12.0 Nernst Equation

Consider an equation $$Aa + Bb \to Cc + Dd\ \ \ \ \ ...(i)$$

Gibbs free energy of the above equation is given by

$$G = \Delta {G^0} + 2.303RT{\log _{10}}\frac{{({a_C}^c \times {a_D}^d)}}{{({a_A}^a \times {a_B}^b)}}...(ii)$$

where $a$ represents the activity of particular reactants and products indicated by the suffix

$\Delta {G^0}$ represents the free energy change of the reaction when the reactants and products are taken in the standard conditions, and called standard free energy change

$G$ represents the free energy change of a cell reaction, related to electrical work that can be obtained from the cell. It is related as $$\Delta {G^0} = - nF{E^0}$$

Substitute these values in equation $i$, we get

$$\begin{equation} \begin{aligned} - nF{E_{cell}} = - nF{E^0} + 2.303RT{\log _{10}}\frac{{({a_C}^c \times {a_D}^d)}}{{({a_A}^a \times {a_B}^b)}}...(iii) \\ nF{E_{cell}} = nF{E^0} - 2.303RT{\log _{10}}\frac{{({a_C}^c \times {a_D}^d)}}{{({a_A}^a \times {a_B}^b)}}...(iv) \\ {E_{cell}} = {E^0} - 2.303\frac{{RT}}{{nF}}{\log _{10}}\frac{{({a_C}^c \times {a_D}^d)}}{{({a_A}^a \times {a_B}^b)}}...(v) \\\end{aligned} \end{equation} $$

This is Nernst equation.

Lets put all the known values like $$\begin{equation} \begin{aligned} R = 8.314J{K^{ - 1}}mo{l^{ - 1}} \\ T = 298K \\ F = 96500C \\\end{aligned} \end{equation} $$

So equation ($v$) reduces to

$$\begin{equation} \begin{aligned} {E_{cell}} = {E^0} - 2.303 \times \frac{{8.314 \times 298}}{{n \times 96500}}{\log _{10}}\frac{{({a_C}^c \times {a_D}^d)}}{{({a_A}^a \times {a_B}^b)}} \\ {E_{cell}} = {E^0} - \frac{{0.0591}}{n} \times {\log _{10}}\frac{{({a_C}^c \times {a_D}^d)}}{{({a_A}^a \times {a_B}^b)}} \\ {E_{cell}} = {E^0} - \frac{{0.0591}}{n} \times {\log _{10}}\frac{{\left[ {\Pr oduct} \right]}}{{\left[ {\operatorname{Re} ac\tan t} \right]}} \\\end{aligned} \end{equation} $$

Now lets find out the potential of single electrode (say anode)

$$M \to {M^{n + }} + n{e^ - }$$...$(i)$

On anode, oxidation occurs so find out the potential for this oxidation reaction by applying Nernst equation,

$${E_{oxd}} = {E_{oxd}}^0 - \frac{{0.0591}}{n}{\log _{10}}\frac{{\left[ {{M^{n + }}} \right]}}{{\left[ M \right]}}$$

${E_{oxd}}$ is the electrode potential of oxidation half cell and ${E_{oxd}}^0$ is standard electrode potential of oxidation half cell $${E_{oxd}} = {E_{oxd}}^0 - \frac{{0.0591}}{n}{\log _{10}}\left[ {{M^{n + }}} \right]$$

Now lets find out the potential of single electrode (say cathode)

$${N^{n + }} + n{e^ - } \to N$\ \ \ ...(ii)$

Lets find out the potential of this by applying nernst equation,

$${E_{red}} = {E_{red}}^0 - \frac{{0.0591}}{n}{\log _{10}}\frac{{\left[ N \right]}}{{\left[ {{N^{n + }}} \right]}}$$

Now consider that $M$ and $N$ form a cell, so combine reaction $(i)$ and $(ii)$, we get

$${N^{n + }} + M \to N + {M^{n + }}$$

$$\begin{equation} \begin{aligned} {E_{red}} + {E_{oxd}} = {E_{red}}^0 - \frac{{0.0591}}{n}{\log _{10}}\frac{{\left[ N \right]}}{{\left[ {{N^{n + }}} \right]}} + {E_{oxd}}^0 - \frac{{0.0591}}{n}{\log _{10}}\frac{{\left[ {{M^{n + }}} \right]}}{{\left[ M \right]}} \\ {E_{cell}} = {E_{red}} + {E_{oxd}} \\ {E_{cell}}^0 = {E_{red}}^0 + {E_{oxd}}^0 \\ {E_{cell}} = {E_{cell}}^0 - \frac{{0.0591}}{n}{\log _{10}}\left\{ {\frac{{\left[ N \right]}}{{\left[ {{N^{n + }}} \right]}} \times \frac{{\left[ {{M^{n + }}} \right]}}{{\left[ M \right]}}} \right\} \\ {E_{cell}} = {E_{cell}}^0 - \frac{{0.0591}}{n}{\log _{10}}\left\{ {\frac{{product}}{{reac\tan t}}} \right\} \\\end{aligned} \end{equation} $$

Note: The concentration of pure solids and liquids are taken as unity.