Nernst Equation

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 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 $-\frac{0.059}{\mathrm{n}} \log \frac{[\mathrm{C}]^{\mathrm{m}}[\mathrm{D}]^{\mathrm{n}}}{[\mathrm{A}]^x[\mathrm{~B}]^{\mathrm{y}}}$

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This Story also Contains

1. Nernst Equation
2. Some Solved Examples
3. Summary

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.

Recommended topic video on(Nernst Equation)

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

Zn(s)+2H+aq⇌Zn₂+aq+H₂(g)Ecell =Ecell ∘−0.0592log⁡[Zn2+]×pH2[H+]2

On addingH2SO4 the[H+] will increase therefore 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

Nernst equation 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