Calculating cell potential (voltaic cells)

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  1. Will a redox reaction occur?
  2. Reduction potential
  3. Will a redox reaction happen spontaneously?
  4. Will a redox reaction happen spontaneously? Worked example 1
  5. Will a redox reaction happen spontaneously? Worked example 2
  6. Cell potential, free energy and equilibrium.
  7. The Nernst equation (non-standard cell potential).
  1. Use a standard potential table1 to predict whether a spontaneous redox reaction will occur.
    1. Will a reaction between Cu (I) and nickel metal occur spontaneously in a redox process?
    2. Will a reaction between Al (III) and scandium metal occur spontaneously in a redox process?
    3. Will a reaction between vanadium metal and Scandium (III) occur spontaneously in a redox process?
  2. Calculate the equilibrium constant K using the cell potential and calculate the cell potential at nonstandard conditions.
    1. The oxidation of zinc metal (Zn) by copper ions (Cu2+) to form Zn2+ ions and Cu (s) happens spontaneously in a redox reaction at standard conditions at 298K. Calculate the equilibrium constant K for this process using the following standard potential data:

      Cu2+ + 2e- β€‰β‡Œβ€‰ \, \rightleftharpoons \, Cu where E0red = +0.34 V
      Zn2+ + 2e- β€‰β‡Œβ€‰ \, \rightleftharpoons \, Zn where E0red = -0.76 V
    2. During the reaction, the reaction quotient Q is measured to be 1.8*108. Use the Nernst equation to find the nonstandard cell potential at these conditions.
      Is the cell potential increasing, decreasing or unchanged during the reaction?
Topic Notes

In this lesson, we will learn:

  • To understand the use of standard potentials in redox reactions.
  • How to predict the spontaneity of a redox reaction using a standard potential table.
  • How to calculate cell potential to prove the spontaneity of a redox reaction.
  • How the cell potential relates to Gibbs free energy.
  • To calculate the cell potential at non-standard conditions with the Nernst equation.


  • We learned in Introduction to electrochemistry that electrochemical cells use redox reactions to convert chemical potential energy into electrical energy. The amount of energy released depends on what is reacting.
    Because of the energy involved when reduction or oxidation takes place, some pairs of substances will react in a redox process spontaneously!
    To find out if a redox reaction will be spontaneous or not, we need to use the table of Standard Reduction Potentials (or just β€˜standard potentials’). Here are some key points about it:
    • The table is a list of substances with the amount of energy (in volts) they release in a reduction half-equation. This means a few things:
      • The substances near the top of the table are excellent oxidizing agents, so they are likely to get reduced in a redox reaction.
      • The substances near the bottom of the table are poor oxidizing agents. On the flip side, these substances are good reducing agents, which means they release lots of energy when they get oxidized themselves. The table always shows reduction reading left to right though, so for oxidation reactions, read the equation right to left.
    • Hydrogen ions being reduced to H2 is defined at zero volts. This is why some of the equations have negative values; it just means the substances are less oxidizing than H + ions.
    • Because some substances (e.g. transition metals) have multiple oxidation states, they will appear multiple times in the table. These are showing different reactions (e.g. an oxidation or reduction from two different states).

  • Because redox reactions involve one reduction and one oxidation (NO EXCEPTIONS), a redox reaction will be built from two half-equations with one from the table being read left to right (reduction) and one going from right to left (oxidation). These two half-cells would be joined together, the half-equation higher on the table will be reduced and the lower half-equation will be oxidized.

  • Whether a redox reaction is spontaneous can also be found mathematically using the values in the table. Using the E0red values, the reaction is spontaneous if E0cell is a positive value. E0cell is worked out by:

    E0cell = E0red (reduction) + E0ox (oxidation)

    Remember, the table ONLY gives E0red values! Find E0red and reverse the sign to get E0ox.


    Will a redox reaction between Bi3+ and In occur spontaneously?
    First, use the table to find an equation with these compounds and their E0red. Below are all the half-equations equations containing the substances in the question, taken from a standard potential table1:

    Bi3+ + 3e- β‡Œ\enspace \rightleftharpoons \enspace BiE0red = +0.20 V
    In+ + e- β‡Œ\enspace \rightleftharpoons \enspace InE0red = -0.14 V
    In3+ + 3e- β‡Œ\enspace \rightleftharpoons \enspace InE0red = -0.34 V

    The second step is to see if one substance can be reduced (left side of the reaction arrow) and if the other can be oxidized (right side of the reaction arrow). For a redox to occur we need both!
    In reading the data, we see:
    • Bi3+ can be reduced into Bi (Bi3+ is on the left of an equation).
    • In+ can be reduced into In, which means In can be oxidized into In+ because In is on the right side of the equation.
    • The reduction of In3+ into In, which means In can be oxidized into In3+.
    We can also see, reading the table from top to bottom (highest E0cell to lowest) that Bi3+ is higher on the table than In. Therefore, the reduction of Bi3+ with the oxidation of In is a spontaneous reaction.

    To find this mathematically, we use the electrode potential expression:

    E0cell = E0red (reduction) + E0ox (oxidation)

    The E0cell must be a positive value for the redox reaction to be spontaneous. We have two possible reactions:
    • Reaction 1: The reduction of Bi3+ to Bi (E0red = +0.20 V) with the oxidation of In to In+ (E0red = -0.14 V)
    • Reaction 2: The reduction of Bi3+ to Bi (E0red +0.20 V) with the oxidation of In to In3+ (E0red = -0.34 V)

    For the oxidation of indium we need to reverse the sign to get E0ox instead of E0red. The calculations are:
    • Reaction 1: E0cell = +0.20 + +0.14 V = +0.34 V Therefore the reaction with Bi3+ and In is spontaneous.
    • Reaction 2: E0cell = +0.20 V + +0.34 V = +0.54V Therefore the reaction with Bi3+ and In is spontaneous.


    Will the oxidation of copper metal (Cu) by zinc ions (Zn2+) happen spontaneously in a redox reaction?

    Use the table1 to find equations with these substances in them:
    Cu+ + e- β‡Œ\enspace \rightleftharpoons \enspace Cu E0red = +0.52 V
    Cu2+ + 2e- β‡Œ\enspace \rightleftharpoons \enspace Cu E0red = +0.34 V
    Zn2+ + 2e- β‡Œ\enspace \rightleftharpoons \enspace Zn E0red = -0.76 V

    Reading the data we see a possibility for two reactions:
    • Reaction 1: The Cu / Cu+ equation is higher up the table than Zn/Zn2+. This means the reduction potential is higher for copper than it is for zinc. Therefore, oxidation of zinc (with the reduction of copper ions) would occur spontaneously but oxidation of copper metal would not occur spontaneously.
    • Reaction 2: The Cu / Cu2+ equation is higher on the table than Zn/Zn2+. Therefore like reaction 1 above the oxidation of copper would not occur spontaneously.

    Mathematically we can calculate the cell potential, E0cell:

    E0cell = E0red (reduction) + E0ox (oxidation)

    For β€œthe oxidation of copper metal (Cu) by zinc ions (Zn2+)” in the question, zinc will be reduced (use the E0red value above) and copper will be oxidized (reverse the sign in the E0red above to get E0ox) so the calculations would be:

    E0cell = E0red (reduction) + E0ox (oxidation)

    Reaction 1: E0cell = -0.76 V + +0.52 V = -0.24 V Therefore the reaction with Cu and Zn2+ is not spontaneous.

    Reaction 2: E0cell = -0.76 V + +0.34 V = -0.42 V Therefore the reaction with Cu and Zn2+ is not spontaneous.

  • We learned in Entropy and Gibbs free energy that Ξ”G\Delta G is a β€˜driving force’ for a reaction. It is a finite source of potential energy that drives a reaction towards equilibrium when Ξ”G\Delta G = 0. In other words, Ξ”G\Delta G pushes a reaction until there is an equal force for the forward and the reverse reaction.
    We saw the equation that relates standard Gibbs free energy change to a reaction’s equilibrium constant, K:

  • Ξ”G=βˆ’RTIn (K)\Delta G = - RTIn \, (K)

    • R = universal gas constant (8.314 J K-1 mol-1)
    • T = temperature (K)
    • K = the equilibrium constant

    There is a similar equation for cell potential and its relation to the standard Gibbs free energy change:

    Ξ”G0=βˆ’nFE0\Delta G^{0} = - nFE^{0}

    • n = the number of moles of electrons being transferred in the reaction.
    • F = the Faraday constant, which is 96485 C mol-1 (Coulombs per mole). This is the amount of charge a mole of electrons carries.
    • E0 = the standard cell potential. This is in volts or J C-1 so the units will cancel to joules.

    Do you notice the similarities? Both are expressions of Ξ”G\Delta G for mixtures at standard conditions showing which direction the reaction favours, in order to reach equilibrium. The only real difference is that the first is for for gases and the second is for aqueous species in a redox cell.
    • A negative E0 yields a positive Ξ”G0\Delta G^{0}, so the mixture will form more reactants in order to reach equilibrium, which has a greater quantity of reactants than products.
    • A positive E0 yields a negative Ξ”G0\Delta G^{0}, so the mixture will form more products in order to reach equilibrium, which has a greater quantity of products than reactants.

    You can therefore treat E0 like Ξ”G\Delta G: a driving force of a reaction towards equilibrium. To avoid having to use the constants, you can equate the two expressions above:

    βˆ’RTln (K)=βˆ’nFE0-RTln \, (K) = -nFE^{0}

    Let’s solve for E0cell using this:

    RTnF\large \frac{RT} {nF}In (K)=E0In \, (K) = E^{0}

    If you are at room temperature of 298 K, this expression for RT/F cancels out as:

    (8.314JK mol) (298 K)n (96485Cmol)\large \frac{(8.314 \frac{J} {K \, mol} ) \, (298 \, K )} {n \, (96485 \frac{C} {mol}) } In (K) In \, (K) \,   \, (8.314 J) (298)n (96485 C)\large \frac{(8.314 \, J) \, (298) } {n \, (96485 \, C) } In (K)=In \, (K) = 0.0257JCn\large \frac{0.0257 \frac{J} {C} } {n} In (K)=E0In \, (K) = E^{0}

    Recall that joules per coulomb is a volt, so:

    E0=E^{0} = 0.0257 Vn\large \frac{0.0257 \, V } {n }In (K)In \, (K)

    This can be used to find KK if you are given the standard cell potential.

  • Like with gas equilibria, we are normally at nonstandard conditions throughout the redox reaction!
    • In redox cells, standard conditions means there is an equal quantity of products and reactants, usually 1M concentration of both. These are the conditions for which we calculate E0E^{0}.
      Redox reactions will only be in this state briefly, probably at the beginning of the reaction when the solutions are prepared.

    Concentrations are changing over time as the system approaches equilibrium, so Ξ”G\Delta G and the cell potential are also both changing. We can use the following equation when not at standard conditions to show how cell potential changes as we approach equilibrium:

    Ξ”G=βˆ’nFE\Delta G = - nFE

    Consider the expression we used in Entropy and Gibbs free energy:

    Ξ”G=Ξ”G0+RTIn (Q)\Delta G = \Delta G^{0} + RTIn \, (Q)

    Because Ξ”G0=βˆ’nFE0\Delta G^{0}= -nFE^{0} and Ξ”G=βˆ’nFE \Delta G = -nFE, we can create the expression:

    βˆ’nFE=βˆ’nFE0+RTln (Q)-nFE = -nFE^{0} + RTln \, (Q)

    Dividing both sides of this expression by βˆ’nF-nF to yield:

    E=E0βˆ’RTnFln (Q)E = E^{0} - \frac{RT}{nF} ln \, (Q)

    This is the Nernst equation used to predict cell potential at nonstandard conditions.
    As shown above, if you are at 298K you can simplify it:

    E=E0+E = E^{0} + (8.314) (298)βˆ’n (96485)\large \frac{(8.314) \, (298) } {-n \, (96485) } In (Q) In \, (Q) \,  E=E0βˆ’ \, E = E^{0} - 0.0257n\large \frac{0.0257} {n} In (Q)In \, (Q)

    This is often derived further using the change of base rule (log10Q= (log_{10} Q = ln(Q)ln(10))\large \frac{ln(Q)} {ln(10)}) to convert ln(Q)ln(Q) into log(Q)log(Q):

    log10Q=log_{10} Q = ln(Q)ln(10)\large \frac{ln(Q)} {ln(10)}  \,  E=E0βˆ’ \, E = E^{0} - 0.0257β€‰βˆ—β€‰In (10)n \large \frac{0.0257 \, * \, In \, (10)} {n} log (Q) log \, (Q) \,  E=E0βˆ’ \, E = E^{0} - 0.0592n\large \frac{0.0592} {n} log (Q)log \, (Q)

    This is the most common form of the Nernst equation.

1 Source for standard potential data: ATKINS, P. W., & DE PAULA, J. (2006).?Atkins' Physical chemistry. Oxford, Oxford University Press.