91Ó°ÊÓ

When an iron nail is dipped into Copper(II) sulfate, iron is oxidized and forms Ferrous sulfate. The Cu metal formed deposits on the iron nail. The Cu metal deposits formed are brownish black in color.

${\text{Fe + CuSO}}_{\text{4}}→{\text{FeSO}}_{\text{4}}\text{+ Cu}$

$\stackrel{0}{\mathrm{Fe}}+\stackrel{+2}{\mathrm{Cu}}\stackrel{-2}{{\mathrm{SO}}_4}→\stackrel{+2}{\mathrm{Fe}}{\stackrel{-2}{\mathrm{SO}}}_4+\stackrel{0}{\mathrm{Cu}}$

As it is evident from the change in oxidation numbers that Fe is oxidized and ${\text{Cu}}^{\text{+ 2}}$ is reduced. Therefore it can be said, that iron has reduced copper and copper has oxidized iron. Thus, Fe is a reducing agent while Cu is an oxidizing agent.

The working principle of a voltaic cell depends on the spontaneity of the redox reaction taking place between the chemical species. If the redox reaction is spontaneous$∆G<0$ the potential difference between the two electrodes generates electrical energy.

In the above situation when an iron nail is dipped into copper sulfate solution, the copper metal deposits on iron nail spontaneously. Thus this reaction can be made into a voltaic cell.

$$\begin{gathered} \underline{{\text{Cu}}^{\text{+ 2}}\left(\text{aq}\right){\text{+ 2e}}^{\text{-}}→\text{Cu}\left(\text{s}\right) \\ \text{Fe}\left(\text{s}\right)→{\text{Fe}}^{\text{+ 2}}\left(\text{aq}\right){\text{+ 2e}}^{\text{-}}} \\ {\text{Cu}}_{\text{(aq)}}^{\text{2 +}}{\text{+ Fe}}_{\text{(s)}}→{\text{Cu}}_{\text{(s)}}{\text{+ Fe}}^{\text{2 +}}\left(\text{aq}\right) \end{gathered}$$

The ${\text{E}}_{\text{cell}}$ can be calculated as the difference in the electrode potentials of the half cell.

${\text{E}}_{\text{cell}}{\text{=E}}_{\text{cathode}}{\text{-E}}_{\text{anode}}$

Let us determine the anode and cathode of the cell. Oxidation occurs at anode and reduction occurs at cathode.

$$\begin{gathered} \text{Anode}  \text{reaction}           {\text{Cu}}^{\text{+ 2}}{\text{+ 2e}}^{\text{-}}→\text{Cu}              {\text{E}}^{\text{0}}\text{= - 0.44} \text{V} \\ \text{Cathode}  \text{reaction}       \text{Fe}→{\text{Fe}}^{\text{+ 2}}{\text{+ 2e}}^{\text{-}}               {\text{E}}^{\text{0}}\text{= 0.38} \text{V} \end{gathered}$$

$\begin{array}{rcl}& & {{\text{E}}^{\text{0}}}_{\text{cell}}\text{=}{{\text{E}}^{\text{0}}}_{\text{cathode}}\text{-}{{\text{E}}^{\text{0}}}_{\text{anode}}\\ & & \text{= 0.34} \text{V -}\left(\text{- 0.44} \text{V}\right)\\ & & \text{= 0.78} \text{V}\\ & & \end{array}$

Hence, the cell potential is $0.78 \text{V}$