chemistry  exam practice help!!
posted by Anya on .
For the cell shown below, which will increase the cell voltage the most?
Fe2+  Fe3+  Cu2+  Cu
A) Halve[Cu2+]
B) Double[Cu2+]
C) Double[Fe2+]
D) Halve[Fe2+]
E) Cut Cu electrode in half
...
Ecell = E0 cell + 0.059/n * log([Cu+]/[Fe2+])
Ecell is directly proportional [Cu+] to and inversely proportional to [Fe2+]..
so i would think the answer would be, B and D both
but there can only be one answer , so i don't really know how else to work it out :/
pls help!

I'm not 100% sure, but I think its like this.

The half rxns are:
Fe^2+ > Fe3+ + e [Reduction]
Cu^2+ + 2e > Cu [Oxidation]
Reduction is increase of electrons while oxidation is decrease.
Reduction=cathode
Oxidation=anode
The equation is Ecell=EcathodeEanode.
So I think the cathode should be higher than anode to gain maximum Ecell and the cathode is Fe so that should be doubled.

Can you help me with question #10 and #25? 
10 is b)
its a 2:1 ratio
so you multipy the rate by 1/2
and 25 is 110.6 kj.
first you find delta H , then delta S , then you put those values into :
delta G = delta H  T*deltaS 
thanks your helping me alot do you know 13?

I know that the rxn is:
Ag^+ + e > Ag
n=M/V
n=0.250/0.6=0.42
to convert from mols to mols of electrons you take into account the ratio. The ratio is 1
so mols of e's=0.42
1 mol e = 96500 C
96500*0.42 = 40530C
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I don't know where I went wrong from this point. There are no answers on the multiple choice like this. 
np. number 13 is like this :
first find moles of Ag which is volume (in L) times molarity. should get 0.15 moles Ag.
to convert to coulombs you multiply it by faraday's constant. (approx 96500 C/mol) . the units will cancel out the moles and you should be left with 1.45 * 10^4 C 
if you have your text on you, you think you can help me out with number 77 on pg 859 ? i have no clue how to do it :S

wow I messed up the mol calculation. I made it n=c/V when its n=c*v
Thanks 
For 77 pg 859
I think first you write up the half rxns.
Zn > Zn^2+ + 2e
Ni > Ni^2+ + 2e
Then I think the E values are in the texbook. The E value that's is greater is the cathode and the lesser one is the anode. From there its:
Ecell=EcathodeEanode
I'm not sure about b) and c)
Even what I just said im not 100%.
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Do you know how to do 18 and 20 are on the old exam? 
sorry for the late reply by the way.