## Textbook Example 6N.1

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

JamieVu_2C
Posts: 108
Joined: Thu Jul 25, 2019 12:16 am

### Textbook Example 6N.1

Calculate the equilibrium constant at 25.00 degreesC for the reaction $AgCl(s)\rightarrow Ag^{+}(aq)+Cl^{-}(aq)$. The equilibrium constant for this reaction is actually the solubility product, Ksp = [Ag+][Cl-], for silver chloride.

The two reduction half reactions are
R: $AgCl(s)+e^{-}\rightarrow Ag(s)+Cl^{-}(aq)$
L: $Ag^{+}(aq)+e^{-}\rightarrow Ag(s)$

Then, the book states that you reverse the second half reaction, which would become an oxidation half reaction: $Ag(s)\rightarrow Ag^{+}(aq)+e^{-}$.

How do you know that AgCl constitutes one half reaction and that Ag alone constitutes the other half reaction? Also, how would you know to reverse the second half reaction for Ag instead of the first half reaction for AgCl?

chemboi
Posts: 101
Joined: Sat Jul 20, 2019 12:16 am

### Re: Textbook Example 6N.1

Unsure about the first part of your question, but as for why you reverse one of them, its to yield the overall net reaction as given in the problem.

Return to “Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)”

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