Question

The standard reduction potential of $A{{g}^{+}}/Ag$ at 298 K is 0.799 volt. Given for AgI , ${{K}_{sp}}=8.7\times {{10}^{-17}}$, evaluate the potential of the $A{{g}^{+}}/Ag$ in a saturated solution of AgI. Also calculate the standard reduction potential of the ${{I}^{-}}/AgI/Ag$ electrode.

Answer 1

Nernst equation is used to calculate cell potential asked in the question. ${{K}_{sp}}$ refers to the solubility product of the salt. It is merely the product of the concentration of ions of salt present in the solution.

Complete step by step solution:
Here, we will first find the potential of $A{{g}^{+}}/Ag$ in the saturated solution of AgI. We will use the Nernst equation.
${{E}_{cell}}={{E}^{o}}_{cell}-\dfrac{RT}{nF}\operatorname{lnQ}$………………….(1)
Here, n=number of electrons involved in the reaction
$F$=Faraday constant=96500 $Cmo{{l}^{-1}}$
$Q$=Equilibrium constant of the reaction
${{E}_{cell}}$=Cell Potential
${{E}^{o}}_{cell}$=Standard cell potential
Here, we don’t have the value of Q. So, we will need to find that first.
We know that cell reaction is $A{{g}^{+}}\to Ag$
So, Equilibrium constant $Q=\left[ \dfrac{Ag}{A{{g}^{+}}} \right]$………………………..(2)
We can find their value from ${{K}_{sp}}$(Solubility Product).
For AgI,
Solubility product ${{K}_{sp}}=\left[ A{{g}^{+}} \right]\left[ {{I}^{-}} \right]$……..(3)
But As $AgI\to A{{g}^{+}}+{{I}^{-}}$, concentration of $A{{g}^{+}}$and${{I}^{-}}$ will be same, So equation (3) can be re-written as
${{K}_{sp}}={{\left[ A{{g}^{+}} \right]}^{2}}$…………(4)
or ${{K}_{sp}}={{\left[ {{I}^{-}} \right]}^{2}}$ but we require $A{{g}^{+}}$, So we will use former equation.
Putting available values in the Equation (4),
$8.7\times {{10}^{-17}}={{\left[ A{{g}^{+}} \right]}^{2}}$
So, we can write that
$\left[ A{{g}^{+}} \right]=\sqrt{8.7\times {{10}^{-17}}}$
Thus,
$\left[ A{{g}^{+}} \right]=9.32\times {{10}^{-9}}M$
Now put this value of concentration in equation (2),
$Q=\left[ \dfrac{Ag}{9.32\times {{10}^{-9}}} \right]$As Ag is metallic we will take its concentration as 1. So,
$Q=\left[ \dfrac{1}{9.32\times {{10}^{-9}}} \right]$………….(5)
Put equation (5) and other constants in equation (1) so will get
${{E}_{cell}}=0.799-\dfrac{8.314\times 298}{96500}\ln \left[ \dfrac{1}{9.32\times {{10}^{-9}}} \right]$
As $A\ln x=2.303A\log x$,
${{E}_{cell}}=0.799-\dfrac{8.314\times 298\times 2.303}{96500}\log \left[ \dfrac{1}{9.32\times {{10}^{-9}}} \right]$
So, we can write that
${{E}_{cell}}=0.799-\dfrac{0.0591}{1}\log \left[ 1.072\times {{10}^{9}} \right]$
Thus, we will get
${{E}_{cell}}=0.799-0.474$
Therefore,
${{E}_{A{{g}^{+}}/Ag}}=0.325volt$
Now, we will find the standard reduction potential of ${{I}^{-}}/AgI/Ag$.
For ${{I}^{-}}/AgI/Ag$ electrode, cell reactions are as below.
At Anode : $Ag\to A{{g}^{+}}+{{e}^{-}}$
At Cathode : $AgI+{{e}^{-}}\to Ag+{{I}^{-}}$
Cell reaction : $AgI\to A{{g}^{+}}+{{I}^{-}}$
Now Applying Nernst equation and applying all the available values, we get
${{E}_{cell}}={{E}^{o}}_{cell}-\dfrac{0.0591}{n}\log \left[ A{{g}^{+}} \right]\left[ {{I}^{-}} \right]$
But at equilibrium ${{E}_{cell}}$=0
So,
$0={{E}^{o}}_{cell}-\dfrac{0.0591}{1}\log \left[ 8.7\times {{10}^{-17}} \right]$
${{E}^{0}}_{cell}=-0.95volt$
Now, we know that
${{E}^{0}}_{cell}={{E}^{0}}_{cathode}+{{E}^{0}}_{anode}$
$-0.95=-0.799+{{E}^{o}}_{Ag/AgI/{{I}^{-}}}$
${{E}^{o}}_{Ag/AgI/{{I}^{-}}}=-0.151volt$
So, potential of the $A{{g}^{+}}/Ag$ in a saturated solution of AgI is ${{E}_{A{{g}^{+}}/Ag}}=0.325volt$.
And Standard potential for ${{I}^{-}}/AgI/Ag$ electrode is ${{E}^{o}}_{Ag/AgI/{{I}^{-}}}=-0.151volt$.

Note: Do not forget to write the true value of the number of electrons involved in the reaction as it often leads to errors when more than one electron are involved in the reaction. Note that in the equilibrium constant, the concentration of Solid metallic atom is considered 1.0