Chemistry Nernst Equation and It's Application
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Topics Covered :

● Nernst Equation
● Equilibrium Constant from Nernst Equation
● Electrochemical Cell and Gibbs Energy of the Reaction

Nernst Equation :

We know that the concentration of all the species involved in the electrode reaction is unity. But this need not be always true. Nernst showed that for the electrode reaction :

`color{red}(M^(n+)(aq) + n e^(-) → M(s))`.

the electrode potential at any concentration measured with respect to standard hydrogen electrode can be represented by :

`color{red}(E_(M^(n+)|M) = E_(M^(n+)|M)^(⊖) - (RT)/(nF) ln \ \ ([M])/([M^(n+)]))`

but concentration of solid `M` is taken as unity and we have

`color{red}(E_(M^(n+) | M) = E_(M^(n+)|M)^(⊖) - (RT)/(n F) ln \ \ 1/([M^(n+) ]))` .........(1)

`color{red}(E_(M^(n+)|M)^(⊖))` has already been defined, `R` is gas constant `(8.314JK^(-1) mol^(-1))`.

`F` is Faraday constant `(96487 C mol^(-1))`,

`T` is temperature in kelvin and `[M^(n+)]` is the concentration of the species `M^(n+)`.

In Daniell cell, the electrode potential for any given concentration of `Cu^(2+)` and `Zn^(2+)` ions, we write

For cathode : `color{red}(E_(Cu^(2+)|Cu) = E_(Cu^(2+)|Cu)^(⊖) - (RT)/(2F) ln \ \ 1/([Cu^(2+) (aq)]))` ...........(2).

For Anode : `color{red}(E_(Zn^(2+)|Zn) = E_(Zn^(2+)|Zn)^(⊖) - (RT)/(2F) ln \ \ 1/([Cu^(2+) (aq)]))` ...............(3).

The cell potential, `color{red}(E_text(cell) = E_(Cu^(2+) |Cu) - E_(Zn^(2+) |Zn))`

` color{red}(= E_(Cu^(2+)|Cu)^(⊖) - (RT)/(2F) ln \ \ 1/([Cu^(2+)(aq)]) - E_(Zn^(2+)|Zn)^(⊖) +(RT)/(2F) ln \ \ 1/([Zn^(2+) (aq)]))`

`color{red}( = E_(Cu^(2+)|Cu)^(⊖) - E_(Zn^(2+)|Zn)^(⊖) - (RT)/(2F) ln \ \ 1/([Cu^(2+) (aq)]) - ln \ \ 1/([Zn^(2+) (aq) ]))`

`color{red}(E_text(cell) = E_text(cell)^(⊖) - (RT)/(2F) ln \ \ ([Zn^(2+)])/([Cu^(2+)]))` ............(4)

`=>` It can be seen that `E_text(cell)` depends on the concentration of both `Cu^(2+)` and `Zn^(2+)` ions.

`=>` It increases with increase in the concentration of `Cu^(2+)` ions and decrease in the concentration of `Zn^(2+)` ions.

By converting the natural logarithm in Eq. (4) to the base 10 and substituting the values of `R`, `F` and `T = 298 K`, it reduces to

`color{red}(E_text(cell) = E_text(cell)^(⊖) - (0.059)/2 log \ ([Zn^(2+)])/([Cu^(2+)]))` ...............(5)

We should use the same number of electrons (n) for both the electrodes and thus for the following cell

`color{red}(Ni(s) | Ni^(2+)(aq) || Ag^(+) (aq) | Ag)`

The cell reaction is `color{red}(Ni (s) + 2 Ag^(+) (aq) → Ni^(2+) (aq) +2 Ag (s))`.

The Nernst equation can be written as

`color{red}(E_text(cell) = E_text(cell)^(⊖) - (RT)/(2F) ln \ \ ([ Ni^(2+)])/([Ag^+]^2))`

For a general electrochemical reaction of the type :

`color{red}(a A + bB overset(n e )→ c C + d D)`

Nernst equation can be written as : `color{red}(E_text(cell) = E_text(cell)^(⊖) - (RT)/(n F) ln Q)`

`color{red}(= E_text(cell)^(⊖) - (RT)/(nF) ln \ \ ([C]^c [D]^d)/([A]^a [B]^b))` ..........(6)
Q 2907234188

Represent the cell in which the following reaction takes place

`Mg(s) + 2Ag^(+) ( 0.0001M) → Mg^(2+) (0.130M) + 2Ag(s)`

Calculate its `E_text(cell)` if `E_text(cell)^(⊖) = 3.17 V`.


The cell can be written as `Mg | Mg^(2+) (0.130M) || Ag^(+) (0.0001M) | Ag`

`E_text(cell) = E_text(cell)^(⊖) - (RT)/(2F) ln \ \ ( [ Mg^(2+)])/([Ag^+]^2)`

` = 3.17 V - (0.059V)/2 log \ \ (0.130)/(0.0001)^2 = 3.17 V - 0.21 V = 2.96 V`.

Equilibrium Constant from Nernst Equation :

If the circuit in Daniell cell (Fig. 3.1) is closed then we note that the reaction

`color{red}(Zn(s) + Cu^(2+)(aq) → Zn^(2+) (aq) + Cu(s))`

takes place.

`=>` As time passes, the concentration of `Zn^(2+)` keeps on increasing while the concentration of `Cu^(2+)` keeps on decreasing.

`=>` At the same time voltage of the cell as read on the voltmeter keeps on decreasing.

`=>` After some time, there is no change in the concentration of `Cu^(2+)` and `Zn^(2+)` ions and at the same time, voltmeter gives zero reading.

`=>` This indicates that equilibrium has been attained.

In this situation the Nernst equation may be written as:

`color{red}(E_text(cell) = 0 = E_text(cell)^(⊖) - (2.303 RT)/(2F) log \ \ ([Zn^(2+)])/([Cu^(2+)]))` or `color{red}(E_text(cell)^(⊖) = (2.303RT)/(2F) log ([Zn^(2+)])/([Cu^(2+)]))`

But at equilibrium `color{red}(([Zn^(2+)])/([Cu^(2+)]) = K_c)` and at `T = 298K` the above equation can be written as

`color{red}(E_text(cell)^(⊖) = (0.059 V)/2 log K_c = 1.1 V \ \ \ \ \ ( E_text(cell)^(⊖) = 1.1V))`.

`color{red}(logK_c = ((1.1V xx 2))/(0.059V) = 37.288)`

`color{red}(K_C = 2xx10^(37) `at `298K)`.

In general,

`color{red}(E_text(cell)^(⊖) = (2.303 RT)/(nF) log K_c)`................(7).

Thus, Eq. (3.14) gives a relationship between equilibrium constant of the reaction and standard potential of the cell in which that reaction takes place.

Thus, equilibrium constants of the reaction, difficult to measure otherwise, can be calculated from the corresponding `E^⊖` value of the cell.
Q 2967434385

Calculate the equilibrium constant of the reaction: `Cu(s) +2Ag^(+) (aq) → Cu^(2+) (aq) +2 Ag (s) \ \ \ \ E_text(cell)^(⊖) = 0.46 V`.


`E_text(cell)^(⊖) = (0.059)/2 log K_c = 0.46 V` or `logK_c = (0.46 V xx 2)/(0.059V) = 15.6`

`K_c = 3.92xx10^(15)`

Electrochemical Cell and Gibbs Energy of the Reaction :

`=>` Electrical work done in one second is equal to electrical potential multiplied by total charge passed.

`=>` If we want to obtain maximum work from a galvanic cell then charge has to be passed reversibly.

`=>` The reversible work done by a galvanic cell is equal to decrease in it's Gibbs energy.

Therefore, if the emf of the cell is `E` and `nF` is the amount of charge passed and `Delta_r G` is the Gibbs energy of the reaction, then

`color{red}(Delta_rG = - nFE_text(cell))` ............(8).

It may be remembered that `E_text(cell)` is an intensive parameter but `Delta_rG` is an extensive thermodynamic property and the value depends on `n`. Thus, if we write the reaction

`color{red}(Zn(s) + Cu^(2+) (aq) → Zn^(2+) (aq) + Cu(s)) `

`color{red}(Delta_rG = -2FE_text(cell))`

but when we write the reaction `color{red}(2 Zn (s) +2 Cu^(2+) (aq) → 2 Zn^(2+) (aq) +2 Cu (s))`.

`color{red}(Delta_rG = - 4 FE_text(cell))`

If the concentration of all the reacting species is unity, then `color{red}(E_text(cell) = E_text(cell)^(⊖))` and we have

`color{red}(Delta_r G^(⊖) = - n F E_text(cell)^(⊖))` ...........(9).

Thus, from the measurement of `color{red}(E_text(cell)^(⊖))` we can obtain an important thermodynamic quantity `color{red}(Delta_r G^(⊖))` standard Gibbs energy of the reaction. From the latter we can calculate equilibrium constant by the equation: `color{red}(Delta_rG^(⊖) = -RT ln K)`.
Q 2937534482

The standard electrode potential for Daniell cell is `1.1V`. Calculate the standard Gibbs energy for the reaction: `Zn(s) +Cu^(2+) (aq) → Zn^(2+) (aq) + Cu(s)`


`Delta_rG^(⊖) = - n F E_text(cell)^(⊖)`

`n` in the above equation is `2, F = 96487 C mol^(–1)` and `E_text(cell)^(⊖) = 1.1 V`

Therefore, `Delta_rG^(⊖) = -2xx1.1V xx 96487 C mol^(-1)`

` = -21227 J mol^(-1)`

` = -21.227 KJ mol^(-1)`