Question

Consider the two redox reactions I and II \[ \text{I. }2\mathrm{Na_2S_2O_3}+ \mathrm{I_2} \rightarrow \mathrm{Na_2S_4O_6}+2\mathrm{NaI} \] \[ \text{II. }2\mathrm{KMnO_4}+10\mathrm{FeSO_4}+3\mathrm{H_2SO_4} \rightarrow 2\mathrm{MnSO_4}+5\mathrm{Fe_2(SO_4)_3}+2\mathrm{K_2SO_4}+3\mathrm{H_2O} \] The ratio of equivalent weights of \(\mathrm{KMnO_4}\) and \(\mathrm{Na_2S_2O_3}\) is \[ (\mathrm{K}=39u;\ \mathrm{Mn}=55u;\ \mathrm{O}=16u;\ \mathrm{Na}=23u;\ \mathrm{S}=32u) \]

• A. \(1:1\)
• B. \(2:1\)
• C. \(1:5\)
• D. \(5:1\)

Hint:

Equivalent weight is given by \[ \boxed{ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}}. } \] For acidic \(\mathrm{KMnO_4}\), \(n=5\), while for \(\mathrm{Na_2S_2O_3}\), \(n=1\).

Answer 1

The Correct Option is C

Step 1: Find the equivalent weight of \(\mathrm{KMnO_4}\). In acidic medium, \[ \mathrm{MnO_4^-}\rightarrow\mathrm{Mn^{2+}} \] involves a change of \[ 5 \] electrons. Molar mass of \(\mathrm{KMnO_4}\) is \[ 39+55+4(16)=158. \] Hence, \[ \text{Equivalent weight} = \frac{158}{5}=31.6. \]

Step 2: Find the equivalent weight of \(\mathrm{Na_2S_2O_3}\). In the reaction, \[ 2\mathrm{S_2O_3^{2-}} \rightarrow \mathrm{S_4O_6^{2-}}+2e^-. \] Thus, one mole of \(\mathrm{Na_2S_2O_3}\) loses \[ 1 \] electron. Its molar mass is \[ 2(23)+2(32)+3(16)=158. \] Therefore, \[ \text{Equivalent weight} = \frac{158}{1}=158. \]

Step 3: Calculate the ratio. \[ \frac{31.6}{158} = \frac15. \] Hence, \[ \boxed{\text{Equivalent weights of } \mathrm{KMnO_4}:\mathrm{Na_2S_2O_3}=1:5.} \] Therefore, the correct option is \(\boxed{(C)}\).