Question

What is the potential of the 1/2 cell `Cu^(2+)+2erightleftharpoonsCu` if `[Cu^(2+)]=0.1M?` `E^@=+0.34V`.

Answer 1

`sf(E=+0.31color(white)(x)V)`

Explanation:

The Nernst Equation for the electrode potential of a 1/2 cell gives us:

`sf(E=E^(@)-(RT)/(zF)ln([[Red]]/([Ox]))`

In reality, activities should be used but molarities are acceptable at low concentrations as `sf(arArr1)`

In this case we have:

`sf(Cu^(2+)+2erightleftharpoonsCu)`

So `sf(z=2)`

At `sf(25^@C)` a convenient form of the equation is:

`sf(E=E^(@)+0.05916/zlog(([Ox])/[[Red]]))`

Putting in the numbers:

`sf(E=+0.34+0.05916/2log(0.1))`

`sf(E=+0.34-0.02958color(white)(x)V)`

`sf(E=+0.31color(white)(x)V)`

This shows that reducing the concentration like this causes a drop in potential.