A solution containing Sn 2+ ions is titrated potentiometrically with Fe 3+. The standard reduction p — Physical Chemistry — Kinetics Chemistry Question
Redox reactions
A solution containing Sn 2+ ions is titrated potentiometrically with Fe 3+. The standard reduction potentials for Sn 4+/2+ and Fe 3+/2+ are given below.
Sn 4+ + 2 e – = Sn 2+ E o = 0.154 V
Fe 3+ + e – = Fe 2+ E o = 0.771 V
Write down the overall reaction and calculate the standard free energy change of the overall reaction.
Model Answer
Sn 2+ + 2 Fe 3+ Sn 4+ + 2 Fe 2+ E o = 0.617 V
G o = – n F E o = – 2 F E o = – 2 × 96485 × 0.617 = – 119 kJ
Determine the equilibrium constant of the reaction.
Model Answer
log K = (n E o) / 0.0592
log K = (2 × 0.617) / 0.0592 20.84
K = 6.92 × 10^20
A volume of 20 cm3 of Sn 2+ solution (c = 0.10 mol dm-3) is titrated with Fe 3+ solution (c = 0.20 mol dm-3). Calculate the voltage of the cell:
i) when 5 cm3 of Fe 3+ solution is added,
ii) at the equivalence point,
iii) when 30 cm3 Fe 3+ of the solution is added.
The saturated calomel electrode (E o SCE = 0.242 V) is used as the reference electrode in the titration.
Model Answer
Before the equivalence point, E of the cell is given by the following equation:
Ecell = E o Sn4+/Sn2+ + (0.0592 / 2) log ([Sn4+]/[Sn2+]) - E o SCE
i. The addition of 5.00 cm3 of Fe3+ solution converts 5.00 / 20.00 of the Sn2+ to Sn4+, thus
[Sn4+]/[Sn2+] = (5.0 / 20.0) / (15.0 / 20.0) = 3.00 / 15.0 = 1/3 (Wait, 5/15 = 1/3)
Ecell = – 0.102 V
ii. At the equivalence point, add the two expressions corresponding to Sn4+ / Sn2+ and Fe3+ / Fe2+ to get
Ecell = (2 E o Sn4+/Sn2+ + E o Fe3+/Fe2+) / 3 - E o SCE
Ecell = (2 × 0.154 + 0.771) / 3 – 0.242 = 0.118 V
iii. Beyond the equivalence point, E of the cell is given by the following equation:
Ecell = E o Fe3+/Fe2+ - 0.0592 log ([Fe2+]/[Fe3+]) - E o SCE
When 30 cm3 of Fe3+ solution is added, 10 cm3 of the solution is in excess, i.e.
[Fe2+]/[Fe3+] = 20.0 / 10.0 = 2.00
Ecell = 0.551 V
One of the important analytical methods for estimation of Cu 2+ is iodometric titration. In this reaction Cu 2+ is reduced to Cu + by I– and the liberated I2 is then titrated with standard Na2S2O3 solution. The redox reaction is as follows:
2 Cu 2+ + 4 I – 2 CuI(s) + I2 (aq)
Electrode potentials of the relevant half-cells are:
Cu 2+ + e – = Cu + E o = 0.153 V
I2 + 2 e – = 2 I– E o = 0.535 V
A consideration of the electrode potentials would indicate that reduction of Cu 2+ by I– is not a spontaneous reaction. However, in the iodometric titration this reaction does take place. Let us try to understand the anomaly:
Cu + has low solubility in water with Ksp = 1.1×10–12 . Calculate the effective E o value for the equilibrium CuI(s) = Cu + + I– .
Model Answer
G o = – R T ln Ksp = 68.27 J mol-1
G o = – n F E o
n = 1
E o = – 0.707 V
Using the result in 9.4 calculate the effective E o value for the reduction of Cu 2+ by I– . What does this value suggest about the spontaneity of the reaction?
Model Answer
Cu + + I – CuI(s) E o = 0.707 V
Cu 2+ + e – Cu + E o = 0.153 V
The overall reaction for reduction of Cu 2+ by I – is
Cu 2+ + I – + e – CuI(s) E o = 0.86 V
The E o value for the reduction of Cu 2+ by I – can now be calculated
2 x (Cu 2+ + I – + e – CuI(s) ) E o = 0.86 V
I2 + 2 e – 2 I – E o = 0.535 V
The overall reaction is
2 Cu 2+ + 4 I – 2 CuI(s) + I2 E o = 0.325 V
The positive value of effective E indicates that the reduction reaction is spontaneous. This has come about since in this reaction, I – is not only a reducing agent, but is also a precipitating agent. Precipitation of Cu + as CuI is the key step of the reaction, as it practically removes the product Cu + from the solution, driving the reaction in the forward direction.
Calculate the equilibrium constant of the reduction reaction in 9.5.
Model Answer
G o = – n F E o
Here n = 1, E o = 0.325V
G o = – 31.3 kJ
G o = – RT ln K
K = 2.9×10^5