Consider the titrtation of 25.0mL of 0.0100M Sn2+ by 0.0500M TI3+ in 1 M HCL, using Pt and Saturated Calomel Electrodes to find the endpoint.
F=96486.7 Cmol
E0 Sn4+/Sn2+ =0.15V
E0 TI3+/TI+ = 1.28
Esce = 0.241V
R=8.3141 J K Mol
(D) calculate E at the following volumes of TI3+ : 1.00mL, 2.50mL, 4.90mL, 5.00mL,5.10mL and 10.0mL
2.4
for the titration of 50.0 mL of 0.0500 M Sn2+ after adding 10, 25 and 40 mL of 0.100 M Tl3+. Both the analyte and the titrant are 1.0 M in HCl.
To calculate E at different volumes of TI3+, we need to use the Nernst equation:
E = E° - (0.0592/n) * log(Q)
where:
E = cell potential at a given volume of TI3+
E° = standard cell potential
n = number of electrons exchanged
Q = reaction quotient
In this case, the half-reactions involved in the cell are:
Sn2+ → Sn4+ + 2e- (Number of electrons exchanged = 2)
TI3+ + e- → TI+ (Number of electrons exchanged = 1)
The overall balanced reaction is:
Sn2+ + 2TI3+ + 2H+ → Sn4+ + 2TI+ + H2O
The reaction quotient, Q, is given by the concentrations of the reactants and products raised to the power of their stoichiometric coefficients.
Now, let's calculate E at the given volumes of TI3+:
1.00 mL:
Sn2+ : 0.0100 M * (25.0 mL - 1.00 mL) / 25.0 mL = 0.0096 M (change in concentration due to adding TI3+)
TI3+ : 0.0500 M * (1.00 mL / 25.0 mL) = 0.0020 M
Sn4+ and TI+ concentrations are initially zero since they haven't been formed yet.
Q = [Sn4+] * [TI+]^2 / [Sn2+]^2
= 0 * (0.0020 M)^2 / (0.0096 M)^2
= 0
E = E° - (0.0592/n) * log(Q)
= E° - (0.0592/2) * log(0)
= E°
E at 1.00 mL of TI3+ is equal to E° (0.15 V).
You can follow the same steps for the other volumes of TI3+ (2.50 mL, 4.90 mL, 5.00 mL, 5.10 mL, and 10.0 mL) by calculating Q and using the Nernst equation as shown above.