Question

For the titration of 35.0 mL of 0.0100 M Sn2 by 0.0500 M Tl3 in 1 M HCl, using Pt and Ag | AgCl electrodes:

(a) What is the balanced titration reaction?

(b) What are the two half-reactions that occur at the indicator electrode?

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

1 mL

3.5 mL

6.9 mL

7 mL

7.1 mL

12 mL

Answer 1

Part a)

35.0 mL of 0.01 M Sn²⁺ is titrated against 0.05 M Tl³⁺ . Here, Tl³⁺ will get reduced and Sn²⁺ will get oxidized according to the following reaction,

Tl³⁺+Sn²⁺   Sn⁴⁺+ Tl⁺

part b)

Here, Ag/AgCl will act as the reference electrode (anode) and Pt will act as the indicator (cathode) electrode

The half cell reactions taking place are,

Tl³⁺+ 2e^-    Tl⁺

Sn²⁺   Sn⁴⁺ +2e^-

Part c)

We have the standard reduction potentials for the half cell reactions

Tl³⁺+ 2e^-    Tl⁺   (E^o = 0.77V)

Sn²⁺ +2e^-     Sn          (E^o= -0.141V)

Sn⁴⁺ +2e^- Sn²⁺      (E^o=0.139V in 1M HCl)

The Nernst equation for the half cell reactions are,

E = E^o (Tl³⁺/Tl⁺) -0.059/2 log[Tl⁺]/[Tl³⁺] = 0.77V -0.059/2 log[Tl⁺]/[Tl³⁺]

E = E^o (Sn⁴⁺/Sn²⁺) -0.059/2 log[Sn²⁺]/[Sn⁴⁺] = 0.139V -0.059/2 log[Sn²⁺]/[Sn⁴⁺]

The standard erduction potential of Ag/AgCl electrode is 0.197 V,

Therefore, before the equillibrium , the cell potential can be calculated using,

Ecell ={ 0.139V -0.059/2 log[Sn²⁺]/[Sn⁴⁺]}-0.197

Ecell = -0.058 -0.059/2 log[Sn²⁺]/[Sn⁴⁺]

Potential at 1.00 mL addition

Initial moles of Sn²⁺ = (0.01M)(35.0mL ) = 0.35 mM

Moles of Tl³⁺ added =(0.05M)(1.00mL ) = 0.05 mM

After the reaction,

[Sn⁴⁺] = Moles of Tl³⁺ added/Total volume = 0.05/(35+1) = 0.0013M

[Sn²⁺] = (Initial moles of Sn²⁺ - Moles of Tl³⁺ added)/Total volume

= 0.0083 M

Therefore, Ecell = -0.058 -(0.059/2) x log[0.0083]/[0.0013] = -0.081V

Potential at 3.5 mL addition

Initial moles of Sn²⁺ = (0.01M)(35.0mL ) = 0.35 mM

Moles of Tl³⁺ added =(0.05M)(3.5mL ) = 0.175 mM

After the reaction,

[Sn⁴⁺] = Moles of Tl³⁺ added/Total volume = 0.175/(35+3.5) = 0.0045M

[Sn²⁺] = (Initial moles of Sn²⁺ - Moles of Tl³⁺ added)/Total volume

= 0.0045 M

Therefore, Ecell = -0.058 -(0.059/2)x log[0.0045]/[0.0045]

Ecell = -0.058 -0= -0.058V

Potential at 6.9 mL addition

Initial moles of Sn²⁺ = (0.01M)(35.0mL ) = 0.35 mM

Moles of Tl³⁺ added =(0.05M)(6.9mL ) = 0.345 mM

After the reaction,

[Sn⁴⁺] = Moles of Tl³⁺ added/Total volume = 0.345/(35+6.9) = 0.0082M

[Sn²⁺] = (Initial moles of Sn²⁺ - Moles of Tl³⁺ added)/41.9

= 0.00011 M

Therefore, Ecell = -0.058 -(0.059/2) x log[0.00011]/[0.0082]

Ecell = -0.027V

Potential at 7 mL addition

Initial moles of Sn²⁺ = (0.01M)(35.0mL ) = 0.35 mM

Moles of Tl³⁺ added =(0.05M)(7mL ) = 0.35 mM

Here, this is the equivalence point,

E = 0.77V -0.059/2 log[Tl⁺]/[Tl³⁺]

E = 0.139V -0.059/2 log[Sn²⁺]/[Sn⁴⁺]

or 2E = 0.909 -0.059/2 x log{[Sn²⁺ ][Tl⁺]/ [Sn⁴⁺][Tl³⁺]}

Also, here [Sn²⁺ ] = [Tl³⁺] and [Tl⁺] = [Sn⁴⁺]

Therefore,

2E = 0.909 -0.059/2 x log1

=0.0909

Therefore E= 0.0909 /2= 0.455V

Cell potential = 0.455- 0.197 = 0.258V

After the equillibrium , the cell potential can be calculated using,

E = 0.77V -0.059/2 log[Tl⁺]/[Tl³⁺] -0.197

or

E = 0.573V -0.059/2 log[Tl⁺]/[Tl³⁺]

Potential at 7.1 mL addition

Initial moles of Sn²⁺ = (0.01M)(35.0mL ) = 0.35 mM

Moles of Tl³⁺ added =(0.05M)(7.1mL ) = 0.355 mM

[Tl⁺] = 0.355/(35+7.1) = 0.0084M

[Tl³⁺] = 0.005/42.1 =0.00011

Therefore, Ecell = 0.573 -0.059/2 log[0.0084]/[0.00011]

Ecell = 0.517V

Potential at 12 mL addition

Initial moles of Sn²⁺ = (0.01M)(35.0mL ) = 0.35 mM

Moles of Tl³⁺ added =(0.05M)(12mL ) = 0.6 mM

[Tl⁺] = 0.6/(35+12) = 0.012M

[Tl³⁺] = 0.25/47 =0.0053 M

Therefore, Ecell = 0.573 -0.059/2 log[0.012]/[0.0053]

Ecell = 0.562V