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Nernst Equation

Given, standa...

Question

Given, standard electrode potentials, $F e^{2 +} + 2 e^{-} \to F e, E^{o} = - 0.440$ V
$F e^{3 +} + 3 e^{-} \to F e, E^{o} = - 0.036$ V

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

+ 0.771

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- 0.771 V

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+ 0.404 V

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- 0.417 V

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Solution

The correct option is D $+ 0.404 V$
$F e^{2 +} + 2 e^{-} \to F e, E^{o} = - 0.440$ ......(1)
$F e^{3 +} + 3 e^{-} \to F e, E^{o} = - 0.036$ .......(2)
The standard electrode potential for $F e^{3 +} + e^{-} \to F e^{2 +}$ is
obtained by substracting the value of the standard electrode potential for first reaction from the standard electrode potential for second reaction.
$- 0.036 - (- 0.440) = 0.404 V$

Hence,option C is correct.

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Similar questions

F e^{2 +} + 2 e^{-}

⟶

F e; E^{o} = - 0.440 V

F e^{3 +} + 3 e^{-}

⟶

F e; E^{o} = - 0.036 V

(E^{o}

F e^{3 +} + e^{-}

\to

F e^{2 +}

{F e}^{3 +} + {3 e}^{-} ⟶ F e; E^{o} = - 0.036 v o l t

{F e}^{2 +} + {2 e}^{-} ⟶ F e; E^{o} = - 0.440 v o l t

E^{o}

{F e}^{3 +} + e^{-} ⟶ {F e}^{2 +}

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:

F e^{3 +} + 3 e ⇌ F e, E^{o}

F e^{3 +} + e ⇌ F e^{2 +}

E^{o}

F e^{2 +} + 2 e ⇌ F e

F e^{2 +} + 2 e \to F e

F e^{3 +} + 3 e \to F e

- 0.440

- 0.036

(E^{0})

F e^{3 +} + e \to F e^{2 +}

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