${{\text{E}}^o }{\text{C}}{{\text{u}}^{{\text{2}} + }}{\text{/Cu = + 0}}{\text{.34 V, E}}_{{\text{F}}{{\text{e}}^{ + {\text{2}}}}/Fe}^o = \,\,{\text{ - 0}}{\text{.44 V}}$
\(Fe\,(s)\,\, \to \,\,F{e^{2 + }}(aq)\,\, + \,\,2{e^ - }\) ....... એનોડ પર
\(C{u^{2 + }}(aq)\, + \,\,2{e^ - }\, \to \,\,Cu\,\,(s)\) ........ કેથોડ પર
આપણે જાણીએ છીએ :
\(\Delta \,{G^o } = \,- \,n\,F\,E_{cell}^o\) ; \(n= 2\)
\(E_{cell}^o = \,\left[ {E_{(C{u^{2 + }}/cu)}^o - E_{(F{e^{2 + }}/Fe)}^o} \right]\)
\( = ( + \,0.34\,V\,)\,- \,( - 0.44\,V)= +0.78\,V\) અને
\(F= 96500\,C\)
\(\therefore \,\Delta \,{G^o } = \,- \,n\,\,F\,E_{cell}^o \,= \, - (2)\,\, \times \,(96500\,C)\,\, \times \,( + \,0\,.78\,V)\)
\( = \,- 150540\,\,CV\, = \, - \,150540\,J\)
\(\,(\;\because \,1\,CV\, = \,1\,J)\)
$mol^{-1}, ᴧ^{0}\, KCl = 150\, S\, cm^{2}\, mol^{-1}$ હોય, તો $ᴧ^{0}\, NaBr$ .............. ${\rm{S}}\,{\rm{c}}{{\rm{m}}^2}{\rm{mo}}{{\rm{l}}^{ - 1}}$ શોધો.
$C{u_{(s)}} + 2A{g^ + }_{(aq)} \to C{u^{2 + }}_{(aq)} + 2A{g_{(s)}}$
$P{b^4} + 2{e^ - } \longrightarrow P{b^{2 + }};\,{E^o} = + 1.67\,V$
$C{e^{4 + }} + {e^ - } \longrightarrow C{e^{3 + }};\,{E^o} = + 1.61\,V$
$B{i^{3 + }} + 3{e^ - } \longrightarrow Bi;\,{E^o} = + 0.20\,V$ આપેલ છે. તો આ ઘટકતી ઓક્સિડેશતકર્તા તરીકેની ક્ષમતા ક્યા ક્રમમાં વધશે?