Ionic conductance of ${\mathrm{H}}^{+}$ and ${\mathrm{SO}}_4^{--}$ are $\mathrm{x}$ and $\mathrm{yS} {\mathrm{cm}}^2{ \mathrm{mol}}^{-1}.$ Hence, equivalent conductivity of ${\mathrm{H}}_2{\mathrm{SO}}_4$ is

From the following molar conductivities at infinite dilution,

${\mathrm{\Lambda }}_{\mathrm{m}}°$ for ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3=858 \mathrm{S}{\mathrm{cm}}^2{ \mathrm{mol}}^{-1}$

${\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}$ for ${\mathrm{NH}}_4\mathrm{OH}=238.3 \mathrm{S}{\mathrm{cm}}^2{ \mathrm{mol}}^{-1}$

${\mathrm{\Lambda }}_{\mathrm{m}}°$ for ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4=238.4 \mathrm{S}{\mathrm{cm}}^2{ \mathrm{mol}}^{-1}$
Calculate ${\mathrm{\Lambda }}_{\mathrm{m}}°$ for $\mathrm{Al}{\left(\mathrm{OH}\right)}_3$

Half cell reactions for some electrodes are given below

I. $\mathrm{A}+{\mathrm{e}}^{-}⟶{\mathrm{A}}^{-}; \mathrm{E}°=0.96 \mathrm{V}$

II. ${\mathrm{B}}^{-}+{\mathrm{e}}^{-}⟶{\mathrm{B}}^{2-}; \mathrm{E}°=-0.12 \mathrm{V}$

III. ${\mathrm{C}}^{+}+{\mathrm{e}}^{-}⟶\mathrm{C}; \mathrm{E}°=+0.18 \mathrm{V}$

IV. ${\mathrm{D}}^{2+}+2{\mathrm{e}}^{-}⟶\mathrm{D}; \mathrm{E}°=-1.12 \mathrm{V}$

Largest potential will be generated in which cell?