2 Questions

[Mr. Study]

Hey guys, these two questions are really bugging me.

1. The overall chemical equation for the reaction that occurs during the electrolysis of concentrated aqueous sodium chloride solutions in the membrane cell is:
                                                                         
What volume of chlorine gas at SLC would be produced every hour, in a cell that has a current of 50,000A passed through it?

2. An electrolytic cell, containing 250 ml of a 0.20M aqueous lead nitrate solution had a current of 1.46 A passed through it for 50 minutes, using carbon electrodes. What is the concentration of lead ions in the solution, after the 50 minutes?

These questions are Multichoice, and from the Lisachem book.

Thanks.

[nisha]

This is how I would do q1:

since Q=It
Therefore, Q=50000*(60*60) as time is in seconds
Q=1.8 *10^8 C

therefore using the equation, Q=n(e-)* F
n(e-)=1.8*10^8/96500
=1865.28 mol

Therefore, n(Cl2(g))=0.5*n(e-)
=932.64mol
(By using the data booklet and obtaining the reduction reaction for chlorine)
v(Cl2)=24.5*932.64mol
=22849.74L

Seems a bit big, but correct me if I'm wrong.

[Hancock]

Q2. In solution we have: H20, Pb^2+, N03 (don't really care about this one)

So, the reaction that occurs in one of the electrodes will be Pb2+(aq) + 2e- makes Pb(s)

n(e-) = Q/F = It / F = (1.46)(50*60) / (96500) = 0.045388 mol

n(Pb) = n(Pb2+ used up) = 0.5n(e-) = 0.022694 mol

n(Pb2+) initially = CV = 0.25*0.2 = 0.05 mol
n(Pb2+) final = 0.05 - 0.022694 = 0.0273 mol

[Pb2+] final = n/V = 0.0273 / 0.250 = 0.1092 M = 0.11 M

Tell me if I made a mistake. Cheers.

[Hancock]

Yes, we did do the 2 separate questions at the same time. *Fist bump*