Introduction

Nernst equation was developed by the German scientist Walther Nernst in 1887. This equation becomes very important in various fields such as the understanding of the electrochemical potential of cells and has multiple applications in electrochemistry and biochemistry.
Nernst's work was totally about electrochemical cells and was limited to electrochemistry. He developed the relationship between the voltage produced by an electrochemical cell and the concentration of particular ions present in it. His observation made him develop his studies in the form of mathematical expression. He developed a mathematical relationship between chemical potential and the concentration of ions.

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The relationship he developed is through the Nernst equation, which determines the redox reaction that occurs inside the electrochemical cell and how their chemical potential and concentration of ions are interconnected. This was a very significant study and formulation of chemistry and physical chemistry. the Nernst equation provides fundamental insights into electrochemical processes, which is important in various fields including the scientific. For reaction prediction, the mean for predicting the equilibrium direction predicts the equilibrium conditions of redox reactions, aiding in understanding the direction and extent of chemical reactions. Nernst equation was the pillar for both the electrochemical theories and the practicals, making it a crucial concept in Chemistry and the field of education and research. Understanding the Nernst equation is important for doing any work with the electrochemical cell, including building any cell such as a galvanic cell, and analyzing any electrolytic cell. The principle of the Nernst equation is used to design the Sensor and detector used for the electrochemical cell which help monitor environmental pollution and industrial processes

Nernst Equation

The Nernst equation relates the concentration of ions to the electrochemical potential of an electrode. It is a key concept in electrochemistry and is used to calculate the electrode potential under non-standard conditions.

This equation gives the relationship between electrode potential and the concentration of ions in the solution. In other words, it shows the electrode potential's dependency on the ions' concentration with which the electrode is reversible.

For a single electrode involving the reduction process,

$\mathrm{M}^{\mathrm{n}+}+\mathrm{ne}^{-} \rightarrow \mathrm{M}(\mathrm{s})$

The reaction quotient Q is defined as $\frac{\mathrm{a}_{\mathrm{M}}}{\left[\mathrm{M}^{+}\right]}$

Now, we learned in thermodynamics that

$\Delta \mathrm{G}=\Delta \mathrm{G}^{\circ}+\mathrm{RT} \ln \mathrm{Q}$ ..(1)

Where $\Delta \mathrm{G}=-\mathrm{nFE}$

and $\Delta \mathrm{G}^{\circ}=-\mathrm{nFE}^{\circ}$

So, substituting these values is (1),

$-\mathrm{nFE}=-\mathrm{nFE}^{\circ}+\mathrm{RT} \ln \mathrm{Q}$

$\Rightarrow \mathrm{E}=\mathrm{E}^{\mathrm{o}}-\frac{\mathrm{RT}}{\mathrm{nF}} \ln Q$

$\Rightarrow \mathrm{E}=\mathrm{E}^{\mathrm{o}}-\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log Q$

This is the Nernst equation which helps us to calculate the non-standard EMF of any Half cell. It can be extended to full of any half cell. It can be extended to full cell which we will be learning later.

Now, at $25^{\circ} \mathrm{C}$ or 298 K
$
\mathrm{E}=\mathrm{E}^{\circ}-\frac{2.303 \times 8.314 \times 298}{\mathrm{n} \times 96500} \log _{10} \frac{[\mathrm{M}]}{\left[\mathrm{M}^{\mathrm{n}+}\right]}
$

$
E=E^{\circ}-\frac{0.059}{\mathrm{n}} \log _{10} \frac{[\mathrm{M}]}{\left.\mathrm{M}^{\mathrm{n}+]}\right]}
$

Here $\mathrm{R}=$ Gas constant
$\mathrm{T}=$ Absolute temperature
$E^{\circ}=$ Standard Emf of the cell
$\mathrm{E}=$ Electrode potential of cell
$\mathrm{F}=$ Faraday number
$\mathrm{n}=$ number of electrons transferred

• If the electrode is solid its activity mass is taken as one.
• For an electrochemical cell having a net reaction:
  $\mathrm{xA}+\mathrm{yB} \xrightarrow{\mathrm{ne}^{-}} \mathrm{mC}+\mathrm{nD}$
  The emf can be calculated asEcell $=\mathrm{E}^{\mathrm{o}}$ cell $-\frac{0.059}{\mathrm{n}} \log \frac{[\mathrm{C}]^{\mathrm{m}}[\mathrm{D}]^{\mathrm{n}}}{[\mathrm{A}]^x[\mathrm{~B}]^{\mathrm{y}}}$

In using the above equation, the following facts should be considered.

• The activity of aq. ions are expressed in terms of their concentration.
• The activity of gases is expressed in terms of their partial pressures.
• The activity of solids is taken to be unity.
• n, the number of electrons transferred should be calculated from the balanced net cell reaction.

EMF of Cells -

It is the potential difference between the two terminals of the cell when no current is drawn from it. It is measured with the help of potentiometer or vacuum tube voltmeter.

Nernst Equation for Full Cell -

$\mathrm{E}_{\mathrm{M}^x+\mid \mathrm{M}}^{\circ}=\mathrm{Q} \quad$ and $\quad \mathrm{E}_{\mathrm{N}^x+\mid \mathrm{N}}^o=\mathrm{P}$

In the full cell both the oxidation and reduction reactions occur simultaneously. Thus, the full cell can be represented as follows:

$M\left|M^{\mathrm{x}+}\right| \mathrm{N}^{\mathrm{x}+} \mid \mathrm{N}$

The electrode potential values for oxidation and reduction are as follows:

$E_{M^x+\mid M}^O=Q \quad$ and $E_{N^{x+} \mid N}^O=P$

At Anode:

$\mathrm{M}(\mathrm{s}) \rightarrow \mathrm{M}^{+\mathrm{x}}(\mathrm{aq})+\mathrm{xe}^{-}$

At Cathode:

$\mathrm{N}^{\mathrm{x}+}(\mathrm{aq})+\mathrm{xe}^{-} \rightarrow \mathrm{N}(\mathrm{s})$

Thus the complete cell reaction is the addition of both anode and cathode reaction. It is given below:

$\mathrm{M}(\mathrm{s})+\mathrm{N}^{\mathrm{x}+}(\mathrm{aq}) \rightarrow \mathrm{M}^{\mathrm{x}+}(\mathrm{aq})+\mathrm{N}(\mathrm{s})$

Thus the reaction quotient(Q) can be given as follows:

$\mathrm{Q}=\frac{\left[\mathrm{M}^{\mathrm{x}+}\right]}{\left[\mathrm{N}^{\mathrm{x}+}\right]}=\frac{\mathrm{c}_1}{\mathrm{c}_2}$
where c₁ and c₂ are the concentrations of M^x+ and N^x+ respectively.

The standard potential of the cell is given as:

$\begin{aligned} \mathrm{E}_{\text {cell }}^{\mathrm{o}} & =\left[\mathrm{E}_{\text {cathode }}^{\mathrm{o}}-\mathrm{E}_{\text {anode }}^{\mathrm{o}}\right] \\ & =\mathrm{P}-\mathrm{Q}\end{aligned}$

At T = 298K, the Nernst equation is given as follows:

$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{0.059}{\mathrm{n}} \log _{10} \mathrm{Q}$
where n is the number of electrons exchanged.

Thus the Nernst equation for the full cell is given as follows:

$\mathrm{E}_{\text {cell }}=(\mathrm{P}-\mathrm{Q})-\frac{0.059}{\mathrm{x}} \log _{10} \frac{\mathrm{c}_1}{\mathrm{c}_2}$

For a better understanding of the topic and to learn more about Nernst Equation with video lesson we provide the link to the

YouTube video:

Some Solved Examples

Example.1

1. In a cell that utilizes the reaction

$\mathrm{Zn}_{(s)}+2{H^{+}}_{(a q)} \rightarrow \mathrm{Zn}^{2+}{ }_{(a q)}+H_{2(g)}$

addition of $\mathrm{H}_2 \mathrm{SO}_4$ to cathode compartment, will

1)lower the E and shift equilibrium to the left

2)lower the E and shift equilibrium to the right

3) (correct)increase the E and shift the equilibrium to the right

4)increase the E and shift the equilibrium to the left

Solution

$\begin{aligned} & Z n_{(s)}+2 H^{+}{ }_{a q} \rightleftharpoons \mathrm{Zn}^{2+}{ }_{a q}+H_{2(g)} \\ & E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.059}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right] \times p_{\mathrm{H}_2}}{\left[\mathrm{H}^{+}\right]^2}\end{aligned}$

On adding$\mathrm{H}_2 \mathrm{SO}_4$ the$\left[H^{+}\right]$ will increase therefore$E_{\text {cell }}$ will also increase and the equilibrium will shift towards the right.

Hence, the answer is the option (3).

Example.2

2. The cell,

$Z n\left|Z n^{2+}(1 M) \| C u^{2+}(1 M)\right| C u\left(E_{\text {cell }}^{\circ}=1.10 \mathrm{~V}\right)$

was allowed to be completely discharged at 298 K .The relative concentration of $\mathrm{Zn}^{2+}$ to $\mathrm{Cu}^{2+}$

$\left(\frac{\left[Z n^{2+}\right]}{\left[C u^{2+}\right]}\right.$is

1)$9.65 \times 10^4$

2)$\operatorname{antilog}(24.08)$

3)$37.3$

4) (correct)$10^{37.3}$

Solution

$\mathrm{Zn}+\mathrm{Cu}^{2+} \rightarrow \mathrm{Zn}^{2+}+\mathrm{Cu}$

From the Nernst equation, we can write

$E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.059}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}$

When the cell is completely discharged, $E_{\text {cell }}=0$

$\begin{aligned} & 0=1.1-\frac{0.059}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]} \\ & \text { or } \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}=\frac{2 \times 1.1}{0.059} \text { or }, \log \frac{\mathrm{Zn}^{2+}}{\mathrm{Cu}^{2+}}=37.3 \\ & \text { or } \frac{\mathrm{Z} n^{2+}}{\mathrm{Cu} u^{2+}}=10^{37.3}\end{aligned}$

Hence, the answer is the option (4).

Example.3

3. For an electrochemical cell:

$\mathrm{Sn}(\mathrm{s})\left|\mathrm{Sn}^{2+}(\mathrm{aq}, 1 \mathrm{M})\right|\left|\mathrm{Pb}^{2+}(\mathrm{aq}, 1 \mathrm{M})\right| \mathrm{Pb}(\mathrm{s})$ the ratio $\frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}$ When this cell attains equilibrium is _____.

$\begin{aligned} & \text { (Given : } \mathrm{E}_{\mathrm{Sn}^2+\mid \mathrm{Sn}}^0=-0.14 \mathrm{~V}, \\ & \left.\mathrm{E}_{\mathrm{Pb}^2+\mid \mathrm{Pb}}^0=-0.13 \mathrm{~V}, \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06\right)\end{aligned}$

1) (correct)2.1544

2)1.11

3)7.15

4)3.14

Solution

As we have learned,

Nernst equation is given as

$\mathrm{E}=\mathrm{E}_{\text {cell }}^0-\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \mathrm{Q}$

Now, the chemical reaction occurring in the cell is given as

$\begin{aligned} & \mathrm{Sn}+\mathrm{Pb}^{2+} \longrightarrow \mathrm{Sn}^{2+}+\mathrm{Pb} \\ & 0=0.01-\frac{0.06}{2} \log \left\{\frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}\right\} \\ & 0.01=\frac{0.06}{2} \log \left[\frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}\right\} \\ & \frac{1}{3}=\log \left[\frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}\right\} \Rightarrow \frac{\left[\mathrm{Sb}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}=10^{1 / 3}=2.1544\end{aligned}$

Hence, the answer is the option (1).

Example.4

4. The potential (in V) of a hydrogen electrode in a solution with pH = 5 at 25^oC is :

1)0.295

2) (correct)-0.295

3)-0.59

4)0.59

Solution

$E=0-0.059 \log \left(\frac{1}{\left[H^{+}\right]}\right)$

$\mathrm{E}=-0.059 \times \mathrm{pH}$
$\mathrm{E}=-0.059 \times 5=-0.295 \mathrm{~V}$

Hence, the answer is the option (2).

Example.5

5. What will be the emf for the given cell

$\mathrm{Pt}\left|\mathrm{H}_2\left(\mathrm{P}_1\right)\right| \mathrm{H}_{\mathrm{aq}}^{+}\left|\mathrm{H}_2\left(\mathrm{P}_2\right)\right| \mathrm{Pt}$

1)$\frac{\mathrm{RT}}{\mathrm{F}} \ln \frac{\mathrm{P}_1}{\mathrm{P}_2}$

2) (correct)$\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\mathrm{P}_1}{\mathrm{P}_2}$

3)$\frac{\mathrm{RT}}{\mathrm{F}} \ln \frac{\mathrm{P}_2}{\mathrm{P}_1}$

4)$\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\mathrm{P}_2}{\mathrm{P}_1}$

Solution

Let's break the reaction into half cells.

Anode: $\mathrm{H}_2\left(\mathrm{P}_1\right) \rightarrow 2 \mathrm{H}^{+}+2 \mathrm{e}^{-}, \mathrm{E}^0=0$

Cathode: $2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2\left(\mathrm{P}_2\right), \mathrm{E}_0=0$

The net cell reaction can be written as

$\mathrm{H}_2\left(\mathrm{P}_1\right) \longrightarrow \mathrm{H}_2\left(\mathrm{P}_2\right)$

According to the Nernst equation, we have

$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^0-\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \left(\frac{\mathrm{P}_2}{\mathrm{P}_1}\right)$

now, since the given cell is a concentration cell, $\mathrm{E}_{\text {cell }}^0=0$

$\mathrm{E}_{\text {cell }}=\frac{-\mathrm{RT}}{2 \mathrm{~F}} \ln \left(\frac{\mathrm{P}_2}{\mathrm{P}_1}\right)=\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \left(\frac{\mathrm{P}_1}{\mathrm{P}_2}\right)$

Hence, the answer is the option (2).

Summary

The nernst equation is a very important topic in electrochemistry as it is the base of various implications in the field of electrochemistry. This equation allows the calculation of electrochemical potential even under non-standard conditions it does not only need the standard condition it Can work even without that. And predict the voltage of the cell with the help of its Concentration, which is very important for making the Batteries and the fuel cells. After understanding the concept of the Nernst equation and cell potential scientists develop a better quality of Batteries with the desired emf. Nernst equation also helps in predicting whether the reaction moving in the forward direction or in the backward direction by calculating cell potential. This concept becomes vital for understanding the mechanism of reaction and optimizing the Chemical process. This also works in the case of the Redox reaction as in the redox reaction it is used to calculate the potential at the equilibrium as well by relating the equilibrium constant to the electrode potential. It has applications in various other fields such as in biochemistry. Its biological application is such that it is used to understand the potential gradients across the cell membrane which is essential for the function of cells and energy production. Nernst equation als help detect corrosion by calculating its chemical potential in different environments.. This is important for materials science and engineering to develop corrosion-resistant materials and coatings. In industrial settings, it helps in controlling and optimizing chemical processes involving electrochemical reactions.