Question

The temperature coefficient of a spontaneous cell is $(\frac{\delta E}{\delta T}{)}_P$. Choose the correct statements.

• A. When $(\frac{\delta E}{\delta T}{)}_P=0$, then $△H=-nFE$
• B. When $(\frac{\delta E}{\delta T}{)}_P<0$, then $△H<nFE$ for endothermic reaction
• C. When $(\frac{\delta E}{\delta T}{)}_P<0$, then $-△H>nFE$ for exothermic reaction
• D. When $(\frac{\delta E}{\delta T}{)}_P=0$, then $△H=nFE$

Answer 1

Answer: (A), (C)

The correct options are
A When $(\frac{\delta E}{\delta T}{)}_P=0$, then $△H=-nFE$
C When $(\frac{\delta E}{\delta T}{)}_P<0$, then $-△H>nFE$ for exothermic reaction

we know,
$△G=△H+T(\frac{\delta G}{\delta T}{)}_P$
$△G=-nFE=△H-nFT(\frac{\delta E}{\delta T}{)}_P$
so, $(\frac{\delta E}{\delta T}{)}_P=\frac{△H+nFE}{nFT}$
(a) When $(\frac{\delta E}{\delta T}{)}_P=0$, then $△H=-nFE$
(b) $(\frac{\delta E}{\delta T}{)}_P<0$ , for endothermic reaction will only be possible when $E_{cell}=-ve$, because for endothermic reactions $△H=+ve$. But as the given cell is spontaneous, $E_{cell}$ has to be positive, so $(\frac{\delta E}{\delta T}{)}_P$ cannot be less than zero.
(c) When $(\frac{\delta E}{\delta T}{)}_P<0$, then $-△H>nFE$, for an exothermic reaction. As for an exothermic reaction, $△H=-ve$ and we have $E_{cell}=+ve$. So, the magnitude of $△H$ has to be greater than that of $nFE$ for $(\frac{\delta E}{\delta T}{)}_P<0$.