10.5: Concentration Cells

The generation of an electrostatic potential difference is dependent on the creation of a difference in chemical potential between two half-cells. One important manner in which this can be created is by creating a concentration difference. Using the Nernst equation, the potential difference for a concentration cell (one in which both half-cells involve the same half-reaction) can be expressed

$$E = -\dfrac{RT}{nF} \ln \dfrac{[\text{oxdizing}]}{[\text{reducing}]} \nonumber$$

Example $\PageIndex{1}$

Calculate the cell potential (at 25 °C) for the concentration cell defined by

$$Cu(s) | Cu^{2+} (aq,\, 0.00100 \,M) || Cu^{2+} (aq,\, 0.100 \,M) | Cu(s) \nonumber$$

Solution

Since the oxidation and reduction half-reactions are the same,

$$E_{cell}^o =0\,V \nonumber$$

The cell potential at 25 °C is calculated using the Nernst equation:

$$E = -\dfrac{RT}{nF} \ln Q \nonumber$$

Substituting the values from the problem:

$$\begin{align*} E_{cell} &= - \dfrac{(8.314 \,J/(mol\,K) (298\,K) }{2(96484\,C)} \ln \left( \dfrac{0.00100\,M}{0.100\,M} \right) \\[4pt] &= 0.059\,V \end{align*}$$