19.5: Cell Potential, Gibbs Energy, And The Equilibrium Constant

The Relationship between Cell Potential & Gibbs Energy

Electrochemical cells convert chemical energy to electrical energy and vice versa. The total amount of energy produced by an electrochemical cell, and thus the amount of energy available to do electrical work, depends on both the cell potential and the total number of electrons that are transferred from the reductant to the oxidant during the course of a reaction. The resulting electric current is measured in coulombs (C), an SI unit that measures the number of electrons passing a given point in 1 s. A coulomb relates energy (in joules) to electrical potential (in volts). Electric current is measured in amperes (A); 1 A is defined as the flow of 1 C/s past a given point (1 C = 1 A·s):

\[\dfrac{\textrm{1 J}}{\textrm{1 V}}=\textrm{1 C}=\mathrm{A\cdot s} \label{20.5.1} \nonumber \]

In chemical reactions, however, we need to relate the coulomb to the charge on a mole of electrons. Multiplying the charge on the electron by Avogadro’s number gives us the charge on 1 mol of electrons, which is called the faraday (F), named after the English physicist and chemist Michael Faraday (1791–1867):

\[\begin{align*} F &=(1.60218\times10^{-19}\textrm{ C})\left(\dfrac{6.02214 \times 10^{23}\, J}{\textrm{1 mol e}^-}\right) \\[4pt] &=9.64833212\times10^4\textrm{ C/mol e}^- \\[4pt] &\simeq 96,485\, J/(\mathrm{V\cdot mol\;e^-})\end{align*} \]

The total charge transferred from the reductant to the oxidant is therefore \(nF\), where \(n\) is the number of moles of electrons.

Michael Faraday (1791–1867)

Faraday was a British physicist and chemist who was arguably one of the greatest experimental scientists in history. The son of a blacksmith, Faraday was self-educated and became an apprentice bookbinder at age 14 before turning to science. His experiments in electricity and magnetism made electricity a routine tool in science and led to both the electric motor and the electric generator. He discovered the phenomenon of electrolysis and laid the foundations of electrochemistry. In fact, most of the specialized terms introduced in this chapter (electrode, anode, cathode, and so forth) are due to Faraday. In addition, he discovered benzene and invented the system of oxidation state numbers that we use today. Faraday is probably best known for “The Chemical History of a Candle,” a series of public lectures on the chemistry and physics of flames.

The maximum amount of work that can be produced by an electrochemical cell (\(w_{max}\)) is equal to the product of the cell potential (\(E^°_{cell}\)) and the total charge transferred during the reaction (\(nF\)):

\[ w_{max} = nFE_{cell} \label{20.5.3} \nonumber \]

Work is expressed as a negative number because work is being done by a system (an electrochemical cell with a positive potential) on its surroundings.

The change in free energy (\(\Delta{G}\)) is also a measure of the maximum amount of work that can be performed during a chemical process (\(ΔG = w_{max}\)). Consequently, there must be a relationship between the potential of an electrochemical cell and \(\Delta{G}\); this relationship is as follows:

\[\Delta{G} = −nFE_{cell} \label{20.5.4} \nonumber \]

A spontaneous redox reaction is therefore characterized by a negative value of \(\Delta{G}\) and a positive value of \(E^°_{cell}\), consistent with our earlier discussions. When both reactants and products are in their standard states, the relationship between ΔG° and \(E^°_{cell}\) is as follows:

\[\Delta{G^°} = −nFE^°_{cell} \label{20.5.5} \]

A spontaneous redox reaction is characterized by a negative value of ΔG°, which corresponds to a positive value of E°cell.

Example \(\PageIndex{1}\)

Suppose you want to prepare elemental bromine from bromide using the dichromate ion as an oxidant. Using the data in Table P2, calculate the free-energy change (ΔG°) for this redox reaction under standard conditions. Is the reaction spontaneous?

Given: redox reaction

Asked for: \(ΔG^o\) for the reaction and spontaneity

Strategy:

1. From the relevant half-reactions and the corresponding values of \(E^o\), write the overall reaction and calculate \(E^°_{cell}\).
2. Determine the number of electrons transferred in the overall reaction. Then use Equation \ref{20.5.5} to calculate \(ΔG^o\). If \(ΔG^o\) is negative, then the reaction is spontaneous.

Solution

A

As always, the first step is to write the relevant half-reactions and use them to obtain the overall reaction and the magnitude of \(E^o\). From Table P2, we can find the reduction and oxidation half-reactions and corresponding \(E^o\) values:

\[\begin{align*} & \textrm{cathode:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{14H^+(aq)}+\mathrm{6e^-}\rightarrow \mathrm{2Cr^{3+}(aq)} +\mathrm{7H_2O(l)} &\quad & E^\circ_{\textrm{cathode}} =\textrm{1.36 V} \\ & \textrm{anode:} &\quad & \mathrm{2Br^{-}(aq)} \rightarrow \mathrm{Br_2(aq)} +\mathrm{2e^-} &\quad & E^\circ_{\textrm{anode}} =\textrm{1.09 V} \end{align*} \nonumber \]

To obtain the overall balanced chemical equation, we must multiply both sides of the oxidation half-reaction by 3 to obtain the same number of electrons as in the reduction half-reaction, remembering that the magnitude of \(E^o\) is not affected:

\[\begin{align*} & \textrm{cathode:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{14H^+(aq)}+\mathrm{6e^-}\rightarrow \mathrm{2Cr^{3+}(aq)} +\mathrm{7H_2O(l)} &\quad & E^\circ_{\textrm{cathode}} =\textrm{1.36 V} \\[4pt] & \textrm{anode:} &\quad & \mathrm{6Br^{-}(aq)} \rightarrow \mathrm{3Br_2(aq)} +\mathrm{6e^-} &\quad & E^\circ_{\textrm{anode}} =\textrm{1.09 V} \\[4pt] \hline & \textrm{overall:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{6Br^{-}(aq)} + \mathrm{14H^+(aq)} \rightarrow \mathrm{2Cr^{3+}(aq)} + \mathrm{3Br_2(aq)} +\mathrm{7H_2O(l)} &\quad & E^\circ_{\textrm{cell}} =\textrm{0.27 V} \end{align*} \nonumber \]

B

We can now calculate ΔG° using Equation \ref{20.5.5}. Because six electrons are transferred in the overall reaction, the value of \(n\) is 6:

\[\begin{align*}\Delta G^\circ &=-(n)(F)(E^\circ_{\textrm{cell}}) \\[4pt] & =-(\textrm{6 mole})[96,485\;\mathrm{J/(V\cdot mol})(\textrm{0.27 V})] \\& =-15.6 \times 10^4\textrm{ J} \\ & =-156\;\mathrm{kJ/mol\;Cr_2O_7^{2-}} \end{align*} \]

Thus \(ΔG^o\) is −168 kJ/mol for the reaction as written, and the reaction is spontaneous.

Exercise \(\PageIndex{1}\)

Use the data in Table P2 to calculate \(ΔG^o\) for the reduction of ferric ion by iodide:

\[\ce{2Fe^{3+}(aq) + 2I^{−}(aq) → 2Fe^{2+}(aq) + I2(s)}\nonumber \]

Is the reaction spontaneous?

Answer

−44 kJ/mol I2; yes