Self-Test 14.12B

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Hannah Chew 2A
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Self-Test 14.12B

Postby Hannah Chew 2A » Sun Feb 18, 2018 4:12 pm

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.

Answer from back of book:
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!!

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Re: Self-Test 14.12B

Postby Chem_Mod » Wed Feb 21, 2018 5:51 pm

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.

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