Question

How many amperes of electricity are needed to electrolyze Cu²⁺ to Cu if 1.21 moles of Cu is produced in 7.4 seconds?

THEN

In an electrolytic cell, the electrolysis of an aqueous zinc solution results in the deposition of zinc at the cathode. If a current of 113.5 A is used, how much time will be needed to plate out 0.338 mol Zn(s)?

Answer 1

A)

A = C/s

then

find total charge required

1 mol of Cu = 2 mol of e- in order to get Cu(s)

1.21 mol of Cu requires = 2x1.21 = 2.42 mol of e-

1 mol of e- = 96500 C according to faraday

so

2.42 mol of e- = 96500*2.42 = 233,530 C present

t = 7.4 s

A = C/s = (233,530)C/7.4 s = 31,558.10 A

I = 31,558.10 A

Q2.

In an electrolytic cell, the electrolysis of an aqueous zinc solution results in the deposition of zinc at the cathode. If a current of 113.5 A is used, how much time will be needed to plate out 0.338 mol Zn(s)?

Zn2+ + 2e- = Zn(s)

I = 113.5 A = 113.5 C/s

find itme for 0.338 mol of zinc to form

1 mol ofZn = 2 mol of e-

0.338 mol of Zn = 2x0.338 = 0.676 mol of e-

1 mol of e- = 96500 C (faraday's)

0.676 mol of e- = 0.676*96500 = 65,234 C

now,

I = 113.5 C/s

I = C/t

t = C/I = (65,234)/(113.5)

t = 574.748 seconds

t = 574.748/60 = 9.579 minutes