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((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((1 mol Au)/(289500 coulombs))((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.