Question

On the basis of the following $E^0$ values, the strongest oxidising agent is
$[Fe(CN{)}_6{]}^{4-}\to [Fe(CN{)}_6{]}^{3-}+e^{-};$
$E^0=-0.35V$
$F{e}^{2+}\to F{e}^{3+}+e^{-};E^0=-0.77V$

• A. $[Fe(CN{)}_6{]}^{4-}$
• B. $F{e}^{2+}$
• C. $F{e}^{3+}$
• D. $[Fe(CN{)}_6{]}^{3-}$

Answer 1

Answer: (C)

Substance which have higher reduction potential are stronger oxidising agent.
$[Fe(CN{)}_6{]}^{4-}\to [Fe(CN{)}_6{]}^{3-}+e^{-},$
$E^0=-0.35V$
$F{e}^{2+}\to F{e}^{3+}+e^{-},E^0=-0.77V$
$\because E_{oxi}^0=-E_{red}^0$
$\therefore [Fe(CN{)}_6{]}^{3-}+e^{-}\to [Fe(CN{)}_6{]}^{4-},E^0=0.35V$
$F{e}^{3+}+e^{-}\to F{e}^{2+}{E}^0=0.77V$
Hence, $F{e}^{3+}$ has maximum tendency to reduced,
so it is the strongest oxidising agent.