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Introduction

The equilibri...

Question

The equilibrium constant of the following redox reaction at $298 K$ is $1 \times 10^{8}$ :

$2 F e_{(a q)}^{3 +} + 2 I_{(a q)}^{-} ⇌ 2 F e_{(a q)}^{2 +} + I_{2 (s)}$ .

If the standard reduction potential of iodine becoming iodide is $+ 0.54 V$ . What is the standard reduction potential of $F e^{3 +} / F e^{2 +}$ ?

+ 1.006 V

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- 1.006 V

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+ 0.8946 V

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+ 0.77 V

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Solution

The correct option is D $+ 0.77 V$
At $E q b^{m}$ , $E_{c e l l} = 0$
Thus, $E_{c e l l}^{o} = \frac{0.0591}{n} log K_{e q}$

As $n$ for the reaction is $2$

$∴ E_{c e l l}^{o} = 0.0591 / 2 log 1 \times 10^{8}$

$E_{c e l l}^{o} = 0.2364 V$
as
$I^{-} ⟶ I_{2}^{o}$ oxidation takes place at anode

$F e^{+ 3} ⟶ F e^{+ 2}$ reduction takes place at cathode

$E_{c e l l}^{o} = E_{c a t h o d e}^{o} - E_{a n o d e}^{o}$

$= E_{F e^{+ 3} / F e^{+ 2}}^{o} - E_{I_{2} / I^{-}}^{o}$

$\Rightarrow 0.2364 = E_{F e^{+ 3} / F e^{+ 2}}^{o} - 0.54$
$= E_{F e^{+ 3} / F e^{+ 2}}^{o} = 0.2364 + 0.54$
$= E_{F e^{+ 3} / F e^{+ 2}}^{o} \approx + 0.77 V$

Thus, standard reduction potential of $F e^{+ 3} ⟶ F e^{+ 2}$ is $0.77 V$

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Similar questions

298 K

1 \times 10^{8}

2 F e^{3 +} (a q .) + 2 I^{-} (a q .) ⇌ 2 F e^{2 +} (a q .) + I_{2} (s)

+ 0.54 V

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

The equilibrium constant for the following reaction is:

$2 F e^{3 +} 3 I^{⊖} ⇌ 2 F e^{2 +} + I_{3}^{⊖}$

0.78 V

0.54 V

F e^{3} + | F e^{2 +} a n d I_{3}^{⊖} | I^{⊖}

(E^{o})

F e^{3 +} \to F e

F e^{2 +} + 2 e 6 - \to F e; E_{F}^{o} e_{F e^{2 +} / F e} = - 0.47 V

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

(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

2 F e^{3 +} + 3 I^{-} ⇌ 2 F e^{2 +} + I_{3}^{-}

0.77

0.54 V

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

{I_{3}}^{-} | I^{-}

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