Class 12 Chemistry

# Electrochemistry

## Nernst Equation

Nernst equation relates the reduction potential of an electrochemical reaction to the standard potential, temperature and activities of the chemical species undergoing reduction and oxidation.

Let us assume the following reaction:

`M^(n+)(aq)+ne^(-)→M(s)`

For this reaction, the electrode potential measured with respect to standard hydrogen electrode can be given by following equation.

E_{(Mn+/M)} = E^{Θ}_{(Mn+/M)} `-(RT)/(nF)ln([M])/([M^(n+)])`

E^{Θ}_{(Mn+/M)} `-(RT)/(nF)ln1/([M^(n+)])`

Since concentration of solid M is taken as unity, hence this equation can be written as follows:

Here, R is gas constant = 8.314 JK^{-1}mol^{-1}

F is Faraday constant = 96487 C mol^{-1}

T is temperature in Kelvin

[M^{n+}] is concentration of the species M^{n+}

In Daniell cell, the electrode potential is given as follows:

*For cathode:*

`E_((Cu^(2+))/(Cu))=E^Θ_ ((Cu^(2+))/(Cu))-(RT)/(2F)ln\1/([Cu^(2+)(aq)])`

*For anode:*

`E_((Zn^(2+))/(Zn))=E^Θ_ ((Zn^(2+))/(Zn))-(RT)/(2F)ln\1/([Zn^(2+)(aq)])`

*Now, cell potential:*

`E_((ce\ll))= E_((Cu^(2+))/(Cu))-E_((Zn^(2+))/(Zn))`

`=E_ ((Cu^(2+))/(Cu))-(RT)/(2F)ln\1/([Cu^(2+)(aq)])^Θ- E_ ((Zn^(2+))/(Zn))+(RT)/(2F)ln\1/([Zn^(2+)(aq)])^Θ`

`=E^Θ_ ((Cu^(2+))/(Cu))= =E^Θ_ ((Zn^(2+))/(Zn))- (RT)/(2F)ln\1/([Cu^(2+)(aq)])- ln\1/([Zn^(2+)(aq)])`

Or, `E_((ce\ll))=E_((ce\ll))-(RT)/(2F)ln([Zn^(2+)])/([Cu^(2+)])^Θ`

By converting the natural logarithm in above equation to the base 10 and substituting the values of R, F and T, the equation is reduced to the following:

Or, `E_((ce\ll))=E_((ce\ll))-(0.059)/2log([Zn^(2+)])/([Cu^(2+)])^Θ`

For a general equation of following type

`aA+bB→cC+dD`

Nernst equation can be written as follows:

`E_((ce\ll))=E_((ce\ll))-(RT)/(nF)ln\Q^Θ`

`=E_((ce\ll))^Θ-(RT)/(nF)ln\([C]^c[D]^d)/([A]^a[B]^b)`

#### Equilibrium constant for Nernst Equation

At equilibrium the electron potential becomes zero. In this situation, the Nernst equation can be written as follows:

`E_((ce\ll))=0`

Or, `0=E_((ce\ll))-(2.303RT)/(2F)-log([Zn^(2+)])/([Cu^(2+)])^Θ`

Or, ` E_((ce\ll))=(2.303RT)/(2F)-log([Zn^(2+)])/([Cu^(2+)])^Θ`

But at equilibrium, `([Zn^(2+)])/([Cu^(2+)])=K_c` and at T = 298 K, the above equation can be written as follows:

`E_((ce\ll))^Θ=(0.059V)/2lo\g K_c=1.1V`

`lo\g K_c=(1.1V\xx2)/(0.059V)=37.288`

Or, `K_c=2xx10^(37)` at 298 K

In general, ` E_((ce\ll))^Θ=(2.303RT)/(nF)lo\g K_c`

### Electrochemical Cell and Gibbs Energy of Reaction

We know that electrical work done in unit time is equal to electrical potential multiplied by total charge passed. In order to obtain maximum work from a galvanic cell, the charge has to be passed reversibly. The reversible work done by a galvanic cell is equal to decrease in its Gibbs energy. Hence, Gibbs energy of reaction is given by following equation:

`Δ_TG=-n\FE_((cell))`

Here, E is the emf and nF is the amount of energy.

Gibbs energy for the following reaction is given below the chemical equation.

`Zn(s)+Cu^(2+)(aq)→Zn^(2+)(aq)+Cu(s)`

`Δ_TG=-2FE_((cell))`

If the reaction is written in a different way (as shown below) then Gibbs energy shall be given accordingly.

`2Zn(s)+2Cu^(2+)(aq)→2Zn^(2+)(aq)+2Cu(s)`

`Δ_TG=-4FE_((cell))`

Daniell Cell

Nernst Equation

Electrolytic Coductance

Electrolytic Cell

InText Solution

NCERT Solution