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Equilibrium Constant from Nernst Equation

Calculate the...

Question

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 $0.77$ and $0.54 V$ respectively for $F e^{3 +} | F e^{2 +}$ and ${I_{3}}^{-} | I^{-}$ couples.

4.26 \times 10^{7}

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5.26 \times 10^{7}

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6.04 \times 10^{7}

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7.26 \times 10^{7}

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Solution

The correct option is C $6.04 \times 10^{7}$
$2 F e^{3 +} + 3 I^{-} ⇌ 2 F e^{2 +} + I_{3}^{-}$

$E_{c e l l}^{\circ} = \frac{R T}{n F} \times l n K$

$⟹ 0.77 - 0.54 = \frac{8.314 \times 298}{2 \times 96500} \times l n K$

$⟹ l n K = 17.917$

$⟹ K = e^{17.917} = 6.04 \times 10^{7}$

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