Match the reactions of H₃PO₃ (molar mass M) in Column I with the corresponding equivalent mass of H₃PO₃ in Column I.

 

(i)

$\begin{array}{l}\mathrm{E}\text{ (oxidation) }=\frac{\mathrm{M}}{2}=\frac{3\mathrm{M}}{6}\\ \mathrm{E}\text{ (reduction) }=\frac{\mathrm{M}}{6}\end{array}$

$\text{ Thus, (i) }\to \text{ (s) }$

$\text{ (ii) }{\mathrm{H}}_3{\mathrm{PO}}_3\equiv 1{\mathrm{OH}}^{-}\equiv 1{\mathrm{H}}^{+}$

$E\left({\mathrm{H}}_3{\mathrm{PO}}_3\right)=M$

$\text{ Thus, (ii) }\to \text{ (p) }$

(iii) ${\mathrm{H}}_3{\mathrm{PO}}_3\equiv 1\mathrm{Ca}(\mathrm{OH}{)}_2\equiv 2{\mathrm{OH}}^{-}\equiv 2{\mathrm{H}}^{+}$

$\therefore 1{\mathrm{H}}^{+}\equiv \frac{{\mathrm{H}}_3{\mathrm{PO}}_3}{2}$

$\mathrm{E}\left({\mathrm{H}}_3{\mathrm{PO}}_3\right)=\frac{\mathrm{M}}{2}$

$\text{ Thus, (iii) }\to \text{ (q) }$

$\underset{+3}{\underset{\uparrow }{{\mathrm{H}}_3{\mathrm{PO}}_3}}⟶\underset{+5}{\underset{\uparrow }{{\mathrm{H}}_3{\mathrm{PO}}_4}}$

$E=\frac{M}{2}$

$\text{ Thus, (iv) }\to \text{ (q) }$