Sciencemadness Discussion Board - Yield 136% while making Fe2O3 by electrolysis

Yield 136% while making Fe2O3 by electrolysis

Diachrynic - 6-2-2018 at 12:44

So I made some Fe₂O₃ but the yield was way to high.

I did wash it, so I can't figure out the reason.

Following setup:

Two sources of iron (steel I suppose) were used as kathode and anode in a solution of 175 g NaCl in 1.3 l H₂O for 320 minutes with a starting power of 2.8 V / 2 A which had risen to 5.1 V / 2 A when I stopped it. Bubbles at the kathode, no bubbles at the anode.
The anode lost 11.08 g of material.

Reactions:

Kathode (-)
2 H₂O + 2 e^- → H₂ + 2 OH^-

Anode (+)
Fe → Fe²⁺ + 2 e^-

The iron undergoes more reactions.

Fe²⁺ + 2 OH^- → Fe(OH)₂
4 Fe(OH)₂ + O₂ → 4 FeO(OH) + 2 H₂O

I washed the stuff that settled to the bottom via repeated dilution and decantation, 5 l each, 2 times (and with a smaller volume a third time).
All that settled on the bottom was first boiled to dryness and then heated with a blowtorch.

2 FeO(OH) → Fe₂O₃ + H₂O
Fe(OH)₂ → FeO + H₂O
4 FeO + O₂ → 2 Fe₂O₃

21.61 g of a lovely rusty coloured powder was obtained.

Now, some calculations.
Assuming 11.08 g pure iron being converted, Faraday's first law tells us our efficiency:

F * n * z = η * I * t
η = F * n * z / (I * t)

Now, we have

F = 26.8 Ah / mol
n = m / M = 11.08 g / (55.85 g / mol) = 0.198 mol
z = 2
I = 2 A
t = 5.33 h

and we get

η = 99.557%

Bit high, but seems fine.

Now, yield.

Overall reaction:
4 Fe + O₂ → 2 Fe₂O₃

So,

n_Iron / n_{Iron(III)-oxide} = 2 / 1
n_{Iron(III)-oxide} = 0.5 * n_Iron
m_{Iron(III)-oxide} = 0.5 * n_Iron * M_{Iron(III)-oxide}

and with

n_Iron = 0.198 mol
M_{Iron(III)-oxide} = 159.697 g / mol

we get

m_{Iron(III)-oxide} = 15.81 g

Now, remember we obtained 21.61 g! This would make 136% yield - impossible. I had some losses anyway.

So, something went wrong.

I repeated the experiment, but I filtered off the precipitate after the second decantation (gravity).
I thought, maybe there was some salt left in the water.

That time, 11.27 g of weight were lost at the anode.
η was 98.33%. While 16.13 g were expected, 21.12 g were obtained after boiling and blowtorching. The losses were bigger due to the filter paper.
Still, 131% yield.

Can someone help me? What is wrong here?

[Edited on 7-2-2018 by Diachrynic]

JJay - 6-2-2018 at 13:08

You probably have a hydrate or some form of iron(III) oxide-hydroxide. I think you can dehydrate it in a crucible to obtain more predictable results.

Diachrynic - 6-2-2018 at 13:13

Okay, I'll repeat the experiment a third time and heat it longer and hotter. Will take some time.

Thanks for the advice!

Pumukli - 7-2-2018 at 00:36

"and we get

η = 99.557%

Bit high, but seems fine."

:-)

I admit that I never done this reaction before, but my gut-feeling is that 99.557% yield in an electrolysis reaction is more than suspect. So no, it's not fine. :-)

May I ask you what kind of water was used for both the dissolution of NaCl and later for the washings?
You did not explicitely write distilled water so it might as well be tap water. Which could lead to co-precipitation of Ca and Mg carbonates/hydroxides/whatnot with the Fe-compounds. It may have caused a 5-15% mass difference.

The proper desiccation (dehydration) of the Fe_xO_y(OH)_z requires fairly high temperature I think. It may contribute to another 15-25% error.

How did you measure the current? What kind of power source did you use?

If you used distilled water and the powder was red hot in the dryer, then I may have to reevaluate my world view. :-)

Anyway, interesting problem.

wg48 - 7-2-2018 at 03:41

Quote: Originally posted by Diachrynic

That time, 11.27 g of weight were lost at the anode.
η was 98.33%. While 16.13 g were expected, 21.12 g were obtained after boiling and blowtorching. The losses were bigger due to the filter paper.
Still, 131% yield.

Can someone help me? What is wrong here?

Your doing the calculation incorrectly. Normally the theoretical yield is the ratio of the actual product to the theoretical maximum possible amount of product.

The theoretical max product of Fe2O3 is 143% wrt iron dissolved, which means your yield is 131/143 = 92% wrt to the theoretical maximum possible Fe2O3 from a given amout of Fe. Percentages are by weight.

Obviously Fe203 is heavy than the iron it contains. When you read this type of stuff or do such calculation its a good idea to always think "wrt what"

Oops: sorry guys, the above is BS, please ignore

[Edited on 7-2-2018 by wg48]

Diachrynic - 7-2-2018 at 06:21

Quote: Originally posted by wg48

Quote: Originally posted by Diachrynic

Your doing the calculation incorrectly. Normally the theoretical yield is the ratio of the actual product to the theoretical maximum possible amount of product.
[Edited on 7-2-2018 by wg48]

21.12 g (obtained mass of oxide) / 16.13 g (theoretical maximum mass of oxide) = 131%

Quote: Originally posted by Pumukli

I admit that I never done this reaction before, but my gut-feeling is that 99.557% yield in an electrolysis reaction is more than suspect. So no, it's not fine. :-)

Yeah, it is kind of suspicious...

Quote: Originally posted by Pumukli

May I ask you what kind of water was used for both the dissolution of NaCl and later for the washings?
You did not explicitely write distilled water so it might as well be tap water. Which could lead to co-precipitation of Ca and Mg carbonates/hydroxides/whatnot with the Fe-compounds. It may have caused a 5-15% mass difference.

Tap water for the dissolution, tap water for the first washing, destilled water for the second one.
I think the water here is soft, but I don't know. At least the toilet almost never has carbonates blocking the way ;D

Quote: Originally posted by Pumukli

The proper desiccation (dehydration) of the Fe_xO_y(OH)_z requires fairly high temperature I think. It may contribute to another 15-25% error.

Did it in a small ceramic bowl from an old chemistry building kit. Ethanol burner. Was kind of slow, so additional heating was done with a small blowtorch (has a nice blue conical flame).

Quote: Originally posted by Pumukli

How did you measure the current? What kind of power source did you use?

Used a DC power supply. Range 0-30 V / 0-5 A.
I used the number straight out of the dial. Probably a multimeter would have been better.

Quote: Originally posted by Pumukli

If you used distilled water and the powder was red hot in the dryer, then I may have to reevaluate my world view. :-)

Anyway, interesting problem.

Not quite red hot. It is funny, how the color changes and gets this lovely rusty red. If you blowtorch it, it actually gets darker, but returns to a fuller red after it cools down.

wg48 - 7-2-2018 at 07:52

Oops: sorry guys, my post was BS, please ignore.
My only excuse apart from being half brain dead is I was in a rush as my dog was nagging me for her walk.

JJay - 7-2-2018 at 08:02

Quote: Originally posted by wg48

Oops: sorry guys, my post was BS, please ignore.
My only excuse apart from being half brain dead is I was in a rush as my dog was nagging me for her walk.

Ok, I'll let it slide this time

wg48 - 7-2-2018 at 08:13

Quote: Originally posted by JJay

Quote: Originally posted by wg48

Oops: sorry guys, my post was BS, please ignore.
My only excuse apart from being half brain dead is I was in a rush as my dog was nagging me for her walk.

Ok, I'll let it slide this time

Oh thankyou JJ your so understanding.

Here is my BS but accurate maths LOL

AJKOER - 7-2-2018 at 16:29

If I am understanding correctly, you employed 175 g NaCl in 1.3 liters of H2O.

With this amount of aqueous NaCl, you create at the electrodes respectively, NaOH (aq) and Cl2 (aq).

Cl2 + H2O = H+ + Cl- + HOCl

The action of HOCl with Fe (electrode) forms eventually FeCl3 (reddish brown).

The action of NaOH on FeCl3:

3 NaOH + FeCl3 = 3 NaCl + Fe(OH)3

2 Fe(OH)3 is best thought of as hydrated ferric oxide, Fe2O3.3H2O .
------------------------------------------------

More detail, the action of Fe and HOCl is likely a fenton-type reaction with hypochlorous acid proceeding as follows:

Fe + 2 HOCl --> Fe(ll) + 2 .OH + 2 Cl-

Fe(ll) + HOCl --> Fe(lll) + .OH + Cl-

.OH + Cl- = OH- + .Cl

.Cl + .Cl --> Cl2

.Cl + Cl- = .Cl2-

.OH + .OH --> H2O2

Fe(ll) + H2O2 --> Fe(lll) + .OH + OH- (limited reaction until pH drops)

Formation of a basic iron chloride is also possible. Interaction with NaOH introduces Fe(OH)2 and Fe(OH)3 products.

Bottom line, not a pure product.

[Edited on 8-2-2018 by AJKOER]

DraconicAcid - 7-2-2018 at 16:48

Quote: Originally posted by AJKOER

If I am understanding correctly, you employed 175 g NaCl in 1.3 liters of H2O.

With this amount of aqueous NaCl, you create at the electrodes respectively, NaOH (aq) and Cl2 (aq).

Cl2 + H2O = H+ + Cl- + HOCl

The action of HOCl with Fe (electrode) forms eventually FeCl3 (reddish brown).

The action of NaOH on FeCl3:

3 NaOH + FeCl3 = 3 NaCl + Fe(OH)3

2 Fe(OH)3 is best thought of as hydrated ferric oxide, Fe2O3.3H2O .

You are not going to oxidize chloride ion to chlorine at an iron electrode, when the iron is much more easily oxidized.

AJKOER - 7-2-2018 at 17:04

With stainless steel, you will observe some chlorine even with 6 V battery. See https://forum.cosmoquest.org/showthread.php?96445-Electrolys...

Diachrynic - 10-2-2018 at 07:00

I did some experiments with stainless steel, I observed no chlorine smell though. In any case, non-stainless steel probably wont generate chlorine due to the potential of E = + 1.31 V IIRC. (And iron having - 0.41 V I think).

Anyway, blew a fuse on my multimeter while measuring the current. Going to take even more time than I tought.

Diachrynic - 11-2-2018 at 02:12

Seems like my multimeter is broken in the 20 A range. Could only measure up to 400 mA. Anyway, in that range we get about 330 mA if we set 300 on the dial. Assuming this tend continues up to 2 A, with 10% more we get 2.2 A, which translates to η = 90.5%. Is that more realistic?