Change in oxidation state of carbon is from $$ - $$ 4 to + 4

In a disproportionation reaction, same substance undergoes oxidation (increase in oxidation number) and reduction (decrease in oxidation number forming two different products.

$$H\mathop N\limits^{ + 5} {O_3},\mathop N\limits^{ + 2} O,\mathop {{N_2}}\limits^0 ,\mathop N\limits^{ - 3} {H_4}Cl$$

Calculate E^o_cell corresponding to each compound undergoing disproportionation reaction. The reaction for which E^o_cell comes out + ve is spontaneous.

HBrO $$ \to $$ Br₂, E^o_HBrO/Br2 = 1.595 V

HBrO $$ \to $$ BrO₃^–, E^o_BrO3^-/HBrO = 1.595 V

E^o_cell for the disproportionation of HBrO,

E^o_cell = E^o_HBrO/Br2 - E^o_BrO3^-/HBrO

= 1.595 - 1.5

= 0.095 V = + ve

So, E^o_cell > 0

So, $$\Delta $$G^o < 0 (Spontaneous)