To find the standard potential of ${\mathrm{M}}^{3+}/\mathrm{M}$ electrode, the following cell is constituted: $\mathrm{Pt}/\mathrm{M}/{\mathrm{M}}^{3+}\mathrm{ }\left(0.001\mathrm{ }\mathrm{mol}\mathrm{ }{\mathrm{L}}^{-1}\right)/{\mathrm{Ag}}^{+}\mathrm{ }\left(0.01\mathrm{ }\mathrm{mol}\mathrm{ }{\mathrm{L}}^{-1}\right)/\mathrm{Ag}$

The emf of the cell is found to be $0.421 \mathrm{volt}$ at $298 \mathrm{K}$. The standard potential of half-reaction ${\mathrm{M}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{M}$ at $298 \mathrm{K}$ will be:

(Given: ${\mathrm{E}}_{\raisebox{1ex}{${\mathrm{Ag}}^{+}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{Ag}$}\right.}^{⊝}$ at $298 \mathrm{K}=0.80 \mathrm{volt}$)

A current of $10.0 \mathrm{A}$ flows for $2.00 \mathrm{h}$ through an electrolytic cell containing a molten salt of metal $\mathrm{X}$. This results in the decomposition of $0.250 \mathrm{mol}$ of metal $\mathrm{X}$ at the cathode. The oxidation state of $\mathrm{X}$ in the molten salt is:

$\left(\mathrm{F}=96,500 \mathrm{C}\right)$

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

Given that,

${\mathrm{E}}_{{\mathrm{O}}_2/{\mathrm{H}}_2\mathrm{O}}^{\mathrm{o}}=+1.23 \mathrm{V};$
${\mathrm{E}}_{{\mathrm{S}}_2{\mathrm{O}}_8^{2-}/{\mathrm{SO}}_4^{2-}}^{\mathrm{o}}=2.05 \mathrm{V}$
${\mathrm{E}}_{{\mathrm{Br}}_2/{\mathrm{Br}}^{-}}^{\mathrm{o}}=+1.09 \mathrm{V};$
${\mathrm{E}}_{{\mathrm{Au}}^{3+}/\mathrm{Au}}^{\mathrm{o}}=1.4 \mathrm{V}$

The strongest oxidizing agent is