Calculate the equilibrium constant of the reaction $F e + C u S O_{4} ⇌ F e S O_{4} + C u$ at $25^{o} C$ Given $E^{0} [F e \to F e^{2 +}) = 0.44 V$ and $E^{0} (C u \to C u^{2 +}) = - 0.337 V$ .

1.28 \times 10^{26}

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1.90 \times 10^{26}

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4.36 \times 10^{26}

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5.12 \times 10^{26}

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Solution

The correct option is B $1.90 \times 10^{26}$
Given,

Reaction: $F e + C u S O_{4} ⇌ F e S O_{4} + C u$

Electrode potential of $F e$ : $E^{0} [F e \to F e^{2 +}) = 0.44 V$

Electrode potential of $C u$ : $E^{0} [F e \to C u^{2 +}) = - 0.337 V$

Number of electrons gained or lost, n = 2

Temperature = $25^{\circ} C = 273 + 25 = 298 K$

In the given reaction, $F e$ undergoes reduction as it gains two electrons while $C u$ undergoes oxidation as it loses two electrons.

The cell potential can be calculated as

${E^{0}}_{c e l l} = {E^{0}}_{r e d u c t i o n} - {E^{0}}_{o x i d a t i o n}$

Substituting the value of electrode potential, we get

${E^{0}}_{c e l l} = 0.44 V - (- 0.337 V) = 0.777 V$

Now, we know that Gibbs free energy can be given as

$Δ G = - n \times F \times Δ E_{0}$ $. . . (i)$

Where n is the number of electrons exchanged, F is Faraday’s constant and $Δ E_{0}$ is the cell potential.

The relation between Gibbs free energy and equilibrium constant can be given as

$Δ G = - R \times T \times ln K$ $. . . (i i)$

Where R is the universal gas constant, T is the thermodynamic temperature and K is the equilibrium constant.

Equating $(i)$ and $(i i)$ we get,

$- n \times F \times Δ E_{0} = - R \times T \times ln K$

K can be given as

$K = e^{\frac{n \times F \times Δ E_{0}}{R \times T}}$

Substituting the respective values in the above equation for this reaction we get,

$K = e^{\frac{2 \times 96485 C \times 0.777 V}{8.3145 J K^{- 1} m o l^{- 1} \times 298 K}}$

Simplifying the above expression we get

$K = e^{\frac{149937.69}{2477.721}}$

$K = e^{60.5143}$

Further solving above by logarithmic values, we get

$\Rightarrow K = 1.90 \times 10^{26}$

Hence, the equilibrium constant for the given reaction $F e + C u S O_{4} ⇌ F e S O_{4} + C u$ will be $K = 1.90 \times 10^{26}$

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Similar questions

F e + C u S O_{4} ⇌ F e S O_{4} + C u

25^{0}

E^{0} (F e / F e^{2 +}) = 0.44 V, E^{0} (C u / C u^{2 +}) = - 0.337 V

F e + C u S O_{4} ⟶ F e S O_{4} + C u

25^{o}

E_{(F e / F e^{2 +})}^{0} = 0.44 V

E_{(C u / C u^{2 +})}^{0} = - 0.337 V

F e + {C u S O}_{4} ⇌ {F e S O}_{4} + C u

E_{F e / F e}^{o} = 0.44 V; E_{C u / {C u}^{2 +}}^{o} = - 0.337 V

M g

F e

C u^{2 +}

M g_{(s)} + C u^{2 +} ⇋ M g^{2 +} + C u_{(s)}; K_{1} = 6 \times 10^{90}

F e_{(s)} + C u^{2 +} ⇋ F e^{2 +} + C u_{(s)}; K_{2} = 3 \times 10^{26}

F e + C u S O_{4} ⇌ F e S O_{4} + C u

25^{0}

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