Respuesta :
Answer: The half reaction occurring at anode is [tex]2I^-(aq.)\rightarrow I_2(s)+2e^-[/tex]
Explanation:
The substance having highest positive [tex]E^o[/tex] potential will always get reduced and will undergo reduction reaction.
For the given chemical equation:
[tex]2ClO_2(g)+2I^-(aq)\rightarrow 2ClO^{-2}(aq.)+I_2(s)[/tex]
The half reaction follows:
Oxidation half reaction: [tex]2I^-(aq.)\rightarrow I_2(s)+2e^-;E^o_{I_2/I^-}=0.53V[/tex]
Reduction half reaction: [tex]ClO_2(g)+e^-\rightarrow ClO_2^-(aq.);E^o_{ClO_2/ClO_2^-}=+0.954V[/tex] ( × 2 )
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
Hence, the half reaction occurring at anode is [tex]2I^-(aq.)\rightarrow I_2(s)+2e^-[/tex]
The half reaction occurring at anode is:
[tex]2I^-(aq)---- > I_2(s)+2e^-[/tex]
Half reaction for the cell:
The substance having highest positive potential will always get reduced and will undergo reduction reaction.
Balanced chemical equation:
[tex]2ClO_2(g)+2I^-(aq)----- > 2ClO^{2-}(aq)+I_2(s)[/tex]
The half reaction follows:
Oxidation half reaction: [tex]2I^-(aq)---- > I_2(s)+2e^-[/tex] , Reduction potential is 0.53V
Reduction half reaction: [tex]ClO_2(g)+e^----- > ClO_2^-[/tex] ( × 2 ), Oxidation potential is +0.954 V
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
Hence, the half reaction occurring at anode is :
[tex]2I^-(aq)---- > I_2(s)+2e^-[/tex]
Find more information about Reduction potential here:
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