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The reduction...

Question

The reduction of a half cell consisting of a Pt electrodes immersed in 1.5 M ${F e}^{2 +}$ and 0.015 M ${F e}^{3 +}$ .Solution at $25 C (E_{{F e}^{3 +} / {F e}^{2 +}} = 0.770 V)$
(A) 0.652 V (B) 0.88 V (C) 0.710 V (D) 0.850 V

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Solution

we have,
$F e^{3 +} + e^{-} \to$ $F e^{2 +}$
$E_{c e l l}$ = $E_{c e l l}^{0}$ - 0.0591 log $\frac{F e^{2 +}}{F e^{3 +}}$

$E_{c e l l}$ = 0.770 - 0.0591 log $\frac{1.5}{0.015}$

$E_{c e l l}$ = 0.770 - 0.1182
$E_{c e l l}$ = 0.652

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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

25^{o}

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

F e^{2 +}

10

F e^{3 +}

E_{F e^{3 +} | F e^{2 +}}^{0} = 0.77 V

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

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

- 0.440

- 0.036

(E^{0})

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

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 +}

Given standard $E^{⊖}$ :

$F e^{3 +} + 3 e^{-} \to F e$ ; $E^{⊖} = - 0.036 V$

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

The $E^{⊖}$ of $F e^{3 +} + e^{-} \to F e^{2 +}$ is:

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