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Application of Electrolysis

Standard elec...

Question

Standard electrode potentials are:
$F e^{2 +} / F e (E^{o} = - 0.44 V), F e^{3 +} / F e^{2 +} (E^{o} = 0.77 V)$
$F e^{2 +}, F e^{3 +}$ and $F e$ blocks are kept together, then:

A

F e^{3 +}

increases

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B

F e^{3 +}

decreases

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C

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

remains unchanged

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D

F e^{2 +}

decreases

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Solution

The correct option is B $F e^{3 +}$ decreases
Conversion of ${F e}^{2 +}$ to $F e$ is non-spontaneous while converting ${F e}^{3 +}$ to ${F e}^{2 +}$ is spontaneous.
So ${F e}^{3 +}$ will be spontaneously converted to ${F e}^{2 +}$ . So, ${F e}^{3 +}$ will decreases
So, option $B$ is correct

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Similar questions

Q. Standard electrode potential of two half-reactions are given below:

F e^{2 +} ⇌ F e; E^{\circ} = - 0.44 V

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

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

F e

are kept together :

Q. Standard electrode potentials of

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

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

- 0.440 V

- 0.036 V

respectively. The standard electrode potential

(E^{o})

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

E_{F e^{2 +} / F e}^{o} = - 0.44 V, E_{F e^{2 +} / F e^{3 +}}^{o} = - 0.77 V

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

F e

solid are kept together, then:

E_{F e 2 + | F e}^{o} = - 0.44 V; E_{F e^{2 +} | F e^{3 +}}^{o} = - 0.77 V

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

F e

solid are kept together, then:

Q. Given standard electrode potentials :

{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

The standard electrode potential

E^{o}

{F e}^{3 +} + e^{-} ⟶ {F e}^{2 +}

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