Question

The standard potential of the following cells is 0.23 V at 15${}^o$C and 0.21 V at 35${}^o$C.
$Pt|H_2\left(g\right)|HCl\left(aq\right)|AgCl\left(s\right)|Ag\left(s\right)$

Calculate $-\Delta H^o$ (in kJ) for the cell reaction by assuming that these quantities remain unchanged in the range 15${}^o$C to 35${}^o$C( to the nearest integer)

Answer 1

The cell notation is $P{t}_{H_2}\left(g\right)|HCl\left(aq\right)|AgC{l}_s|Ag\left(s\right)$
The half cell reactions and the overall cell reactions are given below.
(i) $\frac{1}{2}{H}_2\to H^{+}+eAnode$
$AgCl+e\to Ag+C{l}^{-}Cathode$
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$\frac{1}{2}{H}_2+AgCl\to H^{+}+Ag+C{l}^{-}$
(ii) The standard Gibbs free energy change is
$-\Delta G^o=n{E}^{o}F=1\times 0.23\times 96500=22195J\left(at{15}^{o}C\right)$
$-\Delta G^o=n{E}^{o}F=1\times 0.21\times 96500=20265J\left(at{36}^{o}C\right)$
Also, the relationship between the standard Gibbs free energy change and the standard enthalpy change is
$\Delta G^o=\Delta H^o-T\Delta S^o$
$\therefore -22195=\Delta H^o-288\times \Delta S^o$
$-20265=\Delta H^o-308\times \Delta S^o$
+ - +
------------------------------------------------------------
$\therefore \Delta S^o=-96.50J$
Also, $-22195=\Delta H^o-288\times (-96.5)=-49987J$
$\therefore \Delta H^o=-49.987kJ$
Hence the standard enthalpy change for the reaction is $49.987kJ$.