You visited us 1 times! Enjoying our articles? Unlock Full Access!

Gibb's Energy and Nernst Equation

Given, EFe3...

Question

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

- 0.403 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.385 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.770 V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- 0.270 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $0.770 V$
$F e^{3 +} + 3 e \to F e, {E_{1}}^{o} = - 0.036 V$

$F e^{2 +} + 2 e \to F e, {E_{2}}^{o} = - 0.439 V$

$F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q); {E_{n}}^{o} = ?$
$Δ G$ is an additive property:

$∴ Δ G_{n}^{o} = Δ G_{1}^{o} - Δ G_{2}^{o}$

$Δ G^{o} = n F E^{o}$
Now,

$n_{n} F E_{n}^{o} = n_{1} F E^{o} - n_{2} F E_{2}^{o}$

$∴ E_{n}^{o} = \frac{3 \times (- 0.036) - 2 \times (- 0.439)}{1} = 0.770 V$

Suggest Corrections

Similar questions

E_{F e^{3 +} / F e}^{0} = 0.036 V, E_{F e^{2 +} / F e}^{0} = - 0.439 V

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

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

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

(E^{\circ})

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

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

- 0.440 V

- 0.036 V

E^{0}

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

Join BYJU'S Learning Program