Question

How long must a constant current of 50.0 a be passed through an electrolytic cell containing aqueous cu2+ ions to produce 5.00 moles of copper metal? how long must a constant current of 50.0 a be passed through an electrolytic cell containing aqueous cu2+ ions to produce 5.00 moles of copper metal? 5.36 hours 2.68 hours 0.373 hours 0.187 hours?

Answer 1

When the reaction equation is:

Cu 2+(aq)  +  2 e- → Cu(S)

from the reaction equation, we can see that each 1 mole of Cu 2+ need 2 moles e-

So, the moles of e- when we have 5 moles copper = 5* 2 = 10 moles

by using this formula we  can get the time directly:

 time(per sec) = moles of e- * faraday's constant * current

                        = 10 moles * 96500 coul/mol * 1 sec / 50 coul

                        = 19300 sec 

                        = 19300 /(60*60) = 5.36  Hours 

∴ the answer will be a) 5.36 Hours

Answer 2

Answer: 5.36 hours

Explanation:

$Q=I\times t$

where Q= quantity of electricity in coloumbs

I = current in amperes = 50A

t= time in seconds = ?

$Cu^{2+}+2e^{-1}\rightarrow Cu$

$96500\times 2=193000Coloumb$ of electricity deposits 1 mole of Cu.

1 mole of Copper is produced by = 193000 Coloumb

5 moles of Copper is produced by=$\frac{193000}{1}\times 5=965000C$

$965000=50\times t$

$t=19300s$    (1 hour=3600sec)

$t=5.36hours$