Respuesta :
Answer: The redox reactions that occur spontaneously are reaction b and reaction c.
Explanation:
For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.
Relationship between standard Gibbs free energy and standard electrode potential follows:
[tex]\Delta G^o=-nFE^o_{cell}[/tex]
For a reaction to be spontaneous, the standard electrode potential must be positive.
To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex] .......(1)
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
- For a:
The given chemical equation follows:
[tex]Pb(s)+Mn^{2+}(aq.)\rightarrow Pb^{2+}(aq.)+Mn(s)[/tex]
Oxidation half reaction: [tex]Pb(s)\rightarrow Pb^{2+}(aq.)+2e^-;E^o_{Pb^{2+}/Pb}=-0.13V[/tex]
Reduction half reaction: [tex]Mn^{2+}(aq.)+2e^-\rightarrow Mn(s);E^o_{Mn^{2+}/Mn}=-1.18V[/tex]
Putting values in equation 1, we get:
[tex]E^o_{cell}=-1.18-(-0.13)=-1.05V[/tex]
As, the standard electrode potential is coming out to be negative, this means that standard Gibbs free energy will be positive.
Thus, the reaction will be non-spontaneous.
- For a:
The given chemical equation follows:
[tex]Zn(s)+Fe^{2+}(aq.)\rightarrow Zn^{2+}(aq.)+Fe(s)[/tex]
Oxidation half reaction: [tex]Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-;E^o_{Zn^{2+}/Zn}=-0.76V[/tex]
Reduction half reaction: [tex]Fe^{2+}(aq.)+2e^-\rightarrow Fe(s);E^o_{Fe^{2+}/Fe}=-0.44V[/tex]
Putting values in equation 1, we get:
[tex]E^o_{cell}=-0.44-(-0.76)=0.32V[/tex]
As, the standard electrode potential is coming out to be positive, this means that standard Gibbs free energy will be negative.
Thus, the reaction will be spontaneous.
- For c:
The given chemical equation follows:
[tex]Cr(s)+3Ag^{+}(aq.)\rightarrow Cr^{3+}(aq.)+3Ag(s)[/tex]
Oxidation half reaction: [tex]Cr(s)\rightarrow Cr^{3+}(aq.)+3e^-;E^o_{Cr^{3+}/Cr}=-0.74V[/tex]
Reduction half reaction: [tex]Ag^{+}(aq.)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V[/tex] ( × 3)
Putting values in equation 1, we get:
[tex]E^o_{cell}=0.80-(-0.74)=1.54V[/tex]
As, the standard electrode potential is coming out to be positive, this means that standard Gibbs free energy will be negative.
Thus, the reaction will be spontaneous.
- For d:
The given chemical equation follows:
[tex]Co(s)+Cd^{2+}(aq.)\rightarrow Co^{2+}(aq.)+Cd(s)[/tex]
Oxidation half reaction: [tex]Co(s)\rightarrow Co^{2+}(aq.)+2e^-;E^o_{Co^{2+}/Co}=-0.28V[/tex]
Reduction half reaction: [tex]Cd^{2+}(aq.)+2e^-\rightarrow Cd(s);E^o_{Cd^{2+}/Cd}=-0.40V[/tex]
Putting values in equation 1, we get:
[tex]E^o_{cell}=-0.40-(-0.28)=-0.12V[/tex]
As, the standard electrode potential is coming out to be negative, this means that standard Gibbs free energy will be positive.
Thus, the reaction will be non-spontaneous.
Hence, the redox reactions that occur spontaneously are reaction b and reaction c.
