A. \(2 HNO _{2}+2 HI \rightarrow 2 H _{2} O +2 NO + I _{2}\) B. \(Zn + H _{2} SO _{4} \rightarrow ZnSO _{4}+ H _{2}\) C. \(BaCl _{2}+2 AgNO _{3} \rightarrow 2 AgCl + Ba \left( NO _{3}\right)_{2}\) D. \(4 FeO + O _{2} \rightarrow 2 Fe _{2} O _{3}\)
Correct Answer: C
Explanation
In (A) the oxidation no of iodine changed from \((-1)\) in \(HI\) to \((0)\) in \(HI\), so, it is a redox reaction. - In (B) the oxidation no of zinc changed from (O) in \(Zn\) to (72) in \(ZnSO _{4}\), and hydrogen from \((+1)\) in \(H _{2} SO _{4}\) to (O) in \(H _{2}\). Hence, it is redox reaction. - In (D) iron is oxidized from \((+2)\) state \(( FeO )\) to \((+3)\) state (in \(Fe _{2} O _{3}\) ) making it a redox process. - (C) is a neutralization reaction in which none of the species changed oxidation state.
