Answer: The total reduction potential of the cell is -3.89 V.
Explanation:
We are given:
[tex]Li^++e^-\rightarrow Li[/tex]
The standard reduction potential for this is -3.04 V
[tex]Hg\rightarrow Hg^{2+}+2e^-[/tex]
The standard reduction potential for this is -0.85 V
The cell formed by these half reactions is: [tex]Hg/Hg^{2+}||Li^+/Li[/tex]
The cell potential, [tex]E^o_{cell}=E^o_{oxidation}+E^o_{reduction}[/tex]
[tex]E^o_{cell}=[-0.85+(-3.04)]=-3.89V[/tex]
Hence, the total reduction potential of the cell is -3.89 V.