Answer: The amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours
Explanation:
We are given:
Moles of electron = 1 mole
According to mole concept:
1 mole of an atom contains
number of particles.
We know that:
Charge on 1 electron = ![1.6\times 10^{-19}C](https://tex.z-dn.net/?f=1.6%5Ctimes%2010%5E%7B-19%7DC)
Charge on 1 mole of electrons = ![1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C](https://tex.z-dn.net/?f=1.6%5Ctimes%2010%5E%7B-19%7D%5Ctimes%206.022%5Ctimes%2010%5E%7B23%7D%3D96500C)
![Cu^{2+}+2e^-\rightarrow Cu](https://tex.z-dn.net/?f=Cu%5E%7B2%2B%7D%2B2e%5E-%5Crightarrow%20Cu)
is passed to deposit = 1 mole of copper
63.5 g of copper is deposited by = 193000 C
of copper is deposited by =![\frac{193000}{63.5}\times 14000=42551181 C](https://tex.z-dn.net/?f=%5Cfrac%7B193000%7D%7B63.5%7D%5Ctimes%2014000%3D42551181%20C)
To calculate the time required, we use the equation:
![I=\frac{q}{t}](https://tex.z-dn.net/?f=I%3D%5Cfrac%7Bq%7D%7Bt%7D)
where,
I = current passed = 40.0 A
q = total charge = 42551181 C
t = time required = ?
Putting values in above equation, we get:
![40.0=\frac{42551181 C}{t}\\\\t=1063779sec](https://tex.z-dn.net/?f=40.0%3D%5Cfrac%7B42551181%20C%7D%7Bt%7D%5C%5C%5C%5Ct%3D1063779sec)
Converting this into hours, we use the conversion factor:
1 hr = 3600 seconds
So, ![1063779s\times \frac{1hr}{3600s}=295hr](https://tex.z-dn.net/?f=1063779s%5Ctimes%20%5Cfrac%7B1hr%7D%7B3600s%7D%3D295hr)
Hence, the amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours