Question

Assertion: The standard potential of half-cell is related to that of $A{{g}^{-}}|Ag$.
Reason:
${{E}^{-}}_{A{{g}^{-}}|Ag}={{E}^{-}}_{C{{l}^{-}}|AgCl|Ag}+\frac{RT}{F}\ln {{K}_{sp}}\left( AgCl \right)$
A. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
B. Both Assertion and Reason are true and Reason is not the correct explanation of Assertion.
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true
E. Both Assertion and Reason are false

Answer 1

Standard electrode potential for a half- reaction is basically the potential of a half-reaction which is measured against the SHE (standard hydrogen electrode) under the standard conditions. First write the cathode reaction and then write the anode reaction. Using this write the net reaction and calculate the value of ${{E}_{cell}}$. This will lead you to the answer.

Complete step by step answer:
- First of all as we know that at anode oxidation takes place and at cathode reduction takes place. So we can say that it is a half-cell reaction because here redox reaction takes place, that is oxidation at one electrode and reduction occurs at another electrode. Hence, we can write the anode reaction as:
$Ag+{{e}^{-}}\to A{{g}^{+}}$

We can write the cathode reaction as:
$2C{{l}^{-}}\to C{{l}_{2}}+2{{e}^{-}}$

We can write the net cell reaction as:
$Ag+2C{{l}^{-}}\to C{{l}_{2}}+A{{g}^{+}}$

-The ${{E}_{cell}}$ will be=
${{E}_{cell}}=({{E}^{\circ }}_{C{{l}^{-}}|AgCl}-{{E}^{\circ }}_{A{{g}^{+}}|Ag})-\frac{RT}{F}\ln {{K}_{sp}}$

-The ${{E}_{cell}}$ will be zero
-So, it will become: ${{E}^{\circ }}_{C{{l}^{-}}|AgCl} = {{E}^{\circ }}_{A{{g}^{+}}|Ag}+\frac{RT}{F}\ln {{K}_{sp}}$
So, the correct answer is “Option E”.

Note: - The main factors that affect cell potential are surface area, temperature, and concentration.
- If the value of ${{E}^{\circ }}_{cell}$ is negative, then the reaction is not spontaneous and is found to not occur under standard conditions. Whereas, if the value of ${{E}^{\circ }}_{cell}$ is positive, then the reaction will occur spontaneously.