday.
The reduction process where ${\text{A}}{{\text{g}}^{\text{ + }}}$ gain one electron to become Ag can be represented as below:
${\text{A}}{{\text{g}}^{\text{ + }}}{\text{ + }}{{\text{e}}^ - } \to {\text{Ag}}$
It can be seen that 1 mole of ${\text{A}}{{\text{g}}^{\text{ + }}}$requires 1 electron to get reduced to Ag.
Therefore, n =1. We know that 1 faraday = 96487C
Substituting in the formula, we get,
${\text{q = n x F}} \\ \Rightarrow {\text{q = 1 x 96487}} \\ \Rightarrow {\text{q = 96487 C}} \\$
Thus. 96487 C of charge is required for the reduction of 1 mole of ${\text{A}}{{\text{g}}^{\text{ + }}}$to Ag.

So, the correct option is B.

Additional information: The Faraday constant named after the Michael Faraday is denoted by the symbol F. It is defined as the charge of one mole of electrons.
F = e (1mol) = ${\text{1}}{\text{.6 x 1}}{{\text{0}}^{ - 19}}{\text{ x 6}}{\text{.023 x 1}}{{\text{0}}^{{\text{23}}}} = 96487{\text{ C}}$
Where e is the magnitude of charge equal to ${\text{1}}{\text{.6 x 1}}{{\text{0}}^{ - 19}}$ and 1 mole of electrons has Avogadro number of electrons equal to ${\text{6}}{\text{.023 x 1}}{{\text{0}}^{{\text{23}}}}$.

Note: We need to know the following fundamentals values that are useful to solve this problem:
(i) Magnitude of charge, q = ${\text{1}}{\text{.6 x 1}}{{\text{0}}^{ - 19}}$C
(ii) Avogadro number,${\text{}}{{\text{N}}_{\text{A}}}$= ${\text{6}}{\text{.023 x 1}}{{\text{0}}^{{\text{23}}}}$
(iii) 1 Faraday = 96487 C or 96500 C
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