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Question: The standard electrode potential of a metal ion (\(Ag/A{{g}^{+}}\)) and metal insoluble salt anion (...

The standard electrode potential of a metal ion (Ag/Ag+Ag/A{{g}^{+}}) and metal insoluble salt anion (Ag/AgCl/Cl−^{-}) are related as:
A. EAg−∣Ag−=ECl−/AgCl/Ag−+RTFlnKspE_{A{{g}^{-}}\mid Ag}^{-}=E_{C{{l}^{-}}/AgCl/Ag}^{-}+\dfrac{RT}{F}ln{{K}_{sp}}
B. ECl−∣AgCl∣Ag−=EAg+/Ag−+RTFlnKspE_{C{{l}^{-}}\mid AgCl\mid Ag}^{-}=E_{A{{g}^{+}}/Ag}^{-}+\dfrac{RT}{F}ln{{K}_{sp}}
C. EAg−∣Ag−=ECl−/AgCl/Ag−+RTFln[Cl−]KspE_{A{{g}^{-}}\mid Ag}^{-}=E_{C{{l}^{-}}/AgCl/Ag}^{-}+\dfrac{RT}{F}ln\dfrac{[C{{l}^{-}}]}{{{K}_{sp}}}
D. ECl−∣AgCl∣Ag−=EAg+/Ag−+RTFlnKsp[Cl−]E_{C{{l}^{-}}\mid AgCl\mid Ag}^{-}=E_{A{{g}^{+}}/Ag}^{-}+\dfrac{RT}{F}ln\dfrac{{{K}_{sp}}}{[C{{l}^{-}}]}

Explanation

Solution

In a cell there will be two electrodes called anode and cathode. At anode oxidation chemical reaction occurs and at cathode reduction chemical reaction occurs. Both anode and cathode are going to act as half-cells.

Complete step by step answer:
- In the question it is asked to write the standard electrode potential of the metal ion (Ag/Ag+Ag/A{{g}^{+}}) and metal insoluble salt anion (Ag/AgCl/Cl−^{-}).
- At anode silver metal is going to lose the electrons meaning it undergoes oxidation chemical reaction.
- The oxidation reaction at anode is as follows.
Ag→Ag++Cl−Ag\to A{{g}^{+}}+C{{l}^{-}}
- At cathode reduction chemical reaction is going to occur, at cathode silver chloride is going to accept the electrons and convert into silver metal and chloride anion.
- The reduction in chemical reaction is as follows.
AgCl+e−→Ag+Cl−AgCl+{{e}^{-}}\to Ag+C{{l}^{-}}
- Then we can write the electrode potential of the cell as follows.

Ecell=Ecathode−Eanode Ecell−=ECl−∣AgCl∣Ag−−EAg+/Ag−→(i)  {{E}_{cell}}={{E}_{cathode}}-{{E}_{anode}} \\\ E_{cell}^{-}=E_{C{{l}^{-}}\mid AgCl\mid Ag}^{-}-E_{A{{g}^{+}}/Ag}^{-}\to (i) \\\

- Substitute the cell potential in the below equation to get the electrode potential and it is as follows.

Ecell=Ecell−−RTnFln⁡[Ag+][Cl−](here n=1) 0=Ecell−−RTnFln⁡Ksp Ecell−=RTnFln⁡Ksp→(ii)  {{E}_{cell}}=E_{cell}^{-}-\dfrac{RT}{nF}\ln [A{{g}^{+}}][C{{l}^{-}}](here\text{ }n=1) \\\ 0=E_{cell}^{-}-\dfrac{RT}{nF}\ln {{K}_{sp}} \\\ E_{cell}^{-}=\dfrac{RT}{nF}\ln {{K}_{sp}}\to (ii) \\\

- Substitute (ii) in (i) to get the electrode potential and it is as follows.
ECl−∣AgCl∣Ag−=EAg+/Ag−+RTFlnKspE_{C{{l}^{-}}\mid AgCl\mid Ag}^{-}=E_{A{{g}^{+}}/Ag}^{-}+\dfrac{RT}{F}ln{{K}_{sp}}

Note: We have to consider the cell equation to get the electrode potential of the total cell. The metal which has high potential is going to act as cathode and the electrode which has less potential is going to act as anode (loss electrons or oxidation).