At 25 °C, you conduct a titration of 15.00 mL of a 0.0260 M AgNO3 solution with

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. volume 0.1 mL, 18.40 mL, 30.00 mL , 41.5 mL

Explanation / Answer

For our electrochemical cell: Ecell = Eag/Ag+ - ESCE
ESCE = +0.241V

Eag/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/ [Ag+] - (+0.241 V)

So, we need [Ag+] after the addition of the NaI solution to finish our calculation.
The titration reaction is
Ag++ I- AgI(s)
0.1 mL of NaI addition:

Moles of I- = 0.0001 L X 0.013 M = 1.3 x 10-6 moles
moles of Ag+ = 0.015L X 0.026 = 3.9 x 10-4 moles
moles of Ag+ is very large compared to moles of I- , so we take just molarity of Ag+

Inserting this into our expression above:
Ecell    =    +0.79993 V - 0.05916V log 1 /( 0.026) -    (+0.241    V)   = 0.465 V

18.40 mL of NaI addition:
moles of I- = 0.018 L X 0.013 M = 2.34 x 10-4 moles
moles of Ag+ = 0.015L X 0.026 = 3.9 x 10-4 moles
Moles of Ag+ remaining = 3.9 x 10-4 moles -2.34 x 10-4 moles = 1.56 x 10-4 moles

molarity of [Ag+] = 1.56 x 10-4 moles/0.033L = 4.72 x 10-3 M

Inserting this into our expression above:
Ecell =    +0.79993 V - 0.05916V log 1 /( 0.00472) -    (+0.241    V)   = 0.421 V

30.00 mL of NaI addition:
moles of I- = 0.03 L X 0.013 M = 3.9 x 10-4 moles
moles of Ag+ = 0.015L X 0.026 = 3.9 x 10-4 moles
Moles of Ag+ remaining = 3.9 x 10-4 moles -3.9 x 10-4 moles = 0 moles

From Ksp = [Ag+][I-]
8.3 × 10-17 = x^2 ([Ag+]=[I-] )
[Ag+]= 9.11 x 10-9

Inserting this into our expression above:
Ecell = +0.79993 V - 0.05916V log 1 /( 9.11 x 10-9) - (+0.241    V) = 0.083 V

41.5 mL of NaI addition:
moles of I- = 0.0415 L X 0.013 M = 5.395 x 10-4 moles
moles of Ag+ = 0.015L X 0.026 = 3.9 x 10-4 moles
Moles of I- excess = 5.395 x 10-4 moles -3.9 x 10-4 moles = 1.495x 10-4 moles
molarity of I- = 1.495x 10-4/0.0565 = 2.64 x 10-3 M

From Ksp = [Ag+][I-]
8.3 × 10-17 = 2.64 x 10-3 [Ag+]
[Ag+]= 3.14 x 10-14 M

Inserting this into our expression above:
Ecell =    +0.79993 V - 0.05916V log 1 /(3.14 x 10-14 M) - (+0.241    V)   = -0.24 V

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