Respuesta :
Answer:
[tex]2Cu^++PbO_2+SO_4^{2-}+4H^+\rightarrow 2Cu^{2+}+PbSO_4+2H_2O[/tex]
[tex]E^0=1.53[/tex], the reaction is spontaneous.
Explanation:
Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.
[tex]X\rightarrow X^{n+}+ne^-[/tex]
Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.
[tex]X^{n+}+ne^-\rightarrow X[/tex]
For the given chemical reaction:
[tex]Cu^++PbO_2+SO_4^{2-}\rightarrow PbSO_4+Cu^{2+}[/tex]
The half cell reactions for the above reaction follows:
Oxidation half reaction: [tex]Cu^+\rightarrow Cu^{2+}+e^-[/tex]
Reduction half reaction: [tex]PbO_2+SO_4^{2-}+4H^++2e^-\rightarrow PbSO_4+2H_2O[/tex]
To balance the oxidation half reaction must be multiplied by 2 and thus, the balanced equation is:-
[tex]2Cu^++PbO_2+SO_4^{2-}+4H^++2e^-\rightarrow 2Cu^{2+}+PbSO_4+2H_2O[/tex]
Here [tex]Cu^+[/tex] undergoes oxidation by loss of electrons, thus act as anode. [tex]PbO_2[/tex] undergoes reduction by gain of electrons and thus act as cathode.
[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]
Where both [tex]E^0[/tex] are standard reduction potentials.
[tex]E^0_{[Cu^{2+}/Cu^{+}]}= +0.16V[/tex]
[tex]E^0_{[PbO_2/PbSO_4]}=+1.69V[/tex]
[tex]E^0=E^0_{[PbO_2/PbSO_4]}- E^0_{[Cu^{2+}/Cu^{+}]}[/tex]
[tex]E^0=+1.69- (0.16V)=1.53[/tex]
Since, [tex]E^0>0[/tex], the reaction is spontaneous.
