A) \[Fe+{{H}_{2}}S{{O}_{4}}\,\to \,\,FeS{{O}_{4}}+{{H}_{2}}\]
B) \[Cu+2AgN{{O}_{3}}\,\to \,Cu\,{{(N{{O}_{3}})}_{2}}+2Ag\]
C) \[2KBr+{{I}_{2}}\,\to \,\,2KI+B{{r}_{2}}\]
D) \[CuO+{{H}_{2}}\,\to \,\,Cu+{{H}_{2}}O\]
Correct Answer: C
Solution :
Brown layer is deposited on iron rod because \[Cu\] has greater reduction potential than that of \[F{{e}^{2+}}\].
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