Question

In a chrome-plating apparatus, a power source is used to move electrons through the two half-reactions.

Cr --> Cr³⁺ + 3e^- and

Cr³⁺ + 3e^- --> Cr E= -0.744 V

A. What is the minimum voltage for the power source?

B. While this power source is running, the current runs 5 hours 10 minutes at 1.50 amperes. (1.0 amp = 1.0 coulombs/sec.) Faraday's constant is 96,500 coulombs = 1 mole e^-, How many coulombs of electrons go through the circuit?

C. How many moles of electrons go through the circuit?

D. How many moles of Chromium should be precipitated?

E. How many grams of Chromium could be precipitated?

F. It turns out that the process is only 81.3% efficient. How many grams of Cr are actually precipitated during this process?

Answer 1

Solution:

The computation are expressed as below,

(a) E = E_cathode - E_anode = -0.744 - (-0.42) = -0.342 V

(b) t = 5 hrs 10 mts = 18600 sec

Q = I*t = 1.5*18600 = 27900 C

27900 / 96500 = 0.289 mol e^-

(c) Cr³⁺ + 3e^- ---> Cr

moles of electron = (1 mol / 52 g Cr) *(3 mol e^- / 1 mol) = 0.058 mol e^-

(d) moles of Cr ppt = 3 mol

(e) gram of Cr ppt = 3 mol * 52 g/mol = 156 g of Cr is ppt

(f) for 81.3 % gives the ppt amount = 156*0.813 = 126.83 gm of total amount ppt