For the titration of 30.0 mL of 0.0100 M Sn2 by 0.0500 M Tl3 in 1 M HCl, using P

Question

For the titration of 30.0 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? Answer by adding charges, coefficients, and products to the following.

(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 ? 1ml, 3ml, 5.9ml,6 ml, 6.10ml, 11.0ml.

Explanation / Answer

a) The reaction that takes place is:

Sn2+ + Tl3+ -> Sn4+ + Tl+

This equation is balanced because there is 1 atom of Sn in each side of the equation as well as 1 atom of Tl in each side. The total charge on each side is 5+ too, thus the equation is balanced.

b) We need to write the half reactions, let's start with the oxidation of Sn2+ to Sn4+:

Sn2+ -> Sn4+ + 2e-

We needed to add 2 electrons on the right side to balance the charges (2+ on each side). Also, let's remember that oxidation takes place in the anode.

The half reaction for the reduction of Tl3+ to Tl+:

Tl3+ + 2e- -> Tl+

Again, we added those 2 electrons to balance the charge on each side (1+). Let's remember that reduction takes place in the cathode.

c) The two Nernst Equations for voltage would be:

Ecathode = E0 TI3+/TI+ +(0.059/2)log (Tl3+ / Tl+) - 0.197V

E anode = E0 Sn4+/Sn2+ +(0.059/2)log (Sn4+ / Sn2+) - 0.197V

Where: E0 TI3+/TI+ = 1.28V and E0Sn4+/Sn2+=0.15V.

Note: we divide 0.059 by 2 because 2 is the number of exchanged electrons in each half reaction.

Substituing the values of E0:

Ecathode = 1.28V + (0.059/2)log (Tl3+ / Tl+) - 0.197V

E anode = 0.15V+ (0.059/2)log (Sn4+ / Sn2+) - 0.197V

d) We will construct some ICE tables where we will write the mmoles of each species (mmol = mL*M)

Adding 1mL of Tl3+ 0.05M (V = 30 + 1 = 31mL)

In this case, we only have the Sn4+/Sn2+ pair to calculate E:

E = 0.15V+ (0.059/2)log (Sn4+ / Sn2+) - 0.197V

E = 0.15V+ (0.059/2)log (0.05 / 0.25) - 0.197V

E = -0.0676V

Adding 3mL of Tl3+ 0.05M (V = 30 + 3 = 33mL)

In this case, we only have the Sn4+/Sn2+ pair to calculate E:

E = 0.15V+ (0.059/2)log (Sn4+ / Sn2+) - 0.197V

E = 0.15V+ (0.059/2)log (0.15 / 0.15) - 0.197V

E = -0.047V

Adding 5.9mL of Tl3+ 0.05M (V = 30 + 5.9 = 35.9mL)

In this case, we only have the Sn4+/Sn2+ pair to calculate E:

E = 0.15V+ (0.059/2)log (Sn4+ / Sn2+) - 0.197V

E = 0.15V+ (0.059/2)log (0.295 / 0.005) - 0.197V

E = 0.00524V

adding 6mL of Tl3+ 0.05M (V = 30 + 6 = 36mL)

In this case, we don't have any pair to calculate E but we can calculate E using the following formula:

2E =  E0 TI3+/TI+ + E0 Sn4+/Sn2+ - 2(0.197V)

2E = 1.28V + 0.15V.-2(0.197V)

E = 1.036V

Adding 6.1mL of Tl3+ 0.05M (V = 30 + 6.1 = 36.1mL)

In this case, we only have the Tl3+/Tl+ pair to calculate E:

E = 1.28V + (0.059/2)log (Tl3+ / Tl+) - 0.197V

E = 1.28V + (0.059/2)log (0.005 / 0.3) - 0.197V

E = 1.031V

Adding 11mL of Tl3+ 0.05M (V = 30 + 11 = 41mL)

In this case, we only have the Tl3+/Tl+ pair to calculate E:

E = 1.28V + (0.059/2)log (Tl3+ / Tl+) - 0.197V

E = 1.28V + (0.059/2)log (0.25 / 0.3) - 0.197V

E = 1.081V

Sn2+ Tl3+ -> Tl+ Sn4+ I 0.3 0.05 C 0.05 0.05 0.05 0.05 E 0.25 0 0.05 0.05

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