Question

$E^o$ of an electrode half reaction is related to $\Delta G^o$ by the equation, $E^o=-\Delta G^o/nF$. If the amount of $A{g}^{+}$ in the half reaction $A{g}^{+}+e^{-}\to Ag$ is tripled then:

• A. n is tripled
• B. $\Delta G^o$ increases to three times
• C. $E^o$ reduces to one third
• D. All the above

Answer 1

The correct option is D All the above

If $A{g}^{+}$ is tripled, so Ag and electron will also be tripled as per stoichiometry of the reaction.

Now, $\Delta E^{\circ }=-\frac{\Delta G^{\circ }}{\eta F}$

Since $n=3,\Rightarrow \Delta E^{\circ }=-\frac{\Delta G^{\circ }}{\eta F}\Rightarrow -\Delta G^{\circ }=3\Delta E^{\circ }F$

So, $\Delta E^{\circ }$ is reduced to one-third.

But, $\Delta G^{\circ }$ increases to 3 time.

Therefore, correct option is D.