For the 1/2 cell #sf(MnO_4^(-)+8H^(+)+5erightleftharpoonsMn^(2+)+7H_2O)# the value of #sf(E^(@)=+1.51V)#. What is the value for E if #sf([MnO_4^(-)]=0.0001M)# and #sf([Mn^(2+)]=0.0005M)# ?

Before the calculation it is helpful to make a prediction for the electrode potential using Le Chatelier's Principle:

The 1/2 cell reaction is:

#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(blue)(larr)#

#sf(MnO_4^(-)+8H^++5erightleftharpoonsMn^(2+)+4H_2O)#

#sf(color(red)(0.0001Mcolor(white)(xxxxxxxxx)0.0005M)#

#sf(E^@=+1.51color(white)(x)V)#

You can see that the concentration of #sf(MnO_4^-)# has been reduced relative to the concentration of #sf(Mn^(2+))#.

According to Le Chatelier we would predict that the position of equilibrium will shift to the left to oppose that change, as shown by the blue arrow.

You can see from the 1/2 equation that this will tend to push out more electrons so we would expect the electrode potential to be less positive.

We can calculate this using The Nernst Equation:

#sf(E=E^(@)-(RT)/(zF)ln(([red])/(["ox"]))#

At 298K this simplifies to:

#sf(E=E^@+(0.05916)/(z)log((["ox"])/([red])))#

Where z is the number of moles of electrons transferred.

This becomes:

#sf(E=E^@+(0.05916)/(5)log(([MnO_4^-][H^+]^8)/([Mn^(2+)]))#

Since we are at pH = 0 we can say that #sf([H^+]=1color(white)(x)M)#

This becomes:

#sf(E=E^@+(0.05916)/(5)log(([MnO_4^-])/([Mn^(2+)]))#

Putting in the numbers:

#sf(E=+1.51+(0.05916)/(5)log((0.0001)/(0.0005))#

#sf(E=+1.51-0.00827=+1.50color(white)(x)V)#

As you can see, this is in accordance with our prediction. The potential of the electrode has been made slightly less positive.

Put simply, reducing the concentration of Mn(VII) relative to Mn(II) has made the Mn(VII) slightly less effective as an oxidising agent.