Question

What is the total reduction potential of a cell in which lithium (li) is reduced and mercury (hg) is oxidized?

Answer 1

Answer: The total reduction potential of the cell is -3.89 V.

Step-by-step explanation:

We are given:

• Reduction of lithium follows the reaction:

`Li^++e^-\rightarrow Li`

The standard reduction potential for this is -3.04 V

• Oxidation of mercury follows the reaction:

`Hg\rightarrow Hg^(2+)+2e^-`

The standard reduction potential for this is -0.85 V

The cell formed by these half reactions is:
`Hg/Hg^(2+)||Li^+/Li`

The cell potential,
`E^o_(cell)=E^o_(oxidation)+E^o_(reduction)`

`E^o_(cell)=[-0.85+(-3.04)]=-3.89V`

Hence, the total reduction potential of the cell is -3.89 V.