Question

$\text{Equivalent wt. =} \frac{\text{atomic wt}}{\text{Valency}} \text{OR} \frac{\text{Mol.w.t}}{\text{Valency}}$

Calculate n-f (Both Sides)

① $2Cl^{-} \to Cl_2$

② $Fe_2O_3 \to FeO$

③ $N_3^- \to NH_3$

④ $Fe_2O_3 \to Fe_{0.95}O$

⑤ $Fe_3O_4 \to FeO$

⑥ $H_3PO_2 \to PH_3$

⑦ $SO_2 \to H_2S$

⑧ $SO_3 \to H_2SO_4$

⑨ $C_2O_4^{2-} \to CO_2$

⑩ $H_2O_2 \to H_2O$

⑪ $H_2O_2 \to O_2$

Answer 1

The n-factors (reactant, product) are as follows: ① $2Cl^{-} \to Cl_2$: (1, 2) ② $Fe_2O_3 \to FeO$: (2, 1) ③ $N_3^- \to NH_3$: (8, 8/3) ④ $Fe_2O_3 \to Fe_{0.95}O$: (34/19, 17/20) ⑤ $Fe_3O_4 \to FeO$: (2, 2/3) ⑥ $H_3PO_2 \to PH_3$: (4, 4) ⑦ $SO_2 \to H_2S$: (6, 6) ⑧ $SO_3 \to H_2SO_4$: (2, 2) ⑨ $C_2O_4^{2-} \to CO_2$: (2, 1) ⑩ $H_2O_2 \to H_2O$: (2, 1) ⑪ $H_2O_2 \to O_2$: (2, 2)

Answer 2

The n-factor for each substance is calculated based on the change in oxidation state of the element undergoing redox. For the reactant, we determine the change in oxidation state from its initial state to its final state in the product, multiplied by the number of atoms of that element in the reactant molecule/ion. For the product, we consider the reverse reaction and calculate its n-factor similarly. For non-redox reactions (like $SO_3 \to H_2SO_4$), the n-factor is determined by its acid-base properties (basicity/acidity).