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Equilibrium Constant from Nernst Equation

The equilibri...

Question

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 $0.78 V$ and $0.54 V$ , respectively, for $F e^{3} + | F e^{2 +} a n d I_{3}^{⊖} | I^{⊖}$ couples.

K_{e q} = 10^{8}

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K_{e q} = 10^{7}

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K_{e q} = 10^{5}

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N o n e o f t h e s e

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Solution

The correct option is D $K_{e q} = 10^{8}$
At anode: $3 I^{⊖} \to I_{3}^{⊖} + 2 e^{-}$
At cathode: $2 F e^{3 +} + 2 e^{-} \to 2 F e^{2 +}$
At equilibrium, $E_{c e l l} = 0$
$E_{c e l l}^{⊖} = (E_{r e d}^{⊖})_{c} - (E_{r e d}^{⊖})_{a}$
$= E_{(F e^{3} + | F e^{2 +})}^{⊖} - E_{(I_{3}^{⊖} | 3 I^{⊖})}^{⊖}$
$= 0.78 0.54 = 0.24 V$
$E_{c e l l} = E_{c e l l}^{⊖} - \frac{0.06}{2} l o g K_{e q} (n_{c e l l} = 2)$
$E^{c e l l}$ = $0.03$ log $K_{e q}$
$l o g K_{e q} = \frac{E_{c e l l}^{⊖}}{0.03} = \frac{0.24 V}{0.03 V}$ = 8.0
$K_{e q} = 10^{8}$

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

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

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

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

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

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