Question

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!

Answer 1

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)`