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Electrolytic Cell

Given, Fe3+...

Question

Given, $F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q); E^{\circ} = + 0.77 V$
$A l^{3 +} (a q) + 3 e^{-} \to A l (s); E^{\circ} = - 1.66 V$
$B r_{2} (a q) + 2 e^{-} \to 2 B r^{-}; E^{\circ} = + 1.09 V$

Considering the electrode potentials, which of the following represents the correct order of reducing power?

A

A l < F e^{2} < B r^{-}

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B

F e^{2} < A l < B r^{-}

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C

B r^{-} < F e^{2} < A l

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D

A l < B r^{-} < F e^{2 +}

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Solution

The correct option is C $B r^{-} < F e^{2} < A l$
A negative $E^{o}$ means that the redox couple is a stronger reducing agent. Hence, we can say that Aluminium is strongest reducing agent.

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Q. Based on the date given below, the correct order of reducing power is:

F e^{3 +} (a q) + e \to F e^{2 +} (a q); E^{o} = + 0.77 V

A l^{3 +} (a q) + 3 e \to A l (s); E^{o} = - 1.66 V

B r_{2} (a q) + 2 e \to 2 B^{-} (a q); E^{o} = + 1.08 V

F e^{2 +} (a q) + 2 e^{-} \to F e (s)

E^{⊖} = - 0.44 V

A l^{3 +} + 3 e^{-} \to A l (s)

E^{⊖} = - 1.66 V

B r^{2 +} 2 e^{-} \to 2 B r^{⊖} (a q)

E^{⊖} = - 1.08 V

The decreasing order of reducing power is:

Q. Electrode potential data are given

F e_{(a q)}^{3 +} + e^{-} \to F e_{(a q)}^{2 +}

E^{o} = + 0.77

V

A l_{(a q)}^{3 +} + 3 e^{-} \to A l_{s}; E^{o} = - 1.66

V

B r_{2 (a q)} + 2 e^{-} \to 2 B r_{(a q)}^{-}; E^{o} = + 1.66

V
Based on the given data which statement(s) is/are incorrect?

Q. Which of the following statement(s) is/are correct regarding the given electrode potential data?

F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q); E^{0} = + 0.77 V

A l^{3 +} (a q) + 3 e^{-} \to A l (s); E^{0} = - 1.66 V

B r_{2} (g) + 2 e^{-} \to 2 B r^{-} (a q); E^{0} = + 1.08 V

Q. Given, standard electrode potentials:

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

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

The standard electrode potential

E^{\circ}

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

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