Calculating cell potential (voltaic cells)

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 ‘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 E⁰_red values, the reaction is spontaneous if E⁰_cell is a positive value. E⁰_cell is worked out by:
  
  E⁰_cell = E⁰_red (reduction) + E⁰_ox (oxidation)
  Remember, the table ONLY gives E⁰_red values! Find E⁰_red and reverse the sign to get E⁰_ox.
  
  
  WORKED EXAMPLE 1:¹
  
  Will a redox reaction between Bi³⁺ and In occur spontaneously?
  First, use the table to find an equation with these compounds and their E⁰_red. Below are all the half-equations equations containing the substances in the question, taken from a standard potential table¹:
  
  Bi³⁺ + 3e^- $\enspace \rightleftharpoons \enspace$ BiE⁰_red = +0.20 V
  In⁺ + e^- $\enspace \rightleftharpoons \enspace$ InE⁰_red = -0.14 V
  In³⁺ + 3e^- $\enspace \rightleftharpoons \enspace$ InE⁰_red = -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 E⁰_cell 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:
  
  E⁰_cell = E⁰_red (reduction) + E⁰_ox (oxidation)
  The E⁰_cell 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 (E⁰_red = +0.20 V) with the oxidation of In to In+ (E⁰_red = -0.14 V)
  • Reaction 2: The reduction of Bi³⁺ to Bi (E⁰_red +0.20 V) with the oxidation of In to In³⁺ (E⁰_red = -0.34 V)
  
  For the oxidation of indium we need to reverse the sign to get E⁰_ox instead of E⁰_red. The calculations are:
  • Reaction 1: E⁰_cell = +0.20 + +0.14 V = +0.34 V Therefore the reaction with Bi³⁺ and In is spontaneous.
  • Reaction 2: E⁰_cell = +0.20 V + +0.34 V = +0.54V Therefore the reaction with Bi³⁺ and In is spontaneous.
  
  
  WORKED EXAMPLE 2:¹
  
  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^- $\enspace \rightleftharpoons \enspace$ Cu E⁰_red = +0.52 V
  Cu²⁺ + 2e^- $\enspace \rightleftharpoons \enspace$ Cu E⁰_red = +0.34 V
  Zn²⁺ + 2e^- $\enspace \rightleftharpoons \enspace$ Zn E⁰_red = -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, E⁰_cell:
  
  E⁰_cell = E⁰_red (reduction) + E⁰_ox (oxidation)
  For “the oxidation of copper metal (Cu) by zinc ions (Zn²⁺)” in the question, zinc will be reduced (use the E⁰_red value above) and copper will be oxidized (reverse the sign in the E⁰_red above to get E⁰_ox) so the calculations would be:
  
  E⁰_cell = E⁰_red (reduction) + E⁰_ox (oxidation)
  Reaction 1: E⁰_cell = -0.76 V + +0.52 V = -0.24 V Therefore the reaction with Cu and Zn²⁺ is not spontaneous.
  
  Reaction 2: E⁰_cell = -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 $\Delta G$ is a ‘driving force’ for a reaction. It is a finite source of potential energy that drives a reaction towards equilibrium when $\Delta G$ = 0. In other words, $\Delta G$ 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:

$\Delta G = - RTIn \, (K)$
Where:
• 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:

$\Delta G^{0} = - nFE^{0}$
Where:
• 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 $\Delta G$ 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 $\Delta G^{0}$, 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 $\Delta G^{0}$, 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 $\Delta G$: a driving force of a reaction towards equilibrium. To avoid having to use the constants, you can equate the two expressions above:

$-RTln \, (K) = -nFE^{0}$
Let’s solve for E⁰_cell using this:

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

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

$\large \frac{(8.314 \frac{J} {K \, mol} ) \, (298 \, K )} {n \, (96485 \frac{C} {mol}) }$ $In \, (K) \,$→$\,$ $\large \frac{(8.314 \, J) \, (298) } {n \, (96485 \, C) }$$In \, (K) =$ $\large \frac{0.0257 \frac{J} {C} } {n}$ $In \, (K) = E^{0}$
Recall that joules per coulomb is a volt, so:

$E^{0} =$ $\large \frac{0.0257 \, V } {n }$$In \, (K)$
This can be used to find $K$ 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 $E^{0}$.
  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 $\Delta G$ 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:

$\Delta G = - nFE$
Consider the expression we used in Entropy and Gibbs free energy:

$\Delta G = \Delta G^{0} + RTIn \, (Q)$
Because $\Delta G^{0}= -nFE^{0}$ and $\Delta G = -nFE$, we can create the expression:

$-nFE = -nFE^{0} + RTln \, (Q)$
Dividing both sides of this expression by $-nF$ to yield:

$E = E^{0} - \frac{RT}{nF} ln \, (Q)$
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:

$E = E^{0} +$ $\large \frac{(8.314) \, (298) } {-n \, (96485) }$ $In \, (Q) \,$→$\, E = E^{0} -$$\large \frac{0.0257} {n}$$In \, (Q)$
This is often derived further using the change of base rule $(log_{10} Q =$$\large \frac{ln(Q)} {ln(10)})$ to convert $ln(Q)$ into $log(Q)$:

$log_{10} Q =$ $\large \frac{ln(Q)} {ln(10)}$ $\,$→$\, E = E^{0} -$ $\large \frac{0.0257 \, * \, In \, (10)} {n}$ $log \, (Q) \,$→$\, E = E^{0} -$ $\large \frac{0.0592} {n}$ $log \, (Q)$
This is the most common form of the Nernst equation.