At 25 degree C, you conduct a titration of 15.00 mL of a 0.0560 M AgNO_3 solutio

Question

At 25 degree C, you conduct a titration of 15.00 mL of a 0.0560 M AgNO_3 solution with a 0.0280 M Nal 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 Nal solution? The reduction potential for the saturated calomel electrode is E = 0.241 V. The standard reduction potential for the reaction Ag^+ + e^- rightarrow Ag (s) is E^0 = 0.79993 V. The solubility constant of Agl is K_sp = 8.3 times 10^-17. a) 0.700 mL b) 11.70 mL c) 30.00 mL d) 37.30 mL

Explanation / Answer

when you add 0.028 M 0.7 mL NaI molarity of the I- in the mixture = 0.028 M * 0.7 mL/15.7 mL = 1.25*10^-3 M

concentration of Ag+ that is present in solution = Ksp/[I] = 8*10^-17/ 1.25*10^-3 M = 6.41*10^-14 M

For the reaction : Ag +   + e- --------> Ag(s)

E = E^0 - 0.059/1 ln 1/[Ag+]

E = 0.79993 - 0.059 ln 1/(6.41*10^-14)

or, E = -0.992 V

E(cell) = -0.992 -0.241 =-0.751

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when you add 0.028 M 11.7 mL NaI molarity of the I- in the mixture = 0.028 M * 11.7 mL/26.7 mL = 0.0123 M

concentration of Ag+ that is present in solution = Ksp/[I] = 8*10^-17/ 0.0123 M = 6.52*10^-15 M

For the reaction : Ag +   + e- --------> Ag(s)

E = E^0 - 0.059/1 ln 1/[Ag+]

E = 0.79993 - 0.059 ln 1/(6.52*10^-15)

or, E = 1.182 V

E(cell) = 1.182 -0.241 =0.941

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when you add 0.028 M 30 mL NaI molarity of the I- in the mixture = 0.028 M * 30 mL/45 mL = 0.0187 M

concentration of Ag+ that is present in solution = Ksp/[I] = 8*10^-17/ 0.0187 M = 4.28*10^-15 M

For the reaction : Ag +   + e- --------> Ag(s)

E = E^0 - 0.059/1 ln 1/[Ag+]

E = 0.79993 - 0.059 ln 1/(4.28*10^-15)

or, E = 1.157 V

E(cell) = 1.157 -0.241 = 0.916

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when you add 0.028 M 37.30 mL NaI molarity of the I- in the mixture = 0.028 M * 37.30 mL/52.3 mL = 0.028 M

concentration of Ag+ that is present in solution = Ksp/[I] = 8*10^-17/ 0.028 M = 2.86*10^-15 M

For the reaction : Ag +   + e- --------> Ag(s)

E = E^0 - 0.059/1 ln 1/[Ag+]

E = 0.79993 - 0.059 ln 1/(2.86*10^-15 M)

or, E = 1.13 V

E(cell) = 1.13 -0.241 =0.893 V

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