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Connection between \(E_{cell}\), ∆G, and K

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  • Learning Objectives

    The connection between cell potential, Gibbs energy and constant equilibrium are directly related in the following multi-part equation:

    \[ \Delta G^o= -RT\ln K_{eq} = -nFE^o_{cell} \]

    ∆G: Gibbs Energy

    ∆G is the change of Gibbs (free) energy for a system and ∆G° is the Gibbs energy change for a system under standard conditions (1 atm, 298K). On an energy diagram, ∆G can be represented as:

    Where ∆G is the difference in the energy between reactants and products. In addition ∆G is unaffected by external factors that change the kinetics of the reaction. For example if Ea(activation energy) were to decrease in the presence of a catalyst or the kinetic energy of molecules increases due to a rise in temperature, the ∆G value would remain the same.

    cell: Standard Cell Potential

    cell is the electromotive force (also called cell voltage or cell potential) between two half-cells. The greater the E°cell of a reaction the greater the driving force of electrons through the system, the more likely the reaction will proceed (more spontaneous). E°cell is measured in volts (V). The overall voltage of the cell = the half-cell potential of the reduction reaction + the half-cell potential of the oxidation reaction. To simplify,

    \[E_{cell} = E_{reduction} + E_{oxidation} \tag{1}\]


    \[E_{cell} = E_{cathode} + E_{anode} \tag{2}\]

    The potential of an oxidation reduction (loss of electron) is the negative of the potential for a reduction potential (gain of electron). Most tables only record the standard reduction half-reactions. In other words, most tables only record the standard reduction potential; so in order to find the standard oxidation potential, simply reverse the sing of the standard reduction potential.


    The more positive reduction potential of reduction reactions are more spontaneous. When viewing a cell reduction potential table, the higher the cell is on the table, the higher potential it has as an oxidizing agent.

    Difference between Ecell and Eºcell

    Eº cell is the standard state cell potential, which means that the value was determined under standard states. The standard states include a concentration of 1 Molar (mole per liter) and an atmospheric pressure of 1. Similar to the standard state cell potential, Eºcell, the Ecell is the non-standard state cell potential, which means that it is not determined under a concentration of 1 M and pressure of 1 atm. The two are closely related in the sense that the standard cell potential is used to calculate for the cell potential in many cases.

    \[ E_{cell}= E^o_{cell} -\dfrac{RT}{nF} \ln\; Q \tag{3}\]

    Other simplified forms of the equation that we typically see:

    \[ E_{cell}= E^o_{cell} -\dfrac{0.0257}{n} \ln \; Q \tag{4}\]

    or in terms of \(\log_{10}\) (base 10) instead of the natural logarithm (base e)

    \[ E_{cell}= E^o_{cell} - \dfrac{0.0592}{n} \log_{10}\; Q \tag{5}\]

    Both equations applies when the temperature is 25ºC. Deviations from 25ºC requires the use of the original equation. Essentially, Eº is E at standard conditions

    Example 1

    What is the value of Ecell for the voltaic cell below:

    \[Pt_{(s)}|Fe^{2+}_{(0.1M)},Fe^{3+}_{(0.2M)}||Ag^+_{(0.1M)}|Ag_{(s)} \]


    To use the Nernst equation, we need to establish \(E^o_{cell}\) and the reaction to which the cell diagram corresponds so that the form of the reaction quotient (Q) can be revealed. Once we have determined the form of the Nernst equation, we can insert the concentration of the species.


    \[ E^o_{cell} = E^o_{cathode}- E^o_{anode}\]

    = EoAg/Ag - EoFe3+/Fe2+

    = 0.800V-0.771V = 0.029V

    Now to determine Ecell for the reaction

    Fe2+(0.1M) + Ag+(1.0M) → Fe3+(0.20M) + Ag(s)

    Use the Nernst equation

    Ecell= 0.029V -(0.0592V/1)log [Fe3+]/[Fe2+][Ag]

    =0.029V - 0.0592V*log [0.2]/[0.1]*[1.0]


    K: The Equilibrium Constant

    \(K\) is the equilibrium constant of a general reaction

    \[ aA + bB \leftrightharpoons cC + dD \tag{6} \]

    and is expressed by the reaction quotient:

    \[ K_c= \dfrac{[C]^c[D]^d}{[A]^a[B]^b} \tag{7}\]

    Example 2

    Given \(K = 2.81 \times 10^{-16}\) for a following reaction

    \[Cu^{2+}_{(aq)} + Ag_{(s)} \rightleftharpoons Cu_{(s)} + 2Ag^+\]

    Find ∆G.


    Use the following formula: ∆G=-RTlnK

    = 8.314 x 298 x ln(2.81x10-16) = -8.87x105

    = 8.871 kJ

    The Relationship Between the Three

    The relationship between ∆G, K, and E° cell can be represented by the following diagram.



    • R = 8.314 J mol C-1
    • T = Temp in K
    • n = moles of e- from balanced redox reaction
    • F = Faraday's constant = 96,458 C/mol

    cell can be calculated using the following formula:

    \[E^o_{cell} = E^o_{cathode} – E^o_{anode} = E^o_{Reduction} – E^o_{Oxidation} \tag{8}\]

    Summary Table
    Eo cell \(\Delta G\) Q & K Relationship Reaction Direction Spontaneity (as written)
    >0 - Q<K Forward Spontaneous
    <0 + Q>K Backward Non-spontaneous
    =0 =0 Q=K No Reaction N/A

    Using ∆G=-RT lnK

    Question Find the E° cell for the following coupled half-reactions


    1. Determine the cathode and anode in the reaction

    Zn(s) ↔ Zn2+(aq) + 2e- Anode, Oxidation (since Zn(s) increase oxidation state from 0 to +2)

    Cu2+(aq) + 2e- ↔ Cu(s) Cathode, Reduction (since Cu2+(aq) decreases oxidation state from +2 to 0)

    2. Determine the E° cell values using the standard reduction potential table (Table P1)

    Zn(s) ↔ Zn2+(aq) + 2e- -0.763

    Cu2+(aq) + 2e- ↔ Cu(s) +0.340

    3. Use E° cell = E°cathode - E°anode

    = 0.340 - (-0.763)

    = 1.103 V


    Find ∆G for the following reaction:

    \[2Al_{(s)} +3Br_{2(l)} \rightleftharpoons 2Al^{3+}_{(aq, 1M)} + 6Br^-_{(aq, 1M)}\]


    Step 1: Separate the reaction into its two half reactions

    \[2Al_{(s)} \rightleftharpoons 2Al^{3+}_{(aq)}\]

    \[3Br_{2(l)} \rightleftharpoons 6Br^-_{(aq, 1M)}\]

    Step 2: Balance the half equations using O, H, and charge using e-

    2Al(s) ↔ 2Al3+(aq) +6e-

    6e- + 3Br2(l) ↔ 6Br-(aq)

    Step 3: From the balanced half reactions, we can conclude the number of moles of e- for use later in the calculation of ∆G. Determine the E° values using the standard reduction potentials, using the E° cell table.

    2Al2(s) ↔ Al3+(aq) +6e- -1.676V

    3Br2(l) + 6e- ↔ 6Br-(aq) +1.065V

    Step 4: Determine cell = E°cathode - E°anode.

    = 1.065 - (-1.676)

    = 2.741 V

    Step 5: Once E° cell has be calculated and the number of moles of electrons have been determined, we can use ∆G = -nFE°cell

    = (-6 mol e-)(96458 C/mol e-)(2.741 V)

    = -1586kJ

    This equation can be used to calculate E° cell given K or K given E°cell. If T=298 K, the RT is a constant then the following equation can be used: cell= (0.025693V/n) ln K

    Example 5: Using E° cell=(RT/nF) lnK

    Given the E° cell for the reaction

    \[Cu_{(s)} + 2H^+_{(aq)} \rightleftharpoons Cu^{2+}_{(aq)} +H_{2(g)}\]

    is -0.34V, find the equilibrium constant (K) for the reaction.


    Step 1: Split into two half reaction

    Cu(s) ↔ Cu2+(aq)

    2H+(aq)↔ H2(g)

    Step 2: Balance the half reactions with charges to determine n

    Cu(s) ↔ Cu2+(aq) + 2e-

    2H+(aq) +2e- ↔ H2(g)

    Therefore n=2

    Step 3: From the example above, E° cell = -0.34V

    -0.34 = (0.025693/2) lnK

    K = e(-0.34 x 2/0.025693)

    K = 3.19 x 10-12

    Example 6: Spontaneity

    Given the following reaction determine ∆G, K, and Eocellfor the following reaction at standard conditions? Is this reaction spontaneous?

    \[Mn^{2+}_{(aq)} + K_{(s)} \rightleftharpoons MnO_{2(s)} + K_{(aq)}\]


    Step 1: Separate and balance the half reactions. Label which one is reduction and which one is oxidation. Find the corresponding E° values for the half reactions.

    MnO2(s) + 4H+(aq) + 2e ↔ -Mn2+(aq) + 2H2O(l) Reduction +1.23V

    K+(aq) + e- ↔ K(s) Oxidation -2.92V

    Step 2: Write net balanced reaction in acidic solution, and determine the E° cell.

    E° cell = E°cathode- E°anode = +1.23 - (-2.92) = 4.15

    Mn2+(aq) + 2K+(aq) + 2H2O(l)MnO2(s) + 4H+(aq) + K(s) E° cell = 4.15V

    Step 3: Find ∆G for the reaction.

    Use ∆G = -nFE°cell

    = -2 mol e- x 96458 C x 4.15 = -800.60kJ

    Therefore, since Eocell is positive and ∆G is negative, this reaction is spontaneous.


    1. Find E° cell for \(2Br^-_{(aq)} + I_{2(s)} \rightleftharpoons Br_{2(s)} + 2I^-+{(aq)}\)
    2. Find E° for \(Sn_{(s)} \rightleftharpoons Sn^{2+}_{(aq)} + 2e^-\)
    3. Find E° cell for \(Zn_{(s)} | Zn^{2+}_{(aq)} || Cr^{3+}_{(aq)}, Cr^{2+}_{(aq)}\)

    (See answer key for solutions)

    • If the E° values of the reaction is negative, then the reaction is NOT spontaneous and therefore the reverse reaction is occurring and the electrons are flowing in the opposite direction.
    • These values are tabulated in the standard reduction potential table.

    4. Find ∆G for the following combined half reactions:

    F2(g) + 2e- ↔ 2F-(aq)

    Li+(aq) + e- ↔ Li(s)

    5. Find the equilibrium constant (K) for the following reaction: (Hint: Find E°Cell first!)

    Zn(s) + 2H+(aq) ↔ Zn2+(aq) + H2(g)

    6. Find E°Cell for the given reaction at standard conditions:

    Cu+(aq) + e- ↔ Cu(s)

    I2(s) + 2e- ↔ 2I-(aq)

    (See answer key for solutions)


    1. +0.530 V
    2. +0.137 V
    3. +0.339 V
    4. +1139.68 kJ
    5. 6.4 x 1025
    6. +0.195 V


    1. Brady, James E., Holum, John R. “Chemistry: The Study of Matter and Its Changes”, John Wiley & Sons Inc 1993
    2. Brown, Theodore L., LeMay, H. Eugene Jr. “Chemistry: The Central Science” Third Edition, Prentice-Hall, Inc. Englewood Cliffs, N.J. 07632 1985
    3. Brown, Theodore L., LeMay, H. Eugene Jr., Bursten, Bruce E. “Chemistry: The Central Science” Fifth Edition, Prentice-Hall, Inc. Englewood Cliffs, N.J. 07632 1991
    4. Gesser, Hyman D. “ Descriptive Principles of Chemistry”, C.V. Mosby Company 1974
    5. Harwood, William, Herring, Geoffrey, Madura, Jeffry, and Petrucci, Ralph, General Chemistry: Principles and Modern Applications, Ninth Edition, Upper Saddle River,New Jersey, Pearson Prentice Hall, 2007.

    Contributors and Attributions

    • Deborah S. Gho, Shamsher Singh