# Redox help

#1
So I have to determine whether this reaction is spontaneous:

Cu(s) + Pb2+(aq) ->Cu2+(aq) + Pb(s)

I'm given the following E standards:
Cu2+(aq) + 2e -> Cu(s) +0.34
Pb2+(aq) + 2e -> Pb(s) -0.14

My method:
I know that Pb is the better reducing agent, therefore it will be oxidised:
Pb -> Pb2+ + 2e and so the E standard will also be flipped to +0.14
Now if I add the two equations up, I also add the electrode potentials and I get +0.44

But the answer is that it isn't spontaneous?

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0
7 years ago
#2
(Original post by ps1265A)
So I have to determine whether this reaction is spontaneous:

Cu(s) + Pb2+(aq) ->Cu2+(aq) + Pb(s)

I'm given the following E standards:
Cu2+(aq) + 2e -> Cu(s) +0.34
Pb2+(aq) + 2e -> Pb(s) -0.14

My method:
I know that Pb is the better reducing agent, therefore it will be oxidised:
Pb -> Pb2+ + 2e and so the E standard will also be flipped to +0.14
Now if I add the two equations up, I also add the electrode potentials and I get +0.44

But the answer is that it isn't spontaneous?

Posted from TSR Mobile
You cannot "flip" reduction potentials or they would become oxidation potentials.

Apply

E(cell) = E(red) - E(ox)

to the hypothetical equation.
0
7 years ago
#3
(Original post by charco)
You cannot "flip" reduction potentials or they would become oxidation potentials.

Apply

E(cell) = E(red) - E(ox)

to the hypothetical equation.
I think you mean
E(cell) = E(red) + E(ox)
At least, that was the equation I remember.

Hope this helps
0
7 years ago
#4
(Original post by HenryHiddler)
I think you mean
E(cell) = E(red) + E(ox)
At least, that was the equation I remember.
No, the equation is E(reduced species) - E(oxidised species) = E(cell)

If E(cell) is positive and greater than 0.3V the reaction is spontaneous and likely to go to completion.
0
7 years ago
#5
(Original post by charco)
No, the equation is E(reduced species) - E(oxidised species) = E(cell)

If E(cell) is positive and greater than 0.3V the reaction is spontaneous and likely to go to completion.
What I meant was, if you did 'flip' the sign, you could add the reduction EMF and the hypothetical 'oxidation' EMF to get the right answer.

Hope this helps
0
7 years ago
#6
(Original post by HenryHiddler)
What I meant was, if you did 'flip' the sign, you could add the reduction EMF and the oxidation EMF to get the right answer.
This is why you shouldn't flip the sign. It leads to confusion.

The numbers quoted are standard reduction potentials.

It is best to leave them that way. It is a measure of the tendency of the equilibrium half-equation (written as a reduction by convention) to release electrons.

Zn2+(aq) + 2e <==> Zn(s) Eº = -0.76 V
Cu2+(aq) + 2e <==> Cu(s) Eº = +0.34 V

Hence, the zinc equilibrium tends to the LHS more than the copper equilibrium
In other words Zn(s) is a better reducing agent than Cu(s)
0
7 years ago
#7
(Original post by charco)
This is why you shouldn't flip the sign. It leads to confusion.

The numbers quoted are standard reduction potentials.

It is best to leave them that way. It is a measure of the tendency of the equilibrium half-equation (written as a reduction by convention) to release electrons.

Zn2+(aq) + 2e <==> Zn(s) Eº = -0.76 V
Cu2+(aq) + 2e <==> Cu(s) Eº = +0.34 V

Hence, the zinc equilibrium tends to the LHS more than the copper equilibrium
In other words Zn(s) is a better reducing agent than Cu(s)
Fair enough - personally, I find it easier to flip the signs myself

Hope this helps
0
7 years ago
#8
(Original post by HenryHiddler)
Fair enough - personally, I find it easier to flip the signs myself.
Fine by me
0
#9
(Original post by charco)
Fine by me
So is the answer in the MS wrong?

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#10
(Original post by charco)
Fine by me
So is the answer -0.14 (which is reduction) - 0.34 (which is oxidation?)

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7 years ago
#11
(Original post by ps1265A)
So is the answer -0.14 (which is reduction) - 0.34 (which is oxidation?)

Posted from TSR Mobile
Cu(s) + Pb2+(aq) ->Cu2+(aq) + Pb(s)

So you only have Cu(s) and Pb2+(aq). Therefore, the way you were trying to do it, where Pb(s) is oxidised, is wrong since there is no Pb(s) to react with. The equation you have tells you that Cu(s) is being oxidised. But how could it be, if the Eo(Cu2+/Cu) > Eo(Pb2+/Pb)? Either way you do it, the EMF is negative, so it is not feasible.

Hope this helps
0
#12
(Original post by HenryHiddler)
Cu(s) + Pb2+(aq) ->Cu2+(aq) + Pb(s)

So you only have Cu(s) and Pb2+(aq). Therefore, the way you were trying to do it, where Pb(s) is oxidised, is wrong since there is no Pb(s) to react with. The equation you have tells you that Cu(s) is being oxidised. But how could it be, if the Eo(Cu2+/Cu) > Eo(Pb2+/Pb)? Either way you do it, the EMF is negative, so it is not feasible.
Could you show me the working using the equation Charco mentioned. Because I still don't understand how Cu is being oxidised.

And is Charco saying not to flip the signs?

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#13
(Original post by HenryHiddler)
Cu(s) + Pb2+(aq) ->Cu2+(aq) + Pb(s)

So you only have Cu(s) and Pb2+(aq). Therefore, the way you were trying to do it, where Pb(s) is oxidised, is wrong since there is no Pb(s) to react with. The equation you have tells you that Cu(s) is being oxidised. But how could it be, if the Eo(Cu2+/Cu) > Eo(Pb2+/Pb)? Either way you do it, the EMF is negative, so it is not feasible.
Um, and exactly how is Cu being oxidised?

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7 years ago
#14
(Original post by ps1265A)
Um, and exactly how is Cu being oxidised?

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In the equation you're given, Cu is going to Cu2+

Cu(s) + Pb2+(aq) ->Cu2+(aq) + Pb(s)

Hope this helps
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#15
(Original post by HenryHiddler)
In the equation you're given, Cu is going to Cu2+

Cu(s) + Pb2+(aq) ->Cu2+(aq) + Pb(s)

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7 years ago
#16
We could go about this the GCSE way for a second to keep our feet rooted in reality...
Pb is more reactive than Cu. Pb(s) can therefore displace Cu2+(aq) from its solution.

Now the A-level approach:
Cu2+(aq) + 2e -> Cu(s) +0.34
Pb2+(aq) + 2e -> Pb(s) -0.14

Since the more +ve one is on top, the reaction goes clockwise, i.e. Cu2+ -> Cu (top equation to RHS) and Pb -> Pb2+ (bottom to LHS).
The feasible reaction is therefore Cu2+ + Pb -> Cu + Pb2+, i.e. the opposite of the reaction in the OP, i.e. the given reaction is not feasible.

In the more +ve half cell, reduction occurs and vice versa.
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#17
(null)

This all makes sense, but I need verifying

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7 years ago
#18
Ered - Eox= emf

Reduction reaction would be: Pb2+ + 2e-> Pb
Oxidation reaction would be: Cu > Cu2+ + 2e-

-0.14 - 0.34 = -0.48V
Not feasible (sorry about before, I had a bit of a moment :P)

Hope this helps
0
#19
(Original post by HenryHiddler)
Ered - Eox= emf

Reduction reaction would be: Pb2+ + 2e-> Pb
Oxidation reaction would be: Cu > Cu2+ + 2e-

-0.14 - 0.34 = -0.48V
Not feasible (sorry about before, I had a bit of a moment :P)
Thanks, I sort of understand it. But there's still the contradiction between using E standard values which give me the answer below, and the equation itself which shows that copper is oxidised rather than reduced. Any suggestions?

Ered - Eox
The more negative electrode is Pb, it will be placed on the left, it will be oxidised
The more positive electrode is Cu, it will be placed on the right, it will be reduced
So 0.34 - (-0.14) = 0.48

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7 years ago
#20
(Original post by ps1265A)
Thanks, I sort of understand it. But there's still the contradiction between using E standard values which give me the answer below, and the equation itself which shows that copper is oxidised rather than reduced. Any suggestions?

Ered - Eox
The more negative electrode is Pb, it will be placed on the left, it will be oxidised
The more positive electrode is Cu, it will be placed on the right, it will be reduced
So 0.34 - (-0.14) = 0.48

Posted from TSR Mobile
Well, if you look at the equation, the Cu > Cu2+ + 2e- reaction happens. Ordinarily, what you're saying would happen - the Cu would be oxidised since its reduction potential is greater than the Pb2+ + 2e- reaction. However, from the equation you're given, Cu is oxidised, meaning it is entered into the E(ox) bit. The Pb2+ is reduced in the equation, meaning it is entered into the E(red)
Therefore, since E cell = E(red) +E(ox), then e cell = -0.14 - 0.34 = -0.48

Where you're going wrong is that you're not looking at the species involves - you're right in thinking that if Cu2+ were to react with Pb, THEN Cu2+ would be reduced (as its half cell potential is greater than the half cell potential of Pb2+'s reduction) and the Pb would be oxidised , giving the EMF you mentioned, since E(cell) = E(red) - E(ox) = 0.34 - (-0.14) = 0.48V

However, in the reaction you're given, you only have Cu and Pb2+, where the equation you're given shows Cu is oxidised and the Pb2+ is reduced - this is the opposite reaction to what you're calculating, which means you're getting the wrong answer.
The right answer is E(cell) = E(red) - E(ox) = -0.14 - (0.34) = -0.48V

(Edit: Imagine it as though what you've been calculating js the 'natural' reaction, whereas the one they want is the reverse reaction.

Hope this helps
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