Answer

The standard electrode potential of ${Mg}^{2+} \mid {Mg}$ can be measured with respect to the standard hydrogen electrode, represented by ${Pt_{(s)}}, {H_2(g)}(1 {~atm}) \mid {H_(a q)}^{+}(1 {M})$.

A cell, consisting of ${Mg} \mid {MgSO_4}({aq} 1 {M})$ as the anode and the standard hydrogen electrode as the cathode, is set up.

$ \hspace{2.4cm} {Mg}\left|{Mg}^{2+}({aq}, 1 {M}) \| {H}^{+}({aq}, 1 {M})\right| {H_2}({~g}, 1 \text { bar }), {Pt{(s)}} $

Then, the emf of the cell is measured and this measured emf is the standard electrode potential of the magnesium electrode.

$\begin{array}{ll} & \mathrm{Mg}\left|\mathrm{Mg}^{2+}(1 \mathrm{M})\right|\left|\mathrm{H}^{+}(1 \mathrm{M})\right| \mathrm{H}2,(1 \mathrm{~atm}), \mathrm{Pt} \\ & \mathrm{E}{\text {cell }}^{\mathrm{o}}=\mathrm{E}{\mathrm{H}^{+}, 1 / 2 \mathrm{H}2}-\mathrm{E}{\mathrm{Mg}^{2+}, \mathrm{Mg}} \\ \text { Put } & \mathrm{E}{\mathrm{H}^{+}, 1 / 2 \mathrm{H}2}^o=0 . \\ \text { Hence, } \quad & \mathrm{E}{\mathrm{Mg}^{2+}, \mathrm{Mg}}^{\circ}=-\mathrm{E}_{\text {cell }}^o\end{array}$

Answer

Zinc is more reactive than copper. Hence, it displaces copper from copper sulphate solution as follows :

$ \hspace{2.5cm} \mathrm{Zn}(s)+\mathrm{CuSO}_4(a q) \longrightarrow \mathrm{ZnSO}_4(a q)+\mathrm{Cu}(s) $

Thus, zinc reacts with $\mathrm{CuSO}_4$ solution. Hence, we cannot store copper sulphate solution in a zinc pot.

Answer

Oxidation of ferrous ions means that the following reaction should occur :

$ \hspace{2.5cm} \mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+e^{-} ; \mathrm{E}_{\mathrm{ox}}^{\circ}=-0.77 \mathrm{~V} $

Only those substances can oxidize $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}^{3+}$ which are stronger oxidizing agents and have positive reduction potentials greater than 0.77 V so that EMF of the cell reaction is positive. This is so for elements lying below $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$ in the electrochemical series, e.g., $\mathrm{Br}_2, \mathrm{Cl}_2$ and $\mathrm{F}_2$.