AAMC 11 PS #39 - Redox Reaction

[thomasfx10]

I have a question on REDOX reactions. Question 39 Ask what is the Ecell (V) and is the reaction spontaneous ... There is a picture showing circuit and a salt bridge.

Values given are

Pb2+ + 2 e− -----> Pb(s) E red = -0.127 V
Cu2+ +  2 e− ----> Cu(s) E red = +0.339 V

Answer: Yes it is spontaneous, and E cell = +0.466 V

My question is I know the formula for Ecell = Ecat - Eanode but how do I determine which one is getting oxidized / reduced? If I know that than I can plug and chug since Reduction happens at cathode (Redcat) and Oxidation at Anode (AnOx) ...

In this case Pb was oxidized and Cu was reduced.
0.466V = 0.339V - (-0.127V)


Can I just look at the EV values and see that if Cu is +0.339 it will be reduced? If E Cell is Positive (+) is it spontaneous?

Thanks

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[Bumbl3b33]

I have a question on REDOX reactions. Question 39 Ask what is the Ecell (V) and is the reaction spontaneous ... There is a picture showing circuit and a salt bridge.

Values given are

Pb2+ + 2 e− -----> Pb(s) E red = -0.127 V
Cu2+ +  2 e− ----> Cu(s) E red = +0.339 V

Answer: Yes it is spontaneous, and E cell = +0.466 V

My question is I know the formula for Ecell = Ecat - Eanode but how do I determine which one is getting oxidized / reduced? If I know that than I can plug and chug since Reduction happens at cathode (Redcat) and Oxidation at Anode (AnOx) ...

In this case Pb was oxidized and Cu was reduced.
0.466V = 0.339V - (-0.127V)


Can I just look at the EV values and see that if Cu is +0.339 it will be reduced? If E Cell is Positive (+) is it spontaneous?

Thanks

Honestly, I hate the E=cat-anode equation. Ever since I changed my view to this simplistic paradigm of electrochemistry, I haven't missed a single electrochemistry question. Think of it this way:

Math-wise: You know it's a galvanic cell. If it's a galvanic cell, you know it has to be spontaneous. If it has a -DeltaG, then E has to be positive (DeltaG=-nFE). So, whenever you're given something like this, you want to have a positive E, so what you do is you inverse the negative potential and add the two values.

Theoretically: We are looking at a REDUCTION table. Since +E means spontaneous, that means that Cu would spontaneous reduce, where as Pb wouldn't; if Pb has -0.127V for reduction, then it's oxidation would be +0.127V, meaning that oxidation would be spontaneous.

Thus, 0.466V is the final answer, and yes it's spontaneous. Does that help? I didn't really do a line-by-line on your question because I want you to try this method out; I have introduced it to many people and no one I know has missed electrochemistry questions since learning this, unless it was a stupid error or their part. Of course, the "theoretically....spontaneous" paragraph I wrote is the most important. Give it a shot and let me know if it helps you or if you need clarification!

 

[thomasfx10]

Math-wise: You know it's a galvanic cell. If it's a galvanic cell, you know it has to be spontaneous. If it has a -DeltaG, then E has to be positive (DeltaG=-nFE). So, whenever you're given something like this, you want to have a positive E, so what you do is you inverse the negative potential and add the two values.

Theoretically: We are looking at a REDUCTION table. Since +E means spontaneous, that means that Cu would spontaneous reduce, where as Pb wouldn't; if Pb has -0.127V for reduction, then it's oxidation would be +0.127V, meaning that oxidation would be spontaneous.

Knowing that all galvanic cells are spontaneous helps a lot. I am a little confused on the inversion for the negative potential. Can you give me an example. Also ... so that I am clear ... if I see Cu as Positive (+) than go on the assumption that it will be reduced?

Thanks for your help.

 

[paul411]

Can I just look at the EV values and see that if Cu is +0.339 it will be reduced? If E Cell is Positive (+) is it spontaneous?

Pretty much. Here's a very simple way to figure it out:

Pb2+ + 2 e− -----> Pb(s) E_red = -0.127 V
Cu2+ +  2 e− ----> Cu(s) E_red = +0.339 V

Cu2+ will be reduced because it has a higher potential to be reduced (since its E_red is higher) than Pb2+. Pb will be oxidized.

For MCAT questions like this, just think of E_red as how easily something is reduced (higher = easier).

 

[thomasfx10]

Cu2+ will be reduced because it has a higher potential to be reduced (since its E_red is higher) than Pb2+. Pb will be oxidized.

For MCAT questions like this, just think of E_red as how easily something is reduced (higher = easier).

Wow ... that is easy ... thanks!