Extraction of Metals - Result Question 44
####44. Upon heating with $Cu _2 S$, the reagent(s) that give copper metal is/are
(2014 Adv.)
(a) $CuFeS _2$
(b) $CuO$
(c) $Cu _2 O$
(d) $CuSO _4$
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Solution:
- (b) $4 CuO \xrightarrow{1100^{\circ} C} 2 Cu _2 O+O _2$
$$ 2 Cu _2 O+Cu _2 S \stackrel{\Delta}{\longrightarrow} 6 Cu+SO _2 $$
(c) $Cu _2 S+2 Cu _2 O \xrightarrow{\Delta} 6 Cu+SO _2$
(d) $CuSO _4 \xrightarrow{720^{\circ} C} CuO+SO _2+\frac{1}{2} O _2$
$$ \begin{array}{r} 4 CuO \xrightarrow{1100^{\circ} C} 2 Cu _2 O+O _2 \ 2 Cu _2 O+Cu _2 S \stackrel{\Delta}{\longrightarrow} 6 Cu+SO _2 \end{array} $$
Reaction is believed to proceed as
$$ \begin{aligned} Cu _2 S & \rightleftharpoons 2 Cu^{+}+S^{2-} \ 2 Cu _2 O & \rightleftharpoons 4 Cu^{+}+2 O^{2-} \ S^{2-}+2 O^{2-} & \longmapsto SO _2+6 e^{-} \ 6 Cu^{+}+6 e^{-} & \longrightarrow 6 Cu ; E _{\text {cell }}^{\circ}=0.52 \end{aligned} $$
Here, copper sulphide is reduced to copper metal. Solidified copper has blistered appearance due to evolution of $SO _2$ and thus obtained copper is known as blister copper.
Other compounds which give $Cu$ are
(i) $CuO$ as $4 CuO \xrightarrow{1100^{\circ} C} 2 Cu _2 O+O _2$
$$ 2 Cu _2 O+Cu _2 S \stackrel{\Delta}{\longrightarrow} 6 Cu+SO _2 $$
(ii) $CuSO _4$ as $CuSO _4 \xrightarrow{720^{\circ} C} CuO+SO _2+\frac{1}{2} O _2$
$$ \begin{array}{r} 4 CuO \stackrel{\Delta}{\longrightarrow} 2 Cu _2 O+O _2 \ 2 Cu _2 O+Cu _2 S \stackrel{\Delta}{\longrightarrow} 6 Cu+SO _2 \end{array} $$
While $CuFeS _2$ will not give $Cu$ on heating. The heating in the presence of $O _2$ gives $Cu _2 S$ and $FeS$ with the evolution of $SO _2$.
