Calculate the equilibrium constant for the reaction:

$F e + C u S O_{4} ⟶ F e S O_{4} + C u$

at $25^{o}$ C.

[Given : $E_{(F e / F e^{2 +})}^{0} = 0.44 V$ ; $E_{(C u / C u^{2 +})}^{0} = - 0.337 V$ ].

2.18 \times 10^{26}

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3.18 \times 10^{26}

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1.18 \times 10^{26}

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1.96 \times 10^{26}

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Solution

The correct option is D $1.96 \times 10^{26}$
$F e + {C u S O}_{4} ⟶ {F e S O}_{4} + C u at 25^{\circ} C$

$E^{\circ} F e / F e^{+ 2} = 0.44 V, E_{C u / {C u}^{+ 2}} = - 0.337 V$

$\begin{matrix} Nerst equation E_{cell} = E^{0} - \frac{R T}{n F} ln Q \end{matrix}$

$for equilibrium constant$

$\begin{matrix} E_{cell}^{0} = 0 Q \to K Ecell = - \frac{R T}{n F} ln K ln K = \frac{n F E}{- R T} \end{matrix}$

$\begin{matrix} Ecell & = E_{o x i} - E_{r e d} = 0.44 + 0.337 = 0.777 \end{matrix}$

$\begin{matrix} T & = 273 + 25 = 298 K \end{matrix}$

$= \frac{2 \times 0.777 \times 96500}{0.0821 \times 298}$

$\begin{matrix} ln K = - 6129.41 2.303 log K = 6129.41 log K = - 2661.40 = - 2.66 \times 10^{- 3} \end{matrix}$

$K = 1.96 \times 10^{26}$

Suggest Corrections

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} C

E^{0} [F e \to F e^{2 +}) = 0.44 V

E^{0} (C u \to C u^{2 +}) = - 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

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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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