$[Fe(CN)_6]^{4-} \rightarrow [Fe(CN)_6]^{3-} + e^{-1}\, ;$ $ E^o = -0.35\, V$
$Fe^{2+} \rightarrow Fe^{3+} + e^{-1}\ ;$ $E^o = -0.77\, V$
The substance which has lower reduction potential are stronger reducing agent while the substances which have higher reduction potential are a stronger oxidising agent.
$\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}+\mathrm{e}^{-} \rightarrow\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4} ; \mathrm{E}^{\circ}=0.35 \mathrm{V}$
$\mathrm{Fe}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+} ; \quad \mathrm{E}^{\circ}=0.77 \mathrm{V}$
The reduction potential of $\mathrm{F} \mathrm{e}^{3+/} \mathrm{Fe}^{2+}$ is higher, hence, $\mathrm{Fe}^{3+}$ is a strongest oxidising agent.
$mol^{-1}, ᴧ^{0}\, KCl = 150\, S\, cm^{2}\, mol^{-1}$ હોય, તો $ᴧ^{0}\, NaBr$ .............. ${\rm{S}}\,{\rm{c}}{{\rm{m}}^2}{\rm{mo}}{{\rm{l}}^{ - 1}}$ શોધો.
$Cu ( s )+ Sn ^{2+}( aq ) \rightarrow Cu ^{2+}( aq )+ Sn ( s )$
$\left( E _{ Sn ^{2+} \mid Sn }^{0}=-0.16\, V , E _{ Cu ^{2+} \mid Cu }^{0}=0.34\, V \right.$ Take $F=96500\, C\, mol ^{-1}$ )