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.

__Notes:__- 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 H^{2}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 E
^{0}_{red}values,__the reaction is spontaneous if E__^{0}_{cell}__is a positive value. E__^{0}:_{cell}is worked out byE ^{0}_{cell}= E^{0}_{red}(reduction) + E^{0}_{ox}(oxidation)

Remember, the table ONLY gives E^{0}_{red}values!__Find E__^{0}_{red}and reverse the sign to get E^{0}._{ox}

WORKED EXAMPLE 1:^{1}

Will a redox reaction between Bi^{3+}and In occur spontaneously?

First, use the table to find an equation with these compounds and their E^{0}_{red}. Below are all the half-equations equations containing the substances in the question, taken from a standard potential table^{1}:

Bi^{3+}+ 3e^{-}$\enspace \rightleftharpoons \enspace$ BiE^{0}_{red}= +0.20 V

In^{+}+ e^{-}$\enspace \rightleftharpoons \enspace$ InE^{0}_{red}= -0.14 V

In^{3+}+ 3e^{-}$\enspace \rightleftharpoons \enspace$ InE^{0}_{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:__Bi__^{3+}__can be reduced into Bi__(Bi^{3+}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 In
^{3+}into In, which means__In can be oxidized into In__^{3+}.

^{0}_{cell}to lowest) that__Bi__^{3+}__is higher on the table than In. Therefore, the reduction of Bi__^{3+}__with the oxidation of In is a spontaneous reaction__.

To find this mathematically, we use the electrode potential expression:E ^{0}_{cell}= E^{0}_{red}(reduction) + E^{0}_{ox}(oxidation)

The E^{0}_{cell}must be a positive value for the redox reaction to be spontaneous. We have two possible reactions:- Reaction 1: The reduction of Bi
^{3+}to Bi (E^{0}_{red}= +0.20 V) with the oxidation of In to In+ (E^{0}_{red}= -0.14 V) - Reaction 2: The reduction of Bi
^{3+}to Bi (E^{0}_{red}+0.20 V) with the oxidation of In to In^{3+}(E^{0}_{red}= -0.34 V)

For the oxidation of indium we need to reverse the sign to get E^{0}_{ox}instead of E^{0}_{red}. The calculations are:- Reaction 1: E
^{0}_{cell}= +0.20 + +0.14 V = +0.34 V__Therefore the reaction with Bi__^{3+}__and In is spontaneous__. - Reaction 2: E
^{0}_{cell}= +0.20 V + +0.34 V = +0.54V__Therefore the reaction with Bi__^{3+}__and In is spontaneous__.

WORKED EXAMPLE 2:^{1}

Will the oxidation of copper metal (Cu) by zinc ions (Zn^{2+}) happen spontaneously in a redox reaction?

Use the table^{1}to find equations with these substances in them:

Cu^{+}+ e^{-}$\enspace \rightleftharpoons \enspace$ Cu E^{0}_{red}= +0.52 V

Cu^{2+}+ 2e^{-}$\enspace \rightleftharpoons \enspace$ Cu E^{0}_{red}= +0.34 V

Zn^{2+}+ 2e^{-}$\enspace \rightleftharpoons \enspace$ Zn E^{0}_{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/Zn^{2+}. 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 / Cu
^{2+}equation is higher on the table than Zn/Zn^{2+}. Therefore like reaction 1 above the__oxidation of copper would not occur spontaneously__.

Mathematically we can calculate the cell potential, E^{0}_{cell}:E ^{0}_{cell}= E^{0}_{red}(reduction) + E^{0}_{ox}(oxidation)

For “the oxidation of copper metal (Cu) by zinc ions (Zn^{2+})” in the question, zinc will be reduced (use the E^{0}_{red}value above) and copper will be oxidized (reverse the sign in the E^{0}_{red}above to get E^{0}_{ox}) so the calculations would be:E ^{0}_{cell}= E^{0}_{red}(reduction) + E^{0}_{ox}(oxidation)

Reaction 1: E^{0}_{cell}= -0.76 V + +0.52 V = -0.24 V__Therefore the reaction with Cu and Zn__^{2+}__is not spontaneous__.

Reaction 2: E^{0}_{cell}= -0.76 V + +0.34 V = -0.42 V__Therefore the reaction with Cu and Zn__^{2+}__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: - R = universal gas constant (8.314 J K
^{-1}mol^{-1}) - T = temperature (K)
- K = the equilibrium constant
- 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. - E
^{0}= the standard cell potential. This is in volts or J C^{-1}so the units will cancel to joules. - A negative E
^{0}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 E
^{0}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. - 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.

Where:

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

Where:

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.

You can therefore treat E

^{0}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:

Let’s solve for E

^{0}

_{cell}using this:

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

Recall that joules per coulomb is a volt, so:

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

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:

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

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

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

This is

__the__.

*Nernst equation*used to predict cell potential at nonstandard conditionsAs shown above,

__if you are at 298K you can simplify__it:

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 $log(Q)$:

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.