[maras924]

Hello

I just got a little bit stuck with some homework at Uni.
Problem description:
An aqueous solution contains cadmium ions Cd2+ at a concentration of 2.5 x 10^-3 mole/dm^3
Cadmium hydroxide is to be precipitated from this solution by raising and maintaining the solution PH at 11.2.
I need to calculate:
a)The % recovery that could be achieved by precipitating cadmium as Cd(OH)2 from solution at pH = 11.2
b)If the same amount of Cd^2+ was to be removed by electrolytic deposition, how much electricity would be needed?
c)The electrode potential which would be required to initiate deposition of cadmium in question(b)

My idea for question a) is to use molar mass of cadmium and concentration of the cadmium ions to calculate the mass of cadmium in solution and then convert that mass as percentage of molar mass of cadmium. So I got 0.281 g Cd and 0.25% recovery...
For question b) not sure but I think I need to use 96500C per 1 gram of substance and calculate the electricity per 0.281g?
c) Should I use Nernst equation or relate SEP of Cd which is -0.40V to my solution, my answer was -0.001V.

Thanks for any help
Marek

[Borek]

Your recovery idea is off - use solubility product to calculate how much cadmium will be left for a given concentration of OH^-.

96500 C is not per gram of substance, but per equivalent of substance. Check Faraday's law of electrolysis.

Yes for using Nernst equation. No idea about numbers.

[maras924]

Well, the [OH-] concentration I calculated from ionic product of water was 1.58 x 10^-3 mol.dm^-3. Then I used solubility product equation to get [Cd+] left - I got 1.79 x 10^-9 mol.dm^-3. So how do I get % of recovery?

Cheers
Marek

[Borek]

Assume 1L of solution - what was mass dissolved before, what is mass dissolved after?

[maras924]

Thanks, got it now. I worked it out as 99.99993 % recovery.
About the next task I had, 2 Faraday should be equivalent to 1 mole of Cd2+. So the electricity needed should be 0.99993 x 2?

[Borek]

No, how many moles of Cd were precipitated? Note: you have not listed volume - without volume you can't answer the question, unless you will give the answer as charge per 1L of solution.

[maras924]

I'm sorry but I can't follow what volume do you mean? 1 mole of Cd is 112.4g, then because the solution is 2.5x10^-3 mol.dm^-3 so it contains 0.281g per 1L(1dm^3)... of cadmium. I think that 2.499825x10^-3 moles precipitated...

[maras924]

I'm sorry but I can't follow what volume do you mean? 1 mole of Cd is 112.4g, then because the solution is 2.5x10^-3 mol.dm^-3 so it contains 0.281g per 1L(1dm^3)... of cadmium. I think that 2.499825x10^-3 moles precipitated... Is the formula Q=zmF appropriate to get the charge value? Q=2*2.499825x10^-3 moles * 96500 = 482.466 Coulomb?

[Borek]

As you have already stated - you can calculate mass of cadmium precipitated PER LITRE of solution. That's the volume I am referring to.

Compare:

How many moles of ammonia in 100 mL of 1M solution?

How many moles of ammonia in 1M solution?

Can you give identical answers to both questions? Do you see how it is related to the question you are solving?

[maras924]

For 100 ml of 1M solution you need 0.1 mole of ammonia
In 1M solution there is 1 mole of ammonia

So for my problem the solution with Cadmium will be 0.0025M I hope so it is 0.0025 moles of Cadmium per litre of solution.
How to calculate now the charge? Is it Q=z*m*F; where z is number if electrons taken by Cadmium, m - number of moles, F - Faraday's constant?

[Borek]

So for my problem the solution with Cadmium will be 0.0025M I hope so it is 0.0025 moles of Cadmium per litre of solution.

Just remember you must answer in coulombs/L, not just in coulombs.

Q=z*m*F; where z is number if electrons taken by Cadmium, m - number of moles, F - Faraday's constant?

Yes.

[maras924]

Thanks for help, the last question I got worked out like this: To get an electrode potential to initiate the deposition of cadmium, the Nernst equation is used. Data: Temp. given were 25C. I'm not sure about value 0.999993 which is supposed to be (according to my lecturer)actual concentration of the reactive metal M in the electrode as a mole fraction... Does my calculation look OK?
 
E = -0.40 + (8.314*298/2*96485) x ln(2.50 x 10-3/0.999993) = -0.40 + (2477.572/192970)*0.91629 = -0.40 + (0.012839*0.91629) = -0.388V

[maras924]

Regarding to my last post, Is my calculation for the electrode potential right?