On the basis of $E^{\circ}$ values, the strongest oxidising agent is:

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

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

F e^{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 $F e^{3 +}$
Given reaction:-
$F e^{+ 2} ⟶ F e^{+ 3} + e^{-}$ $e^{o} = - 0.77 V$
The reduction potential of $F e^{+ 2} = -$ Oxidation potential of $F e^{+ 3}$
$∴$ Oxidation potential of $F e^{+ 3}$ is $0.77 V$ .
By $[F e (C N)_{6}]^{4 -} ⟶ [F e (C N)_{6}]^{- 3} + e^{-}$ $e^{o} = - 0.35 V$
$∴$ Oxidation potential of $[F e (C N)_{6}]^{- 3}$ is $0.35 V$ .
Thus, the oxidation potential of $F e^{3 +}$ is highest.

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Similar questions

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)_{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}

[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

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 $E^{⊖}$ :

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

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

The $E^{⊖}$ of $F e^{3 +} + e^{-} \to F e^{2 +}$ is:

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