At 25 °C, you conduct a titration of 15.00 mL of a 0.0260 M AgNO3 solution with a 0.0130 M NaI solution within the following cell:

Question

At 25 °C, you conduct a titration of 15.00 mL of a 0.0260 M AgNO3 solution with a 0.0130 M NaI solution within the following cell: 
Saturated Calomel Electrode || Titration Solution | Ag (s)

For the cell as written, what is the voltage after the addition of the following volume of NaI solution? The reduction potential for the saturated calomel electrode is E = 0.241 V. The standard reduction potential for the reaction 
Ag+ +e- --> Ag(s)
is E0 = 0.79993 V. The solubility constant of AgI is Ksp = 8.3 × 10-17.

a) 0.5 mL ; b) 17.0 mL ; c) 30.0 mL ; d) 39.7 mL
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This is what I did:
AgNO3 + NaI --> AgI + NaNO3
For the half-reaction: NaI --> Na+ + I- 
I set ICE table and got concentration of [I-] = 0.0128 M
Then I found the concentration of [Ag+]:
[Ag+] = Ksp/[I-] = 6.48*10^(-15) M
Then I found cell potential of Ag electrode:
E = 0.799 - 0.05916 log(1/[Ag+]) = -0.0404 V

And now I don't know how to proceed . How do I calculate the voltage after the addition of certain volumes of NaI solution?

Thank you for the help.

DrBob222 · Answer

I think you need to redo the ICE table and I don't understand what (or why) you did with the NaI.
The (AgNO3) for the first addition is
0.0260 x (15.00/15.50) = 0.02516 M.
The NaI is reduced likewise as
0.0130 x (0.5/15.5) = 0.0004194 and you should check these numbers and add or subtract digits as appropriate.
..........AgNO3 + NaI ==> AgI + NaNO3
I......0.02516.....0.......0......0
add...........0.0004194............
C...-0.0004194..-0.0004194..0.0004194
E...0.02474........0.......0.0004194

I don't think you need Ksp for this one or the 17.0 mL addition since the Ag^+ from the AgNO3 far more than that from the solubility of AgI. I think you will need it for the 30.0 mL and 39.7 mL.

When you find those concentrations, you substitute it into the formula you have shown and calculate voltage. Remember, however, that the voltage you are calculating is with reference to the H electrode so when you finish, correct for voltage with reference calomel (SCE); that's the voltage the problem is asking for.