Posted by Anya on Thursday, December 13, 2012 at 11:00am.
I'm not 100% sure, but I think its like this.
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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=Ecathode-Eanode.
So I think the cathode should be higher than anode to gain maximum Ecell and the cathode is Fe so that should be doubled.
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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 thats is greater is the cathode and the lesser one is the anode. From there its:
Ecell=Ecathode-Eanode
I'm not sure about b) and c)
Even what I just said im not 100%.
================
Do you know how to do 18 and 20 are on the old exam?
sorry for the late reply by the way.
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