91Ó°ÊÓ

Change from liquid to vapour phase involves heat of vaporisation, which requires energy to be consumed or released. Chemical energy is the energy associated with a molecule's chemical bonds.

Given,

$\begin{array}{l}{\mathrm{H}}_{2\left(\mathrm{g}\right)}{+2\mathrm{Ag}}_{\left(\mathrm{aq}\right)}^{+}→{2\mathrm{H}}_{\left(\mathrm{aq}\right)}^{+}+2{\mathrm{Ag}}_{\left(\mathrm{s}\right)}\\ {\mathrm{P}}_{{\mathrm{H}}_2}=1.00\mathrm{atm};\\ \left[{\mathrm{Ag}}^{+}\right]=0.100\mathrm{M}\end{array}$

Subtract the anode's reduction potential from the cathodes to get the ${\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}$

role="math" localid="1663832506705" $\begin{array}{r}{\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}{=\mathrm{E}}_{\mathrm{red}}^{\mathrm{o}}{-\mathrm{E}}_{\mathrm{ox}}^{\mathrm{o}}\\ =0.80-0\\ =0.80\mathrm{V}\end{array}$

Using the Nernst equation, determine the reaction's ${\mathrm{E}}_{\mathrm{cell}}{=\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{\mathrm{RT}}{\mathrm{nF}}×\mathrm{lnQ}$

$\begin{array}{c}{\mathrm{E}}_{\mathrm{cell}}{=\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{\mathrm{RT}}{\mathrm{nF}}×\mathrm{lnQ}\\ \mathrm{lnQ}=\frac{-\left({\mathrm{E}}_{\mathrm{cell}}{-\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}\right)×\mathrm{nF}}{\mathrm{RT}}\\ =\frac{-(0.915\mathrm{V}-0.80\mathrm{V})×2×96500\mathrm{C}}{8.314×298\mathrm{K}}\\ =-8.953\\ \mathrm{Q}={\mathrm{e}}^{-8.953}\\ =1.29×{10}^{-4}\end{array}$

Using the Q equation, calculate the ${\mathrm{H}}^{+}$concentration:$\mathrm{Q}=\frac{{\left[{\mathrm{H}}^{+}\right]}^2}{{\mathrm{P}}_{{\mathrm{H}}_2}×{\left[{\mathrm{Ag}}^{+}\right]}^2}$

$\begin{array}{c}\mathrm{Q}=\frac{{\left[{\mathrm{H}}^{+}\right]}^2}{{\mathrm{P}}_{{\mathrm{H}}_2}×{\left[{\mathrm{Ag}}^{+}\right]}^2}\\ 1.29×{10}^{-4}=\frac{{\left[{\mathrm{H}}^{+}\right]}^2}{{\mathrm{P}}_{{\mathrm{H}}_2}×{\left[{\mathrm{Ag}}^{+}\right]}^2}\\ =\frac{{\left[{\mathrm{H}}^{+}\right]}^2}{1.00\mathrm{atm}×[0.100\mathrm{M}{]}^2}\\ =\frac{{\left[{\mathrm{H}}^{+}\right]}^2}{0.01}\\ {\left[{\mathrm{H}}^{+}\right]}^2=1.29×{10}^{-4}×0.01\\ \left[{\mathrm{H}}^{+}\right]=\sqrt{1.29×{10}^{-4}×0.01}\\ =1.14×{10}^{-3}\mathrm{M}\end{array}$

Using, $\left[{\mathrm{H}}^{+}\right]$, calculate the $\mathrm{pH}$

$\begin{array}{c}\mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]\\ =2.94\end{array}$

Therefore, the$\mathrm{pH}=2.94$