Question

1. Write the balanced oxidation and reduction half reactions for the reaction of (As_2O_3)^3- and I_2. What ion is oxidized and what ion is reduced?

2. Assuming that 5 mL of 0.006M arsenite solution is subjected to coulometric titration using a current of 30 mamps. Calculate the time that the current should be applied to reach the endpoint of the titration.

Answer 1

1) The equation will be

As₂O₃+ I₂ --> AsO4^3- + I-

SO here the oxidized species will be As2O3 and reduced one will be I-

the reduction is said to take place if atom accept electron (I2 (oxidation state of I =0 , I- = -1)

The oxidation is said to take place if atom loses electron and gains positive charge

As in As2O3 = +3

As in  AsO4^3- = +5

So oxidation of Arsenic.

The half reactions will be ( lets say in acidic medium)

Oxidation

As₂O₃   --> AsO4^3-

Let us balance the As first

As₂O₃   --> 2AsO4^3-

Now balance the oxygen on the two side by adding water

As₂O₃ + 5H2O --> 2AsO4^3-

Now balance the hydrogen by adding H+

As₂O₃ + 5H2O --> 2AsO4^3- + 10H+

Now balance charge by adding electrons

As₂O₃ + 5H2O --> 2AsO4^3- + + 4e + 10H+ [Balanced oxidation half cell]

Let us consider the reduction half cell

I2 --> I-

Balance atoms and charge

I2 + 2e --> 2I-   [Balanced reduction half reaction]

Complete balanced equation will be

As₂O₃ + 2 I₂ + 5 H2O --> 2 AsO4^3- + 10 H⁺ + 4 I^-

2) The moles of aresenite solution at equivalence point or end point will be equal to

Moles = Current X time / number of electons X faraday constant

Moles = Molarity X volume = 0.006 x 5 / 1000 = 0.03 X 10^-3 moles

So time =Molex X number of electrons X Faraday constant / current

Time = 0.03 X 10^-3 X 4 X 96500 / 30 = 0.386 seconds