Answer: The standard potential of the cell is 0.77 V
Explanation:
We know that:
[tex]E^o_{Ni^{2+}/Ni}=-0.25V\\E^o_{Cu^{+}/Cu}=0.52V[/tex]
The substance having highest positive [tex]E^o[/tex] reduction potential will always get reduced and will undergo reduction reaction.
The half reaction follows:
Oxidation half reaction: [tex]Ni(s)\rightarrow Ni^{2+}(aq)+2e^-[/tex]
Reduction half reaction: [tex]Cu^{+}(aq)+e^-\rightarrow Cu(s)[/tex] ( × 2)
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]
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
Putting values in above equation follows:
[tex]E^o_{cell}=0.52-(-0.25)=0.77V[/tex]
Hence, the standard potential of the cell is 0.77 V
