Calculate the degree of dissociation of acetic acid at $298 \mathrm{K}$, given that ${\mathrm{\Lambda }}_{\mathrm{m} }\left({\mathrm{CH}}_3\mathrm{COOH}\right) =11.7 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1}$;$\mathrm{\Lambda }{°}_{\mathrm{m}} \left({\mathrm{CH}}_3{\mathrm{COO}}^{-}\right)=40.9 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1}$; $\mathrm{\Lambda }{°}_{\mathrm{m}} \left({\mathrm{H}}^{+}\right)=349.1 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1}$.
 

In the electrolysis of aqueous sodium bromide, there are two possible anodic reactions.
$$\begin{gathered} 2 {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{O}}_2\left(\mathrm{g}\right)+4{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+4{\mathrm{e}}^{-}, {\mathrm{E}}^{°}=0.23 \mathrm{V} \\ 2 {\mathrm{Br}}^{-}\left(\mathrm{aq}\right)\to {\mathrm{Br}}_2\left(\mathrm{g}\right)+2{\mathrm{e}}^{-}, {\mathrm{E}}^{°}=1.08 \mathrm{V} \end{gathered}$$

Which reaction occurs at the anode and why?
 

Explain why electrolysis of aqueous solution of $\mathrm{NaCl}$gives ${\mathrm{H}}_2$ at cathode and ${\mathrm{Cl}}_2$ at the anode? Write the overall reaction. Given, $\mathrm{E}{°}_{({\mathrm{Na}}^{+}\overline{)\mathrm{Na})}}=-2.71 \mathrm{V}, \mathrm{E}{°}_{({\mathrm{H}}_2\mathrm{O}\overline{){\mathrm{H}}_2)}}=-0.83 \mathrm{V}$; $\mathrm{E}{°}_{({\mathrm{Cl}}_2\overline{){\mathrm{Cl}}^{-})}}=+1.36 \mathrm{V}, \mathrm{E}{°}_{({\mathrm{H}}^{+}, {\mathrm{H}}_2\overline{){\mathrm{H}}_2\mathrm{O})}}=+1.23 \mathrm{V}$.