## Concentration effect on E

$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$

Ryan Yoon 1L
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Joined: Mon Jul 01, 2019 12:15 am

### Concentration effect on E

Why is that if you increase the concentration of product, E is smaller than E standard?

Rita Chen 1B
Posts: 112
Joined: Sat Jul 20, 2019 12:15 am

### Re: Concentration effect on E

I think it depends on the reaction? I could be wrong, but in Dr. Lavelle's example today, the increasing in Mn concentration, caused the forward reaction to be favored, making E increase since the E standard was positive. I think it depends what you increase in concentration.

Alice Chang 2H
Posts: 101
Joined: Fri Aug 30, 2019 12:18 am

### Re: Concentration effect on E

I think it's similar to the concepts of Le Chatelier's Principles.

This website talks a bit about it with a worked example: https://www.mikeblaber.org/oldwine/chm1046/notes/Electro/Conc/Conc.htm

PranaviKolla2B
Posts: 114
Joined: Fri Aug 30, 2019 12:17 am

### Re: Concentration effect on E

Could someone explain what exactly E is?

Emily Chirila 2E
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Joined: Sat Jul 20, 2019 12:16 am

### Re: Concentration effect on E

If you increase the concentration of the product > 1.0M, we get a larger value of Q (CONCENTRATIONproducts/CONTENTRATIONreactants) which is used in the equation:

E= E(standard conditions) - (RT/nF(lnQ))

Taking the ln of a Q that is more than 1 gives a positive number for lnQ, which is being subtracted from E(standard conditions), thus giving a smaller Ecell value in comparison to E(standard conditions)

If you were increasing concentration of reactants > 1.0M, you would end up with a Q that is less than 1, resulting in a negative value for lnQ. This changes the subtraction to addition and you end up getting a value for Ecell that is larger than E(standard conditions)

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