Question

The chlorate ion can be disproportionate in basic solution according to reaction,
 $ 2ClO_{3}^{-}\text{ }\rightleftharpoons \text{ }ClO_{2}^{-}\text{ + }ClO_{4}^{-} $
What is the equilibrium concentration of perchlorate ions from a solution initially at 0.1 M in chlorate ions at 298 K?
Given that: $ E_{ClO_{4}^{-}\text{ }\!\!|\!\!\text{ }ClO_{3}^{-}}^{O} $ = 0.36 V and $ E_{ClO_{3}^{-}\text{ }\!\!|\!\!\text{ }ClO_{2}^{-}}^{O} $ = 0.33V at 298 K.
(A) 0.019M
(B) 0.024M
(C) 0.1M
(D) 0.19M

Answer 1

Hint: The disproportionation of chlorate ion ( $ ClO_{3}^{-}\text{ } $ ) forms Chlorite ( $ \text{ }ClO_{2}^{-} $ )and perchlorate ( $ \text{ }ClO_{4}^{-} $ ) ions. To find the equilibrium concentration we will use the Nernst equation first and then calculate the equilibrium constant. $ ClO_{3}^{-}\text{ } $ is oxidized to form $ \text{ }ClO_{4}^{-} $ and it is reduced to form $ \text{ }ClO_{2}^{-} $ .

Complete Step By Step Answer:
We’ll first see the oxidation of $ ClO_{3}^{-}\text{ } $ : It loses electrons to form perchlorate ions.
(i) $ ClO_{3}^{-}{{\text{ }}_{\left( aq \right)}}\text{ + }{{\text{H}}_{2}}{{O}_{\left( l \right)}}\text{ }\to \text{ }ClO{{_{4}^{-}}_{\left( aq \right)}}\text{ + 2}{{\text{H}}^{+}}_{_{\left( aq \right)}\text{ }}\text{ + 2}{{\text{e}}^{-}} $
Now, let’s see the reduction of $ ClO_{3}^{-}\text{ } $ It gains the electrons to form chlorite ions.
(ii) $ ClO_{3}^{-}{{\text{ }}_{\left( aq \right)}}\text{+ 2}{{\text{e}}^{-}}\text{+ 2}{{\text{H}}^{+}}_{_{\left( aq \right)}\text{ }}\to \text{ 2}ClO{{_{2}^{-}}_{\left( aq \right)}}\text{ + }{{\text{H}}_{2}}{{O}_{\left( l \right)}} $
Adding the equation (i) and (ii) we get the resultant disproportionation equation as,
 $ 2ClO{{_{3}^{-}}_{\left( aq \right)}}\text{ }\rightleftharpoons \text{ }ClO{{_{2}^{-}}_{_{\left( aq \right)}}}\text{ + }ClO{{_{4}^{-}}_{_{\left( aq \right)}}} $

Emf of the anode	$ E_{ClO_{4}^{-}\text{ }\!\!|\!\!\text{ }ClO_{3}^{-}}^{O} $ = 0.36V
Emf of the cathode.	$ E_{ClO_{3}^{-}\text{ }\!\!|\!\!\text{ }ClO_{2}^{-}}^{O} $ = 0.33V

We know the EMF of the cell,
 $ \Rightarrow E_{cell}^{o}\text{ = E}_{cathode}^{o}\text{ - E}_{anode}^{o} $
 $ =\text{ }(0.33\text{ - 0}\text{.36) V} $
 $ =\text{ - 0}\text{.03 V} $
Now, using Nernst equation,
 $ \Rightarrow EMF\text{ = E}_{cell}^{o}\text{ - }\dfrac{0.059}{n}\text{ log K} $
But the equation is in equilibrium so EMF will be zero.
Substituting 0 for EMF in the above equation we get,
 $ 0\text{ = E}_{cell}^{o}\text{ - }\dfrac{0.059}{n}\text{ log K} $
Bringing $ \dfrac{0.059}{n}\log \text{K} $ to the LHS, we get
 $ \Rightarrow \dfrac{0.059}{n}\text{ log K = E}_{cell}^{o} $
Arrange the above equation with $ E_{cell}^{o} $ in the LHS,
 $ \Rightarrow E_{cell}^{o}\text{ = }\dfrac{0.059}{n}\text{ log K} $
 $ \Rightarrow E_{cell}^{o}\text{ = }\dfrac{RT}{nF}\text{ ln K} $
n=2 as there are two ions formed and $ E_{cell}^{o}\text{ = -0}\text{.03 V} $
Putting the above values in the equation we get,
 $ \Rightarrow -0.03\text{ = }\dfrac{RT}{2F}\text{ ln K} $
 $ We\text{ get }\dfrac{RT}{F}\text{ = 0}\text{.06} $ ,
 $ \therefore \text{ - 0}\text{.03 = }\dfrac{0.06}{2}\text{ ln K} $
 $ \therefore \text{ K = 0}\text{.1} $
Now, assume that the concentration of each reactant and product is,
 $ 2ClO_{3}^{-}\text{ }\rightleftharpoons \text{ }ClO_{2}^{-}\text{ + }ClO_{4}^{-} $

$ 2ClO_{3}^{-}\text{ } $	$ ClO_{2}^{-} $	$ ClO_{4}^{-} $
$ 0.1\text{ - 2x } $	x	x

Thus, $ K\text{ }=\text{ }\dfrac{concentration\text{ }of\text{ }products}{concentration\text{ of reactants}} $
 $ \Rightarrow K\text{ = }\dfrac{{{x}^{2}}}{{{(0.1-2x)}^{2}}} $
 $ \Rightarrow \text{0}\text{.1 = }\dfrac{{{x}^{2}}}{{{(0.1-2x)}^{2}}} $
 $ \Rightarrow \dfrac{1}{10}\text{ = }\dfrac{{{x}^{2}}}{{{(0.1-2x)}^{2}}} $
By cross multiplying,
 $ {{(0.1-2x)}^{2}}\text{ = 10}{{x}^{2}} $
Calculating the square of $ {{\left( 0.1\text{ - 2x} \right)}^{2}} $ by using the identity $ {{\left( a-b \right)}^{2}}\text{ = }{{\text{a}}^{2}}\text{ - 2ab + }{{\text{b}}^{2}} $
 $ 0.01\text{ - 0}\text{.4x + 4}{{\text{x}}^{2}}\text{ = 10}{{\text{x}}^{2}} $
 $ \Rightarrow 3.16x\text{ = 0}\text{.1 - 2x} $
 $ \Rightarrow 5.16x\text{ }=\text{ }0.1 $
 $ \Rightarrow x\text{ = }\dfrac{0.1}{5.16} $
 $ \therefore x\text{ = 0}\text{.0193 M} $
Final answer is Option A= 0.019 M.

Note:
Calculating the Nernst equation of various concentrations of cells will help you to understand the equation questions easily. Also, always keep in mind that if the compound loses an electron, reduction occurs, and if it gains electrons oxidation takes place. Simultaneous loss and gain of electrons takes place in the redox reactions.