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19.6: Electrochemistry and Thermodynamics

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    In section 19.3.2 we related free energy to cell potential with the following equation, which will allow us to relate thermodynamics to electrochemistry.

    \[\Delta G = -nFE_{Cell}\]

    We also know from section 18.6.3 where we discussed two meanings for "spontaneity".

    • Spontaneous in the sense that it is product favored
      \(\Delta G^o_{rxn}\)<<0 (it is very negative) and \(K \gg 1\) (it is a very large number)
      (once equilibrium is reached there will be lots of products)
    • Spontaneous in the sense that it is reactant loaded
      \(\Delta G_{rxn} < 0\) (Forward reaction is spontaneous) and Q<K
      (the reaction will make products, even if not many)

    We also know that from section that at equilibrium Q=K and \(\Delta G_{rxn} \) = 0, so substituting these equilibrium conditions into eq. 18.5.11 gives

    \[ \Delta G^o_{rxn} + RTlnK_{eq} = 0 \\ \color{blue}{ rearranging \; \; gives }\\ \\ \color{red}{ \Delta G^o_{rxn} = -RTlnK } \\ \color{blue}{ or} \\ \\ \color{red}{ \large{K =e^{\frac{-\Delta G^o_{rxn}}{RT}}}} \]

    (note: the above are different forms of the same equation)

    We will start by looking at systems that are at equilibrium, and see that we can determine K from standard state reduction potentials.

    Equilibrium Constants and Standard Cell Potentials

    We can not get the equilbrium constant from a table of standard reduction potentials. Noting from

    \[\Delta{G°} = −RT \ln K \; \; and \; \; \Delta{G°}= −nFE^°_{cell}\]

    Gives the relationship between the standard free-energy change and the standard cell potential:

    \[−nFE^°_{cell} = −RT \ln K \]

    Rearranging this equation,

    \[E^\circ_{\textrm{cell}}= \left( \dfrac{RT}{nF} \right) \ln K \]

    or in the exponential form

    \[\large{K =e^{\frac{nFE^o_{rxn}}{RT}}}\]

    Thus K is exponentially related to \(E^°_{cell}\) and large equilibrium constants correspond to large positive values of \(E^°_{cell}\).

    Example \(\PageIndex{2}\)

    Use the data in the References Table of Standard Potentials to calculate the equilibrium constant at standard conditions for the reaction of metallic lead with PbO2 in the presence of sulfate ions to give PbSO4 under standard conditions. (This reaction occurs when a car battery is discharged.) Report your answer to two significant figures.

    Given: redox reaction

    Asked for: \(K\)


    1. Write the relevant half-reactions and potentials. From these, obtain the overall reaction and E°cell.
    2. Determine the number of electrons transferred in the overall reaction. Use Equation 19.5.6 to solve for K.


    A. The relevant half-reactions and potentials from Table P2 are as follows:

    \(\begin{align} & \textrm {cathode:}
    & & \mathrm{PbO_2(s)}+\mathrm{SO_4^{2-}(aq)}+\mathrm{4H^+(aq)}+\mathrm{2e^-}\rightarrow\mathrm{PbSO_4(s)}+\mathrm{2H_2O(l)}
    & & E^\circ_\textrm{cathode}=\textrm{1.69 V} \nonumber \\
    & \textrm{anode:}
    & & \mathrm{Pb(s)}+\mathrm{SO_4^{2-}(aq)}\rightarrow\mathrm{PbSO_4(s)}+\mathrm{2e^-}
    & & E^\circ_\textrm{anode}=-\textrm{0.36 V} \nonumber \\ \nonumber
    & \textrm {overall:}
    & & \mathrm{Pb(s)}+\mathrm{PbO_2(s)}+\mathrm{2SO_4^{2-}(aq)}+\mathrm{4H^+(aq)}\rightarrow\mathrm{2PbSO_4(s)}+\mathrm{2H_2O(l)}
    & & E^\circ_\textrm{cell}=\textrm{2.05 V} \end{align}\)

    B. Two electrons are transferred in the overall reaction, so n = 2. Solving Equation \(\ref{20.5.13}\) for log K and inserting the values of n and E°,

    \[\begin{align} \nonumber & =e^{\frac{nFE^o_{rxn}}{RT}} \nonumber \\ \nonumber & =e^{\frac{2(96,500\frac{J}{V\cdot mol\, e^{-}})2.05V}{8.314\frac{J}{mol\cdot K}(298K)}} \nonumber \\ & =2.3\times10^{69} \end{align}\]

    Thus the equilibrium lies far to the right, meaning the reaction will proceed towards products.

    Exercise \(\PageIndex{3}\)

    Use the data in the Reference Table of Standard Potentials, calculate the equilibrium constant for the reaction of Sn2+(aq) with oxygen to produce Sn4+(aq) and water under standard conditions. Report your answer to two significant figures. The reaction is as follows:

    \[2Sn^{2+}_{(aq)} + O_{2(g)} + 4H^+_{(aq)} \rightleftharpoons 2Sn^{4+}_{(aq)} + 2H_2O_{(l)}\]

    \(1.2 \times 10^{73}\)

    Figure \(\Pa

    This page titled 19.6: Electrochemistry and Thermodynamics is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Robert Belford.

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