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

Respuesta :

Answer:

D) 5.36 hours

Explanation:

According to mole concept:

1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of particles.

We know that:

Charge on 1 electron = [tex]1.6\times 10^{-19}C[/tex]

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 = [tex]10\times 1.6\times 10^{-19}\times 6.022\times 10^{23}=9.6352\times 10^5C[/tex]

To calculate the time required, we use the equation:

[tex]I=\frac{q}{t}[/tex]

where,

I = current passed = 50.0 A

q = total charge = [tex]9.6352\times 10^5C[/tex]

t = time required = ?

Putting values in above equation, we get:

[tex]50.0A=\frac{9.6352\times 10^5C}{t}\\\\t=\frac{9.6352\times 10^5C}{50.0A}=19270.4s[/tex]

Converting this into hours, we use the conversion factor:

1 hr = 3600 seconds

So, [tex]19270.4s\times \frac{1hr}{3600s}=5.36hr[/tex]

Hence, the amount of time needed is 5.36 hrs.

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