At 25 degree C. you conduct a titration of 15.00 mL of a 0.0500 M AgNO_3 solutio

Question

At 25 degree C. you conduct a titration of 15.00 mL of a 0.0500 M AgNO_3 solution with a 0.0250 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 degree = 0.79993 V. The solubility constant of Agl is K_sp = 8 3 times 10^-17. a) 0.700 mL b) 16.30 mL c) 30.00mL d) 43.20 mL

Explanation / Answer

Saturated Calomel Electrode || Titration Solution | Ag (s)

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.

For electrochemical cell: E cell = E Ag/Ag+ - ESCE

ESCE = +0.241V

E Ag/Ag+ are described by the Nernst equation:

E Ag/Ag+ = +0.79993 V - 0.05916V log 1/ [Ag+]

Combining the two:

E cell = +0.79993 V - 0.05916V log 1/[Ag+] - (+0.241 V)

The titration reaction is Ag+ + I- AgI (s)

The equivalence point occurs at:

15.00 mL Ag+ x 0.0500 mol Ag+ /L x 1 mol I-/1 mol Ag + x 1 L / 0.0250 mol = 30 ml

a) 0.700 ml NaI – Given volume is before the equivalence point-

The (AgNO3) for the first addition is

0.0500 x (15.00/15.70) = 0.0478 M.

The NaI is reduced as

0.0250 x (0.700/15.700) = 0.0011

AgNO3 + NaI ==> AgI + NaNO3

I......0.0478.....0.......0......0

C...-0.0011 -0.0011 0.0011

E...0.0467.......0.......0.0011

E = 0.79993 - 0.05916 log (1/[Ag+]) - (+0.241 V)

E = 0.79993 - 0.05916 log (1/[0.0467]) - (+0.241 V)

E= 0.79993 – (0.0787V) – (0.241 V) = 0.48023V

b) 16.30 ml (Given volume is also before the equivalence point

The (AgNO3) for the first addition is

0.0500 x (15.00/31.30) = 0.024 M.

The NaI is reduced as 0.0250 x (16.30/31.30) = 0.013M

AgNO3 + NaI ==> AgI + NaNO3

I......0.024.....0.......0......0

C...-0.013 -0.013 0.013

E...0.011........0.......0.013

E = 0.79993 - 0.05916 log (1/[Ag+]) - (+0.241 V)

E = 0.79993 - 0.05916 log (1/[0.011]) - (+0.241 V)

E= 0.79993 – (0.11587V) – (0.241 V)

E = 0.44313 V

c) 30.0 ml -

Given volume is at the equivalence point. Here voltage does not depend on volume and concentration.

The equivalence volume of NaI is 30mL.     That would mean 100% is already precipated.[Ag+] = 0

E cell = +0.79993 V - 0.05916V log 1/[Ag+] - (+0.241 V)

E cell = +0.79993 V - 0.05916V log 1/[0] - (+0.241 V)

E cell = +0.79993 V - (+0.241 V) = 0.55893 V

d) 43.20ml

Given volume is after the equivalence point, therefore, prior to returning to equilibrium, all of the Ag+ has been converted to AgI(s) and some excess I- remains.

We have 43.2 – 30.00 =13.2 mL excess I- solution,

Therefore the concentration = (13.2mL)(0.0250 M I-) /(43.2 + 15.00) mL = 0.00567 M I-

From the Ksp expression-

Ksp = [Ag+][I-]

[Ag+] =Ksp/[I-] = 8.3 × 10^-17/0.00567 = 1.46 x10^-14

Inserting this into our expression above:

E cell = +0.79993 V - 0.05916V /1 log 1 /1.46 x 10^-14 - (+0.241 V)

E cell = +0.79993 V – (0.8379V) – (+0.241 V) = - 0.278V

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