Question

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 for the potential of the cell, consisting of the Nernst equation for each half-reaction minus the potential of the SCÉ (0.241 V) Sn log 2+ 0.05916 A E0.139- 0.241 4+ Sn 4+ Sn 0.05916 |в Е%3 0.139 log 0.241 Scroll down to answer part D of this question. 2+ Sn C E0.139-0.05916 log | 0.241 _ [A] 4+

Answer 1

-Answer Givens - Page No-o A Titration of 25.0ml of 0.0000 H snit in im Hel with 0.0500 M Agt to give sn2t and Agst, (a) Balanced bitration reaction : sg"+ng* = sn ago (B) Two half cell reaction are ; shu + + 200 st E, 0,139 Agot + deo Agt 6 - 1990 (C) Two Neinst quations are : 77 E- 0.139. 0.05916 -0.241 Ec 0:139 - Dior go log for en -0241 = 1.90 - 0.05906 long I forgot ow E: 1.90 -0.05916 2 241 so the Answer is (A) & (E) O Ceu voltage (E) : (a) Ime of Agt added Holes of snut present = 0.0 x 25 = 0.25 mmol Holes of Agt added = 0.05H XIml = 0.05 mmol Fan 4+ ] remained = 0.2126 = 0.00477 [3n24] formed = 0.05/26 = 0.0019 M E = [0.139 -0.0592/2 109 (0.601910.0077) -0.241 = -0.084V

Page No - (b) 2.5 me Agt added Holes of snut present = 0.0x25 = 0.25 mmol moles of Agt added = 0.05 H * 25 ml = 0.125 mmol (sr44] - [szt] E = 0.139 -0.241 - -0.102 v (c) 4.8 ml Ag+ added Holes of snut present = 0.01x25=0.25 mmol Holes of Agt added = 0.05 M *4.8 ml = 0.24 mmol (54+] semained = 0.01/29.8 = 0.000335 M (3n+t] formed = 0.24/ 29.8 = 0.00805 M E = (0.139 -0.0592/2 109 (0.00805/ 0.000335) - -0.143 v (d) 5 ml Agt added Holes of snut present = 0.01x25 = 0.25 mmol Holes of Agt added = 0.05 MX sme = 0.25 mmol E- (0.139 +1.90)/2] -0.241 = 0.7785V (e) 5.10 me Agt added Moles of snut present - 0.01*25 = 0.25 mmol Holes of Agt odded = 0.05 MX5.1m1 = 0.285 mmol {agt ] remained = 0.005/30.1 = 0.00017 M (Ag3] formed - 0.25/ 50.1 = 0.0083 M E = (1.90 -0.0592/2 103 10.00014 10.0083)] -0.241 = 1176 v

Page No - . (f) lome Agt Added Holes ' present SnYX = 0.01x25= 0.25 mmol Holes of Ag* added = 0.05m xiom -0.5 mmol E = 1.90 -0.241 z 1.659