Potential:

Let's recapitulate Nernst equation which you have studied in the previous unit. You have seen previous while a metal is immersed in a solution containing its own ions, say zinc rod in zinc sulphate solution, a potential difference is established between the metal and the solution.

Mⁿ⁺ + ne- → Mº                                ... (2.1)

A potential difference E for an electrode reaction could be given through Nernst equation:

E=Eº - (RT/nF) ln (1/a_Mⁿ⁺)             ... (2.2)

while R is gas constant, F is Faraday Constant, T is absolute temperature, n is the valency of ions,  a_Mⁿ⁺  is the activity of ions in the solution, and E⁰ is a constant depending upon the metal.

In this equation no term for elemental metal is included in the logarithmic term because it is a pure solid and its activity is one. Eq. 2.2 could be simplified through introducing the known values of F, R, and T, and converting natural logarithms to base 10 by multiplying by 2.3; it then becomes:

E=Eº - (0.0591/n) log (1/a_Mⁿ⁺)             ... (2.3)

For most reasons in quantitative analysis, it is sufficiently accurate to replace an a_Mⁿ⁺ with [Mⁿ⁺], the concentration of the metal ions in mol/dm³.

E=Eº - (0.0591/n) log (1/[Mⁿ⁺])             ...(2.4)

This is the simplest form of Nernst equation where Eo is standard electrode potential of the metal, which is constant. Thus, the electrode potential varies linearly with the logarithm of the reciprocal of the metal ion concentration. This equation can be used to calculate potential of half-cell if we know the concentration of ions involved in half- cell reaction or calculate the concentration of ions in the solution if we know the potential and standard potential of the half-cell shown above. To further understand this try out following SAQs.