Chemistry - Electrochemistry Question with Solution | TestHub

For feasibility, check , ${{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cathode}\left(\mathrm{reduction}\right)}^{\mathrm{o}}-{\mathrm{E}}_{\mathrm{anode}\left(\mathrm{oxidation}\right)}^{\mathrm{o}}>0$

The ${\mathrm{E}}^{\mathrm{o}}$ values when metal acts as anode and ${\mathrm{NO}}_3^{-}$ reaction is cathodic reaction.

For Vanadium metal, ${\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}=0.97+1.19=2.16\mathrm{V}$

For Iron metal, ${\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}=0.97+0.04=1.01\mathrm{V}$

For silver metal, ${\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}=0.97-0.80=0.17\mathrm{V}$

For Gold metal, ${\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}=0.97-1.140=-0.17\mathrm{V}$

For electrodes having oxidation potential greater than $-0.97 \mathrm{V}$, ${{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}>0$

$\therefore \mathrm{Ag}, \mathrm{Fe}$ and $\mathrm{V}$ can be oxidised