The equilibrium constant for the reaction is:
$2 F e^{3 +} + 3 I^{⊖} ⇌ F e^{3 +} + C e^{3 +}$
Given :
$E_{C e^{4 +} 1 C e^{3 +}}^{⊖}$ = 1.44 V
$E_{F e^{3 +} 1 F e^{2 +}}^{⊖}$ = 0.68 V.

7.236 \times 10^{12}

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7.386 \times 10^{12}

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7.158 \times 10^{12}

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Solution

The correct option is A $7.236 \times 10^{12}$
For the cell reaction :
$F e^{2 +} + C e^{4 +} ⇌ F e^{3 +} + C e^{3 +}$
Half cell reaction are :
At equilibrium,
$E_{c e l l}^{⊖} = \frac{0.0591}{n} l o g k_{e q}$ (because at equilibrium, $E_{c e l l} = 0)$ .
+ 0.76 = $\frac{0.0591}{1} l o g K_{e q}$
$l o g K_{e q} = \frac{0.76}{0.0591} = 12.895 \Rightarrow K_{e q} = 7.236 \times 10^{12}$

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Similar questions

F e^{2 +} + C e^{4 +} ⇌ F e^{3 +} + C e^{3 +}

E_{{C e}^{4 +} / C e^{3 +}}^{\circ} = 1.44 V; E_{F e^{2 +} / F e^{3 +}}^{\circ} = 0.68 V

F e^{2 +} + C e^{4 +} | F e^{3 +} + C e^{3 +}

E^{0} C e^{4 +} / C e^{3 +} = 1.44

E^{0} F e^{3 +} / F e^{2 +} = 0.68 V]

F e^{2 +} + C e^{4 +} ⇌ F e^{3 +} + C e^{3 +}

E_{C e^{4 +} / C e^{3 +}}^{o} = 1.44 V; E_{F e^{3 +} / F e^{2 +}}^{o} = 0.68 V

F e^{2 +} + C e^{4 +} ⇌ F e^{3 +} + C e^{3 +}

E_{C e^{4 +} / C e^{3 +}}^{0} = 1.44 V; E_{F e^{3 +} / F e^{2 +}}^{0} = 0.68 V]

F e^{2 +} + C e^{4 +} ⇌ F e^{3 +} + C e^{3 +}

E_{C e^{4 +} | C e^{3 +}}^{o} = 1.44 V; E_{F e^{3 +} | F e^{2 +}}^{o} = 0.68 V

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