Maximum Useful Work from a Fuel Cell

[salman213]

What is the maximum useful work which can be obtained from a hydrogen-oxygen fuel cell that produces 1.96 kg of water at 25°C?



How do we approach this problem?

I guess -Ecell x charge = w?

 

[salman213]

2H2(g) -----> 4H+ + 4e
O2(g) + 4H+ + 4e -------> 2H2O(g)


E cell = 1.2 V


anyhelp for the rest

 

[ShawnD]

It looks like you're on the right track. The reduction table I have says the voltage is 1.23 while yours says 1.2, so that's good.

What you're supposed to do is figure out how many moles of water is 1.96kg

From there you use the mole ratio between water and elections to see how many moles of electrons is that. The way you wrote it, it looks like 4 electrons makes 2 waters.

Then how many coulombs of electrons is that. The conversion rate between moles and coulombs is called Faraday's Number, = 96,485.3383 coulomb/mole

1 Volt = 1 Joule / 1 Coulomb

 

[arivero]

what about the maximum power?

 

[ShawnD]

what about the maximum power?

Impossible to tell. You would need to know the internal resistance of the cell.

 

[salman213]

can you check this:
I need to find charge
.....(1 mol)
1960 g x --------- x 2 moles e = 217.536 moles e
.....(18.02g)

217.536 moles e x 96485 C/mole e = 20988960.96 C


w = -1.23 x 20988960.96 =-25816422 J


according to my assignment online that is wrong :( HELPP

 

[salman213]

Hint: Note that work is conventionally defined as being done ON the system. Here, however, you are asked for the (useful) work that can be obtained FROM the system.

 

[chemisttree]

Maximum useful work always makes me think about the definition of Gibbs energy.

Do not approach this problem by counting charges. It will lead you nowhere.

 

[salman213]

G = -nFEcell


(2)(96485)(1.2) = work
?

 

[chemisttree]

The change in Gibbs free energy, ΔG, in a reaction is a very useful parameter. It can be thought of as the maximum amount of work obtainable from a reaction.

try ΔG = ΔH - TΔS for the reaction:

H_2 + 1/2O_2 \rightarrow H_2O

 

[salman213]

so just do that for this question, i don't have to take into account that mass of 1.96 kg?

 

[chemisttree]

You might be able to find molar gibbs energy in a table somewhere. From there convert the molar energy into the energy contained in 1.96 Kg.

 

[salman213]

-237.192 kJ/mol is the delta G formation

so ur saying just multiply that by the following

1.96 kg = 1960 g x 1mol / 18.02 g x-237.192 kJ/mol = -25798 kJ

??/

 

[salman213]

If i use

G = -nFEcell
(2)(96485)(1.23) = work
then multiply by 1960/18.02 it gives me

-25816 kJ

so which should i put I only have 1 try left but those answers seem to be slightly different!

-(2)(96485)(1.23)(1960)/(18.02) = -25816 kJ

or

1960 g x 1mol / 18.02 g x-237.192 kJ/mol = -25798 kJ

also: I am guessing i put positive of those values

 

[chemisttree]

If i use

G = -nFEcell
(2)(96485)(1.23) = work
then multiply by 1960/18.02 it gives me

-25816 kJ

so which should i put I only have 1 try left but those answers seem to be slightly different!

-(2)(96485)(1.23)(1960)/(18.02) = -25816 kJ

This is the answer you obtained in post #6, isn't it? Was it correct?

or

1960 g x 1mol / 18.02 g x-237.192 kJ/mol = -25798 kJ

also: I am guessing i put positive of those values[/b]

These answers are very close to each other. Does your assignment require you to use significant figures?
How many significant figures for 1.2 V?
For 1.23 V?
For 1.96 Kg?

 

[salman213]

i never treid any of these answers unfortunately, I was doing it wrong the first few times. But it doesn't really require sig figs because I remember I put in for an answer 2104 mols for another question and it showed me that I was correct and the correct answer was 2104.23 mols

but I am still hesitant as to what I should put :(

 

[unhip_crayon]

If i use

G = -nFEcell
(2)(96485)(1.23) = work
then multiply by 1960/18.02 it gives me

-25816 kJ

so which should i put I only have 1 try left but those answers seem to be slightly different!

-(2)(96485)(1.23)(1960)/(18.02) = -25816 kJ

or

1960 g x 1mol / 18.02 g x-237.192 kJ/mol = -25798 kJ

also: I am guessing i put positive of those values

It would be positive. So, the answer is, in your case, +25816 kJ