## Knowing If K>1 in Redox Reactions

Morgan Carrington 2H
Posts: 54
Joined: Wed Nov 14, 2018 12:22 am

### Knowing If K>1 in Redox Reactions

I've been seeing while doing problems to find the standard cell potentials that it also sometimes asks if K > 1 (like 6M.13), but I'm not sure how to go about figuring out this part of the question. Can someone give guidance about how to go about thinking about/solving these kinds of questions?

For reference, here's 6M.13a:

Identify the reactions with K > 1 in the following list and, for each such reaction, identify the oxidizing agent and calculate the standard cell potential.

(a) Cl2(g) + 2 Br− (aq) → 2 Cl− (aq) + Br2(l)

Daniel Honeychurch1C
Posts: 109
Joined: Thu Jul 11, 2019 12:15 am

### Re: Knowing If K>1 in Redox Reactions

When K > 1, the reaction is spontaneous because this makes delta G negative. So when the question asks to find if K > 1, it wants you to find the sign of delta G and thus if the reaction is spontaneous.

Luc Zelissen 1K
Posts: 57
Joined: Mon Jun 17, 2019 7:23 am

### Re: Knowing If K>1 in Redox Reactions

Daniel Honeychurch1C wrote:When K > 1, the reaction is spontaneous because this makes delta G negative. So when the question asks to find if K > 1, it wants you to find the sign of delta G and thus if the reaction is spontaneous.

You can easily find the sign of delta G, by finding the of the cell. You can use the equation . If the E cell is negative, then delta G will be positive. If the E cell is positive, then delta G will be negative.

Abigail Sanders 1E
Posts: 112
Joined: Wed Sep 11, 2019 12:16 am

### Re: Knowing If K>1 in Redox Reactions

I believe in the textbook it states that K>1 when E°cell >0. This makes sense because at equilibrium E°cell= RT/nF * ln K so, when K>1, ln k is a positive number and E°cell will be greater than 0. Furthermore, the sign of K, or K>1 can also be determined from delta G as Delta G = -RTlnK at equilibrium. Thus, when standard free energy (delta G nought) is less than zero, the reaction is spontaneous and K >1.