For the cell shown below, which will increase the cell voltage the most?

Question

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!

Answer 1

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?

Answer 2

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

Answer 3

thanks your helping me alot do you know 13?

Answer 4

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.

Answer 5

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

Answer 6

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

Answer 7

wow I messed up the mol calculation. I made it n=c/V when its n=c*v

Thanks

Answer 8

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=Ecathode-Eanode

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?

Answer 9

sorry for the late reply by the way.