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Nernst Equation

The redox pot...

Question

The redox potential of $F e^{3 +} | F e^{2 +}$ electrode at pH = 4 is:

0.8 V

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0.5 V

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0.2 V

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0.1 V

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Solution

The correct option is D 0.5 V
At pH = 4,
$E_{c e l l} = E_{c e l l}^{⊖} - \frac{0.06}{1} l o g \frac{[F e^{2 +}]}{[F e^{3 +}]}$
$= 0.80 - 0.06 l o g \frac{0.01}{10^{- 37} / (10^{- 10})^{3}}$
$= 0.80 - 0.06 l o g \frac{10^{- 2} \times 10^{- 30}}{10^{- 37}}$
$0.80 - 0.06 l o g 10^{5}$
$= 0.80 - 0.06 \times 5 = 0.5 V$

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Similar questions

F e^{3 +} | F e^{2 +}

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 +}

F e^{2 +} + 2 e^{-} \to F e, E^{o} = - 0.440

F e^{3 +} + 3 e^{-} \to F e, E^{o} = - 0.036

(E^{o})

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

25^{o}

F e^{3 +} | F e^{2 +}

F e^{2 +}

10

F e^{3 +}

E_{F e^{3 +} | F e^{2 +}}^{0} = 0.77 V

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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