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Equivalent, Molar Conductivity and Cell Constant

Given, E0Fe3+...

Question

Given, $E_{F e^{3 +} / F e}^{0} = 0.036 V, E_{F e^{2 +} / F e}^{0} = - 0.439 V$ . The value of standard electrode potential for the change, $F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q)$ will be :

A

- 0.403 V

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B

0.385 V

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C

0.986 V

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D

- 0.270 V

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Solution

The correct option is C $0.986 V$
$F e^{3 +} + 3 e^{-} \to F e, E_{1}^{0} = 0.036 V$
$F e^{2 +} + 2 e^{-} \to F e, E_{2}^{0} = - 0.439 V$
$F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q); E_{n}^{0} = ?$
$Δ G$ is an additive property:
$Δ G_{n}^{0} = Δ G_{1}^{0} + Δ G_{2}^{0}$
Here,
$∴ Δ G_{n}^{0} = Δ G_{1}^{0} - Δ G_{2}^{0}$
$Δ G^{0} = - n F E^{0}$
Now,
$- n_{n} F E_{n}^{0} = - n_{1} F E_{1}^{0} + n_{2} F E_{2}^{0}$
$∴ E_{n}^{0} = \frac{3 \times (0.036) - 2 \times (- 0.439)}{1} = 0.986 V$

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Similar questions

Q. Standard electrode potentials of

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

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

- 0.440

- 0.036

V respectively. The standard electrode potential

(E^{0})

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

Q. Given, standard electrode potentials,

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

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

The standard electrode potential

(E^{o})

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

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 +}

Q. Given standard electrode potentials,

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

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

The standard electrode potential

(E^{\circ})

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

Q. Given standard electrode potential

F e^{2 +} + 2 e^{-} \to F e; E^{0} = - 0.440 V

F e^{3 +} + 3 e^{-} \to F e; E^{0} = - 0.036

V. Calculate the electrode potential of cell.

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