Given that :
$F{e^{2 + }} + 2{e^ - } \to Fe;$ ${E^o}_{F{e^{2 + }}/Fe} = - 0.47\,V$
$F{e^{3 + }} + {e^ - } \to F{e^{2 + }};$ ${E^o}_{F{e^{3 + }}/F{e^{2 + }}} = + 0.77\,V$
$(i)\,F{e^{2 + }}\, + \,2{e^ - }\, \to \,Fe;$ ${E^o}\, = \, - 0.47\,\,V;$
$(ii)\,F{e^{3 + }}\, + \,{e^ - }\, \to \,F{e^{2 + }};$ ${E^o}\, = \, + 0.77\,\,V;$
$(iii)\,F{e^{3 + }}\, + \,3{e^ - }\, \to Fe$
$(i)\,\Delta {G^o}\, = \, - nF{E^o}\, = \, - \,2\,( - 0.47)F\, = \,0.94\,F$
$(ii)\,\Delta {G^o}\, = \, - nF{E^o}\, = \, - \,1\,( + 0.77)F\, = \, - 0.77\,F$
$(iii)$ on adding $:\,\Delta {G^o}\, = \, + \,0.17\,F$
$\Delta {G^o}\, = \, - nF{E^o}\,{E^o}$ for
$(F{e^{3 + }} \to Fe)\, = \,\frac{{\Delta {G^o}}}{{ - nF}}\,$ $ = \,\frac{{0.17F}}{{ - 3F}}\, = \, - \,0.057\,V$
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$MnO _{4}^{-}+8 H ^{+}+5 e ^{-} \rightarrow Mn ^{2+}+4 H _{2} O$,
$E^{o} _{ Mn ^{2+} / MnO _{4}^{-}}=-1.510 \,V$
$\frac{1}{2} O _{2}+2 H ^{+}+2 e ^{-} \rightarrow H _{2} O$,
$E _{ O _{2} / H _{2} O }^{o}=+1.223 \,V$
Will the permanganate ion, $MnO _{4}^{-}$liberate $O _{2}$ from water in the presence of an acid ?
(ii) Silver metal does not react with $1 M$ zinc nitrate solution
(iii) Zinc metal dissolves in $1M$ copper sulphate solution and copper metal gets depositedHence the order of decreasing strength of the three metals as reducing agents will be