Nernst equation

Nernst equation

Nernst equation is the one which relates the cell potential and the concentration of the species involved in an electrochemical reaction. Let us consider an electrochemical cell for which the overall redox reaction is,

xA + yB ↔ lC + mD

The reaction quotient Q for the above reaction is given below

We have already learnt that,

∆G = ∆G’ + RT lnQ                 .....(9.29)

The Gibbs free energy can be related to the cell emf as follows [ ∴ equation (9.24) and (9.25) ]

∆G = - nFE_cell; ∆G^o= - nFE_cell^o

Substitute these values and Q from (9.28) in the equation (9.29)

(9.29) ⇒  - nFE_cell = - nFE^o_cell + RT ln { [C]^l[D]^m / [A]^x[B]^y }       .....(9.30)

Divide the whole equation (9.30) by (-nF)

The above equation (9.31) is called the Nernst equation

At 25º C (298K), the above equation (9.31) becomes,

Let us calculate the emf of the following cell at 25^oC using Nernst equation.

Cu (s) | Cu²⁺ (0.25 aq, M) | | Fe³⁺ (0.005 aq M) | Fe²⁺ (0.1 aq M) | Pt (s)

Given : ( Eº )_Fe3+|Fe2+ = 0.77V and ( Eº ) _{Cu2+ | Cu} = 0.34 V

Half reactions are

Cu (s) → Cu²⁺ (aq) + 2e^-                        ..... (1)

2 Fe³⁺ → (aq) + 2e- → 2Fe²⁺ (aq)              ..... (2)

the overall reaction is

Cu (s) + 2 Fe³⁺ (aq) → Cu²⁺ (aq) + 2 Fe²⁺ (aq), and n = 2

Apply Nernst equation at . 25ºC .

Given standard reduction potential of Cu²⁺ | Cu is 0.34V

(E^o_ox )_Cu|Cu2+ = - 0.34V

(E^o_red) _{Fe3+|Fe 2+} = 0.77V

Eº_cell = - 0.34 + 0.77

Eº_cell = 0.43V

Evaluate yourself

The electrochemical cell reaction of the Daniel cell is

Zn (s) + Cu²⁺ (aq) → Zn²⁺ (aq)+Cu (s)

What is the change in the cell voltage on increasing the ion concentration in the anode compartment by a factor 10?