16.0 Concentration Cells

It is the limited form of galvanic cell which has two equivalent half cells of the same composition differ only in the concentrations. OR

It can be defined as the cells in which emf arise due to transfer of matter from one half cell to other which are differ in the concentration.

There are two types of the concentration cells:

$(i)$ Electrode Concentration Cell

$(ii)$ Electrolyte Concentration Cell

In this, two similar electrodes of different concentration are dipped in the same solution .

For example, two hydrogen electrodes at different pressures are dipped in the same solution of hydrogen ions. Here potential difference is developed between two electrodes at different concentrations dipped in the same electrolyte.

$$\begin{equation} \begin{aligned} (Pt,{H_2}(pressure\ {P_1}))/Anode/{H^ + }/({H_2}(pressure\ {P_2})Pt)/Cathode) \\ Pt;{H_2}({P_1})/{H^ + }ions(solution)/({H_2}({P_2});Pt \\ \\\end{aligned} \end{equation} $$

Lets write the reaction:

At Anode (i.e LHE): $${H_2}({P_1}) \to 2{H^ + } + 2{e^ - }$$

At cathode (i.e RHE): $$2{H^ + } + 2{e^ - } \to {H_2}({P_2})$$

Overall reaction: $${H_2}({P_1}) \to {H_2}({P_2})$$

Using Nernst equation, we can find out ${E_{cell}}$ as $$\begin{equation} \begin{aligned} {E_{cell}} = 0 - \frac{{0.059}}{2}\log \frac{{{P_2}}}{{{P_1}}} \\ {E_{cell}} = - 0.0295\log \frac{{{P_2}}}{{{P_1}}} \\ {E_{cell}} = 0.0295\log \frac{{{P_1}}}{{{P_2}}} \\\end{aligned} \end{equation} $$

In this, identical electrodes are dipped in the same electrolyte of different concentrations.

Here electrical energy source is the tendency of the electrolyte to diffuse from a solution of higher concentration to lower concentration. With the decrease in the time or (with the expiry of the time), two concentrations become equal, so in the beginning, emf is the maximum and gradually decreases to zero.

$$M/{M^{n + }}({C_1})//{M^{n + }}({C_2})/M$$

The emf of the cell is given by the following expression at ${25^0}C$.

$$\begin{equation} \begin{aligned} {E_{cell}} = 0 - \frac{{0.059}}{n}\log \frac{{{C_1}}}{{{C_2}}} \\ {E_{cell}} = \frac{{0.059}}{n}\log \frac{{{C_2}}}{{{C_1}}} \\\end{aligned} \end{equation} $$

Take an example of copper:

$$(Cu/C{u^{2 + }}({C_1}))/Anode//C{u^{2 + }}({C_2})/Cu)/Cathode$$

At anode: $$Cu \to C{u^{2 + }}({C_1}) + 2{e^ - }$$

At cathode: $$C{u^{2 + }}({C_2}) + 2{e^ - } \to Cu$$

Overall Reaction: $$C{u^{2 + }}({C_2}) \to C{u^{2 + }}({C_1})$$

Here value of $n$ is $2$.

Using Nernst Equation, we can find ${E_{cell}}$ as

$$\begin{equation} \begin{aligned} {E_{cell}} = 0 - \frac{{0.059}}{2}\log \frac{{{C_1}}}{{{C_2}}} \\ {E_{cell}} = - 0.0295\log \frac{{{C_1}}}{{{C_2}}} \\ {E_{cell}} = 0.0295\log \frac{{{C_1}}}{{{C_2}}} \\\end{aligned} \end{equation} $$

The concentration cells are used to determine the solubility of sparingly soluble salt, and also in the valency of the cation of electrolyte and in case of transition point between two allotropes of metal which are used as electrodes etc.

And for every reaction, ${E_{cell}}^0$ is equal to $0$.