## Reaction quotient

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

Shanzey
Posts: 120
Joined: Wed Sep 18, 2019 12:20 am

### Reaction quotient

How does the reaction quotient affect our value for E in the Nernst equation? Does having a larger reaction quotient result in a smaller or larger value for E? Also, what is going on conceptually when this occurs?

Uisa_Manumaleuna_3E
Posts: 60
Joined: Wed Sep 21, 2016 2:56 pm

### Re: Reaction quotient

I might be wrong about this but I think have a larger reaction quotient actually makes the E smaller, considering the Nernst equation. We know that E is equal to the E standard minus (RT/nF)*(lnQ). So if the lnQ was to grow in number, the E standard would keep subtracting larger and larger values.

I think that's what would possibly happen, but again I'm not 100 percent sure.

Suraj Doshi 2G
Posts: 100
Joined: Fri Aug 02, 2019 12:15 am

### Re: Reaction quotient

An increase in the reaction quotient will lead to E becoming smaller which, I believe means that the more product we have, the cell potential would be decreasing.

Jocelyn Thorp 1A
Posts: 103
Joined: Wed Sep 18, 2019 12:20 am
when you have a larger Q, you have more products and less reactants. this makes for a low potential difference mathematically because you're subtracting the log of Q. Thinking of it in terms of electrical potential, since your reaction is going from left to right, anode to cathode, it would make sense that having more products would mean a less potential difference: all the electrons have moved from the anode to cathode already, and the potential difference thrives on having a lot of electrons in the anode and not many in the cathode.

Let me know if this doesn't make sense, sometimes my logic only makes sense to me.

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