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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

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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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