Q21.58CP - Chemistry: Molecular Nature Of Matter And Change

Question

If the ${\mathbf{E}}_{\mathbf{cell}}$of the following cell is $\mathbf{0}\mathbf{.}\mathbf{915}\mathbf{ }\mathbf{V}$. what is the $\mathbf{pH}$ in the anode compartment? 

$\mathbf{Pt}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\left|}{\mathbf{H}}_{\mathbf{2}}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{atm}\mathbf{)}\mathbf{\right|}{\mathbf{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}{\mathbf{Ag}}^{\mathbf{+}}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{M}\mathbf{)}\mathbf{\mid }\mathbf{Ag}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$

Step-by-Step Solution

Answer

The $\mathrm{pH}=2.94$.

Step 1: Define energy

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.

Step 2: Explanation

Given,

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

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

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

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

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

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

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

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 $\mathrm{pH}=2.94$