2.5 The Nernst Equation | Electrochemistry | Chemistry 12

Introduction to the Nernst Equation

The Nernst equation is a fundamental mathematical expression in electrochemistry that quantifies the relationship between the concentration of ions involved in a redox reaction and the actual electrode potential ( $E$ ) under non-standard conditions. It allows calculation of the electrode potential when ion concentrations deviate from the standard $1 M$ (or $1 m o l / d m^{3}$ ) and the temperature is not necessarily $298 K$ .

The electrode potential is crucially influenced by the concentrations of the species participating in the redox half-reaction. The Nernst equation connects the standard electrode potential ( $E^{\circ}$ ), measured under standard conditions ( $1 M$ concentration for ions, $1 a t m$ pressure for gases, $298 K$ temperature), to the non-standard potential ( $E$ ).

Electrode Potential→

Mathematical Formulations

The general form of the Nernst equation is:

$E = E^{\circ} + \frac{R T}{n F} ln \frac{[ oxidized species ]}{[ reduced species ]}$

This equation can also be expressed using the common logarithm ( $lo g_{10}$ ):

$E = E^{\circ} + \frac{2.303 R T}{n F} lo g \frac{[ oxidized species ]}{[ reduced species ]}$

Remember the conversion: $ln x = 2.303 lo g x$

At standard temperature ( $T = 298 K$ or $2 5^{\circ} C$ ), the term $\frac{2.303 R T}{F}$ simplifies to a constant:

$\frac{2.303 \times R \times T}{F} = \frac{2.303 \times 8.314 J K ^{- 1} m o l ^{- 1} \times 298 K}{9.648 \times 1 0 ^{4} C m o l ^{- 1}} \approx 0.059 J C^{- 1} = 0.059 V$

Thus, at $298 K$ , the Nernst equation simplifies to:

$E = E^{\circ} + \frac{0.059}{n} lo g \frac{[ oxidized species ]}{[ reduced species ]}$

Components of the Nernst Equation

Define each term in the Nernst equation:

$E$ : The actual electrode potential under non-standard conditions (in volts, V).
$E^{\circ}$ : The standard electrode potential (in volts, V), measured at $1 M$ concentrations, $1 a t m$ pressure for gases, and $298 K$ .
$R$ : The universal gas constant, with a value of $8.314 J m o l^{- 1} K^{- 1}$ .
$T$ : The absolute temperature in Kelvin ( $K$ ).
$n$ : The number of electrons transferred in the balanced half-reaction from the reduced species to the oxidized species.
$F$ : Faraday's constant, which is the charge of one mole of electrons. Its value is $9.648 \times 1 0^{4} C m o l^{- 1}$ (coulombs per mole).
$[oxidized species]$ : The concentration of the species with the higher oxidation state (typically in $m o l / L$ or $M$ ).
$[reduced species]$ : The concentration of the species with the lower oxidation state (typically in $m o l / L$ or $M$ ).

Note: For pure solids or liquids, their concentrations are considered constant and are omitted from the Nernst equation (effectively treated as 1).

Electrode Potentials and Concentration

The electrode potential measures how easily a species gains or loses electrons. This ease is directly tied to the availability of the species, hence its concentration.

Consider the standard electrode potential for the copper half-reaction:

$C u^{2 +} + 2 e^{-} \to Cu E^{\circ} = + 0.34 V$

This $E^{\circ}$ value is valid when the concentration of $C u^{2 +}$ ions is $1 m o l / d m^{3}$ .

Effect of Concentration Changes (Le Chatelier's Principle)

If the concentration of $C u^{2 +}$ ions is decreased (e.g., by adding water), according to Le Chatelier's Principle, the equilibrium position shifts to counteract this change. In this case, the equilibrium will shift to the left to increase the $C u^{2 +}$ concentration again. A shift to the left implies that the reduction process (gaining electrons) becomes less favorable.

Increase in ion concentration: If the concentration of ions (e.g., $C u^{2 +}$ ions for copper electrodes) increases, there are more ions available to gain electrons. This makes reduction easier, and thus the electrode potential becomes more positive.
Decrease in ion concentration: If the concentration of ions decreases, there are fewer ions available to accept electrons. This makes reduction harder, and thus the electrode potential becomes less positive (or more negative).

The Nernst equation provides a quantitative tool to predict these precise changes in electrode potential.

Worked Examples

Example: Calculating Electrode Potential for $C u^{2 +} / Cu$ System

Problem: The standard electrode potential of the $C u^{2 +} / Cu$ system is $+ 0.34 V$ . What is the electrode potential of a solution containing $0.5 M$ $C u^{2 +}$ ions?

Solution:

Write the given values:
- Half-reaction: $C u^{2 +} + 2 e^{-} \to Cu$
- Standard electrode potential, $E^{\circ} = + 0.34 V$
- Number of electrons transferred, $n = 2$
- Concentration of oxidized species, $[C u^{2 +}] = 0.5 M$
- Concentration of reduced species, $[Cu] = 1$ (for a pure solid, concentration is constant)
- Temperature is assumed to be $298 K$ , allowing use of the simplified Nernst equation.
Apply the Nernst equation: The simplified Nernst equation at $298 K$ is: $E = E^{\circ} + \frac{0.059}{n} lo g \frac{[ oxidized species ]}{[ reduced species ]}$ For the $C u^{2 +} / Cu$ system, this becomes: $E_{C u^{2 +} / Cu} = E_{C u^{2 +} / Cu}^{\circ} + \frac{0.059}{n} lo g \frac{[ C u ^{2 +} ]}{[ Cu ]}$
Show calculation with proper formatting: $E_{C u^{2 +} / Cu} = 0.34 V + \frac{0.059}{2} lo g \frac{0.5}{1}$ $E_{C u^{2 +} / Cu} = 0.34 + 0.0295 \times lo g (0.5)$ $E_{C u^{2 +} / Cu} = 0.34 + 0.0295 \times (- 0.301)$ $E_{C u^{2 +} / Cu} = 0.34 - 0.0089$ $E_{C u^{2 +} / Cu} = 0.331 V$

As predicted by Le Chatelier's Principle, decreasing the $C u^{2 +}$ concentration from $1 M$ to $0.5 M$ makes the reduction less favorable, resulting in a less positive electrode potential ( $+ 0.331 V$ compared to $+ 0.34 V$ ).

Example: Calculating Electrode Potential for $F e^{3 +} / F e^{2 +}$ System

Problem: The standard electrode potential of $F e^{3 +} / F e^{2 +}$ is $0.77 V$ . What is the electrode potential of the system containing $1.0 m o l / d m^{3}$ of $F e^{3 +}$ and $0.2 m o l / d m^{3}$ of $F e^{2 +}$ ions?

Solution:

Half-reaction: $F e^{3 +} + e^{-} \to F e^{2 +}$
Given values: $E^{\circ} = 0.77 V$ , $n = 1$ , $[oxidized species] = [F e^{3 +}] = 1.0 M$ , $[reduced species] = [F e^{2 +}] = 0.2 M$ .
Apply Nernst equation: $E = E^{\circ} + \frac{0.059}{n} lo g \frac{[ F e ^{3 +} ]}{[ F e ^{2 +} ]}$ $E = 0.77 V + \frac{0.059}{1} lo g \frac{1.0}{0.2}$ $E = 0.77 + 0.059 \times lo g (5)$ $E = 0.77 + 0.059 \times 0.699$ $E = 0.77 + 0.0412$ $E = 0.811 V$

Worked Examples

Problem: The standard electrode potential of the

C u^{2 +} / Cu

system is

+ 0.34 V

. What is the electrode potential of a solution containing

0.5 M

C u^{2 +}

ions?

Solution:

Write the given values:

Half-reaction: $C u^{2 +} + 2 e^{-} \to Cu$
Standard electrode potential, $E^{\circ} = + 0.34 V$
Number of electrons transferred, $n = 2$
Concentration of oxidized species, $[C u^{2 +}] = 0.5 M$
Concentration of reduced species, $[Cu] = 1$ (for a pure solid, concentration is constant)
Temperature is assumed to be $298 K$ , allowing use of the simplified Nernst equation.

Apply the Nernst equation: The simplified Nernst equation at $298 K$ is: $E = E^{\circ} + \frac{0.059}{n} lo g \frac{[ oxidized species ]}{[ reduced species ]}$ For the $C u^{2 +} / Cu$ system, this becomes: $E_{C u^{2 +} / Cu} = E_{C u^{2 +} / Cu}^{\circ} + \frac{0.059}{n} lo g \frac{[ C u ^{2 +} ]}{[ Cu ]}$

Show calculation with proper formatting: $E_{C u^{2 +} / Cu} = 0.34 V + \frac{0.059}{2} lo g \frac{0.5}{1}$ $E_{C u^{2 +} / Cu} = 0.34 + 0.0295 \times lo g (0.5)$ $E_{C u^{2 +} / Cu} = 0.34 + 0.0295 \times (- 0.301)$ $E_{C u^{2 +} / Cu} = 0.34 - 0.0089$ $E_{C u^{2 +} / Cu} = 0.331 V$

As predicted by Le Chatelier's Principle, decreasing the

C u^{2 +}

concentration from

1 M

0.5 M

makes the reduction less favorable, resulting in a less positive electrode potential (

+ 0.331 V

compared to

+ 0.34 V

Problem: The standard electrode potential of

F e^{3 +} / F e^{2 +}

0.77 V

. What is the electrode potential of the system containing

1.0 m o l / d m^{3}

F e^{3 +}

and

0.2 m o l / d m^{3}

F e^{2 +}

ions?

Solution:

Half-reaction: $F e^{3 +} + e^{-} \to F e^{2 +}$

Given values: $E^{\circ} = 0.77 V$ , $n = 1$ , $[oxidized species] = [F e^{3 +}] = 1.0 M$ , $[reduced species] = [F e^{2 +}] = 0.2 M$ .

Apply Nernst equation: $E = E^{\circ} + \frac{0.059}{n} lo g \frac{[ F e ^{3 +} ]}{[ F e ^{2 +} ]}$ $E = 0.77 V + \frac{0.059}{1} lo g \frac{1.0}{0.2}$ $E = 0.77 + 0.059 \times lo g (5)$ $E = 0.77 + 0.059 \times 0.699$ $E = 0.77 + 0.0412$ $E = 0.811 V$