**mail@learnmetallurgy.com**

**The Nernst equation gives the relationship between cell potential at a given T to standard potential and to the activities/concentration of chemical species involved. Hence enabling us to determine cell potential under non-standard conditions.**

Lets consider a general cell reaction :

\[ M^+ + ne^- \rightleftarrows M \]

So according to Van't Hoff equation:
\[ \Delta G = \Delta G^o + RTlnK \]
\[\Delta G = \Delta G^o + RTln \frac{Products}{Reactants} \tag{1}\]
Also work done by a electrochemical cell is equal to the amount of electrical energy which it generates unser fixed fixed conditions of T,P and C. Hence the work perform be the cell will be equal to the decrease in the free energy of system. So..
\[ \boxed{\Delta G = - nEF \quad and \quad \Delta G^o = - nE^oF } \tag{2}\]
Here ,

**\( \Delta G = \)** Gibbs free energy (J/mol)

**\( \Delta G^o = \)** Standard gibbs free energy (J/mol)

**\( \Delta E = \)** Cell potential (V)

**\( \Delta E^o = \)** Standard cell potential (V)

**\( \Delta n = \)** No. of electron involved

**\( \Delta F = \)** Faraday's constant (1F = 96500 C/mol)

Now on putting **equ (2)** in **equ (1)**, we get
\[ -nFE = -nFE^o + RTln \left[ \frac{Products}{Reactants}\right] \]
\[ \boxed{E = E^o - \frac{RT}{nF} ln\left[ \frac{Products}{Reactants}\right]} \]
At standard condition. i.e., T=298K and P = 1atm
\[ \boxed{E = E^o - \frac{0.0591}{n} log \left[ \frac{Products}{Reactants}\right]} \tag{3}\]
The above equation (Equ 3) is known as nernst equation. Let's take a cell reaction \( M^+ + ne^- \rightleftarrows M \) and write nernst equation for that..
\[ E = E^o - \frac{0.0591}{n} \times log \left[ \frac{1}{M^+}\right] \]
Application of Nernst equation :

**(1) **Calculation of cell potential of a cell at non-standard condition.

**(2) **To find equlibrium constant

**(3) **Find out the concentration of a chemical species if concentration of other species is given

**(4) **Used to determine PH of a solution

Practice Question : Performing electrolysis of Cd at 527

Show Answer

Answer : a= 0.0549

**Solution** \[ \Delta G = -nEF = RTlnK \] \[ -2*(-0.1)*96500 = 8.314*800 ln \frac{1}{a^{Cd-alloy}} \]