How do I calculate the standard potential of a chemical reaction?

I have a standard hydrogen gas electrode which is connected to a H+/H2 electrode with PH2= 1.00 and [H+] = 1.00*10-7mol l-1. How should I calculate the cell potential? Use Nerst formula and if so, how?

Thanks!

1 Answer
Mar 1, 2018

You can do it like this:

Explanation:

This is an example of a concentration cell. In this device the 1/2 cells are made of the same material but the concentrations differ.

This sets up a potential difference between the two half cells causing electron flow.

This example uses the hydrogen electrode:

Alevelchem.com

In the standard hydrogen electrode #sf(p_(H_2)=1color(white)(x)"Atm")# and #sf([H^+]=1color(white)(x)"mol/l")#. It operates at #sf(25^@C)#

The other electrode is the same but #sf([H^+]=10^(-7)color(white)(x)"mol/l")#

The cell reaction is:

#sf(2H^++H_2rightleftharpoonsH_2+2H^(+))#

#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxx))(color(red)(rarr)#

#sf(1Mcolor(white)(xxxxxxx)10^(-7)M)#

According to Le Chatelier's Principle we would expect the position of equilibrium to shift from left to right.

The cell diagram is:

#sf(Pt|H_2|2H^(+)(1M)||2H^(+)(10^(-7)M)|H_2|Pt)#

#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(red)(larr)#

#sf(color(white)(xxxxxx)"electron flow")#

Electrons will flow from the #sf(10^(-7)M)# half cell into the external circuit and enter the standard hydrogen electrode where #sf(H^+)# ions are reduced to hydrogen. This makes the SHE the cathode. The other half cell is the anode.

To find #sf(E_(cell))# we need to use The Nernst Equation:

#sf(E_(cell)=E_(cell)^@-(RT)/(zF)logQ)#

At #sf(25^@C)# this simplifies to:

#sf(E_(cell)=E_(cell)^@-0.0591/(z)logQ)#

#sf(Q)# is the reaction quotient so this becomes:

#sf(E_(cell)=E_(cell)^@-0.0591/(2)log(([H^+]^(2)"R")/([H^+]^(2)"L")))#

In a concentration cell #sf(E_(cell)^@# is zero so:

#sf(E_(cell)=0-0.0591/(2)log[10^(-14)])#

#sf(E_(cell)=-0.029655xx-14=+0.414 color(white)(x)V)#