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Electrode Potential

The equilibri...

Question

The equilibrium constant of the following redox reaction at $298 K$ is $1 \times 10^{8}$
$2 F e^{3 +} (a q .) + 2 I^{-} (a q .) ⇌ 2 F e^{2 +} (a q .) + I_{2} (s)$
If the standard reduction potential of iodine becoming iodides 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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+ 0.77 V

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- 0.77 V

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- 0.652 V

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Solution

The correct option is C $+ 0.77 V$
$E^{\circ} = \frac{0.059}{n} {log}_{10} K$
$= \frac{0.059}{2} {log}_{10} 10^{8} = 0.236$
$E_{C e l l}^{\circ} = E_{R e d u c e d s p e c i e s}^{\circ} - E_{O x i d i s e d s p e c i e s}^{\circ}$
$0.236 = E_{F e^{3 +} / F e^{2 +}}^{\circ} - 0.54$
$E_{F e^{3 +} / F e^{2 +}}^{\circ} = 0.77 V$ .

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Similar questions

298 K

1 \times 10^{8}

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

+ 0.54 V

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

E_{F e^{3 +} / F e^{2 +}}^{\circ} = + 0.77 V; E_{S n^{2 +} / S n}^{\circ} = - 0.14 V

S n_{(s)} + 2 F e_{(a q)}^{3 +} \to 2 F e_{(a q)}^{2 +} + S n_{(a q)}^{2 +}

E^{0}

E_{F e^{3 +} / F e^{2 +}}^{0}

E_{S n^{2 +} / S n}^{0}

S n (s) + 2 F e^{3 +} (a q) \to 2 F e^{2 +} (a q) + S n^{2 +} (a q)

E^{0}

E^{0_{F e 3 + / F e 2 +}} = + 0.77 V

E_{S n^{2 +} / S n}^{0} = - 0.14 V

S n (s)

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

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^{⊖}

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