Question

For the titration of 40.0 mL of 0.0100 M Sn2 by 0.0500 M Tl3 in 1 M HCl, using Pt and Ag | AgCl electrodes:

(c) What are the two Nernst equations for the cell voltage? The potential for the Ag | AgCl electrode is 0.197 V.]

(d) What is the value of E at the following volumes of added Tl3 ?

1.0mL 4.0mL 7.9mL 8.0mL 8.1mL 13mL

Answer 1

Titration

(a) Nernst equations,

E = 0.15 - 0.0592/2 log([Sn2+]/[Sn4+]) - 0.197

and,

E = 0.77 - 0.0592/2 log([Tl+]/[Tl3+]) - 0.197

(d) value of E

1.0 ml Tl3+ added

initial Sn2+ = 0.01 M x 40 ml = 0.40 mmol

added Tl3+ = 0.05 M x 1 ml = 0.05 mmol

Sn2+ remained = 0.35 mmol

Sn4+ formed = 0.05 mmol

Using first Nernst equation,

E = 0.15 - 0.0592/2 log(0.35/0.05) - 0.197

   = -0.072 V

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4.0 ml Tl3+ added

initial Sn2+ = 0.01 M x 40 ml = 0.40 mmol

added Tl3+ = 0.05 M x 4 ml = 0.20 mmol

Sn2+ remained = 0.20 mmol

Sn4+ formed = 0.20 mmol

half-equivalence point

Using first Nernst equation,

E = 0.15 - 0.0592/2 log(0.20/0.20) - 0.197

   = -0.047 V

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7.9 Tl3+ added

initial Sn2+ = 0.01 M x 40 ml = 0.40 mmol

added Tl3+ = 0.05 M x 7.9 = 0.395 mmol

Sn2+ remained = 0.005 mmol

Sn4+ formed = 0.395 mmol

Using first Nernst equation,

E = 0.15 - 0.0592/2 log(0.005/0.395) - 0.197

   = 0.0092 V

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8.0 ml Tl3+ added

initial Sn2+ = 0.01 M x 40 ml = 0.40 mmol

added Tl3+ = 0.05 M x 8 ml = 0.40 mmol

Equivalence point

E = (0.15 + 0.77)/2 - 0.197

   = 0.263 V

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8.1 ml Tl3+ added

initial Sn2+ = 0.01 M x 40 ml = 0.40 mmol

added Tl3+ = 0.05 M x 8.1 ml = 0.405 mmol

Tl3+ remained = 0.005 mmol

Tl+ formed = 0.40 mmol

Using second Nernst equation,

E = 0.77 - 0.0592/2 log(0.40/0.005) - 0.197

   = 0.517 V

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13 ml Tl3+ added

initial Sn2+ = 0.01 M x 40 ml = 0.40 mmol

added Tl3+ = 0.05 M x 13 ml = 0.65 mmol

Tl3+ remained = 0.25 mmol

Tl+ formed = 0.40 mmol

Using second Nernst equation,

E = 0.77 - 0.0592/2 log(0.40/0.25) - 0.197

   = 0.567 V