Calculate the reduction potential at $25^{o}$ C for $F e^{3 +} | F e^{2 +}$ electrode, if the concentration of $F e^{2 +}$ ion is $10$ times higher than $F e^{3 +}$ ion.

[Given $E_{F e^{3 +} | F e^{2 +}}^{0} = 0.77 V$ ].

- 0.71 V

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

0.71 V

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

- 0.34 V

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

0.34 V

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

Open in App

Solution

The correct option is D $0.71 V$
The reduction of $F e^{3 +}$ can be given as:

$F e^{3 +} + e^{-} \to F e^{2 +}$
We have, the expression for reduction potential as:

$E_{F e^{3 +} | F e^{2 +}} = E_{F e^{3 +} | F e^{2 +}}^{0} - \frac{0.0591}{n} l o g \frac{[F e^{2 +}]}{[F e^{3 +}]}$

Now concentration of $F e^{2 +}$ ion is 10 times higher than $F e^{3 +}$ ion.

$∴ E_{F e^{3 +} | F e^{2 +}} = 0.77 V - \frac{0.0591}{1} l o g \frac{10 [F e^{3 +}]}{[F e^{3 +}]}$

$E_{F e^{3 +} | F e^{2 +}} = 0.71 V$

Hence, option $B$ is correct.

Suggest Corrections

Similar questions

25^{o} C

P t (s) | F e^{2 +} (a q, 0.1 M), F e^{3 +} (a q, 0.01 M); E_{F e^{3 +} / F e^{2 +}}^{0} = + 0.77 V

P t / F e^{2 -}, F e^{3 +}

P t / S n^{4 +}, S n^{2 +}

25^{o}

S n^{4 +} + 2 F e^{2 +} \to S n^{2 +} + 2 F e^{3 +}

F e^{2 +} + C e^{4 +} | F e^{3 +} + C e^{3 +}

E^{0} C e^{4 +} / C e^{3 +} = 1.44

E^{0} F e^{3 +} / F e^{2 +} = 0.68 V]

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 +}

(E^{o})

F e^{3 +} \to F e

F e^{2 +} + 2 e^{-} \to F e; E_{F e^{2 +} / F e}^{o} = - 0.47 V

F e^{3 +} + e^{-} \to F e^{2 +}; E_{F e^{3} / F e^{2 +}}^{o} = + 0.77 V

Join BYJU'S Learning Program