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

Answer:

D) 5.36 hours

Explanation:

According to mole concept:

1 mole of an atom contains $6.022\times 10^{23}$ number of particles.

We know that:

Charge on 1 electron = $1.6\times 10^{-19}C$

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 = $10\times 1.6\times 10^{-19}\times 6.022\times 10^{23}=9.6352\times 10^5C$

To calculate the time required, we use the equation:

$I=\frac{q}{t}$

where,

I = current passed = 50.0 A

q = total charge = $9.6352\times 10^5C$

t = time required = ?

Putting values in above equation, we get:

$50.0A=\frac{9.6352\times 10^5C}{t}\\\\t=\frac{9.6352\times 10^5C}{50.0A}=19270.4s$

Converting this into hours, we use the conversion factor:

1 hr = 3600 seconds

So, $19270.4s\times \frac{1hr}{3600s}=5.36hr$

Hence, the amount of time needed is 5.36 hrs.