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.

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"