Sapling Learning 25 ec, you conduct atitration of 15.00 mL of a o.0500 MAgNO, so

Question

Sapling Learning 25 ec, you conduct atitration of 15.00 mL of a o.0500 MAgNO, solution with a 0.0250 M Nal solution within the following cell Saturated Calomel Electrode Il Titration solution I Ag (s) the cell is the voltage after the addition of the following ofNal solution? The uction potential for the saturated calomel electrode E- 0.241 V. The standard reduction potential for the reaction Ag(s) Ag e is EP 0.79993 v. The solubility constant of Agl is Ksp 8.3 x109. c) 30.00 mL a) 0.700 mL. Number d) 43.20 mL b) 16.30 mL o Previous 8 Give up & view soluson checkAnswer o Next aExe you ge in your O eTextboo o Help with o Web Help o Technical

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+ is described by the Nernst equation:

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

Combining the two:

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

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) 0.5 ml NaI –

Given volume is before the equivalence point-

The (AgNO3) for the first addition is

0.0500 x (15.00/15.50) = 0.0484 M.

The NaI is reduced likewise as

0.0130 x (0.5/15.5) = 0.000806

.AgNO3 + NaI ==> AgI + NaNO3

I......0.0484.....0.......0......0

add...........0.000806............

C...-0.000806 -0.000806 0.000806

E...0.047594........0.......0.000806

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

E = 0.79993 - 0.05916 log(1/[0.047594]) - (+0.241 V)

E= 0.79993 – (0.0782V) – (0.241 V) = 0.481V

b) 16.30 ml (Given volume is also before the equivalence point), so calculate as in a)

c) 30ml

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

d) 43.2ml

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:

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

1 3.99x10-14 -14.164 x

Ecell = +0.79993 V – (0.8379V) – (+0.241 V) = -0.278V

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