Calculate the emf for the given cell at ${25}^{°}\mathrm{C}$: $\mathrm{Cr}\overline{){\mathrm{Cr}}^{3+}}(0.1 \mathrm{M})\parallel {\mathrm{Fe}}^{2+}(0.01\mathrm{M})|\mathrm{Fe}$ [Given, $\mathrm{E}{°}_{({\mathrm{Cr}}^{3+}\overline{)\mathrm{Cr})}} =-0.74 \mathrm{V}, \mathrm{E}{°}_{\left({\mathrm{Fe}}^{2+}\overline{)\mathrm{Fe}}\right)}=-0.44 \mathrm{V}$].
 

The conductivity of two electrolytic solutions $\mathrm{A}$ and $\mathrm{B}$ each having a concentration of$0.1 \mathrm{M}$are $8.5\times {10}^{-2} \mathrm{S} {\mathrm{cm}}^{-1}$ and $4.1\times {10}^{-4} \mathrm{S} {\mathrm{cm}}^{-1 }$respectively. Which among the two offers less resistance to the flow of current ? 

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, $\mathrm{E}{°}_{({\mathrm{Na}}^{+}\overline{)\mathrm{Na})}}=-2.71 \mathrm{V}, \mathrm{E}{°}_{({\mathrm{H}}_2\mathrm{O}\overline{){\mathrm{H}}_2)}}=-0.83 \mathrm{V}$; $\mathrm{E}{°}_{({\mathrm{Cl}}_2\overline{){\mathrm{Cl}}^{-})}}=+1.36 \mathrm{V}, \mathrm{E}{°}_{({\mathrm{H}}^{+}, {\mathrm{H}}_2\overline{){\mathrm{H}}_2\mathrm{O})}}=+1.23 \mathrm{V}$.