Skip to main content
Chemistry LibreTexts

10.5: Concentration Cells

  • Page ID
    84750
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    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:

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


    This page titled 10.5: Concentration Cells is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Patrick Fleming.

    • Was this article helpful?