${E^o}_{\left( {{I_2}/{I^ - }} \right)},{E^ \circ }_{\left( {Br^-/B{r_2}} \right)}$ and ${E^ o }_{\left( {{Fe}/{Fe^{2+} }} \right)}$
are respectively $+ 0.54\ V$, $-1.09\ V$ and $0.44\ V$. On the basis of above data which of the following process is non-spontaneous
$\text{E}_{\text{cell }}^{o}=\left( \text{E}_{\text{Rp}}^{{}^o } \right)\text{ cathode }-{{\left( \text{E}_{\text{RP}}^{{}^o } \right)}_{\text{anode }}}$
$\because \,\,\,\text{E}_{\text{cell}}^{o}$ is negative hence this reaction is non-spontaneous
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| Compound | $\ce{NH_3}$ | $\ce{PH_3}$ | $\ce{AsH_3}$ | $\ce{SbH_3}$ |
| $\Delta_{\text{diss}}(\text{E}-\text{H})/\text{kJ mol}^{-1}$ | $389$ | $322$ | $297$ | $255$ |
(Density of water $= 1\,gm/ml$ )
$F{e^{ + 3}}\left( {aq} \right) + {e^ - } \to F{e^{2 + }}\left( {aq} \right);{E^o} = + 0.77\,V$
$A{l^{ + 3}}\left( {aq} \right) + 3{e^ - } \to Al\left( s \right);{E^o} = - 1.66\,V$
$B{r_2}\left( {aq} \right) + 2{e^ - } \to 2B{r^ - };{E^o} = + 1.09\,V$
Considering the electrode potentials, which of the following represents the correct order of reducing power?
