Question

$\Delta H^{\ominus }$ for the cell reaction assuming that these quantities remain unchanged in the range ${15}^{\circ }C$ to ${35}^{\circ }$ C is:

• A. $-99.97kJmo{l}^{-1}$
• B. $99.97kJmo{l}^{-1}$
• C. $-9.99kJmo{l}^{-1}$
• D. $9.99kJmo{l}^{-1}$

Answer 1

Answer: (A)

$-99.97kJmo{l}^{-1}$

At anode : $H_2\left(g\right)\to 2H^{\oplus }\left(aq\right)+2e^{-}\left(oxidation\right)$
At cathode : $2AgBr\left(s\right)+2e^{-}\to 2Ag\left(s\right)+\begin{array}{c}2B{r}^{\ominus }\left(aq\right)\\ \left(Reduction\right)\end{array}$
$\Delta G^{\ominus }at{15}^{\circ }C=-nF{E}_{cell}^{\ominus }$
$=-2\times 96500Cmo{l}^{-1}\times 0.23V$
$=-44390Jmo{l}^{-1}$
$\Delta G^{\ominus }at{35}^{\circ }C=-nF{E}_{cell}^{\ominus }$
$=-2\times 96500Cmo{l}^{-1}\times 0.21V$
$=-40530Jmo{l}^{-1}$
Now, $\Delta G^{\ominus }at{15}^{\circ }C=\Delta H^{\ominus }-T\Delta S^{\ominus }$
$-44390Jmo{l}^{-1}=\Delta H^{\ominus }-\left(288K\right)\Delta S^{\ominus }$ ....(i)
and $\Delta G^{\ominus }at{35}^{\circ }C=\Delta H^{\ominus }-\left(308K\right)\Delta S^{\ominus }$
$-40530Jmo{l}^{-1}=\Delta H^{\ominus }-\left(308K\right)\Delta S^{\ominus }$ .... (ii)
(Assuming $\Delta H$ and $\Delta S$ at 298 K are temperature independent, i.e., $\Delta H$ at ${15}^{\circ }C=\Delta H^{\ominus }$ and $\Delta H$ at ${35}^{\circ }C=\Delta H^{\ominus }$).
Solving for $\Delta H^{\ominus }$ and $\Delta S^{\ominus }$ from equation (i) and (ii) we get,
$-44390+288\Delta S^{\ominus }=\Delta H^{\ominus }$ .... (iii)
$-40530+308\Delta S^{\ominus }=\Delta H^{\ominus }$ .... (iv)
Equating equations (iii) and (iv), we get
$\Delta S^{\ominus }=-193J{K}^{-1}mo{l}^{-1}$
Substituting the value of $\Delta S^{\ominus }$ in either equation (iii) or (iv),
$-44390J{K}^{-1}+288+288\times (-193J{K}^{-1})=\Delta H^{\ominus }$
$\therefore \Delta H^{\ominus }=-99974J{K}^{-1}=-99.974kJmo{l}^{-1}$