Question

Which of the following is correct relation?

• A. $-\Delta S={\left(\frac{\partial E}{\partial T}\right)}_P\times n$
• B. ${\left(\frac{\partial E}{\partial T}\right)}_P=\frac{\partial \left(\Delta S\right)}{\partial T}$
• C. $\Delta S={\left(\frac{\partial E}{\partial T}\right)}_p\times nF$
• D. $\left(\frac{\partial E}{\partial T}\right)=\frac{\Delta G-\Delta E}{T}$

Answer 1

The correct option is C $\Delta S={\left(\frac{\partial E}{\partial T}\right)}_p\times nF$

As we know,
$\Delta G_{system}=\Delta H-T\Delta S_{total}$......$\left(i\right)$
And, $\Delta G=-nFE$................$\left(ii\right)$
According to Gibb's Helmholtz equation, decrease in free energy , $\Delta G$ at constant pressure would be,
$-\Delta G=-\Delta H-T{\left(\frac{\delta \left(\Delta G\right)}{\delta T}\right)}_p$.....$\left(iii\right)$
Where $-\Delta H$ is the decrease in enthalpy of the cell at constant pressure.
Substituting the value of $-\Delta G$ from equation $\left(ii\right)$ in eqation $\left(iii\right)$, we get,
$nEF=-\Delta H-T{\left(\frac{\delta (-nEF)}{\delta T}\right)}_p$
$\Rightarrow nEF=-\Delta H+TnF{\left(\frac{\delta E}{\delta T}\right)}_p$
$\Rightarrow -nEF=\Delta H-TnF{\left(\frac{\delta E}{\delta T}\right)}_p$
$\Rightarrow \Delta G=\Delta H-TnF{\left(\frac{\delta E}{\delta T}\right)}_p$......$\left(iv\right)$
so, compairing equation $\left(i\right)\text{and}\left(iv\right)$ we know that,
$\Rightarrow \Delta S={\left(\frac{\partial E}{\partial T}\right)}_p\times nF$