Question

Write down the reduction reaction between Fe(CN)63- and Fe(CN)64-; find the standard electrode potential of this half reaction.

Calculate the ionic strength of 0.1 KCl and 0.1 M Na2SO4, respectively.

Answer 1

Solution:

Hydrogen peroxide in basic medium reduces Fe (CN)6^3- to Fe(CN)6^4-.

The net reaction are:

Fe(CN)6^3- + H2O2 + alkaline medium = Fe(CN)6^4- + O2

Here, in Fe(CN)6^3-, the oxidation number of Fe = +3 and it is changed into +2 in Fe(CN)6^4-

Therefore process in reduction.

The half cell reaction is:

Fe(CN)6^3- = Fe(CN)6^4-

The standard half cell potential in electrochemical series for:

E°(Fe^3+/Fe^2+) =  0.771 V

Part B)

Ionic strength (I) for binary electrolyte is calculated as:

I = 1/2 (m1z1^2 + m2z2^2)

Where, m1 =concentration of cation

m2= concentration of anion

z1 =charge on cation

z2 =charge on anion

For 0.1M KCl:

m1 = 0.1, m2= 0.1,z1 = +1, z2 =-1

Then, I = 1/2 [0.1 (+1)^2 + 0.1 (-1)^2] = 1/2 x 0.2 = 0.1

Ionic strength =0.10

For 0.1M Na2SO40:

Na2SO4 is dissociated as:

Na2SO4 = 2Na+ + SO4^2-

Then, m1 = 0.1 x 2 =0.2, z1 =+1

m2 =0.1 ,z2 = -2

Hence,

I = 1/2 [0.2(+1)^2 + 0.1(-2)^2] = 1/2 x 0.6 =0.3

Ionic strength =0.30