On the basis of the following E values- the strongest oxidizing agent is- -Fe CN 6-4- -Fe CN 6-3-e-1-E0-0-35V Fe2- Fe3-e-1-E0-0-77V

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Electrochemical Series

On the basis ...

Question

On the basis of the following E values, the strongest oxidizing agent is:

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

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

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

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E^{o}

{[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

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

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}

In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of:

(a) Na₄[Fe(CN)₆] (b) Fe₄[Fe(CN)₆]₃ (c) Fe₂[Fe(CN)₆] (d) Fe₃[Fe(CN)₆]₄

Given standard electrode potentials:

$F e^{3 +} + 3 e^{-} \to F e; E^{0} = - 0.036 V$

$F e^{2 +} + 2 e^{-} \to F e; E^{0} = - 0.440 V$

The standard electrode potential $E^{o}$ for $F e^{3 +} + e^{-} \to F e^{2 +}$ is:

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