# Redox help

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ps1265A

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#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?

Posted from TSR Mobile

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

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charco

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#2

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#2

(Original post by

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

**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

Apply

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

to the hypothetical equation.

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HenryHiddler

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#3

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#3

(Original post by

You cannot "flip" reduction potentials or they would become oxidation potentials.

Apply

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

to the hypothetical equation.

**charco**)You cannot "flip" reduction potentials or they would become oxidation potentials.

Apply

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

to the hypothetical equation.

E(cell) = E(red)

**+**E(ox)

At least, that was the equation I remember.

Hope this helps

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charco

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#4

(Original post by

I think you mean

E(cell) = E(red)

At least, that was the equation I remember.

**HenryHiddler**)I think you mean

E(cell) = E(red)

**+**E(ox)At least, that was the equation I remember.

If E(cell) is positive and greater than 0.3V the reaction is spontaneous and likely to go to completion.

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HenryHiddler

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#5

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#5

(Original post by

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.

**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.

Hope this helps

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charco

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#6

(Original post by

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.

**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.

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.

Zn

^{2+}(aq) + 2e <==> Zn(s) Eº = -0.76 V

Cu

^{2+}(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)

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HenryHiddler

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#7

(Original post by

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.

Zn

Cu

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)

**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.

Zn

^{2+}(aq) + 2e <==> Zn(s) Eº = -0.76 VCu

^{2+}(aq) + 2e <==> Cu(s) Eº = +0.34 VHence, 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)

Hope this helps

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#8

(Original post by

Fair enough - personally, I find it easier to flip the signs myself.

**HenryHiddler**)Fair enough - personally, I find it easier to flip the signs myself.

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ps1265A

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ps1265A

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#10

(Original post by

Fine by me

**charco**)Fine by me

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#11

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#11

(Original post by

So is the answer -0.14 (which is reduction) - 0.34 (which is oxidation?)

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**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 Pb

^{2+}

_{(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 E

^{o}(Cu

^{2+}/Cu) > E

^{o}(Pb

^{2+}/Pb)? Either way you do it, the EMF is negative, so it is not feasible.

Hope this helps

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#12

(Original post by

So you only have Cu(s) and Pb

**HenryHiddler**)**Cu(s) + Pb2+(aq)**->Cu2+(aq) + Pb(s)So you only have Cu(s) and Pb

^{2+}_{(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 E^{o}(Cu^{2+}/Cu) > E^{o}(Pb^{2+}/Pb)? Either way you do it, the EMF is negative, so it is not feasible.And is Charco saying not to flip the signs?

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#13

**HenryHiddler**)

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

So you only have Cu(s) and Pb

^{2+}

_{(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 E

^{o}(Cu

^{2+}/Cu) > E

^{o}(Pb

^{2+}/Pb)? Either way you do it, the EMF is negative, so it is not feasible.

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#14

^{2+}

**Cu(s)**+ Pb2+(aq) ->

**Cu2+(aq)**+ Pb(s)

Hope this helps

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#15

(Original post by

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

**HenryHiddler**)In the equation you're given, Cu is going to Cu

^{2+}**Cu(s)**+ Pb2+(aq) ->**Cu2+(aq)**+ Pb(s)Posted from TSR Mobile

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Pigster

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#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.

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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#18

E

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

_{red}- E_{ox}= emfReduction 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

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#19

(Original post by

E

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)

**HenryHiddler**)E

_{red}- E_{ox}= emfReduction 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)

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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#20

(Original post by

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

**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

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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