Formula: Ecell = Eocell - 0.059 V/n * Log(1/[Ag+]) = Eocell + 0.059 V/n * Log[Ag+]
Here, n = no. of electrons tranferred in the cell reaction = 1
Now, Ksp = [Ag+][Cl-], then [Ag+] = Ksp/[Cl-]
(a) [Cl-] = 25 mL * 0.15 M/(25+50) mL = 0.05 M
i.e. E = 0.799 V + 0.059 V/1 * Log(1.8*10-8/0.05) = 0.419 V
(b) [Cl-] = 50 mL * 0.15 M/(50+50) mL = 0.075 M
i.e. E = 0.799 V + 0.059 V/1 * Log(1.8*10-8/0.075) = 0.408 V
0. (15 points) An analytical chemist conducted the titration of 50.00 mL of 0.100 M Ag...
please answer the following question 2. Calculate E for the titration of 48.00 mL of 0.110 M Ag when 50.00 mL of 0.140 M Cl" is added. Silver wire is the indicator electrode and the saturated Ag-AgCl electrode is the reference electrode. E=0.197 V for the saturated Ag-AgCl electrode, E.= 0.7993 V and Ksp = 1.8 10- for AgCl.
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A chemist titrates 25.0 mL of 0.00100 M I- solution with 0.000500 M Ag+ . It takes 50.0 mL of titrant to reach the equivalence point. The Ksp= 8.3 x 10-17 for AgI solid. a) What is the [I-] value when the added titrant volume is 48.0 mL? b) What is the pAg+ value when the added titrant volume is 48.0 mL?
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At 25 C, a titration of 15.00 mL of a 0.0440 M A^NO, solution with a 0.0220 M Nal solution is conducted within the cell SCE | titration solution | Ag(s) For the cell as written, what is the potential after the addition of each 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 _» Ag(s) is E = 0.79993 V. The...
Мар You are performing a titration of 25.0 mL of 0.0100 M Sn4 in 1 M HCI with 0.0500 M Ag* to give Sn2 anu 3+ Ags using a Pt indicator electrode and a saturated calomel electrode (SCE) as the reference electrode. A) Write the balanced titration reaction B) Complete the two half-reactions that occur at the indicator electrode (shown is their corresponding reduction potential) Sn E 0.139 V 3+ Ag® Eo 1.90 V C) Choose the two different expressions...