MCQ 19

[mustafa]

Instead of feeding reactants u feed water into the cell and reverse the terminals.

Of course, by changing reactants you have an entirely new cell with a new set of equations.
This is not 'recharging', recharging is the reverse reaction, i.e. carbon dioxide + water -->

oxygen + fuel. As far as I know, this is a very very very unlikely energetically unfavoured
reaction. If someone have gotten that to work they might have just solved both the fuel as
well as the global warming crisis.


Umm, recharging a fuel
cell is simple.... The reaction is simply electrolysis of water into hydrogen and oxygen....
Which is a prevalent and much practiced reaction. Fuel cells CAN be recharged, albeit it being unprofitable to do so.

[Mao]

The problem I have with recharging is:

In recharging, one would simply reverse the direction of electrical current by giving an EMF source. This is the case with secondary cells.

For fuel cells, this would not be possible as a fuel cell has pumps and valves to continuously supply fuel/oxygen and continuously remove products. Furthermore, this is only really applicable to H2/O2 fuel cells, methanol injection/ethanol injection can't work as their products include CO2, and it is extremely difficult to turn CO2 back into fuel, and even then you cannot control what kind of fuel you get.

In this sense, 'recharging' a fuel cell is simply electrolysis of water. I agree it is a much practiced reaction in the industry, but their setup is much different from a fuel cell. For starters they don't use porous Pt-coated electrodes, or phosphoric acid electrolyte, or anything like that.

So I still think that fuel cells cannot be recharged in the sense that it is impossible from the design, it cannot be generalized for all fuel cells, and it achieves nothing.

[masonnnn]

in all honesty i found this question easy... fuel cells aren't 'rechargeable' in a reversible sense like secondary cells... the other two were ridiculous and the one Mao's said was kind of...obvious.
:p

[Slumdawg]

I went with B. And I just thought... hmm this doesn't seem right. So 10 seconds before, I just went screw it. I'm changing to C. THANK GOD I DID. Thanks Mao for reassuring me that I made the right move. So lucky

What do you think of the exam Mao? Sorry if you've already answered this. Easy, medium or hard? I thought it was medium but I dunno, not the most devoted chem student so my judgment is probably off..

[3Xamz]

OMFG. I picked C, rubbed it out and picked A.

A) "Reactants are within the cell" I'm like everything else is wrong so this could be read as the whole system including the two bottles of reactants were referred to as the cell.

I read C, but then went wait a minute "reductant to anode" that's the same thing! Because the reductant is at the anode. WTF?

[Slumdawg]

OMFG. I picked C, rubbed it out and picked A.

A) "Reactants are within the cell" I'm like everything else is wrong so this could be read as the whole system including the two bottles of reactants were referred to as the cell.

I read C, but then went wait a minute "reductant to anode" that's the same thing! Because the reductant is at the anode. WTF?

I remembered that in fuel cells reactants need to constantly be supplied. So I knew A was probably wrong. I get what you mean about C, but something was telling me to go with that... I was tossing up between B and C in the end.

[Intronaut]

C definitely made the most sense to me at the time. It will be interesting to see how it is handled.

[kyzoo]

I don't see what's so ambiguous about this question

A. Reactants are not stored in the fuel cell
B. Reactino products are not continously removed from secondary cells
D. Fuel cells can't be recharged. You have a constant inflow of reactants and the products need ot escape somehow 0.o

A, B and D are eliminated, hence C

[m@tty]

surely 19 is not C, since when did stupid electrons decide to go to the anode??? and Mao, the anode and the reluctant are the same, its not like a dissolved reluctant is donating electrons to the non reactive anode, the anode is the reluctant, this makes no sense??

Dude the reductant is the chemical species which loses electrons. The electrons then go to the anode and pass through the anode (this is the actual electrode. It is separate from the reductant.).

Hence C.

[stonecold]

surely 19 is not C, since when did stupid electrons decide to go to the anode??? and Mao, the anode and the reluctant are the same, its not like a dissolved reluctant is donating electrons to the non reactive anode, the anode is the reluctant, this makes no sense??

Dude the reductant is the chemical species which loses electrons. The electrons then go to the anode and pass through the anode (this is the actual electrode. It is separate from the reductant.).

Hence C.

I agree.  I cannot believe this question caused so many problems.  I thought it was really simple.  Conceptually, option C is correct, and the other 3 alternatives were ludicrous, so you could have got it purely by elimination.

Why you would recharge a fuel cell, defies logic.

[Studyinghard]

surely 19 is not C, since when did stupid electrons decide to go to the anode??? and Mao, the anode and the reluctant are the same, its not like a dissolved reluctant is donating electrons to the non reactive anode, the anode is the reluctant, this makes no sense??

Dude the reductant is the chemical species which loses electrons. The electrons then go to the anode and pass through the anode (this is the actual electrode. It is separate from the reductant.).

Hence C.

I agree.  I cannot believe this question caused so many problems.  I thought it was really simple.  Conceptually, option C is correct, and the other 3 alternatives were ludicrous, so you could have got it purely by elimination.

Why you would recharge a fuel cell, defies logic.

This question has caused so many problems because clearly its ambiguous and retarded. Its not often that a lot of people are divided amongst answers. Now just because you and who ever else got it right you think its straight forward and happy days. So please lets not talk about logic and ludicrous answers because I can tell you right now that it will lead to a combustion reaction with H2 being produced in this thread with no escape. 

[m@tty]

Instead of simply attacking us that have clear understanding of the question why don't you actually (try) and substantiate your argument. Ad hominem attacks only serve to turn everyone against you.

Yes, I agree that it tested correct understanding of the common terms "anode" and "reductant". Even I was confused at first and couldn't figure the right answer, simply because I had not made the proper distinction between anode and reductant.

It is in no way ambiguous. I guarantee you that the question will stand. Good test of superficial understanding of terms, I think.

[Studyinghard]

First of all, my post wasn't an attack of any means. If you bothered to read my post, it wasnt in any way an attack on stonecold or anyone who got it right. I didn't use any sort of language that was derogatory towards anyone.

Secondly, I am sure 99% of the people who picked C worked on the basis of elimination, so dont give me all this hoo hah that the answers were OBVIOUS. Going from Mao's explanation that electrons go from the surface of the electrode to the centre, pardon my language, is complete BS. I haven't bothered to read over the chapter again but I am almost certain that is not mentioned anywhere in the textbook.

Thirdly I couldn't give two poodles if everyone turned against me. I stand for what I believe in.

 

[D27RII]

no, in this specifc example teh anode and reductant are not different, they are BOTH FE(s), hence it option C makes no sense, electrons dont flow from Fe(s) to Fe(s), that makes no sense, how is that an acceptable answer???

[m@tty]

The reductant is the chemical species that loses electrons. Only some of the Fe atoms are losing electrons at any one time. They then become ions and NO LONGER PART OF THE ANODE. The electrons then travel to/up the remaining Fe atoms in the anode. The atoms that are not being oxidised are not reductants at that time.

Hence they are clearly different - and C is correct.