From which of the following reaction is blister copper obtained?

C u_{2} S + F e O ⟶ 2 C u + F e O

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C u_{2} S + F e S ⟶ 2 C u + F e S_{2}

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C u_{2} S + 2 C u_{2} O ⟶ 6 C u + S O_{2}

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C u^{2 +} + F e ⟶ F e^{2 +} + C u

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Solution

The correct option is B $C u_{2} S + 2 C u_{2} O ⟶ 6 C u + S O_{2}$
The formed molten copper when cooled gives off dissolved $S O_{2}$ . The $S O_{2}$ gas escaping in the form of bubbles leaves the surface of the metal full of cavities, which gives the metal a blistered appearance. This is why the metal obtained is blister copper.
Reaction:
$C u_{2} S + 2 C u_{2} O ⟶ 6 C u + S O_{2}$

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Similar questions

F e^{2 +} + 2 e^{-} \to F e

E^{\circ} = - 0.44 V

C u^{2 +} + 2 e^{-} \to C u

E^{\circ} = + 0.34 V

E^{\circ}

F e + C u^{2 +} ⇋ F e^{2} + C u

C u_{2} S + 2 C u_{2} O \to 6 C u + S o_{2}

(Δ G^{\circ})

2 C u^{+} \to C u + C u^{2 +}

E_{C u^{+} / C u}^{\circ} = 0.52 V, E_{C u^{2 +} / C u^{+}}^{\circ} = 0.16 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}

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