Question

1) At 25

Answer 1

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/1) * log 1/ [Ag+]}

Combining the two:

Ecell = (+0.79993 V) - {(0.05916V/1) * log 1/ [Ag+]} - 0.241

So, we need [Ag+] after the addition of the NaI solution to finish our calculation. The titration reaction is:

                                               Ag++ I-?AgI(s)

The equivalence point occurs at:

moles Ag+ = 0.015*0.0340 = 0.00051

moles Ag+ = MI-*VI-

VI- = 0.00051 / 0.0170 M = 0.030 L = 30 ml

a) VI-= 0.7 ml

moles of I- = 0.0007 * 0.0170 M = 0.000012

We�ll have that all I- will deplete and we�ll have Ag+ left:

moles Ag+ left = 0.00051 - 0.000012 = 0.000498 moles

[Ag+] = 0.000498 moles / (0.015 + 0.0007) = 0.032 M

E = (+0.79993 V) - {0.05916 V * log 1/ 0.032} - 0.241 = 0.79993 - 0.329 = 0.47093 V

b) VI-= 19.70 ml

moles of I- = 0.0197 * 0.0170 M = 0.0003349

We�ll have that all I- will deplete and we�ll have Ag+ left:

moles Ag+ left = 0.00051 - 0.00003349 = 0.000477 moles

[Ag+] = 0.000477 moles / (0.015 + 0.0197) = 0. 0137 M

E = (+0.79993 V) - {0.05916 V * log 1/ 0.0137} - 0.241 = 0.79993 - 0.351= 0.449 V

c) VI-= 30 ml

This is just in the equivalence point, so there won�t be any Ag+ or I- in solution

d) VI-= 44.90 ml

Our 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.

What is the I- concentration? We have 44.90 � 30.00 = 14.90 mL excess I-

solution, therefore the concentration must be:

(14.90 mL)(0.0170 M I-) / (44.90 + 15.00) mL= 0.00423 M I-

From the Ksp expression:

Ksp= [Ag+][I-]

[Ag+] = Ksp/[I-] = 8.3 � 10-17/ 0.00423 = 1.945 *10-14 M

Inserting this into our expression above:

Ecell= (+0.79993 V) - {0.05916 V * log 1/ 1.945 *10-14} - 0.241

Ecell= 0.79993 - 1.052 = -0.252 V