<u>Answer:</u> The redox reactions that occur spontaneously are Reaction W and Reaction Z.
<u>Explanation:</u>
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:
![\Delta G^o=-nFE^o_{cell}](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D-nFE%5Eo_%7Bcell%7D)
For a reaction to be spontaneous, the standard electrode potential must be positive.
To calculate the
of the reaction, we use the equation:
.......(1)
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
The chemical reaction follows:
![2Al(s)+3Pb^{2+}(aq.)\rightarrow 2Al^{3+}(aq.)+3Pb(s)](https://tex.z-dn.net/?f=2Al%28s%29%2B3Pb%5E%7B2%2B%7D%28aq.%29%5Crightarrow%202Al%5E%7B3%2B%7D%28aq.%29%2B3Pb%28s%29)
We know that:
![E^o_{Al^{3+}/Al}=-1.66V\\E^o_{Pb^{2+}/Pb}=-0.13V](https://tex.z-dn.net/?f=E%5Eo_%7BAl%5E%7B3%2B%7D%2FAl%7D%3D-1.66V%5C%5CE%5Eo_%7BPb%5E%7B2%2B%7D%2FPb%7D%3D-0.13V)
Calculating the
using equation 1, we get:
![E^o_{cell}=-0.13-(-1.66)=1.53V](https://tex.z-dn.net/?f=E%5Eo_%7Bcell%7D%3D-0.13-%28-1.66%29%3D1.53V)
As, the standard electrode potential is coming out to be positive. So, the reaction is spontaneous.
The chemical reaction follows:
![Fe(s)+Cr^{3+}(aq.)\rightarrow Fe^{3+}(aq.)+Cr(s)](https://tex.z-dn.net/?f=Fe%28s%29%2BCr%5E%7B3%2B%7D%28aq.%29%5Crightarrow%20Fe%5E%7B3%2B%7D%28aq.%29%2BCr%28s%29)
We know that:
![E^o_{Fe^{3+}/Fe}=0.77V\\E^o_{Cr^{3+}/Cr}=-0.74V](https://tex.z-dn.net/?f=E%5Eo_%7BFe%5E%7B3%2B%7D%2FFe%7D%3D0.77V%5C%5CE%5Eo_%7BCr%5E%7B3%2B%7D%2FCr%7D%3D-0.74V)
Calculating the
using equation 1, we get:
![E^o_{cell}=-0.74-(0.77)=-1.51V](https://tex.z-dn.net/?f=E%5Eo_%7Bcell%7D%3D-0.74-%280.77%29%3D-1.51V)
As, the standard electrode potential is coming out to be negative. So, the reaction is not spontaneous.
The chemical reaction follows:
![Zn(s)+Ca^{2+}(aq.)\rightarrow Zn^{2+}(aq.)+Ca(s)](https://tex.z-dn.net/?f=Zn%28s%29%2BCa%5E%7B2%2B%7D%28aq.%29%5Crightarrow%20Zn%5E%7B2%2B%7D%28aq.%29%2BCa%28s%29)
We know that:
![E^o_{Ca^{2+}/Ca}=-2.87V\\E^o_{Zn^{2+}/Zn}=-0.76V](https://tex.z-dn.net/?f=E%5Eo_%7BCa%5E%7B2%2B%7D%2FCa%7D%3D-2.87V%5C%5CE%5Eo_%7BZn%5E%7B2%2B%7D%2FZn%7D%3D-0.76V)
Calculating the
using equation 1, we get:
![E^o_{cell}=-2.87-(-0.76)=-2.11V](https://tex.z-dn.net/?f=E%5Eo_%7Bcell%7D%3D-2.87-%28-0.76%29%3D-2.11V)
As, the standard electrode potential is coming out to be negative. So, the reaction is not spontaneous.
The chemical reaction follows:
![Co(s)+2Cu^{+}(aq.)\rightarrow Co^{2+}(aq.)+2Cu(s)](https://tex.z-dn.net/?f=Co%28s%29%2B2Cu%5E%7B%2B%7D%28aq.%29%5Crightarrow%20Co%5E%7B2%2B%7D%28aq.%29%2B2Cu%28s%29)
We know that:
![E^o_{Cu^{+}/Cu}=0.34V\\E^o_{Co^{2+}/Co}=-0.28V](https://tex.z-dn.net/?f=E%5Eo_%7BCu%5E%7B%2B%7D%2FCu%7D%3D0.34V%5C%5CE%5Eo_%7BCo%5E%7B2%2B%7D%2FCo%7D%3D-0.28V)
Calculating the
using equation 1, we get:
![E^o_{cell}=0.34-(-0.28)=0.62V](https://tex.z-dn.net/?f=E%5Eo_%7Bcell%7D%3D0.34-%28-0.28%29%3D0.62V)
As, the standard electrode potential is coming out to be positive. So, the reaction is spontaneous.
Hence, the redox reactions that occur spontaneously are Reaction W and Reaction Z.