Question

16-3. Consider the titration of 25.0 mL of 0.010 0 M Sn21 by 0.050 0 M Tl31 in 1 M HCl, using Pt and saturated calomel electrodes to find the end point. (a) Write a balanced titration reaction. (b) Write two different half-reactions for the indicator electrode. (c) Write two different Nernst equations for the cell voltage. (d) Calculate E at the following volumes of Tl31: 1.00, 2.50, 4.90, 5.00, 5.10, and 10.0 mL. Sketch the titration curve. How do we know which ion gets reduced and which one gets oxidized

Answer 1

In an oxidation/reduction reaction electrons are transferred from one reactant to another.

b)

Sn⁴⁺ + 2e^- --------> Sn²⁺ E^o = 0.154 [ 0.14 in 1 M HCl ]

Tl³⁺ + 2e^- -------> Tl⁺ E^o = 1.25 [ 0.77 in 1 M HC l ]

b)

Balanced titration reation is

Sn⁴⁺ + Tl³⁺ -------> Sn²⁺ + Tl⁺

c)

E = E^o - 0.0592 / n [Product] / [Reactant]

E^o = the standard electrode potential

n = number of moles of electrons that appears in the half-reaction for the electrode process as written

Sn⁴⁺ + 2e^- --------> Sn²⁺ E^o = 0.154 [ 0.14 in 1 M HCl ]

E = 0.14 - 0.0592 / 2 [Sn²⁺] / [ Sn⁴⁺]

Tl³⁺ + 2e^- -------> Tl⁺ E^o = 1.25 [ 0.77 in 1 M HC l ]

E = 0.77 - 0.0592 / 2 [Tl⁺] / [ Tl³⁺]

d)

The Equivalence point volume of Ti3+ is 5.00 mL

V1M1 = V2M2

25.00 x 0.0100 = 0.0500 x V2

V2 = 5.00 mL

i) When 1.00 mL of 0.0500 M Ti³⁺ is added

When Ti³⁺ is added, it is essentially completely consumed by reaction with Sn²⁺

Intial number of moles of Sn²⁺ is

= 25.0 mL x 0.0100 mM/ mL

= 0.25 mM

Number of moles of Ti3+ in 1.00mL is 1.00 mL x 0.0500 mM / mL = 0.05 mM

after adding 1.00ml of Ti3+ the number of moles of Sn2+ is

= 0.25 mM - 0.05 mM

= 0.20mM

The concentration of Sn2+

= 0.20 mM / 26.00 mM

The concentration of Sn2+ 0.00769 M

Now the concentration of Sn4+

= 0.05 mM / 26 mM

= 0.0019 M

E^o = 0.14 - 0.0592 / 2 log(0.00769  / 0.0019)

E_system = 0.122 V

ii) When 2.50 mL of 0.0500 M Ti3+ is added

Intial number of moles of Sn2+ is 0.25 mM

Number of moles of Ti3+ in 2.50mL is 2.50 mL x 0.0500 mM / mL = 0.125 mM

the number of moles of Sn2+ is

= 0.25 mM - 0.125 mM

= 0.125mM

The concentration of Sn2+

= 0.125 mM / 27.50 mM

The concentration of Sn2+ 0.004545 M

Now the concentration of Sn4+

= 0.05 mM / 27.50 mM

= 0.004545 M

Eo = 0.14 - 0.0592 / 2 log(0.004545 / 0.004545)

E_system = 0.14 V

iii)

Like wise when 4.90 mL of 0.0500 M Ti3+ is added

E_system = 0.145 V

iv)

We know that 5.00 mL is equivalence point volume

[Sn⁴⁺] = [Tl³⁺] and [Sn²⁺] = [Tl⁺]

Adding two equivalence point we have

2ESn₂₊ = 2Eo - 0.059 [Sn²⁺] / [ Sn⁴⁺]

2E_Ti+ = 2Eo - 0.0592 [Tl⁺] / [ Tl³⁺]

2+2 E_system = 2Eo_Sn2+ + 2EoTi₊

E_system = 2Eo_Sn2+ + 2EoTi₊ / 2+2

E_system = 2[0.14 + 0.17] / 4

E_system = 0.155 V

After the equivalence point:

Beyond the equivalence point, Sn2+ has been essentially completely converted to Sn4+. Additional Ti3+ has no Sn2+ with which to react, hence it is straightforward to calculate Esystem from the concentrations of Ti3+ and Ti+.

5.10 mL of Ti3+

E = 0.77 - 0.0592 / 2 [Tl⁺] / [ Tl³⁺]

E = 0.77 - 0.0592 / 2 log(5.00 / 0.10)

E_system = 0.719 V

10.00 mL of Ti3+

Which is 100% beyond equivalence point.So,

E = 0.77 - 0.0592 / 2 log( 1/ 1)

E_system= 0.77V

Titration curve: