Calculate the equilibrium constant for the reaction $F e + C u S O_{4} ⇌ F e S O_{4} + C u$ at $25^{0}$ C.

Open in App

Solution

$F e + C u {S O}_{4} ⇌ F e {S O}_{4} + C u$
$K_{C} = \frac{[F e {S O}_{4}] [C u]}{[F e] [C u {S O}_{4}]}$

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}

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

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

Join BYJU'S Learning Program