Question

Three electrolytic cells are connected in a series. The electrolytes in the cells are aqueous copper(II) sulfate, gold(III) sulfate, and silver nitrate. A current of 2.26 A is applied and after some time 1.71 g Cu is deposited.

(a) How long was the current applied?

(b) What mass of gold was deposited?

(c) What mass of silver was deposited?

Answer 1

a)

Electrolysis equation is:

Cu2+ + 2e- ------> Cu

1 mol of Cu requires 2 mol of electron

1 mol of electron = 96485 C

So,1 mol of Cu requires 192970 C

let us calculate mol of element deposited:

molar mass of Cu = 63.55 g/mol

use:

number of mol of Cu, n = mass of Cu/molar mass of Cu

= 1.71/63.55

= 2.691*10^-2 mol

total charge = mol of element deposited * charge required for 1 mol

= 2.691*10^-2*1.93*10^5

= 5.192*10^3 C

use:

time = Q/i

= 5.192*10^3/2.26

= 2.298*10^3 seconds

= 2.298*10^3/60 min

= 38.29 min

Answer: 38.3 min

b)

Electrolysis equation is:

Au3+ + 3e- ------> Au

1 mol of Au requires 3 mol of electron

1 mol of electron = 96485 C

So,1 mol of Au requires 289455 C

let us calculate the charge passed:

t = 38.29 min = 38.29*60 s = 2.297*10^3 s

time, t = 2.297*10^3s

Q = I*t

= 2.26A * 2.297*10^3s

= 5.191*10^3 C

mol of Au plated = 5.191*10^3/289455 = 1.793*10^-2 mol

Molar mass of Au = 1.97*10^2 g/mol

mass of Au = number of mol * molar mass

= 1.793*10^-2 * 1.97*10^2

= 3.533 g

Answer: 3.53 g

c)

Electrolysis equation is:

Ag1+ + 1e- ------> Ag

1 mol of Ag requires 1 mol of electron

1 mol of electron = 96485 C

So,1 mol of Ag requires 96485 C

let us calculate the charge passed:

t = 38.29 min = 38.29*60 s = 2.297*10^3 s

time, t = 2.297*10^3s

Q = I*t

= 2.26A * 2.297*10^3s

= 5.191*10^3 C

mol of Ag plated = 5.191*10^3/96485 = 5.38*10^-2 mol

Molar mass of Ag = 1.079*10^2 g/mol

mass of Ag = number of mol * molar mass

= 5.38*10^-2 * 1.079*10^2

= 5.805 g

Answer: 5.80 g