Question

The standard electrode potential of a metal ion $(\left. {Ag} \right|A{g^ + })$ and metal insoluble salt anion $(\left. {Ag} \right|AgCl\left| {C{l^ - }} \right.)$ are related as:
A. $E_{A{g^ + }\left| {Ag} \right.}^ - \left| = \right.E_{C{l^ - }\left| {AgCl\left| {Ag} \right.} \right.}^ - + \dfrac{{RT}}{F}\ln {K_{sp}}$
B. $E_{C{l^ - }\left| {AgCl\left| {Ag} \right.} \right.}^ - = E_{A{g^ + }\left| {Ag} \right.}^ - + \dfrac{{RT}}{F}\ln {K_{sp}}$
C. $E_{A{g^ + }\left| {Ag} \right.}^ - = E_{C{l^ - }\left| {AgCl\left| {Ag} \right.} \right.}^ - + \dfrac{{RT}}{F}\ln \dfrac{{[C{l^ - }]}}{{{K_{sp}}}}$
D. $E_{C{l^ - }\left| {AgCl\left| {Ag} \right.} \right.}^ - = E_{A{g^ + }\left| {Ag} \right.}^ - + \dfrac{{RT}}{F}\ln \dfrac{{{K_{sp}}}}{{[C{l^ - }]}}$

Answer 1

This kind of electrode functions as a redox electrode and the equilibrium is attained between silver metal and its salt. The Nernst equation shows the dependence of the potential of silver-silver chloride on the activity of chloride ions. Further, we know that at equilibrium, the ionic product is equivalent to the solubility product of a cell.

Complete step-by-step answer:
Metal-metal insoluble salt electrodes such as silver-silver chloride have the metal in contact with one of its sparingly soluble salt and a solution containing the ion present in the salt other than the metal. The silver-silver chloride electrode is represented as $C{l^ - }\left| {AgCl\left| {Ag} \right.} \right.$ is made by dipping a silver wire into a solution containing chloride ions resulting in the deposition of silver chloride on the silver wire. Its reaction can be written as:
At Cathode: $AgCl \to Ag + C{l^ - }$
At anode: $Ag \to A{g^ + } + {e^ - }$
So, the overall equation for the reaction
$Ag(s) + C{l^ - } \rightleftharpoons AgCl(s) + {e^ - }$
When Q is the ionic product of the cell and ${K_{sp}}$ is the solubility product constant, at equilibrium when the solution is saturated, electrode potential of the cell becomes zero and ionic product becomes equal to solubility product.
${E_{cell}} = 0 \Rightarrow {Q_{cell}} = {K_{sp}}$ i.e. the system is at equilibrium and there is no shift to either left or right. Or we can also say that there will be no evolution of the reaction either forward or backward. We can define the cell potential of this reaction with the help of Nernst equation.
${E_{cell}} = E_{cell}^o - \dfrac{{RT}}{F}\ln {K_{sp}}$
Where, ${K_{sp}}$ is the solubility product constant, which indicates how far the dissolution proceeds at saturation or equilibrium.
For this reaction, we can write the electrode potential difference as
$E_{cell}^o = E_{Ag\left| {AgCl\left| {C{l^ - }} \right.} \right.}^o - E_{A{g^ + }\left| {Ag} \right.}^o$
$0 = E_{cell}^o - \dfrac{{RT}}{F}\ln {K_{sp}}$
Putting the value of $E_{cell}^o$ in the above equation, we get
$E_{C{l^ - }\left| {AgCl\left| {Ag} \right.} \right.}^ - = E_{A{g^ + }\left| {Ag} \right.}^ - + \dfrac{{RT}}{F}\ln {K_{sp}}$
Hence, the correct option is (B).

Note: Metal insoluble salt anion electrode can be of mercury-amalgam too. But we generally use silver-silver chloride one because it can be easily prepared in the laboratory and it is reliable due to its non-toxicity as compared to calomel electrode i.e. mercury one.