Self-Test 14.12B

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

Hannah Chew 2A
Posts: 76
Joined: Fri Sep 29, 2017 7:05 am
Been upvoted: 2 times

Self-Test 14.12B

Calculate the potential of a cell constructed with two silver electrodes. The electrolyte in one compartment is 1.0 m AgNO3(aq). In the other compartment, NaOH has been added to a AgNO3 solution until the pH 12.5 at 298 K.

At pH 12.5, pOH = 1.5 and [OH-] =10 ^ -1.5 = 0.032 mol/L; [Ag+] = (Ksp/[OH-]) = (1.5 x 10 ^-8/(0.032) = 4.7 x 10^-7 mol/L. E = -(0.025693 V/1) ln(4.7 x10^-7/1.0)= 0.37 V

1) Where does the Ksp = 1/5x10^-8 come from?
2) What exactly is occurring with the OH- and how does it affect the compartment? I'm not sure what exactly is happening in this reaction.

Thank you!!

Chem_Mod
Posts: 17949
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 406 times

Re: Self-Test 14.12B

The Ksp is a unique value for this specific reaction, and you would not be expected to know this number.

The addition of OH- will change the pH of the solution.

You are combining two delta G equations here to solve for E: G=-RTlnK and G=-nFE.

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

Who is online

Users browsing this forum: No registered users and 1 guest