Respuesta :
Answer:
For a: The reaction is not spontaneous.
For b: The reaction is spontaneous.
For c: The reaction is not spontaneous.
For d: The reaction is spontaneous.
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 chemical reaction follows:
[tex]Fe(s)+Mn^{2+}(aq.)\rightarrow Fe^{2+}(aq.)+Mn(s)[/tex]
We know that:
[tex]E^o_{Fe^{2+}/Fe}=-0.44V\\E^o_{Mn^{2+}/Mn}=-1.18V[/tex]
Calculating the [tex]E^o_{cell}[/tex] using equation 1, we get:
[tex]E^o_{cell}=-1.18-(-0.44)=-0.74V[/tex]
As, the standard electrode potential is coming out to be negative. So, the reaction is not spontaneous.
- For b:
The chemical reaction follows:
[tex]Fe(s)+2Ag^{+}(aq.)\rightarrow Fe^{2+}(aq.)+2Ag(s)[/tex]
We know that:
[tex]E^o_{Fe^{2+}/Fe}=-0.44V\\E^o_{Ag^{+}/Ag}=0.80V[/tex]
Calculating the [tex]E^o_{cell}[/tex] using equation 1, we get:
[tex]E^o_{cell}=0.80-(-0.44)=1.24V[/tex]
As, the standard electrode potential is coming out to be positive. So, the reaction is spontaneous.
- For c:
The chemical reaction follows:
[tex]Zn(s)+Mg^{2+}(aq.)\rightarrow Zn^{2+}(aq.)+Mg(s)[/tex]
We know that:
[tex]E^o_{Zn^{2+}/Zn}=-0.76V\\E^o_{Mg^{2+}/Mg}=-2.37V[/tex]
Calculating the [tex]E^o_{cell}[/tex] using equation 1, we get:
[tex]E^o_{cell}=-2.37-(-0.76)=-1.61V[/tex]
As, the standard electrode potential is coming out to be negative. So, the reaction is not spontaneous.
- For d:
The chemical reaction follows:
[tex]2Al(s)+3Pb^{2+}(aq.)\rightarrow 2Al^{3+}(aq.)+3Pb(s)[/tex]
We know that:
[tex]E^o_{Al^{3+}/Al}=-1.66V\\E^o_{Pb^{2+}/Pb}=-0.13V[/tex]
Calculating the [tex]E^o_{cell}[/tex] using equation 1, we get:
[tex]E^o_{cell}=-0.13-(-1.66)=1.53V[/tex]
As, the standard electrode potential is coming out to be positive. So, the reaction is spontaneous.
