Answer:
Explanation:
reduction potential of NO gas is highly unfavourable . It is - 1.7 V . So it is highly unlikely to be reduced to NOā» . On the other hand it is easily oxidised .
Half cell reaction of given cell
At anode ( where oxidation occurs )
NOā» ā NO + e ( reduction potential is - 1.7 V )
At cathode ( where reduction takes place )
Brā + 2e ā2 Brā» Ā ( reduction potential is 1.09 V )
(NOā» ā NO + e ) x 2
Brā + 2e ā2 Brā»
--------------------------------
Brā + 2NOā» ā 2NO +2 Brā»
Ecell = Ecathode - E anode
= 1.09 - 2 x ( - 1.7)
4.5 V