EMF vs Ecell

[globy321]

What is the difference between electromotive force [EMF=E(reduction)+E(oxidation)] and E(cell)=E(cathode)-E(anode)? When do you use these formulas?

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[V5RED]

Those two formulae mean the same thing. You can use either one since they are the same thing.

Recall that oxidation=anode and reduction=cathode. Also recall that going from reduction potential to oxidation potential just means you multiply the reduction potential by negative one.

EMF=Ecell
Ereduction=Ecathode
Eoxidation=-(Eanode)

What material are you studying from?

 

[globy321]

Those two formulae mean the same thing. You can use either one since they are the same thing.

Recall that oxidation=anode and reduction=cathode.
EMF=Ecell
Ereduction=Ecathode
Eoxidation=-(Eanode)

What material are you studying from?

I am using Kaplan right now.
when using the above formulas, does E(oxidation)and E(reduction) have to be positive before you use the formula?
In an electrolytic cell how do you determine whether lead (Pb) or copper (Cu) is being reduced?

 

[Courtjester8]

Oxidation is loss of electrons, Reduction is a gain of electrons. At the anode (the site of oxidation) the neutral metal will be losing electrons and converting to the cation. At the cathode (the site of reduction) the metal ions in solution will be gaining electrons and attaching to the metal cathode. If this isn't enough, we'll need to know the half cell potentials for copper and lead to tell you more. In fact, if you can't get it from that, type all the information on the page about the cell just to make sure we have a complete picture.


cj8

EDIT: also important is realizing that an electrolytic cell is nonspontaneous unlike a voltaic (galvanic) cell.

 

[V5RED]

I am using Kaplan right now. when using the above formulas, does E(oxidation)and E(reduction) have to be positive before you use the formula? In an electrolytic cell how do you determine whether lead (Pb) or copper (Cu) is being reduced?

In a galvanic cell, Ecell is always greater than or equal to zero. When you use the formula, if you get a negative number then you either entered the wrong numbers or you assumed the reaction moved in the direction opposite to which it moves.

As to an electrolytic cell, you would need more information than you have just given me.

If everything is at standard state, then you need the reduction potentials of copper and lead. Reduction will occur in the cell with a lower reduction potential because it is an electrolytic cell.

If they are not at standard state, then you need the concentrations as well as the equilibrium constant for the reaction. You would also still need the standard reduction potentials.

How far along are you in your prep, and have you actually taken general chemistry? I am not trying to be mean, but you have made quite a few threads asking pretty basic questions that would usually be able to be answered by reading your textbook.

 

[globy321]

Oxidation is loss of electrons, Reduction is a gain of electrons. At the anode (the site of oxidation) the neutral metal will be losing electrons and converting to the cation. At the cathode (the site of reduction) the metal ions in solution will be gaining electrons and attaching to the metal cathode. If this isn't enough, we'll need to know the half cell potentials for copper and lead to tell you more. In fact, if you can't get it from that, type all the information on the page about the cell just to make sure we have a complete picture.


cj8

The standard potential of lead is given as -0.127V and copper as +0.339V. Since copper has the higher standard potential it's being reduced. is this right? if lead had a standard potential of -0.400 then would lead be reduced?

 

[V5RED]

The standard potential of lead is given as -0.127V and copper as +0.339V. Since copper has the higher standard potential it's being reduced. is this right? if lead head a standard potential of -0.400 then would lead be reduced?

No.

In an electrolytic cell at standard state, the side with the lower potential gets reduced. Electrolytic cells drive unfavorable reactions, so in the electrolytic cell you asked about, lead would be reduced.

If it were a galvanic cell, then copper would be reduced, but that is not what you asked about.

Changing lead's potential from -127V to -400V would not change which electrode experiences oxidation since it would still be a lower potential than copper.

 

[globy321]

No.

In an electrolytic cell at standard state, the side with the lower potential gets reduced. Electrolytic cells drive unfavorable reactions, so in the electrolytic cell you asked about, lead would be reduced.

If it were a galvanic cell, then copper would be reduced, but that is not what you asked about.

Changing lead's potential from -127V to -400V would not change which electrode experiences oxidation since it would still be a lower potential than copper.

Sorry I meant the question was for a galvanic cell.
So for a glavanic cell, the one which has the higher standard potential would be reduced and the one which has the lower standard potential would be oxidized.
For an electrolytic cell it would be the opposite. The one with the lower potential would be reduced and the one with the higher potential would be oxidized. Is that right?

 

[V5RED]

Sorry I meant the question was for a galvanic cell.
So for a glavanic cell, the one which has the higher standard potential would be reduced and the one which has the lower standard potential would be oxidized.
For an electrolytic cell it would be the opposite. The one with the lower potential would be reduced and the one with the higher potential would be oxidized. Is that right?

At standard state, yes this is true.

At other concentrations it might not be.

 

[globy321]

At standard state, yes this is true.

At other concentrations it might not be.

Can you give me an example when the above would not be true?

 

[V5RED]

Can you give me an example when the above would not be true?

Standard potentials refer to the potentials when the concentrations of reactant and product are both 1M.

Look at the Nernst equations to determine the potentials at nonstandard conditions.

Estd=RT/nF*lnK
Eactual=Estd-RT/nF*lnQ

If Q<1, then Eactual>Estd because the natural log of a number less than 1 is a negative number, so subtracting a negative value from Estd makes Eactual>Estd

If Q=1, then Eactual=Estd because ln1=0, so Eactual=Estd-Rt/nF*lnQ=Estd-0=Estd

If K>Q>1, then 0<Eactual<Estd

If Q=K, then Eactual=0 because RT/nF*lnK=Estd, so Eactual=Estd-Estd=0

If Q>K, then Eactual<0 because RT/nF*ln(#>K) will be larger than Estd, so Estd-(#greater than Estd)=a negative value. In this instance, the reaction would move in reverse, and the side with the larger Estd would be getting oxidized because the concentrations were beyond the equilibrium concentrations for the reaction we would expect from looking at standard reduction potentials.