Question

$△G=△H-T△S$ and $△G=△H+T{\left(\frac{d\left(△G\right)}{dT}\right]}_p$ then $\left(\frac{dEcell}{dT}\right)$ is

• A. $\frac{△S}{nF}$
• B. $\frac{nE}{△S}$
• C. $-nF{E}^{cell}$
• D. $+nE{F}_{cell}$

Answer 1

The correct option is A $\frac{△S}{nF}$

$\Delta G=\Delta H-T\Delta S⟶\left(i\right)$

$\Delta G=\Delta H+{T\left[\frac{d\left(\Delta G\right)}{dT}\right]}_P$

$=\Delta H+T.(-nF\left[\frac{d{E}_{cell}}{dT}\right])\{since\Delta G=-nF{E}_{cell}\Rightarrow {\left[\frac{d\left(\Delta G\right)}{dT}\right]}_P=-nF{\left[\frac{d{E}_{cell}}{dT}\right]}_P$

$=\Delta H.T.n.F.{\left[\frac{d{E}_{cell}}{dT}\right]}_P⟶\left(ii\right)$

Comparing (i) with (ii) we will get,

$\Delta S=\frac{d{E}_{cell}}{dT}\times nF$

$\Rightarrow \frac{d{E}_{cell}}{dT}=\frac{\Delta S}{nF}$