[poli275]

Hi guys, I have a question on polarograph which involves calculation. It will be great if anyone of you guys can point me to the correct track in getting the answer.

The question is as follow:
Iron(III) is polarographically reduced to Iron(II) at potentails more negative than about +0.4V vs SCE, and is further reduced to Iron metal at -1.5V vs SCE. Iron(II) is also reduced to Iron metal at -1.5V. A polarogram is run using a dropping mercury electrode on a solution containing both Fe3+ and Fe2+. A current is recorded at zero applied volts, and it's magnitude is 12.5uA. A wave is also recorded with E1/2 equal to -1.5V vs SCE, and its height is 30uA. Calculate the relative concentrations of Fe3+ and Fe2+ in solution.

I gone through several books I can get from my uni library but they provide no relevant examples. I hope someone here can point me in the right direction to solve this problem, thanks

[Pradeep]

Poli,

I answering your question assuming a dropping mercury electrode is used. The current at 0v (because it is negaive than 0.4 V) is corresponding the faltue of Fe3+(aq)  Fe2+(HG) wave.

The current at -1.5 V is due to reduction of Fe2+  Fe(Hg) and  Fe3+ (aq)  Fe2+(Hg).

The current due to Fe2+(aq)  Fe(Hg) is 12.5 -30 uA. = 17.5 uA

[Fe2+]/[Fe3+] = 12.5/17.5 = 1.4

[poli275]

Hi and thanks for the reply. Well the answer is 5:1, which your working seems incorrect. Anyway, I will try it again using your method, thanks

[Borek]

How did you get 5:1?

Strange.

First things first - current is given by the Ilkovič equation:

i = k n C

(k is a constant that combines an electrode characteristic and a diffusion coefficient, n is a number of electrons, C is a concentration)

At 0V we observe current 12.5 μA from a one electron reaction:

Fe³⁺(aq) -> Fe²⁺(aq)

30 μA (at -1.5V) is a sum of two reactions:

Fe³⁺(aq) -> Fe(Hg) (three electrons)

and

Fe²⁺(aq) -> Fe(Hg) (two electrons)

This yields two equations for current:

12.5 = k [Fe³⁺]

and

30 = k (2[Fe²⁺]+3[Fe³⁺])

k will cancel out in the end. Actually it should be slightly different for each ion, as it is a function of diffusion coefficient.

Trick is, when solving it yields a negative concentration of Fe²⁺.

Perhaps I need more coffee, but I don't see what could went wrong.

[poli275]

Hey man I apologise for my previous post, I read the wrong answer and your working is perfectly correct. Thanks again for your help, have a nice day ahea dof you

[Pradeep]

Poli,

Was I correct?

Glad to hear that.

[Pradeep]

Poli,

I carefully read the problem. This should be the correct way to work out.

I answering your question assuming a dropping mercury electrode is used. The current at 0v (because it is negaive than 0.4 V) is corresponding the faltue of Fe3+(aq)    Fe2+(HG) wave.

The current at -1.5 V is due to reduction of Fe2+    Fe(Hg) and  Fe3+ (aq)    Fe2+(Hg).

The current due to Fe2+(aq)    Fe(Hg) is 12.5 -30 uA. = 17.5 uA

[Fe3+]/[Fe2+] = 12.5 / ((30-12.5)/2)

[Fe3+]/[Fe2+] = 12.5/8.75 = 1.43

[Borek]

Either I am missing something or there is something wrong with the question and the answer.

The current at 0v (because it is negaive than 0.4 V) is corresponding the faltue of Fe3+(aq)    Fe2+(HG) wave.

Agreed. Just the product is not Fe²⁺(Hg) but Fe²⁺(aq)

The current at -1.5 V is due to reduction of Fe2+    Fe(Hg) and  Fe3+ (aq)    Fe2+(Hg).

At -1.5V both Fe³⁺ and Fe²⁺ get reduced to Fe(0). What you wrote seems to be ambiguous to me - what is the final product? Fe(Hg), or Fe(Hg) and Fe²⁺(aq)? Judging from

[Fe3+]/[Fe2+] = 12.5 / ((30-12.5)/2)

looks like you are assuming Fe³⁺ gets reduced to Fe²⁺ which doesn't get reduced further. How come Fe²⁺ that was present in the solution gets reduced to Fe(0), but Fe²⁺ that is product of Fe³⁺ reduction doesn't get reduced?

The current due to Fe2+(aq)    Fe(Hg) is 12.5 -30 uA. = 17.5 uA

This is wrong (and not only because 12.5-30=-17.5, not 17.5). To reduce Fe³⁺ you need three times the charge necessary to reduce Fe³⁺ to Fe²⁺, as this is now a three electron process. So the current used for reduction of Fe²⁺ (initially present in the solution, not the one coming from Fe³⁺ reduction) is not 30-12.5 but 30-3x12.5. Unfortunately, this yields a negative current for the reduction of Fe²⁺ present initially.

Besides, because of small inconsistencies your posts are hard to follow:

12.5 -30 uA. = 17.5 uA

No, -17.5.

12.5/17.5 = 1.4

No, 0.71.

[Pradeep]

Sir,

Thank you very much for the advice. I accept my week point. I am so sorry. I had to double check all these. I beleave that I have to learn a lot of things form the experiences you have. I may little bit good at chemistry, but my teaching skills are looks like poor!!!!!!!!!!!!!!

I try to improve it  , and thank for giving a nice and detailed constructive comment.

Just the product is not Fe2+(Hg) but Fe2+(aq)

Yes. You may be correct. The solubility of ions in mucury is less, than the solubility of water. But the Fe2+ formed at the surface of Hg drop may be adhered to negatively polarized Hg surface and may be removed from the aquas layer  with the detachment of the drop.

At -1.5V both Fe3+ and Fe2+ get reduced to Fe(0).

At -1.5 Fe3+(aq) get reduced to Fe2+(aq) only. If both of those process happening the polorographic wave should be splitted in to two.  Because energy required for Fe2+ Fe(0) and Fe3+   Fe(0) are different values. (Practical values are normally -1.3 and -1.73 respectively) Since Poli sayed about one  wave  (not splitted waves) I consider the potential was not enough to reduce Fe3+ Fe(0). Hence reduction of Fe2+ Fe(0) only will happen.

At -1.5V both Fe3+ and Fe2+ get reduced to Fe(0). What you wrote seems to be ambiguous to me - what is the final product? Fe(Hg), or Fe(Hg) and Fe2+(aq)? Judging from

The product formed at working electrode should be Fe²⁺_(aq) and Fe_(Hg)

looks like you are assuming Fe3+ gets reduced to Fe2+ which doesn't get reduced further. How come Fe2+ that was present in the solution gets reduced to Fe(0), but Fe2+ that is product of Fe3+ reduction doesn't get reduced?

Polorography is known as a non destructive technique. (Want you have suggested is happening on bulk electrolysis, but not in polarography) Although it yields a minute current, one (or fairly large finite number of ) scan/s doesn't consume a considerable amount of the analytes in sample.  That is why we get the same polorographic wave height upon repeating linear scans. (I am not talking about cyclic scans.) If polarography consumes the sample you should see a decay of wave height upon repeating linear scans.

As same as the consumption of reactants, Polagraphy doesnt produce a considerable amount product either.  Therefore any practical number of linear scans doesn't change the Fe2+(aq) concentration of the sample. All Fe2+ that is product of Fe3+ reduction and Fe2+ in the sample are participating for the reduction. But the contribution of Fe2+ that is product of Fe3+ reduction to the total Fe2+ is truly negligible amount.

Therefore         12.5 = k [Fe3+]                    Equation 1
                      30   = k ( [Fe³⁺] + [2Fe²⁺] )     Equation 2  


At -1.5 also Fe3+_(aq) Fe2+ _{(lets say aq)} is happening. Since the process is controlled my mass transfer the current produced by this reduction is remaining at 12.5.


I hope now you can solve these two equations.

I hope this post has clarified your doubts.

I specially thank to Bork for his nice comments on my previous post.

Best Regards,
Pradeep

[Borek]

Polorography is known as a non destructive technique. (Want you have suggested is happening on bulk electrolysis, but not in polarography)

You are right about bulk solution - concentration of Fe²⁺ doesn't change. But when Fe³⁺ gets reduced produced Fe²⁺ is already on the electrode, not even in the diffusion layer. It doesn't have a chance to "run" away, it should be reduced further. As the potential is low enough, reduction takes place immediately, and no matter how many steps are required, if they are fast enough, we should observe a complete reduction of Fe³⁺ to Fe(0).

I will try to consult a colleague (and a former boss) of mine - he is an electrochemistry professor at Warsaw University. I must admit I have not dealt with such problems since late eighties, when we were both working in the Polarography Lab, so I can be wrong.

[poli275]

I will consult my lecturer too, and I shall have a full worked out and verified answer by saturday. Great discussion we having here.

[Borek]

So, I consulted Ziggy and he confirmed I am right - Fe³⁺ gets reduced to Fe(0).

However, he also pointed out that the second wave is measured not from zero, but from the background given by the first wave, so the total current of the second wave is not 30, but 30+12.5, and equations I wrote earlier take form

42.5 = k*(2*[Fe2+]+3*[Fe3+])
12.5 = k*1*[Fe3+]

which yields - surprise - ratio of concentrations 1:5.

I would not miss it seeing the current/potential curve, I hate plots that are just "described". One picture is worth a million words.

[Pradeep]

Still it is not clear for me how, two electrochemical reduction process with considerably different two E1/2 values can be merged in to one wave.

[Borek]

Still it is not clear for me how, two electrochemical reduction process with considerably different two E1/2 values can be merged in to one wave.

Not clear to me what the problem is, I guess you are struggling with the fact both Fe³⁺ gets reduced to Fe(0), not just to Fe²⁺.

Let's assume we are at -1.5V.

We start with a single Fe²⁺ cation. It gets to the electrode, it gets reduced. End of the story.

Now take a single Fe³⁺ cation. When it touches the electrode it gets reduced to Fe²⁺. What can happen then? If the electrode potential is not low enough to reduce it further, it will diffuse away. But we are at -1.5V now, so Fe²⁺ can be reduced, it can't just diffuse away - it is already in contact with the electrode, it has to accept electrons.

[Pradeep]

Anyhow, This is a good conversation.