Question

Calculate the equilibrium constant for Zn + HgO --> ZnO + Hg (Eo cell = +1.35 V) assuming standard conditions?

When I use the equation, Eo cell= (0.0592V x mole-/n) x log(K), I keep getting 6.37x10^22, but that answer is wrong. Please explain.

Answer 1

`sf(K=4.84xx10^(45))`

Explanation:

If 1 Coulomb of charge is moved through a potential difference of 1 Volt then 1 Joule of work is done.

In a cell:

If n moles of electrons are moved through a potential difference of E volts then nFE Joules of work is done.

This is the maximum amount of work you can get from the cell and is equal to The Free Energy change.

Under standard conditions this gives:

`sf(DeltaG^@=-nFE^@)`

`:.` `sf(DeltaG^(@)=-2xx9.65xx10^(4)xx1.35=-22.055xx10^(4)color(white)(x)J)`

The relationship between `sf(DeltaG)` and `sf(K)` is:

`sf(DeltaG=DeltaG^(@)+RTlnK)`

At equilibrium `sf(DeltaG=0)` so we can say:

`sf(DeltaG^(@)=-RTlnK)`

`:.` `sf(lnK=(DeltaG^(@))/(-RT))`

`:.` `sf(lnK=(-26.055xx10^(4))/(-8.31xx298)=105.193)`

From which:

`sf(K=4.84xx10^45)`

This number is so large that we can say that the reaction goes to completion.