A technician plates a faucet with 2.68 g of Cr metal by electrolysis of aqueous Cr2 (SO4) 3. If 15.2 min is allowed for the plating, what current is needed? Use 96500 C/mol e - for the Faraday constant. Enter a number to 2 decimal places.

16.36A

Explanation:

We'll begin by writing a balanced dissociation equation of aqueous Cr2 (SO4) 3. This is illustrated below:

Cr2 (SO4) 3 - > 2Cr^3 + 3 (SO4) ^2-

From the above, we can see that Cr is trivalent.

Next, let us determine the number of faraday needed to deposit metallic Cr. This is illustrated below:

Cr^3 + 3e - - > Cr

From the above equation, 3 faradays are needed to deposit metallic Cr

1 faraday = 96500C

Therefore, 3 faraday = 3 x 96500C = 289500C.

Molar Mass of Cr = 52g/mol

Now let us determine the quantity of electricity needed for 2.68g of Cr metal

This is shown below:

52g of Cr required 289500C.

Therefore, 2.68g of Cr will require = (2.68 x 289500) / 52 = 14920.38C

Now, with this quantity of electricity (i. e 14920.38C), we can easily calculate the current needed for the process. This is illustrated below:

Q (quantity of electricity) = 14920.38C

t (time) = 15.2mins = 15.2 x 60 = 912secs

I (current) = ?

Apply the equation Q = It

Q = It

14920.38 = I x 912

Divide both side by 912

I = 14920.38/912

I = 16.36A

Therefore, a current of 16.36A is needed for the process.