Question

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

Answer 1

We can easily find the solubility product (Ksp) from the standard electrode potential values by using the Nernst equation, which is given by,${{E}_{cell}}\text{ }=\text{ }{{E}^{0}}_{cell}-\text{ }0.0591nlog\text{ }K$. By applying the equilibrium conditions, the answer to this question can be approached.

Complete Solution :
We have studied about the cell reactions in our lower classes of physical chemistry and now we shall see the calculation of emf of the cell.
- The basic cell reaction here in the above question is as shown below:
$AgI\left( s \right)\rightleftharpoons A{{g}^{+}}\left( aq \right)\text{ }+\text{ }{{I}^{-}}\left( aq \right)$

The expression of the solubility product that is${{K}_{sp}}$ is given by:
${{K}_{sp}}\text{ }=\text{ }[A{{g}^{+}}][{{I}^{-}}]$
where $[A{{g}^{+}}]$ is the concentration of silver ions and $[{{I}^{-}}]$ 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 :
$AgI\left( s \right)\text{ }+\text{ }{{e}^{-}}\to ~\text{ }Ag\left( s \right)\text{ }+\text{ }{{I}^{-}}~$E0 = -0.15 V
Half cell reaction at anode :
$Ag\left( s \right)\text{ }\to ~\text{ }A{{g}^{+}}+\text{ }{{e}^{-}}$ E0 = 0.80 V

We know that the expression of E cell is given by:
${{E}^{0}}_{cell}=\text{ }{{E}^{0}}_{cathode}-\text{ }{{E}^{0}}_{anode}$
From the above half cell reactions, we know that ${{E}^{0}}_{cathode}=\text{ }-0.15\text{ }V$and ${{E}^{0}}_{anode}=\text{ }0.80V$
${{E}^{0}}_{cell}=\text{ }-0.15\text{ }-\text{ }0.80\text{ }V\text{ }=\text{ }-0.95\text{ }V.$

Now, using the Nernst equation:
${{E}_{cell}}\text{ }=\text{ }{{E}^{0}}_{cell}-\text{ }0.0591nlog\text{ }K$ ----- (i)
where, n is the number of electrons , here in this case n=1
At equilibrium ${{E}_{cell}}=\text{ }0$ (because at equilibrium emf of cathode becomes equal to emf of anode)
So, equation (i) becomes:
$0\text{ }=\text{ }\left( -0.95 \right)\text{ }-\text{ }0.0591(1)log\text{ }{{K}_{sp}}$
We need to find the value of the solubility product, which is${{K}_{sp}}$.

Therefore, by rearranging we will get,
$0.95-0.0591=\text{ }log\text{ }{{K}_{sp}}$
$\Rightarrow log\text{ }{{K}_{sp}}=\text{ }-16.07$

Thus, by taking antilog of above equation, we have,
${{K}_{sp}}=8.51\times {{10}^{-17}}{{(mol/L)}^{2}}$
Therefore, the answer is $8.51\times {{10}^{-17}}{{(mol/L)}^{2}}$

Note: The principle of solubility product ${{K}_{sp}}$ is mainly used in the determination of solubility of particular compounds in a particular solvent. Mostly the 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 occurs. 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 ${{O}_{sp}}$, is the ionic product.