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Chemistry LibreTexts

17.12: Galvanic Cells and Free Energy

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    We see in the section on the Electromotive Force of Galvanic Cells that the emf of a galvanic cell can tell us whether the cell reaction is spontaneous. In other sections we show that the free-energy change ΔG of a chemical process also indicates whether that process is spontaneous. It is quite reasonable, then, to expect some relationship between ΔG and E, and indeed one exists.

    In the section on Free Energy we stated that the free-energy change corresponds to the maximum quantity of useful work which can be obtained when a chemical reaction occurs. In other words,

    \[\triangle G = –w_{max}\]

    where the minus sign is necessary because the free energy decreases as the chemical system does useful work on its surroundings. If we are referring to a redox reaction, that work can be obtained in electrical form by means of an appropriate galvanic cell. It can be measured readily, because when a quantity of charge Q moves through a potential difference ΔV, the work done is given by

    \[w = Q \triangle V\]

    Thus if one coulomb passes through a potential difference of one volt, the work done is

    \[w = 1 \text{ C} \times 1 \text{ V} = 1 \cancel{\text{As}} × 1 \text{ J} \cancel{\text{A}^{-1}\text{ s}^{-1}} = 1 \text{ J}\]

    Now suppose we construct a Zn-Cu cell of the type described earlier:

    \[\ce{Zn} \mid \ce{Zn^2+ (1 M)} \parallel  \ce{Cu^2+ (1 M)} \mid \ce{Cu}\]

    and suppose we make the cell large enough that the concentrations of Cu2+ and Zn2+ will not change significantly even though 1 mol Zn is oxidized to 1 mol Zn2+ according to the cell reaction

    \[\ce{Zn (s) + Cu^2+ (aq) -> Zn^2+ (aq) + Cu(s)} \label{5}\]

    If this cell is discharged through a large enough resistance, the potential difference will have its maximum value, namely, the cell emf, E°; so if we know how much charge is transferred, we can calculate the electrical work done. For the oxidation of 1 mol Zn [that is, for the occurrence of 1 mol of reaction \(\ref{5}\)], there must be 2 mol e transferred according to the half-equation

    \[\ce{Zn (s) -> Zn^2+ (aq) + 2}e^–\]

    Therefore the quantity of electrical charge transferred per mole of reaction is

    \[Q_m = \ce{2 * F = 2 * 9.649 × 10^4 C mol^{-1} = 1.930 * 10^5 C mol^{-1}}\]

    (The Faraday constant, F, is the quantity of charge per mole of electrons. It has the value 96,485 C/mol.) The maximum useful work per mole of reaction which the cell can perform while discharging is thus

    \[w_{max} = Q_{m}E^{o} = \ce{2FE^{o} = 2 * 9.649 * 10^{4} C mol^{-1} * 1.10 V = 212} \frac{\text{ kJ}}{\text{ mol}}\]

    The standard molar free energy change for the cell reaction is thus

    \[\Delta G_{m}^{o} =  – w_{max} = – 2FE^{o} = – 212 \frac{\text{ kJ}}{\text{ mol}}\]

    A similar argument can be applied to any cell in which the reactants and products are all at their standard concentrations or pressures. If the standard emf of such a cell is E°, while ΔGm° is the standard molar free energy change for the cell reaction,these two quantities are related by the equation

    \[\Delta G_{m}^{o} = – zFE^{\circ} \]

    where z (a dimensionless number) corresponds to the number of moles of electrons transferred per mole of cell reaction.

    A similar relationship holds even when reactants and products are not at standard concentrations and pressure:

    \[\Delta G_{m} = – zFE \]

    This connection between cell emf and free-energy change provides a means of measuring ΔGm, directly, rather than by determining ΔHm, and ΔSm, and then combining them.


    Example  \(\PageIndex{1}\) : The emf of a Cell

    \[\ce{Pt, H_{2} (1 atm)} \mid \ce{H^{+} (1 M)} \mid \ce{O_{2} (1 atm), Pt}\]

    is 1.229 V at 298.15 K. Calculate ΔGf° for liquid water at this temperature.

    Solution: The half-equations for the cell are

                         \(\ce{2H2 (g) -> 4H+ (aq) + 4}e^{-}\)

                    \(\ce{4}e^{-} \ce{+ 4H+ (aq) + O2 (g) -> 2H2O (l)}\)

    so that the cell reaction is

    \[\ce{2 H2 (g) + O_{2} (g) -> 2 H_{2}O (l)} \quad \text{1 atm, 298.15 K} \]

    Since there are 4 mol e transferred per mol cell reaction, z = 4 and

    \[\Delta G_{m} = - zFE = - 4 \times \frac{9.649 * 10^{4} \text{C}}{ \text{1 mol}} \times 1.229 \text{V} = - 474.3 \frac{ \text{ kJ}}{ \text{ mol}} \]

    The reaction produces liquid water at standard pressure and the desired temperature, but 1 mol reaction produces 1 mol 2H2O(l), that is, 2 mol H2O. Therefore

    \[\triangle G_{f}^{o}[\text{H}_2 \text{O} (l) \text{, 298 K}] = \frac{1}{2} \triangle G_{m} = – 237.2 \frac{\text{ kJ}}{ \text{ mol}}\]