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?

A) 0.187 hours
B) 0.373 hours
C) 2.68 hours
D) 5.36 hours

Answer 1

D) 5.36 hours

Step-by-step explanation:

According to mole concept:

1 mole of an atom contains
`6.022* 10^(23)` number of particles.

We know that:

Charge on 1 electron =
`1.6* 10^(-19)C`

Also, copper will produce 2 electrons. So, out of 5 moles of copper, 10 moles of electrons will be produced.

So,

Charge on 10 mole of electrons =
`10* 1.6* 10^(-19)* 6.022* 10^(23)=9.6352* 10^5C`

To calculate the time required, we use the equation:

`I=(q)/(t)`

where,

I = current passed = 50.0 A

q = total charge =
`9.6352* 10^5C`

t = time required = ?

Putting values in above equation, we get:

`50.0A=(9.6352* 10^5C)/(t)\\\\t=(9.6352* 10^5C)/(50.0A)=19270.4s`

Converting this into hours, we use the conversion factor:

1 hr = 3600 seconds

So,
`19270.4s* (1hr)/(3600s)=5.36hr`

Hence, the amount of time needed is 5.36 hrs.