What is the four-electron reduced form of $\mathrm{O}_2$ ?

How many electrons are needed to reduce $\mathrm{N}_2$ to $\mathrm{NH}_3 ?$

The equivalent weight of $$\mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3(\mathrm{Gram}$$ molecular weight $$=\mathrm{M})$$ in the given reaction is

$$\mathrm{I}_2+2 \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3=2 \mathrm{NaI}+\mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6$$

The equivalent weight of KIO$$_3$$ in the given reaction is (M = molecular mass):

$$\mathrm{2Cr{(OH)_3} + 4O{H^ - } + KI{O_3} \to 2CrO_4^{2 - } + 5{H_2}O + KI}$$