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

1 Answer
Dec 1, 2016

Answer:

#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.