[linds]

Hello,
I'm having trouble figuring out where to start in solving the following problem.  Any help would be appreciated!

Both Al and Zn are used in air batteries, those that use oxygen from the air as an oxidizing agent.  These reactions take place in out time, or the other:
4Al(s) + 3O2(g)    2Al2O3(s)
2Zn(s) + O2(g)     2ZnO(s)

a) Calculate  :delta:G for these reactions at 25C. 
b) what is the maximum number of kilojoules that can be extracted from one pound (454g) of each metal? 
c) what is the voltage, E0cell of each battery?


I think part A is this:
 :delta: G = 2(-1582) - 4(0) + 3(0) = -3164
 :delta: G = 2(0) + 0 - 2(-318.2) = 636.4
Total  :delta: G = -2528.6

Can that be right?  And how to I get part B?

Thanks so much for any help you can offer.
Lindsey

[MrKeller]

First, you don't add the two  :delta: G values together. Each reaction has its own value and this is what your question is asking for. If you looked up the correct :delta: G_f values then your part A is fine.

As to part B, just use dimensional analysis:
1 pound * (454 g/pound) * (1 mol/molar mass of metal) * (# of kJ/mol)

the last one is the :delta: G you calculated for each reaction. The value of :delta: G is the value of the maximum amount of work you can get out of a process.

For part C you need to use the fact that :delta: G = -nFE to find the E0cell. Find n by figuring out how many electrons need to be transferred to reduce oxygen to O2- and to oxidize the metal to the correct oxidation state.

[linds]

Thank you so much, this was a great *delete me*  I think I have it right, but if you have time to check it that'd be great.

Part A:
 :delta: G for Al2O3 = -3164 kJ/mol
 :delta: G for ZnO = 636.4 kJ/mol

Part B:  1 lb = (454g/1lb) * (1mol/MW) * (kJ/mol)
So for Al2O3,   454/101.96 * x kJ/mol = 1lb
x = .225kJ/mol Al2O3

For ZnO,    454/81.41 * x kJ/mol = 1lb
x = 0.180kJ/mol
(the only thing I'm unsure of is if I'm using the proper MW for the second part- it's supposed to be for ZnO, right?)

Part C:    :delta: G = -nFEcell
For Al2O3. -3164 kJ/mol = -(3)(96.5)(Ecell)
Ecell = 10.929V

For ZnO,  636.4 = -(-2)(96.5)(Ecell)
Ecell = 32.974V


Does this look about right?  Thanks!

[MrKeller]

Part A:
 :delta: G for Al2O3 = -3164 kJ/mol
 :delta: G for ZnO = 636.4 kJ/mol

These could be right for all I know. I don't have time to look up the values and do the arithmetic.

Part B:  1 lb = (454g/1lb) * (1mol/MW) * (kJ/mol)
So for Al2O3,   454/101.96 * x kJ/mol = 1lb
x = .225kJ/mol Al2O3

For ZnO,    454/81.41 * x kJ/mol = 1lb
x = 0.180kJ/mol
(the only thing I'm unsure of is if I'm using the proper MW for the second part- it's supposed to be for ZnO, right?)

Assuming your :delta: G values are OK this looks fine except that you need to use the MW of the metal, not the metal oxide. It makes more sense to know the work for the amount of starting material than to know it for the amount of product.

Part C:    :delta: G = -nFEcell
For Al2O3. -3164 kJ/mol = -(3)(96.5)(Ecell)
Ecell = 10.929V

For ZnO,  636.4 = -(-2)(96.5)(Ecell)
Ecell = 32.974V

I think some small corrections are in order:
 :delta: G = -nFEcell
 For Al2O3: 3.164e6 J/mol = -(12 mol e-)(96,500C/mol e-)Ecell

Basically, pay attention to units (use J not kJ because 1 V = 1 J/C) and be sure to get the correct number of moles of electrons. I get 2.72 V for the Al reaction. This seems reasonable to me.