In which of the following reactions, the underlined substance has been reduced?

Question

In which of the following reactions, the underlined substance has been reduced? (A)  underlineCO + CuOarrow underlineCO2 + Cu (B)  underlineCuO + 2 HClarrow underlineCuCl2 + H2O (C)  underline4H2O (g) + 3 Fe arrow underline4H2 (g) + Fe3O4 (D)  underlineC + 4 HNO3 arrow underlineCO2 + 2H2O + 4 NO2

Tardigrade Admin · Accepted Answer

a) overset+2C O longrightarrow overset+4CO 2 b) overset+2CuO longrightarrow overset+2C uCl 2 c) overset+1H 2 O longrightarrow overset0 H 2 d) overset0 C longrightarrow overset+4C O 2 Only in (c), O.N. of hydrogen decreases from +1 to 0 and hence H 2 O gets reduced to H 2

Q. In which of the following reactions, the underlined substance has been reduced?

$\underline{CO} + C u O \to \underline{C O_{2}} + C u$

$\underline{C u O} + 2 H Cl \to \underline{C u C l_{2}} + H_{2} O$

$\underline{4 H_{2} O} (g) + 3 F e \to \underline{4 H_{2}} (g) + F e_{3} O_{4}$

$\underline{C} + 4 H N O_{3} \to \underline{C O_{2}} + 2 H_{2} O + 4 N O_{2}$

a) $C + 2 O ⟶ C + 4 O_{2}$
b) $C + 2 u O ⟶ C + 2 u C l_{2}$
c) $H + 1_{2} O ⟶ H_{2} 0$
d) $C 0 ⟶ C + 4 O_{2}$
Only in (c), O.N. of hydrogen decreases from +1 to 0 and hence $H_{2} O$ gets reduced to $H_{2}$

Q. In which of the following reactions, the underlined substance has been reduced?

CO​+CuO→CO2​​+Cu

CuO​+2HCl→CuCl2​​+H2​O

4H2​O​(g)+3Fe→4H2​​(g)+Fe3​O4​

C​+4HNO3​→CO2​​+2H2​O+4NO2​

a) C+2O⟶C+4O2​ b) C+2uO⟶C+2uCl2​ c) H+12​O⟶H2​0​ d) C0⟶C+4O2​ Only in (c), O.N. of hydrogen decreases from +1 to 0 and hence H2​O gets reduced to H2​

$\underline{CO} + C u O \to \underline{C O_{2}} + C u$

$\underline{C u O} + 2 H Cl \to \underline{C u C l_{2}} + H_{2} O$

$\underline{4 H_{2} O} (g) + 3 F e \to \underline{4 H_{2}} (g) + F e_{3} O_{4}$

$\underline{C} + 4 H N O_{3} \to \underline{C O_{2}} + 2 H_{2} O + 4 N O_{2}$

a) $C + 2 O ⟶ C + 4 O_{2}$
b) $C + 2 u O ⟶ C + 2 u C l_{2}$
c) $H + 1_{2} O ⟶ H_{2} 0$
d) $C 0 ⟶ C + 4 O_{2}$
Only in (c), O.N. of hydrogen decreases from +1 to 0 and hence $H_{2} O$ gets reduced to $H_{2}$