Skip to main content
Chemistry LibreTexts

6. Potential of an Electrochemical Cell

  • Page ID
    81963
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Potassium dichromate (K2Cr2O7) reacts with Fe(II) to produce Cr(III) and Fe(III). What is the standard state potential and K for this reaction?

    The first step in solving this is to identify the two appropriate half reactions that make up the cell. From a table of Eo values we find the following two reactions:

    \[\mathrm{Fe^{3+}(aq) + e^- = Fe^{2+}(aq) \hspace{40px} E^o = 0.77\: V}\]

    \[\ce{Cr2O7^2- (aq) + 14H+(aq) + 6e-} = \ce{2Cr^3+(aq) + 7H2O} \hspace{40px} \mathrm{E^o = 1.33\: V}\]

    An examination of the Eo values indicates that the iron reaction proceeds as an oxidation and is the anode and the chromium reaction proceeds as a reduction and is the cathode.

    Standard State Potential

    The following equation is used when calculating the standard state potential of an electrochemical cell.

    \[\mathrm{E^o_{CELL} = E^o_{CAT} - E^o_{AN}}\]

    When using this equation, the Eo values are used directly from the table without changing the sign and without using any coefficients in the balanced overall electrochemical reaction. So even though the anodic reaction is reversed in direction from that in the table, the values in the table are all ranked relative to each other and the minus sign in the equation is accounting for the fact that the anodic reaction occurs in a reverse direction. There is no use of coefficients because Eo values are measured for standard state starting conditions (i.e., all reactants and products beginning at 1 Molar). With the pair in this problem, inequivalent molar amounts of the iron and chromium species will be present when the system achieves equilibrium because six mole equivalents of the iron are used up for each mole equivalent of the chromium.

    One final point to note is that EoCELL will always be positive.

    \[\mathrm{E^o_{CELL} = 1.33 - 0.77 = 0.56\: V}\]

    Equilibrium Constant

    Earlier in the unit we had developed the following equation relating Eo to the equilibrium constant.

    \[\mathrm{E^o = \dfrac{0.059}{n}\log ⁡K}\]

    This expression can be rewritten as follows:

    \[\mathrm{K = 10^{(n)(E^o)/0.059}}\]

    n is the number of electrons that are transferred in the balanced electrochemical reaction (6 in this example). Putting numbers in and evaluating this term gives the following:

    \[\mathrm{K = 10^{(6)(0.56)/0.059} = 8.9 \times 10^{56}}\]

    Note that this is an exceptionally large equilibrium constant meaning that the reaction goes nearly toward completion leaving only tiny amounts of reactants at equilibrium.

    Cell Potential with the Non-standard State Conditions Given in the Problem

    In this example, one half-cell is made up with 1.50 M potassium dichromate and 0.30 M chromium(III)nitrate hexahydrate in 1.00 M nitric acid and the other half cell is made up with 0.050 M iron(III)chloride hexahydrate and 0.10 M iron(II)chloride tetrahydrate?

    Using the shorthand notation for an electrochemical cell, we could write the above cell as follows:

    \[\mathrm{Pt \:|\: Cr_2O_7^{2-} (1.50\: M),\: Cr^{3+} (0.30\: M),\: H^+ (1.00\: M) \:||\: Fe^{2+} (0.050\: M),\: Fe^{3+} (0.10\: M) \:|\: Pt}\]

    There are two ways to solve this problem. One is to use the Nernst equation in the form we previously defined.

    \[\mathrm{E_{CELL}= E_{CELL}^0 - \dfrac{0.059}{n} \log Q}\]

    When using this method, we first need the complete electrochemical reaction for the cell. Note that n = 6 for this reaction as six electrons were needed to balance out the two half reactions.

    \[\ce{Cr2O7^2- (aq) + 14H+(aq) + 6Fe^2+(aq)} = \ce{2Cr^3+(aq) + 6Fe^3+(aq) + 7H2O}\]

    We can then write the term for Q:

    \[\mathrm{Q= \dfrac{[Cr^{3+}]^2 [Fe^{3+}]^6}{[Cr_2 O_7^{2-}][H^+ ]^{14} [Fe^{2+}]^6}}\]

    \[\mathrm{Q= \dfrac{(0.30)^2 (0.050)^6}{(1.50)(1.00)^{14} (0.10)^6} =9.4 \times 10^{-4}}\]

    We can now substitute values into the Nernst equation and solve for ECELL.

    \[\mathrm{E_{CELL}= 0.56 -\dfrac{0.059}{6} \log⁡(9.4 \times 10^{-4})=0.56+0.03=0.59 \:V}\]

    Instead of using the Nernst equation, is also possible to use the following equation to calculate the cell potential.

    \[\mathrm{E_{CELL} = E_{CAT} - E_{AN}}\]

    When using this equation, each of the half reactions is written and evaluated as a reduction. The minus sign accounts for the fact that the anodic reaction is reversed in the complete, balanced electrochemical reaction. When evaluating each of the two terms, use the associated value of n for each of the individual half reactions.

    \[\mathrm{E_{CAT}= E_{CAT}^0 - \dfrac{0.059}{n}\log Q_{CAT}}\]

    \[\mathrm{Q_{CAT}=\dfrac{[Cr^{3+}]^2}{[Cr_2 O_7^{2-} ][H^+ ]^{14}} =\dfrac{(0.30)^2}{(1.50)(1.00)^{14}}=0.060}\]

    The number of electrons in the cathode reaction is six so n = 6.

    \[\mathrm{E_{CAT}= 1.33-\dfrac{0.059}{6} \log⁡(0.060)=1.33+0.01=1.34\: V}\]

    The number of electrons in the anode reaction is one so n = 1.

    \[\mathrm{E_{AN}= E_{AN}^0 - \dfrac{0.059}{n} \log Q_{AN}}\]

    \[\mathrm{Q_{AN}=\dfrac{[Fe^{2+}]}{[Fe^{3+}]} =\dfrac{(0.10)}{(0.050)}=2.0}\]

    \[\mathrm{E_{AN}= 0.77-\dfrac{0.059}{1} \log⁡(2.0)=0.77-0.02=0.75\: V}\]

    Now we can evaluate the cell potential.

    \[\mathrm{E_{CELL} = E_{CAT} - E_{AN} = 1.342 - 0.75 = 0.59\: V}\]

    As expected, the use of these two possible methods provides the exact same value for the cell potential.

    What is the cell potential if the chromium half-cell were operated at a pH of 7 instead of using 1 M nitric acid?

    The point of this problem is to realize that some electrochemical reactions are critically dependent on the pH. The particular reaction in this problem has a stoichiometric coefficient of 14 for the H+, meaning that the reaction will be highly dependent on pH. Since only the cathodic half reaction depends on pH, we can evaluate the overall cell potential using the second of the two methods from above.

    \[\mathrm{E_{CAT}= E_{CAT}^0 - \dfrac{0.059}{n} \log Q_{CAT}}\]

    \[\mathrm{Q_{CAT}=\dfrac{[Cr^{3+}]^2}{[Cr_2 O_7^{2-}][H^+]^{14}} =\dfrac{(0.30)^2}{(1.50)(1.00 \times 10^{-7})^{14}} =6.0 \times 10^{96}}\]

    \[\mathrm{E_{CAT}= 1.33 - \dfrac{0.059}{6} \log⁡(6.0 \times 10^{96} )=1.33- -.95=0.38\: V}\]

    Remember that:

    \[\mathrm{E_{CELL} = E_{CAT} - E_{AN}}\]

    and that EAN calculate previously was 0.75 V.

    \[\mathrm{E_{CELL} = 0.38 - 0.75 = -0.37\: V}\]

    It is important to note that ECELL in this case is negative. The cell potential is positive for a reaction that proceeds in the forward direction toward products. The fact that this value is negative means that the reaction actually goes in the reverse direction under the conditions provided. Looking again at the overall reaction for this cell provided below, with the concentration of H+ so low (1.00 x 10-7 M) and the coefficient of 14 for the H+, it is reasonable to think that the overall reaction will actually proceed toward reactants to establish equilibrium than proceeding toward products.

    \[\ce{Cr2O7^2- (aq) + 14H+(aq) + 6Fe^2+(aq)} = \ce{2Cr^3+(aq) + 6Fe^3+(aq) + 7H2O}\]

    Similar calculations at pH 1 ([H+] = 0.10 M) and pH 3 ([H+] = 0.0010 M) give cell potentials of 0.46 V and 0.18 V, respectively. The importance of pH on this particular electrochemical cell is apparent from these data.


    This page titled 6. Potential of an Electrochemical Cell is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Thomas Wenzel.

    • Was this article helpful?