Cell potential with different volumes Please Help!!!!

Question

At 25 ∘C, a titration of 15.00 mL of a 0.0360 M AgNO3 solution with a 0.0180 M NaI solution is conducted within the cell

SCE || titration solution | Ag(s)SCE || titration solution | Ag(s)

For the cell as written, what is the potential after the addition of each volume of NaINaI 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)  𝐸∘=0.79993 V. The solubility constant of AgI is 𝐾sp=8.3×10−17.

at 7.00mL

12.00mL

30mL

if someone could walk me through one of these I would really appreciate it so I can know how to do the others

Answer 1

Let's solve for #1: V= 7.0 ml

Ag⁺+e−→Ag(s)

The cell of the electrode:

E∘=E_R^∘−E_O^∘

=0.799−0.241

=0.558V

Remember the Nernst equation? Now we are going to use it again. However, before that let's figure out what's going on with the reaction between AgNO₃ + NaI = AgI + NaNO₃

number of moles for AgNO3 = 0.015L x 0.036 M = 5.4 x 10^-4 moles

[I^-] = 0.018 M x 7.0 ml = 1.26 x 10^-4 moles

[Ag+]= K_sp/ [I^-] = 6.59 x 10^-13 M

From Nernst equation (remember?)

E_cell=0.558 V-(0.059/1)log(1/.6.59x10^-13)

E_cell=0.558V -0.059xlog1.517x10^-16

E_cell= -0.161 V

In the same way solve for other volumes of iodine, just double check calculations.

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