Question

What mass of gold is produced when 23.8 a of current are passed through a gold solution for 36.0 min ?

Answer 1

Answer:-

Solution:- The equation for the conversion of gold ions to gold metal is as follows:

$Au^3^+(aq)+3e^-\rightarrow Au(s)$

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 = $23.8 ampere*36.0 min(\frac{60 sec}{1 min})$

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:

$51408 coulomb(\frac{1 mol Au}{289500 coulombs})(\frac{196.967 g}{1 mol})$

= 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.