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Question: How much charge in Faraday is required for the reduction of 1 mole of \[{\text{A}}{{\text{g}}^{\text...

How much charge in Faraday is required for the reduction of 1 mole of Ag + {\text{A}}{{\text{g}}^{\text{ + }}}to Ag?
(A) 19.29 x 104{\text{19}}{\text{.29 x 1}}{{\text{0}}^{\text{4}}} C
(B) 96487 C
(C) 38.59 x 104{\text{38}}{\text{.59 x 1}}{{\text{0}}^{\text{4}}} C
(D) 4824 C

Explanation

Solution

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 q = n x F{\text{q = n x F}}.

Complete step by step answer: It is given that 1 mole of Ag + {\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 q = n x F{\text{q = n x F}}
Where n is the number of electrons and F is faraday.
The reduction process where Ag + {\text{A}}{{\text{g}}^{\text{ + }}} gain one electron to become Ag can be represented as below:
Ag +  + e−→Ag{\text{A}}{{\text{g}}^{\text{ + }}}{\text{ + }}{{\text{e}}^ - } \to {\text{Ag}}
It can be seen that 1 mole of Ag + {\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,
q = n x F ⇒q = 1 x 96487 ⇒q = 96487 C  {\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 Ag + {\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) = 1.6 x 10−19 x 6.023 x 1023=96487 C{\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 1.6 x 10−19{\text{1}}{\text{.6 x 1}}{{\text{0}}^{ - 19}} and 1 mole of electrons has Avogadro number of electrons equal to 6.023 x 1023{\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 = 1.6 x 10−19{\text{1}}{\text{.6 x 1}}{{\text{0}}^{ - 19}}C
(ii) Avogadro number,NA{\text{}}{{\text{N}}_{\text{A}}}= 6.023 x 1023{\text{6}}{\text{.023 x 1}}{{\text{0}}^{{\text{23}}}}
(iii) 1 Faraday = 96487 C or 96500 C