Respuesta :
Answer: [tex]Cu+Au^{3+}\rightarrow Cu^{2+}+Au[/tex]
Explanation: when [tex]E^0[/tex]= +ve, reaction is spontaneous
[tex]E^0[/tex]= -ve, reaction is non spontaneous
[tex]E^0[/tex]= 0, reaction is in equilibrium
1. [tex]Ni+Zn^{2+}\rightarrow Ni^{2+}+Zn[/tex]
Here Ni undergoes oxidation by loss of electrons, thus act as anode. zinc 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_{[Zn^{2+}/Zn]}= -0.76V[/tex]
[tex]E^0_{[Ni^{2+}/Ni]}= -0.26V[/tex]
[tex]E^0=E^0_{[Zn^{2+}/Zn]}- E^0_{[Ni^{2+}/Ni]}[/tex]
[tex]E^0=-0.76V- (-0.26V)=-0.50V[/tex]
As the standard emf comes to be negative, the reaction is non spontaneous.
2) [tex]Mg+Li^{+}\rightarrow Mg^{2+}+Li[/tex]
Here Mg undergoes oxidation by loss of electrons, thus act as anode. Lithium undergoes reduction by gain of electrons and thus act as cathode.
[tex]E^0_{[Mg^{2+}/Mg]}= -2.37V[/tex]
[tex]E^0_{[Li^{+}/Li]}= -3.05V[/tex]
[tex]E^0=E^0_{[Li^{+}/Li]}- E^0_{[Mg^{2+}/Mg]}[/tex]
[tex]E^0=-3.05- (-2.37V)=-0.68V[/tex]
As the standard emf comes to be negative, the reaction is non spontaneous.
3) [tex]Cu+Au^{3+}\rightarrow Cu^{2+}+Au[/tex]
Here Cu undergoes oxidation by loss of electrons, thus act as anode. Gold undergoes reduction by gain of electrons and thus act as cathode.
[tex]E^0_{[Au^{3+}/Au]}= +1.40V[/tex]
[tex]E^0_{[Cu^{2+}/Cu]}= 0.34V[/tex]
[tex]E^0=E^0_{[Au^{3+}/Au]}- E^0_{[Cu^{2+}/Cu]}[/tex]
[tex]E^0=1.40- (+0.34V)=1.06V[/tex]
As the standard emf comes to be positive, the reaction is spontaneous.
4) [tex]Zn+Ca^{2+}\rightarrow Zn^{2+}+Ca[/tex]
Here Zn undergoes oxidation by loss of electrons, thus act as anode. Calcium undergoes reduction by gain of electrons and thus act as cathode.
[tex]E^0_{[Zn^{2+}/Zn]}= -0.76V[/tex]
[tex]E^0_{[Ca^{2+}/Ca]}=-2.87V[/tex]
[tex]E^0=E^0_{[Ca^{2+}/Ca}- E^0_{[Zn^{2+}/Zn]}[/tex]
[tex]E^0=-2.87V- (-0.76V)=-2.11V[/tex]
As the standard emf comes to be negative, the reaction is non spontaneous.
Answer : The spontaneous redox reaction will be:
[tex]Cu+Au^{3+}\rightarrow Cu^{2+}+Au[/tex]
Explanation :
The given reactions are the examples of single displacement reactions.
Single displacement reactions : It is defined as the reactions in which most reactive metal displaces the least reactive metal from its chemical reaction. The reactivity of metals is given by the series known as reactivity series.
That means the metals that lie above in the series can replace the metals which lie low in the reactivity series.
For the given options:
Option A :
[tex]Ni+Zn^{2+}\rightarrow Ni^{2+}+Zn[/tex]
Nickel lies low in the series than zinc and hence is less reactive. So, it will not displace zinc. Thus, this is a non-spontaneous reaction.
Option B :
[tex]Mg+Li^+\rightarrow Mg^{2+}+Li[/tex]
Magnesium lies low in the series than lithium and hence is less reactive. So, it will not displace lithium. Thus, this is a non-spontaneous reaction.
Option C :
[tex]Cu+Au^{3+}\rightarrow Cu^{2+}+Au[/tex]
Copper lies above in the series than gold and hence is more reactive. So, it will easily displace gold. Thus, this is a spontaneous reaction.
Option D :
[tex]Zn+Ca^{2+}\rightarrow Zn^{2+}+Ca[/tex]
Zinc lies low in the series than calcium and hence is less reactive. So, it will not displace calcium. Thus, this is a non-spontaneous reaction.
Hence, the spontaneous redox reaction will be:
[tex]Cu+Au^{3+}\rightarrow Cu^{2+}+Au[/tex]
