Question

Calculate the solubility product constant of AgI from the following values of standard electrode potentials.
${ E }_{ { Ag }^{ + }/Ag }^{ 0 }$ = 0.80 volt and ${ E }_{ { I/Ag }I/Ag }^{ 0 }$ = -0.15 volt at${ 25 }^{ 0 }C$.

Answer 1

Hint: ${ AgI }_{ (s) }\rightarrow { Ag }_{ (aq) }^{ + }+{ I }_{ (aq) }^{ - }$ is the cell reaction for the above question. We can easily find the solubility product $\left( { K }_{ sp } \right) $ from the standard electrode potential values by using the Nernst equation, which is given by :
${ E }_{ cell }={ E }_{ cell }^{ 0 }-\dfrac { 0.0591 }{ n } log\quad K$ . At equilibrium value of ${ E }_{ cell }$ is zero and ${ E }_{ cell }^{ 0 }$ is given by ${ E }_{ cell }^{ 0 }={ E }_{ cathode }^{ 0 }-{ E }_{ anode }^{ 0 }$ . Cathode is the half cell in which oxidation takes place and anode is the half cell in which reduction takes place.

Complete step by step answer:
Solubility product $\left( { K }_{ sp } \right) $ is defined as the equilibrium constant for the dissolution of a solid substance into an aqueous solution.
${ AgI }_{ (s) }\rightarrow { Ag }_{ (aq) }^{ + }+{ I }_{ (aq) }^{ - }$
Above given is the cell reaction as per the question.
And expression of Ksp is given by:
${ K }_{ sp }=\left[ { Ag }^{ + } \right] \left[ { I }^{ - } \right] $
Where $\left[ { Ag }^{ + } \right] $ is the concentration of silver ions and $\left[ { I }^{ - } \right] $ is the concentration of iodide ions.
To calculate the ${ K }_{ sp }$ of AgI from the given standard reduction potentials, first of all we need to write the half cell reactions.

Half cell reaction at cathode :
${ AgI }_{ (s) }+{ e }^{ - }\rightarrow { Ag }_{ (s) }+{ I }^{ - }$ ${ E }^{ 0 }$= -0.15 V

Half cell reaction at anode :
${ Ag }_{ (S) }\rightarrow { Ag }_{ (aq) }^{ + }+e^{ - }$ ${ E }^{ 0 }$= 0.80 V
We know that the expression of ${ E }_{ cell }^{ 0 }$ is given by :
${ E }_{ cell }^{ 0 }={ E }_{ cathode }^{ 0 }-{ E }_{ anode }^{ 0 }$
from the above half cell reactions, we know that ${ E }_{ cathode }^{ 0 }$= -0.15 V and ${ E }_{ anode }^{ 0 }$ = 0.80 V
${ E }_{ cell }^{ 0 }$ = -0.15 - 0.80 V = -0.95 V.

Now, using the Nernst equation:
${ E }_{ cell }={ E }_{ cell }^{ 0 }-\dfrac { 0.0591 }{ n } log\quad K$ ----- (i)
where n is the number of electrons , here in this case n=1
At equilibrium ${ E }_{ cell }^{ 0 }$ = 0 (because at equilibrium ${ E }_{ cathode }^{ 0 }$ becomes equal to ${ E }_{ anode }^{ 0 }$)
So, equation (i) becomes,
$0=\left( -0.95 \right) -\dfrac { 0.0591 }{ 1 } log\quad K$

We need to find the value of the solubility product, which is${ K }_{ sp }$.
Therefore, by rearranging we will get ,
$\dfrac { 0.95 }{ -0.0591 } =log\quad K$
log ${ K }_{ sp }$ = -16.07
${ K }_{ sp }={ 8.51 }\times { 10 }^{ -17 }{ (mol/L) }^{ 2 }$
Therefore, the answer is ${ 8.51 }\times { 10 }^{ -17 }{ (mol/L) }^{ 2 }$

Note: The principle of ${ K }_{ sp }$ solubility product is mainly used in the determination of solubility of particular compounds in a particular solvent. Mostly common ion effect comes into picture for this.When, the ${ Q }_{ sp }$<${ K }_{ sp }$ solution will be unsaturated in which more solute can be dissolved. I.e. No precipitation. When ${ Q }_{ sp }$=${ K }_{ sp }$ then the solution is saturated in which no more solute can be dissolved but no precipitation is formed. When ${ Q }_{ sp }$>${ K }_{ sp }$ then the solution will be supersaturated and precipitation takes place. Here${ Q }_{ sp }$ is the ionic product.