Nernst Equation - Practice Questions & MCQ

This equation gives the relationship between electrode potential and the concentration of ions in the solution. In other words, it shows the dependency of electrode potential on the concentration of the ions 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 learnt in thermodynamics that 

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

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

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

So, substituting these values is (1),

$\begin{aligned} & -\mathrm{nFE}=-\mathrm{nFE}^{\mathrm{O}}+\mathrm{RT} \ln \mathrm{Q} \\ & \Rightarrow \mathrm{E}=\mathrm{E}^{\mathrm{o}}-\frac{\mathrm{RT}}{\mathrm{nF}} \ln Q \\ & \Rightarrow \mathrm{E}=\mathrm{E}^{\mathrm{o}}-\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log Q\end{aligned}$

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

$\mathrm{E}=\mathrm{E}^{\circ}-\frac{2.303 \times 8.314 \times 298}{\mathrm{n} \times 96500} \log _{10} \frac{[\mathrm{M}]}{\left[\mathrm{M}^{\mathrm{n}+}\right]}$

$
\mathrm{E}=\mathrm{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
$\mathrm{E}^{\circ}=$ 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 net reaction:
  $\mathrm{xA}+\mathrm{yB} \xrightarrow{\mathrm{ne}^{-}} \mathrm{mC}+\mathrm{nD}$
  The emf can be calculated as

Ecell $=\mathrm{E}^{\mathrm{o}}$ 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 kept in mind.

• The activity of aq. ions are expressed in terms of their concentration.
• 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.

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

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

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

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

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

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 given as 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 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, 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:

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

If the circuit in Daniell cell is closed then we note that the reaction
$\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$
takes place and as time passes, the concentration of Zn²⁺ keeps on increasing while the concentration of Cu²⁺ keeps on decreasing. At the same time, the voltage of the cell as read on the voltmeter keeps on decreasing. After some time, we shall note that there is no change in the concentration of Cu²⁺ and Zn²⁺ 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:

$
\begin{aligned}
& \mathrm{E}_{\text {(cell) }}=0=\mathrm{E}_{(\text {cell })}^{\ominus}-\frac{2.303 \mathrm{RT}}{2 \mathrm{~F}} \log \frac{\left[\mathrm{Zn}^2\right]}{\left[\mathrm{Cu}^2\right]} \\
& \text { or } \mathrm{E}_{(\text {cell })}^0=\frac{2.303 \mathrm{RT}}{2 \mathrm{~F}} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^2+\right]}
\end{aligned}
$

But at equilibrium,
$\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}=\mathrm{K}_{\mathrm{c}}$ for the above reaction and at $\mathrm{T}=298 \mathrm{~K}$ the above equation can be written as

$
\begin{aligned}
& \mathrm{E}_{(\text {cell })}^{\ominus}=\frac{0.059 \mathrm{~V}}{2} \log \mathrm{~K}_{\mathrm{C}}=1.1 \mathrm{~V} \quad\left(\mathrm{E}_{\text {cell }}^{\Theta}=1.1 \mathrm{~V}\right) \\
& \log \mathrm{K}_{\mathrm{C}}=\frac{(1.1 \mathrm{~V} \times 2)}{0.059 \mathrm{~V}}=37.288 \simeq 37.3 \\
& \mathrm{~K}_{\mathrm{C}}=2 \times 10^{37} \text { at } 298 \mathrm{~K}
\end{aligned}
$
 

In general,

$
\mathrm{E}_{(\text {cell })}^{\ominus}=\frac{2.303 \mathrm{RT}}{n \mathrm{~F}} \log \mathrm{~K}_{\mathrm{C}}
$
 

Alternatively 

$\Delta \mathrm{G}=\Delta \mathrm{G}^{\circ}+\mathrm{RT} \ln \mathrm{Q}$

At equilibrium, $\Delta \mathrm{G}=0$ and $\mathrm{Q}=\mathrm{K}$, so.

$0=\Delta \mathrm{G}^{\mathrm{o}}+\mathrm{RT} \ln \mathrm{K}$

$\Delta \mathrm{G}^{\circ}=-\mathrm{RT} \ln \mathrm{K}$

$\Delta \mathrm{E}^{\mathrm{o}}=\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \ln \mathrm{K}$

$\ln \mathrm{K}=\frac{\mathrm{nE} \mathrm{E}^{\mathrm{o}}}{0.059}$

Thus, the above equation gives a relationship between the 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^o value of the cell.