Genchem galvanic cell, quick help please

[wall1two]

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In a voltaic cell, one half-cell contains
a cobalt electrode immersed in a
1.0 M Co2+ solution. In the other half-
cell, a lead electrode is immersed in a
1.0 M Pb2+ solution. The value for E°
is 0.15 V in this cell. In a similar vol-
taic cell, a cobalt electrode is immersed
in a 0.0010 M Co2+ solution and a lead
electrode is immersed in a 0.10 M Pb2+
solution. Which statement below accu-
rately describes the second voltaic cell?
Co2+ 1 2e → Co E° -0.28 V
Pb2+ 1 2e → Pb E° -0.13 V
(A) The cell will have a lower E.
(B) The cell will have a higher E.
(C) The cell will have an identical
E, since E is unaffected by
concentration.
(D) Cobalt will become the cathode.
(E) The cathode will dissolve.

ok the funny thing is i dont know why i got it wrong in the first place, if lead is being reduced and you are lowering the concentration of your oxidation product then the reaction would be shifting right= higher E
(not given but Pb^2+ + Co --> Pb + Co^2+ ) right? so lowering prod = up E and correct i think. well somehow i messed this up and got A as my answer. the back of the book has the same answer as what i think is correct (B) but it says:

"Remember, the substance with the most negative reduction potential is the most easily oxidized, making it the anode In this problem, lead is the anode and cobalt the cathode"
wtf how can lead be the anode if it has the more positive reduction potential?
I understand why the answer is B but only when pb is the cathode. arnt the lead cations traveling to the cathode and becoming reduced and plated out (solid on the right hand side) my test is soon and i spent way too much time googling and chading and searching about this problem. any help would be great.

 

[qkchen]

looks like they made a typo in their answer explanation. where did you see this question from?

In a voltaic cell, one half-cell contains
a cobalt electrode immersed in a
1.0 M Co2+ solution. In the other half-
cell, a lead electrode is immersed in a
1.0 M Pb2+ solution. The value for E°
is 0.15 V in this cell. In a similar vol-
taic cell, a cobalt electrode is immersed
in a 0.0010 M Co2+ solution and a lead
electrode is immersed in a 0.10 M Pb2+
solution. Which statement below accu-
rately describes the second voltaic cell?
Co2+ 1 2e → Co E° -0.28 V
Pb2+ 1 2e → Pb E° -0.13 V
(A) The cell will have a lower E.
(B) The cell will have a higher E.
(C) The cell will have an identical
E, since E is unaffected by
concentration.
(D) Cobalt will become the cathode.
(E) The cathode will dissolve.

ok the funny thing is i dont know why i got it wrong in the first place, if lead is being reduced and you are lowering the concentration of your oxidation product then the reaction would be shifting right= higher E
(not given but Pb^2+ + Co --> Pb + Co^2+ ) right? so lowering prod = up E and correct i think. well somehow i messed this up and got A as my answer. the back of the book has the same answer as what i think is correct (B) but it says:

"Remember, the substance with the most negative reduction potential is the most easily oxidized, making it the anode In this problem, lead is the anode and cobalt the cathode"
wtf how can lead be the anode if it has the more positive reduction potential?
I understand why the answer is B but only when pb is the cathode. arnt the lead cations traveling to the cathode and becoming reduced and plated out (solid on the right hand side) my test is soon and i spent way too much time googling and chading and searching about this problem. any help would be great.

 

[wall1two]

"mastering AP Chemistry" from Peterson's. ok book but not pleased about this question

 

[wall1two]

another question also says which of the following is a conjugate acid-base pair, and both are options

"NH4+ / NH3"

"HClO4 / ClO4- "

and the answer is the ammonium one. both of them are conjugate acid/base pairs, i don't understand, or this book is trash

 

[qkchen]

this book would be the ultimate source for testing your knowledge, as it forces you to pick out mistakes in their questions 😛

another question also says which of the following is a conjugate acid-base pair, and both are options

"NH4+ / NH3"

"HClO4 / ClO4- "

and the answer is the ammonium one. both of them are conjugate acid/base pairs, i don't understand, or this book is trash

 

[RCLEE]

Some questions are just wrong. You just have to accept that. I've seen several PAT question on the angle wrong on Kaplan as well as their chemistry questions.

How about using the nernst equation?

I think that since this is not a normal chemical reaction that the le chatelier's principle would apply since this involves both the oxidation and reduction potential on both sides. Since the reduction potential has been reduced, the E itself would be reduced too. The same thing is true when the oxidation only is reduced. At the same (equivalent) concentration then the reduction potential remain constant.

 

[Teeth12345]

Hey Wall,
I just did this and got (A) as my answer.
I'm using the Nernst Eqn. I understand how to set up the cell and which one is the anode/cathode but what I don't get is, if you're decreasing the concentration and shifting the reaction to the right, wouldn't Q increase?
So then E=Eo - (.059)/n x logQ leads to a smaller E?

Using different logic, I ignored LeChatelier and treated each different cell as two separate systems already at equilibrium. Then, in that case I understand why it's (B) because your Q is smaller from the start and logQ is negative.

I don't know which one of my thought processes is correct - could someone please explain why E gets larger? That is, if you can even understand my convoluted thought process 🙂

 

[wall1two]

if the cell voltage originally is .15 (positive) like they said then the equation for the redox reaction has to be

Pb2+ + Co(s) ---> Pb(s) + Co2+.

E= Enot - .0592/n* log(Q) then cobalt ion is on top of the Q expression and pb is on the bottom. and according to destroyer decreasing the concentration of the products at a given point (Q value) will give a more positive E value at that point in time (log of a number less than 1 is negative, so subtracting a negative number makes the E at the time more positive) so i think the book is right, but their reasoning is wrong(the say pb is the anode) and i shouldn't have gotten the wrong answer in the first place yesterday was a looong day. i have a feeling im either going to have 10 cell questions or none

 

[recyrb]

Hey Wall,
I just did this and got (A) as my answer.
I'm using the Nernst Eqn. I understand how to set up the cell and which one is the anode/cathode but what I don't get is, if you're decreasing the concentration and shifting the reaction to the right, wouldn't Q increase?
So then E=Eo - (.059)/n x logQ leads to a smaller E?

Using different logic, I ignored LeChatelier and treated each different cell as two separate systems already at equilibrium. Then, in that case I understand why it's (B) because your Q is smaller from the start and logQ is negative.

I don't know which one of my thought processes is correct - could someone please explain why E gets larger? That is, if you can even understand my convoluted thought process 🙂

I have not mastered anything, but my thought process is, that if you are decreasing the product concentration, then the reaction is going to move to the right. Which is the same way a reaction moves if it is spontaneous. Q does not increase if the products are decreased, i think, because isn't K < Q mean that the reactants increase-- the 'arrow' less than sign, points towards the Left. If products/reatants is less than 1 then you increase the energy of the system, which in my brain is analogous to the reaction becoming 'more spontaneous' thats my best guess

wall pretty sure thats just a bad explanation, dont sweat it, how many chemistry books are you going to dominate lol? making me feel bad

 

[wall1two]

rcb check your inbox i shot some questions at ya

 

[johnnyt09]

It would have a higher E value.