Question

A 20 cm³ aliquot of a 0.0577 mol dm^-3 vanadium(V) solution wasreduced by iodide ions to an unknown oxidation state (x). The liberated iodine (I₂) was titrated with 10.2 cm³ of a 0.0986 mol dm^-3 Na₂S₂O₃ solution.

Calculate the experimental value for the unknown oxidation state x (to two decimal places).

Answer 1

Solution :-

Balanced reaction equation for the I2 with Na2S2O3 is as follows

I2   +   2 Na2S2O3   -----> 2 NaI   +   NaS4O6

So lets first calculate the moles of the Na2S2O3

Moles = molarity * volume

Moles of Na2S2O3 = 0.0986 mol /L * 0.0102 L = 0.000986 mol Na2S2O3

Now using the mole rati lets calculate the moles of the I2 liberated

Moles of I2 liberated = 0.000986 mol Na2S2O3 * 1 mol I2 / 2 mol Na2S2O3

                                   = 0.000493 mol I2

Moles of iodine unknown ion = 0.000493 mol I2 * 2 mol iodine ion = 0.000986 mol iodine ions

Lets calculate the moles of the Vadium (v) ion

Moles of Vanadium ion = 0.0577 mol per L * 0.020 L = 0.001154 mol

Now lets calculate the total electrons needed to reduce this moles of vanadium ion

0.001154 molVanadium (V) ion * 5 mol e- / 1mol ion = 0.00577 mol electrons

So the number of electrons given by the iodine = 0.00577

So change in the electrons of each iodine ion = 0.00577 mol / 0.000986 mol = 5.85

So the oxidation state of the each unknown iodine ion is 5.85 or w can round it to 6.00