ing Sapling Learning Map di For the titration of 30.0 ml of 0.0100 M sn? by 0050
ID: 1068529 • Letter: I
Question
ing Sapling Learning Map di For the titration of 30.0 ml of 0.0100 M sn? by 00500 M Tpr using Pt and Ag l Agca electrodes: in 1 M HCI, (a) What is the balanced titration reaction? b) What are the two half-reactions that occur at the ind electrode? Answer by adding charges, coefficients, and products to the following. Fe sn' Eo 0.139 V 0.77 V (c) What are the two Nernst equations for the cell voltage? The potential for the Ag l Agol electrode is o.197 v Sn (Scroll down for more questions ll (i) E-0.77-0.05916 log Isn' 0.197 IV (II) E 0.139 0.05916 log Sn -0.197 O Previous ® Give up 8 view Solution a Try Again 0 Next ExitExplanation / Answer
Titration
(a) balanced equation,
Sn2+ + Tl3+ <==> Sn4+ + Tl+
(b) Sn4+ + 2e- <==> Sn2+ Eo = 0.139 V
Tl3+ + 2e- <==> Tl+ Eo = 0.770 V
(c) Nernst equations
E = [0.139 - 0.0592/2 log([Sn2+]/[Sn4+]) - 0.197
E = [0.770 - 0.0592/2 log([Tl+]/[Tl3+]) - 0.197
(d) Addition
(i) 1 ml Tl3+ added
moles of Sn2+ = 0.01 M x 30 ml = 0.3 mmol
Tl3+ added = 0.05 M x 1 ml = 0.05 mmol
[Sn4+] formed = 0.05 mmol/31 ml = 0.0016 M
[Sn2+] remined = 0.25 mmol/31 ml = 0.0081 M
E = [0.139 - 0.0592/2 log(0.0081/0.0016)] - 0.197
= -0.080 V
(ii) 3 ml Tl3+ added
moles of Sn2+ = 0.01 M x 30 ml = 0.3 mmol
Tl3+ added = 0.05 M x 3 ml = 0.15 mmol
half equivalence point
E = 0.139 - 0.197 = -0.058 V
(iii) 5 ml Tl3+ added
moles of Sn2+ = 0.01 M x 30 ml = 0.3 mmol
Tl3+ added = 0.05 M x 5 ml = 0.25 mmol
[Sn4+] formed = 0.25 mmol/35 ml = 0.0071 M
[Sn2+] remined = 0.05 mmol/35 ml = 0.00143 M
E = [0.139 - 0.0592/2 log(0.00143/0.0071)] - 0.197
= -0.0374 V
(iv) 5.9 ml Tl3+ added
moles of Sn2+ = 0.01 M x 30 ml = 0.3 mmol
Tl3+ added = 0.05 M x 5.9 ml = 0.295 mmol
[Sn4+] formed = 0.295 mmol/35.9 ml = 0.00822 M
[Sn2+] remained = 0.005 mmol/35.9 ml = 0.0004 M
E = [0.139 - 0.0592/2 log(0.0004/0.00822)] - 0.197
= -0.019 V
(v) 6 ml Tl3+ added
moles of Sn2+ = 0.01 M x 30 ml = 0.3 mmol
Tl3+ added = 0.05 M x 6 ml = 0.3 mmol
Equivalence point
E = [0.139 + 0.77]/2 - 0.197
= 0.2575 V
(vi) 11 ml Tl3+ added
moles of Sn2+ = 0.01 M x 30 ml = 0.3 mmol
Tl3+ added = 0.05 M x 11 ml = 0.55 mmol
[Tl+] formed = 0.3 mmol/41 ml = 0.0134 M
[Tl3+] remained = 0.25 mmol/41 ml = 0.0061 M
E = [0.77 - 0.0592/2 log(0.0134/0.0061)] - 0.197
= 0.563 V
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