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.
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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 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?
sorry for the late reply by the way.
To determine which choice will increase the cell voltage the most, we can use the Nernst equation. The Nernst equation is given as:
Ecell = E°cell + (0.059/n) * log([Cu2+]/[Fe2+])
Where:
- Ecell is the cell voltage
- E°cell is the standard cell voltage
- [Cu2+] is the concentration of Cu2+
- [Fe2+] is the concentration of Fe2+
- n is the number of electrons transferred in the balanced equation
Based on the equation, we can see that increasing [Cu2+] or decreasing [Fe2+] will result in an increase in cell voltage.
Let's consider the given choices:
A) Halve[Cu2+]:
If we halve the concentration of Cu2+, the value in the log term would decrease. However, since there is a negative sign in front of the log term, halving [Cu2+] would result in a smaller negative value, ultimately increasing the cell voltage.
B) Double[Cu2+]:
If we double the concentration of Cu2+, the value in the log term would increase. Since there is a positive coefficient (0.059/n) in front of the log term, doubling [Cu2+] would result in a larger positive value, ultimately increasing the cell voltage.
C) Double[Fe2+]:
If we double the concentration of Fe2+, the value in the log term would decrease. Doubling [Fe2+] would result in a smaller negative value in the log term, which would decrease the cell voltage.
D) Halve[Fe2+]:
If we halve the concentration of Fe2+, the value in the log term would increase. However, since there is a negative coefficient (0.059/n) in front of the log term, halving [Fe2+] would result in a larger negative value, ultimately decreasing the cell voltage.
E) Cut Cu electrode in half:
Reducing the surface area of the Cu electrode by cutting it in half would not affect the concentrations of Cu2+ or Fe2+. Therefore, it would not have any impact on the cell voltage.
From analyzing the choices, we can see that the option that will increase the cell voltage the most is B) Double[Cu2+]. By doubling the concentration of Cu2+, the value in the log term increases, resulting in a larger positive value and a higher cell voltage.