A 1.0 L solution of CaF2 was electrolyzed for 14.7 h to give 99.47 g of calcium. Assuming the minimum voltage needed was avalible, what amperage would be needed to complete the electrolysis in the given time

Respuesta :

Answer: The amperage needed to complete the electrolysis in the given time is 9.08 A

Explanation:

We are given:

Moles of electron = 1 mole

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]

Charge on 1 mole of electrons = [tex]1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C[/tex]

moles of calcium =[tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{99.47g}{40g/mol}=2.49moles[/tex]

[tex]Ca^{2+}+2e^-\rightarrow Ca[/tex]

[tex]2\times 96500=193000C[/tex] is required to deposit = 1 mole

Thus 1 mole is deposited by = 193000 C

2.49 moles will be deposited by =[tex]\frac{193000}{1}\times 2.49=480570C[/tex]

To calculate the time required, we use the equation:

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

where,

I = current passed = ?

q = total charge = [tex]480570C[/tex]

t = time required = 14.7 hours =52920 s  ( 1hour=3600s)

Putting values in above equation, we get:

[tex]I=\frac{480570C}{52920s}\\\\I=9.08A[/tex]

Hence, the amperage needed to complete the electrolysis in the given time is 9.08 A

The amount of current needed to complete the electrolysis in the given time is 9.08 A.

What information do we have?

Moles of electron = 1 mole

According to mole concept:

1 mole of an atom contains a number of particles.

We know that:

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

Charge on 1 mole of electrons =  [tex]1.6*10^{-19} C *6.022*10^{-23} =96500 C[/tex]

The moles of calcium = given mass / Molar mass

Moles of calcium = 2.49 moles

Thus 1 mole is deposited by = 193000 C

2.49 moles will be deposited by =193000 * 2.49 = 480570 C

Calculation for time required:

where,

I = current passed = ?

q = total charge = 480570 C

t = time required = 14.7 hours =52920 s  ( 1hour=3600s)

On subsituting the values:

I= 480570 C / 52920 s = 9.08 A

Hence, the Current needed to complete the electrolysis in the given time is 9.08 A

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