## Calculating K for reaction using half reactions [ENDORSED]

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

Julia_Gordon_2A
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Joined: Wed Sep 21, 2016 2:57 pm

### Calculating K for reaction using half reactions

In the textbook example 14.8 on page 586, the question asks to calculate the equilibrium constant at 25 degrees C for AgCl(s)--->Ag+(aq)+Cl-(aq). The two half reactions that are required to use are
R: AgCl(s) + e- ---->Ag(s) +Cl-(aq) E=+0.22V
L: Ag+(aq) +e- -----Ag(s) E=+0.80V

It says that the second equation is reversed. Therefore, wouldn't it make that equation the anode? I thought that the more positive E value was the cathode? Could someone please explain how to know which equation to reverse and how the value of E ties into that?

Thank you!

Beata_Vayngortin_3L
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### Re: Calculating K for reaction using half reactions

you flip the equation where flipping the E value will give you the largest most positive E value for your overall redox reaction.

Beata_Vayngortin_3L
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### Re: Calculating K for reaction using half reactions  [ENDORSED]

you flip the equation where flipping the E value will give you the largest most positive E value for your overall redox reaction.

Julia_Gordon_2A
Posts: 30
Joined: Wed Sep 21, 2016 2:57 pm

### Re: Calculating K for reaction using half reactions

But this would mean that Ecell=0.22-0.80=-0.58V

Why is the negative value accepted in this case?

Zulfiqar Lokhandwala 1H
Posts: 10
Joined: Fri Sep 25, 2015 3:00 am

### Re: Calculating K for reaction using half reactions

In the explanation for the question it says that we should expect E to be negative because the K to be very small, and in order of that to happen the the value of E(naught) must be negative

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