Question
On the basis of the following $${E^ \circ }$$ values, the strongest oxidising agent is
$${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }} \to {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }} + {e^ - };$$ $${E^ \circ } = - 0.35\,V$$
$$F{e^{2 + }} \to F{e^{3 + }} + {e^ - };$$ $${E^ \circ } = - 0.77\,V$$
A.
$${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }}$$
B.
$$F{e^{2 + }}$$
C.
$$F{e^{3 + }}$$
D.
$${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }}$$
Answer :
$$F{e^{3 + }}$$
Solution :
Substance which have higher reduction potential are stronger oxidising agent.
$${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }} \to {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }} + {e^ - },$$ $${E^ \circ } = - 0.35\,V$$
$$\eqalign{
& F{e^{2 + }} \to F{e^{3 + }} + {e^ - },\,{E^ \circ } = - 0.77\,V \cr
& \because \,\,E_{oxi}^ \circ = - E_{red}^ \circ \cr} $$
$$\therefore \,{\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }} + {e^ - } \to {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }},$$ $${E^ \circ } = 0.35\,V$$
$$F{e^{3 + }} + {e^ - } \to F{e^{2 + }},\,\,{E^ \circ } = 0.77\,V$$
Hence, $$F{e^{3 + }}$$ has maximum tendency to reduced, so it is the strongest oxidising agent.