A) \[CuS{{O}_{4}}\]
B) \[{{(N{{H}_{4}})}_{2}}S{{O}_{4}}\]
C) \[ZnS{{O}_{4}}\]
D) \[FeS{{O}_{4}}\]
Correct Answer: A
Solution :
Iodine being a strong reducing agent reduce \[C{{u}^{2+}}\] ions to \[C{{u}^{+}}\] ions and itself gets oxidised to iodine. \[\underset{\text{Reduced}}{\overset{+2}{\mathop{2CuS{{O}_{4}}}}}\,+\underset{\text{Oxidised}}{\overset{-1}{\mathop{4KI}}}\,\to \overset{+1}{\mathop{C{{u}_{2}}{{I}_{2}}}}\,+\overset{0}{\mathop{{{I}_{2}}}}\,+2{{K}_{2}}S{{O}_{4}}\]
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