Respuesta :
Answer:-
Solution:- The equation for the conversion of gold ions to gold metal is as follows:
[tex]Au^3^+(aq)+3e^-\rightarrow Au(s)[/tex]
From this equation, 1 mol of Gold metal is deposited by 3 mole of electrons.
We know that one mol of electron carries one Faraday that is 96500 Coulombs.
So. 3 moles of electrons would be = 3(96500) = 289500 Coulombs
We could calculate the total Coulombs by using the formula:
q = i*t
where, q is the charge in coulombs, i is the current in ampere and t is time in seconds.
q = [tex]23.8 ampere*36.0 min(\frac{60 sec}{1 min})[/tex]
q = 51408 ampere per second
Since, 1 ampere per second = 1 coulomb
So, 51408 ampere per second is same as 51408 coulombs.
Let's use this value of q and the info from balanced equation and molar mass(196.967 g epr mol) of gold to calculate the mass of it being deposited while the current is passed as:
[tex]51408 coulomb(\frac{1 mol Au}{289500 coulombs})(\frac{196.967 g}{1 mol})[/tex]
= 35.0 g Au
So, 35.0 g of gold are deposited on passing 23.8 ampere of current for 36.0 min to a gold solution.
