Question

A 3.50 amp power supply is used to deposit chromium from a solution of CrCl3. How long will it take to deposit 100.0 grams of chromium?

Answer 1

Step-by-step explanation:

The given data is as follows.

Current (I) = 3.50 amp, Mass deposited = 100.0 g

Molar mass of Cr = 52 g

It is known that 1 faraday of electricity will deposit 1 mole of chromium. As 1 faraday means 96500 C and 1 mole of Cr means 52 g.

Therefore, 100 g of Cr will be deposited by "z" grams of electricity.

`z * 52 g = 96500 * 100 g`

z =
`(96500 * 100 g)/(52 g)`

= 185576.9 C

As we know that, Q = I × t

Hence, putting the given values into the above equation as follows.

Q = I × t

185576.9 C =
`3.50 amp * t`

t = 53021.9 sec

Thus, we can conclude that 100 g of Cr will be deposited in 53021.9 sec.