So, \(Fe \rightarrow \mathrm{Fe}^{2+}+2 \mathrm{e}^{-}, \mathrm{E}^{\circ}=+0.441 \mathrm{V} \ldots .\) (i)
and \(E_{F e^{3}+/ F_{e}^{2}}^{0}=0.771 \mathrm{V}\)
So, \(Fe ^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}, \mathrm{E}^{\circ}=0.771 \mathrm{V} \ldots \ldots\) (ii)
Cell reaction (i) \(Fe \rightarrow \mathrm{Fe}^{2+}+2 \mathrm{e}^{-}, \quad \mathrm{E}^{\circ}=0.441 \mathrm{V}\)
(ii) \(2 \mathrm{Fe}^{3+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Fe}^{2,+} \quad \mathrm{E}^{\circ}=+0.771 \mathrm{V}\)
\(\mathrm{Fe}^{2+}+2 \mathrm{Fe}^{3+} \rightarrow 3 \mathrm{Fe}^{2+}, \quad E_{\mathrm{cell}}^{0}=1.212 \mathrm{V}\)
So, on the basis of cell reaction following half-cell reactions are written
At anode:
\(\mathrm{Fe} \rightarrow \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \quad\) (oxidation)
At cathode:
\(2 \mathrm{Fe}^{3+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Fe}^{2+}(\text { reduction })\)
\(\mathrm{So}\)
\(\mathrm{E}_{\mathrm{cell}}^{0}=\mathrm{E}_{\text {cathode }}^{0}-\mathrm{E}_{\text {anode }}^{0}\)
\(=E_{F e^{3+} / / F e^{2 +}}^{0}-E_{F e^{2+}{ / F e}}^{0}\)
\((0+.771)-(-0.441)=+1.212 \mathrm{V}\)
$E_{{A^{3 + }}/A}^o = 1.50\,\,V\,,$ $E_{{B^{2 + }}/B}^o = 0.3\,\,V,$
$E_{{C^{3 + }}/C}^o = - \,0.74\,\,V,$ $E_{{D^{2 + }}/D}^o = - \,2.37\,\,V.$
યોગ્ય ક્રમ જેમાં કઈ વિવિધ ધાતુઓ કેથોડ પર જમા થાય છે
$F{e^2}+ \left( {aq} \right) + A{g^ + }\left( {aq} \right) \to F{e^{3 + }}\left( {aq} \right) + Ag\left( s \right)$
$E_{Ag^+/Ag}^o = xV$, $E_{F{e^{2 + }}/Fe}^o = yV$, $E_{F{e^{3 + }}/Fe}^o = zV$