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Electrochemical Series

What is the e...

Question

What is the equilibrium constant for the recation, $2 F e^{3 +} \to 3 I^{-} ⇌ 2 F e^{2 +}, I_{3}^{-}$ . The standard reduction potentials in acidic conditions are $0.77$ V and $0.54$ V respectively for $F e^{3 +} | F e^{2 +}$ and $I_{3}^{-} | I^{-}$ couples?

A

4.25 \times 10^{7}

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B

7.05 \times 10^{5}

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C

6.25 \times 10^{5}

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D

6.25 \times 10^{7}

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Solution

The correct option is D $6.25 \times 10^{7}$
$2 F e^{3 +} + 3 I^{⊝} ⇋ 2 F e^{2 +} + I_{3}^{⊝}$
$n = 2$
$log K_{e q} = \frac{n}{0.0591} E_{c e l l}^{o}$
$E_{c e l l}^{o} = E_{F e^{3 +} / F e^{2 +}} - E_{I_{3}^{⊝} / I^{⊝}}$
$E_{c e l l}^{o} = 0.77 - 0.54 = 0.23$ $V$
$log K_{e q} = \frac{2}{0.0591} \times 0.23 = 7.78$
$K_{e q} = 6.2 \times 10^{7}$

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Similar questions

Q. Calculate the equilibrium constant for the reaction,

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

. The standard reduction potentials in acidic conditions are

0.77

0.54 V

respectively for

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

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

The equilibrium constant for the following reaction is:

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

The standard reduction potential in acidic conditions is

0.78 V

0.54 V

, respectively, for

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

Q. The standard electrode potential (reduction) of

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

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

are 0.77 V and 0.15 v respectively at

25^{o}

C The standard EMF of the reaction

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

Q. Question below refers to the overall reaction and half-reactions with standard reduction potentials below.

2 F e^{2 +} + C l_{2} \to 2 F e^{3 +} + 2 C l

?

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

E_{r e d}^{o}

C l_{2} + 2 e^{-} \to 2 C l^{-}

E_{r e d}^{o}

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Q. The equilibrium constant of the following redox reaction at

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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

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