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, thanks.
To calculate E at the given volumes of TI3+, you will need to use the Nernst equation. The Nernst equation relates the electrode potential (E) of a half-reaction to the concentration of reactants and products involved.
The Nernst equation is given as:
E = E0 - (RT/nF) * ln(Q)
Where:
E is the electrode potential
E0 is the standard electrode potential
R is the ideal gas constant (8.3141 J K^(-1) mol^(-1))
T is the temperature in Kelvin
n is the number of moles of electrons transferred
F is Faraday's constant (96486.7 Cmol^(-1))
ln is the natural logarithm
Q is the reaction quotient, which is the ratio of products to reactants, each raised to their stoichiometric coefficient.
Let's calculate E at the given volumes of TI3+ using the Nernst equation:
For 1.00 mL of TI3+:
Q = (concentration of Sn2+)/(concentration of Sn4+) = (0.0100 M)/(0.0500 M) = 0.200
E = 1.28 V - (8.3141 J K^(-1) mol^(-1))*(298 K)/(2*96486.7 Cmol^(-1)) * ln(0.200) - 0.241 V
For 2.50 mL of TI3+:
Q = (concentration of Sn2+)/(concentration of Sn4+) = (0.0100 M)/(0.0500 M) = 0.200
E = 1.28 V - (8.3141 J K^(-1) mol^(-1))*(298 K)/(2*96486.7 Cmol^(-1)) * ln(0.200) - 0.241 V
For 4.90 mL of TI3+:
Q = (concentration of Sn2+)/(concentration of Sn4+) = (0.0100 M)/(0.0500 M) = 0.200
E = 1.28 V - (8.3141 J K^(-1) mol^(-1))*(298 K)/(2*96486.7 Cmol^(-1)) * ln(0.200) - 0.241 V
For 5.00 mL of TI3+:
Q = (concentration of Sn2+)/(concentration of Sn4+) = (0.0100 M)/(0.0500 M) = 0.200
E = 1.28 V - (8.3141 J K^(-1) mol^(-1))*(298 K)/(2*96486.7 Cmol^(-1)) * ln(0.200) - 0.241 V
For 5.10 mL of TI3+:
Q = (concentration of Sn2+)/(concentration of Sn4+) = (0.0100 M)/(0.0500 M) = 0.200
E = 1.28 V - (8.3141 J K^(-1) mol^(-1))*(298 K)/(2*96486.7 Cmol^(-1)) * ln(0.200) - 0.241 V
For 10.0 mL of TI3+:
Q = (concentration of Sn2+)/(concentration of Sn4+) = (0.0100 M)/(0.0500 M) = 0.200
E = 1.28 V - (8.3141 J K^(-1) mol^(-1))*(298 K)/(2*96486.7 Cmol^(-1)) * ln(0.200) - 0.241 V
By plugging in the values into the equation, you can calculate E at each volume of TI3+.