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
(a) Write the balanced reaction.
Sn4+ + 2e- <--> Sn2+
TI3+ <--> TI+ + 2e-
Sn2+ + Tl3+- -> Sn4+ + Tl+
(b) Write two different half reactions for the net cell reaction.
I'm assuming this means
Sn4+ + 2e- <--> Sn2+ E0= 0.15V
TI3+ <--> TI+ + 2e- E0= 1.28 V
(c) Write the two nernst equations for the net cell reaction
Prior to the equivalence point, the equation would be;
E=1.28+RT/F ln(TI3+/TI+)-SCE (0.241)
Where TI3+/TI+ = Conc. products/reactants.
After the equivalence point, reaction then becomes,
E=0.15 + RT/F ln(Sn4+/Sn2+) - SCE
(D) calculate E at the following volumes of TI3+ : 1.00mL, 2.50mL, 4.90mL, 5.00mL,5.10mL and 10.0mL
This is the section I am having difficulty with, if anyone could help, thank
you!
Sure, I'd be happy to help! Let's calculate E at the given volumes of TI3+:
1.00 mL:
E = 1.28 + (8.3141 J K mol / 96486.7 Cmol) * ln(0.0500 / 1.00) - 0.241
E = 1.28 + (8.3141 J K mol / 96486.7 Cmol) * ln(0.0500) - 0.241
E = 1.28 - 0.381 - 0.241
E = 0.658 V (approx.)
2.50 mL:
E = 1.28 + (8.3141 J K mol / 96486.7 Cmol) * ln(0.0500 / 2.50) - 0.241
E = 1.28 + (8.3141 J K mol / 96486.7 Cmol) * ln(0.0200) - 0.241
E = 1.28 - 0.912 - 0.241
E = 0.127 V (approx.)
4.90 mL:
E = 1.28 + (8.3141 J K mol / 96486.7 Cmol) * ln(0.0500 / 4.90) - 0.241
E = 1.28 + (8.3141 J K mol / 96486.7 Cmol) * ln(0.0102) - 0.241
E = 1.28 - 1.339 - 0.241
E = -0.301 V (approx.)
5.00 mL:
E = 1.28 + (8.3141 J K mol / 96486.7 Cmol) * ln(0.0500 / 5.00) - 0.241
E = 1.28 + (8.3141 J K mol / 96486.7 Cmol) * ln(0.0100) - 0.241
E = 1.28 - 1.628 - 0.241
E = -0.589 V (approx.)
5.10 mL:
E = 1.28 + (8.3141 J K mol / 96486.7 Cmol) * ln(0.0500 / 5.10) - 0.241
E = 1.28 + (8.3141 J K mol / 96486.7 Cmol) * ln(0.0098) - 0.241
E = 1.28 - 1.651 - 0.241
E = -0.611 V (approx.)
10.0 mL:
E = 1.28 + (8.3141 J K mol / 96486.7 Cmol) * ln(0.0500 / 10.0) - 0.241
E = 1.28 + (8.3141 J K mol / 96486.7 Cmol) * ln(0.0050) - 0.241
E = 1.28 - 2.335 - 0.241
E = -1.296 V (approx.)
I hope this helps! Let me know if you have any more questions or need further assistance.
To calculate E at different volumes of TI3+, you can use the Nernst equation. The Nernst equation relates the electrode potential to the concentrations of reactants and products in the electrochemical cell.
The Nernst equation is given by:
E = E0 - (RT/nF) * ln(Q)
Where:
E is the cell potential
E0 is the standard electrode potential
R is the gas constant (8.3141 J K^(-1) mol^(-1))
T is the temperature in Kelvin
n is the number of electrons transferred in the balanced reaction
F is the Faraday constant (96486.7 Cmol)
Q is the reaction quotient
At each volume of TI3+, you need to calculate the concentration of TI3+ and TI+ and use them to calculate the reaction quotient Q. Then, plug the values into the Nernst equation to find E.
For example, let's calculate E at 1.00 mL of TI3+:
TI3+ concentration = (0.0500 M) * (1.00 mL / total volume)
TI+ concentration = (0.0500 M) * (1.00 mL / total volume)
Calculate Q = (TI+ concentration)^2 / (TI3+ concentration)
Then, plug the values into the Nernst equation to find E.
Repeat the same process for the other volumes of TI3+ given.
To calculate E at different volumes of TI3+, you need to use the Nernst equation:
E = E0 - (RT/F) * ln(Q)
where:
- E is the cell potential
- E0 is the standard cell potential
- R is the gas constant (8.3141 J K Mol)
- T is the temperature in Kelvin
- F is Faraday's constant (96486.7 Cmol)
- Q is the reaction quotient
For the given reaction:
Sn2+ + Tl3+- -> Sn4+ + Tl+
(a) At volume 1.00 mL of TI3+:
- The concentration of TI3+ is 0.0500 M * (1.00 mL / total volume)
- The concentration of Sn2+ is 0.0100 M - (0.0500 M * (1.00 mL / total volume))
- Calculate Q using the concentrations of each species
- Substitute the values into the Nernst equation to calculate E
(b) Repeat the same process for volumes 2.50 mL, 4.90 mL, 5.00 mL, 5.10 mL, and 10.0 mL.
Note: Remember to convert the given volumes to the total volume, which may require additional information.