The electode potential of anode at standard condition is called standard potential.

Calculate the standard cell potential (in V) of the cell in which the following reaction takes place:

${\mathrm{F}\mathrm{e}}^{2+}\mathrm{ }\left(\mathrm{a}\mathrm{q}\right)+\mathrm{A}{\mathrm{g}}^{+}\left(\mathrm{a}\mathrm{q}\right)\to \mathrm{F}{\mathrm{e}}^{3+}\left(\mathrm{a}\mathrm{q}\right)+\mathrm{A}\mathrm{g}\left(\mathrm{s}\right)$

Given that
${\mathrm{E}}_{{\mathrm{Ag}}^{+}/\mathrm{A}\mathrm{g}}^{\mathrm{o}}=\mathrm{x}\mathrm{ }\mathrm{V}$
${\mathrm{E}}_{{\mathrm{F}\mathrm{e}}^{2+}/\mathrm{F}\mathrm{e}}^{\mathrm{o}}=\mathrm{y}\mathrm{ }\mathrm{V}$
${\mathrm{E}}_{{\mathrm{F}\mathrm{e}}^{3+}/\mathrm{F}\mathrm{e}}^{\mathrm{o}}=\mathrm{z}\mathrm{ }\mathrm{V}$

Calculate the E.M.F. of the cell,

$\mathrm{Mg}\left(\mathrm{s}\right)|{\mathrm{Mg}}^{2+} (0.1\mathrm{M}\left)\right|\left|{\mathrm{Ag}}^{+}\left(1\times {10}^{-3}\right)\right|\mathrm{Ag}\left(\mathrm{s}\right)$

${\mathrm{E}}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}^{°} =+0.8\mathrm{V}, {\mathrm{E}}_{{\mathrm{Mg}}^{2+}/\mathrm{Mg}}^{°}=-2.37\mathrm{V}$

What happens to the E.M.F. if the concentration of ${\mathrm{Ag}}^{+}$ is decreased to $1\times {10}^{-4} \mathrm{M}$ ? [given log 5 = 0.6990]

Represent a cell consisting of ${\mathrm{Mg}}^{2+}$ | $\mathrm{Mg}$ half cell and ${\mathrm{Ag}}^{+}$ | $\mathrm{Ag}$ half cell and write the cell reaction.

$(\mathrm{E}{°}_{\mathrm{Ag}}=0.799 \mathrm{V}, \mathrm{E}{°}_{\mathrm{Mg}}=-2.37 \mathrm{V})$

The standard reduction potential for ${\mathrm{Cu}}^{2+}/\mathrm{Cu}$ is $+\mathrm{0.34 }\mathrm{V}.$ Calculate the reduction potential at $\mathrm{pH}=14$ for the above couple.

$\left({\mathrm{K}}_{\mathrm{sp}} \mathrm{of} \mathrm{Cu}{\left(\mathrm{OH}\right)}_2=1\times {10}^{-19}\right)$

Give reason for ${\mathrm{E}}^{\circ }$ value for $({\mathrm{Mn}}^{+2}/\mathrm{Mn})$ is negative whereas for $({\mathrm{Cu}}^{+2}/\mathrm{Cu})$ is positive.