On the basis of the following $E^{o}$ values, the strongest oxidising agent is:
${[F e {(C N)}_{6}]}^{4 -} ⟶ {[F e {(C N)}_{6}]}^{3 +} + e^{-}, E^{o} = - 0.35 V$
${F e}^{2 +} \to {F e}^{3 +} + e^{- 1}$ ;
${F e}^{2 +} \to {F e}^{3 +} + e^{- 1}; E^{o} = - 0.77 V$

{F e}^{3 +}

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{[F e {(C N)}_{6}]}^{3 -}

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{[F e {(C N)}_{6}]}^{4 -}

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{F e}^{2 +}

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Solution

The correct option is A ${F e}^{3 +}$
For the strongest oxidizing agent, the oxidizing potential should be least. Here, the oxidizing potential of ${F e}^{+ 2}$ is less than that of ${[{F e (C N)}_{6}]}^{4 -}$ . Therefore, ${F e}^{+ 2}$ is stronger oxidizing agent than ${[{F e (C N)}_{6}]}^{4 -}$ . Also, the stronger oxidizing agent should easily reduce itself. Here, ${F e}^{+ 3}$ is easily reduced than ${F e}^{+ 2}$ . Therefore, among all the four, ${F e}^{+ 3}$ is the stronger oxidizing agent.

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Similar questions

E^{\circ}

[F e (C N)_{6}]^{4 -} \to [F e (C N)_{6}]^{3 -} + e^{-} E^{\circ} = - 0.35 V

F e^{2 +} \to F e^{3 +} + e^{-}; E^{\circ} = - 0.77 V

[F e {(C N)}_{6}]^{4 -} \to [F e {(C N)}_{6}]^{3 -} + e^{- 1}, E^{0} = - 0.35 V

F e^{2 +} \to F e^{3 +} + e^{- 1}, E^{0} = - 0.77 V

E^{\circ}

[F e (C N)_{4}]^{4^{-}} \to [F e (C N)_{6}]^{3 -} + e^{-}

\begin{matrix} E^{\circ} = - 0.35 V F e^{2 +} \to F e^{3 +} + e^{-} & E^{\circ} = - 0.77 V \end{matrix}

(E^{o})

F e^{3 +} \to F e

F e^{2 +} + 2 e 6 - \to F e; E_{F}^{o} e_{F e^{2 +} / F e} = - 0.47 V

F e^{3 +} + e^{-} \to F e^{2 +}; E_{F e^{3 +} / F e^{2 +}}^{o} = + 0.77 V

(E^{o})

F e^{3 +} \to F e

F e^{2 +} + 2 e^{-} \to F e; E_{F e^{2 +} / F e}^{o} = - 0.47 V

F e^{3 +} + e^{-} \to F e^{2 +}; E_{F e^{3} / F e^{2 +}}^{o} = + 0.77 V

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