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

Question

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

Vedantu Content Team · Accepted Answer

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 = 96487CSubstituting 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

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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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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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