Relationship between Electrode Potential, Gibb’s Energy and Equilibrium Constant

An electrochemical cell is an apparatus used to generate electricity from a spontaneous redox reaction or, conversely, that uses electricity to drive a non-spontaneous redox reaction.

An electrochemical process is driven by an oxidation-reduction (redox) reaction in which electrons flow from one chemical substance to another. A redox reaction occurs when electrons are transferred from an oxidised substance to one that is being reduced. The substance that loses electrons and is oxidised in the electrochemical process is known as the reductant; however, the oxidant is the species that gains electrons and is reduced in the process. Because it is impossible to have a reduction without oxidation and vice versa, a redox reaction can be described as two half-reactions, one representing the oxidation process and one the reduction process. Let’s learn how this process is carried and establish a relationship between Electrode Potential, Gibb’s Energy, and Equilibrium Constant.

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 is the amount of energy available to do electrical work. This energy depends on both the cell potential and the total number of electrons transferred from the reductant to the oxidant during a reaction.

The resulting electric current is measured in coulombs \(({\rm{C}})\) It is the SI unit of charge that measures the number of electrons passing through a given point in \(1\,{\rm{s}}\) A coulomb is related to energy (in joules) and electrical potential (in volts) as per the given relation.

\({\rm{1}}\,{\rm{C}}\,{\rm{ = }}\,\frac{{1\,{\rm{J}}}}{{1{\rm{V}}}} = \,1{\rm{A}}{\rm{.s}}\)

Electric current is measured in amperes \({\rm{(A)}}\,{\rm{;1}}\,{\rm{A}}\,\) is defined as the flow of \(1\,{\rm{C}}\) of charge through a point in \(1\) second.

Multiplying the charge on the electron \((1.60218 \times {10^{ – 19}}{\rm{C)}}\) by Avogadro’s number \({\rm{(6}}{\rm{.02214}} \times {\rm{1}}{{\rm{0}}^{23}})\) gives us the charge on 1 mol of electrons, which is called the faraday \(({\rm{F}})\) named after the English physicist and chemist Michael Faraday \((1791\, – \,1867)\)
\(1{\rm{C}}\,{\rm{ = }}\,\frac{{1{\rm{J}}}}{{1{\rm{V}}}} = 1{\rm{A}}{\rm{.s}}\)

\({\rm{1F}}\,{\rm{ = }}\,{\rm{96,485C/mol}}\)

Therefore, the total charge transferred from the reductant to the oxidant is \({\rm{nF}}\) where n is the number of moles of electrons.

\({{\rm{E}}^ \circ }_{{\rm{cell}}}\): Standard Cell Potential

\({{\rm{E}}^ \circ }_{{\rm{cell}}}\) is the cell potential or the electromotive force (also called cell voltage) between two half-cells. The greater the \({{\rm{E}}^0}_{{\rm{cell}}}\) of a reaction, the more likely the reaction will proceed (more spontaneous). \({{\rm{E}}^0}_{{\rm{cell}}}\)  is measured in volts \(({\rm{V}})\)

The overall voltage of the cell \( = \) the half-cell potential of the reduction reaction + the half-cell potential of the oxidation reaction. To simplify,

\({{\rm{E}}^0}_{{\rm{cell}}} = {{\rm{E}}^0}_{{\rm{reduction}}} – {{\rm{E}}^0}_{{\rm{oxidation}}}\)

The negative of the potential for a reduction potential (gain of electrons) is the potential of an oxidation-reduction (loss of electron).

The more positive reduction potential of reduction reactions is more spontaneous. In a cell reduction potential table, the higher the cell is on the table, the higher potential it has as an oxidising agent.

\({\rm{\Delta G}}\): Gibbs Energy

The change in Gibbs (free) energy for a system is represented by \({\rm{\Delta G}}\) and \({\rm{\Delta }}{{\rm{G}}^0}\)is the change in Gibbs energy for a system under standard conditions \((1\,{\rm{atm,}}\,{\rm{298\, K)}}\) On an energy diagram \({\rm{\Delta G}}\) can be represented as follows:

Here, \({\rm{\Delta G}}\) is the difference in the energy between the reactants and the products. \({\rm{\Delta G}}\) is unaffected by external factors such as a change in temperature or the presence of a catalyst that changes the kinetics of the reaction. 

K: The Equilibrium Constant

\({\rm{K}}\) represents the equilibrium constant of a general reaction. It expresses the ratio between products and reactants of a reaction at equilibrium. For example, let us consider a chemical reaction that can be represented as-

The equilibrium constant for the above reaction can be given by-

\({\rm{K}}\,{\rm{ = }}\,\frac{{{{[{\rm{C}}]}^{\rm{c}}}{{[{\rm{D}}]}^{\rm{d}}}}}{{{{[{\rm{A}}]}^{\rm{a}}}{{[{\rm{B}}]}^{\rm{b}}}}}\)

Difference between \({{\rm{E}}_{{\rm{cell}}}}\) and \({{\rm{E}}^0}_{{\rm{cell}}}\)

\({{\rm{E}}^0}_{{\rm{cell}}}\) is the standard state cell potential, whose value is determined under standard states that include a concentration of 1 Molar (mole per liter) and an atmospheric pressure of 1 atm. However,\({{\rm{E}}_{{\rm{cell}}}}\) is the non-standard cell potential, which is not determined under a concentration of 1 Molar and pressure of \(1\,{\rm{atm}}\). The two are closely related, as shown below.

\({{\rm{E}}_{{\rm{cell}}}} = {{\rm{E}}^0}_{{\rm{cell}}} – \frac{{{\rm{RT}}}}{{{\rm{nF}}}}\ln {\rm{K}}\)

Other simplified forms of the equation that we typically see:

\({{\rm{E}}_{{\rm{cell}}}} = {{\rm{E}}^0}_{{\rm{cell}}} – \frac{{0.0257}}{{\rm{n}}}\ln {\rm{K}}\)

or in terms of \({\log _{10}}\)

\({{\rm{E}}_{{\rm{cell}}}} = {{\rm{E}}^0}_{{\rm{cell}}} – \frac{{0.0592}}{{\rm{n}}}{\log _{10}}{\rm{K}}\)

Relationship Between the Three

The work done due to electrical energy is equal to the product of charge and e.m.f. of the cell \(\left( {{{\rm{E}}_{{\rm{cell}}}}} \right)\)

\({\rm{W}} = ({\rm{electrical}}\,{\rm{work}}\,{\rm{done}}) =  – {{\rm{E}}_{{\rm{cell}}}} \times {\rm{q}}\)

The negative sign indicates that work is done by the system present in the surrounding. If n is the number of moles of electrons transferred in the redox reaction of the cell, then the quantity of electric charge that flows through the external circuit is given by the equation.

\({\rm{q}}\,{\rm{ = }}\,{\rm{nF}}\)

\(\therefore \,{\rm{W}}\,{\rm{ = }}\,{\rm{(electrical work done)}}\,{\rm{ = }}\, – {\rm{nF}}\,{{\rm{E}}_{{\rm{cell}}}}\)

We know that:

\(\Delta {\rm{G}}\, = \, – {\rm{W}}({\rm{non}}\, – \,{\rm{expansion)}}\)

As in this case, the non-expansion work done is electrical work done. Therefore,

\(\Delta {\rm{G}}{\mkern 1mu} {\rm{ = }}{\mkern 1mu} {\rm{W}}({\rm{electrical}}{\mkern 1mu} {\rm{work}}{\mkern 1mu} {\rm{done}}){\rm{ = }}{\mkern 1mu} \, – {\rm{nF}}{\mkern 1mu} {{\rm{E}}_{{\rm{cell}}}}\)

For comparing different cells, standard cell potentials \(({{\rm{E}}^0}_{{\rm{cell}}})\)

The corresponding energy charge is called standard energy \((\Delta {{\rm{G}}^0})\) Thus we can write-

\(\Delta {{\rm{G}}^0} =  – {\rm{nF}}{{\rm{E}}^0}_{{\rm{cell}}}\)

The relationship between \(\Delta {\rm{G}},{\mkern 1mu} {\rm{K}}{\mkern 1mu}\) and \({{\rm{E}}^0}_{{\rm{cell}}}\) can be represented by the following diagram.

Where,

1. \({\rm{R}} = 8.314\,{\rm{J}}\,{\rm{mol}}\,{{\rm{C}}^{ – 1}}\)
2. \({\rm{T = }}\,{\rm{Temp}}\,{\rm{in}}\,{\rm{K}}\)
3. \({\rm{n}}\,{\rm{ = }}\) moles of \({{\rm{e}}^ – }\) from balanced redox reaction
4. \({\rm{F}}\, = \) Faraday’s constant \( = \,96,\,485\,{\rm{C/mol}}\)

The connection between cell potential, Gibbs energy, and equilibrium constants are directly related in the following equation:

\(\Delta {{\rm{G}}^0} =  – {\rm{RT}}\ln {\rm{K = }}\,{\rm{ – nF}}\,{{\rm{E}}^0}_{{\rm{cell}}}\)

Solved Examples Questions

Q.1. Given \({\rm{K}}\, = \,2.81\, \times \,{10^{ – 16}}\)  for the following reaction:

Find \(\Delta {{\rm{G}}^0}\)
Ans: Use the following formula:
\(\Delta {{\rm{G}}^0} = \, – \,{\rm{RT}}\,{\rm{lnK}}\)
\({\rm{ = }}\,{\mkern 1mu} {\rm{ – }}\left( {{\rm{8}}.{\rm{314}}} \right)\left( {{\rm{298K}}} \right){\rm{ln}}\,{\rm{K}}\left( {{\rm{2}}.{\rm{81}} \times {\rm{1}}{{\rm{0}}^{ – 16}}} \right)\)
\( =  – 8.87 \times {10^5}{\rm{J/mol}}\)

Q.2. Given the \({{\rm{E}}^{\rm{0}}}_{{\rm{cell}}}\) for the reaction

is \({\rm{ – 0}}{\rm{.34}}\,{\rm{V}}\) find the equilibrium constant \(\left( {\rm{K}} \right)\) for the reaction.
Solution:
Step 1: Split into two half-reactions.

Step 2: Balance the half-reactions with charges to determine \({\rm{n}}\).

Therefore \({\rm{n = 2}}\)
Step 3: Substitute the values in the following equation

\({\rm{ – RTlnK = }}\,{\rm{ – nF}}{{\rm{E}}^{\rm{0}}}_{{\rm{cell}}}\)

\({{\rm{E}}^{\rm{0}}}_{{\rm{cell}}}{\rm{ = }}\frac{{{\rm{RTlnK}}}}{{{\rm{nF}}}}\)

\({\rm{ – 0}}{\rm{.34V = }}\frac{{\left( {{\rm{0}}{\rm{.0257}}} \right){\rm{lnK}}}}{{\rm{2}}}\)

\({\rm{K = exp}}\left( {\frac{{{\rm{ – 0}}{\rm{.34V\; \times 2}}}}{{{\rm{0}}{\rm{.0257}}}}} \right){\rm{ = \;3}}{\rm{.19\; \times 1}}{{\rm{0}}^{{\rm{ – 12}}}}\)

Summary

A coulomb \(({\rm{C}})\)  relates electrical potential (volts) and energy (joules). The current generated from a redox reaction is measured in amperes \(({\rm{A}})\) The product of the cell potential and the total charge is the maximum amount of energy available to do work, which is related to the change in energy \((\Delta {\rm{G}})\) that occurs during an electrochemical process.\(\Delta {\rm{G}}\) for the overall reaction is proportional to both the cell potential and the number of electrons \(({\rm{n}})\) transferred. The equilibrium constant \({\rm{K}}\)  is related to \(\Delta {\rm{G}}\)  hence, \({{\rm{E}}^0}_{{\rm{cell}}}\) and \({\rm{K}}\) are also related. Large equilibrium constants correspond to large positive values of \({{\rm{E}}^0}_{{\rm{cell}}}\)

FAQs on Relationship b/w Electrode Potential, Gibb’s Energy, Equilibrium Constant

Q.1. What is the relationship between Gibbs energy and equilibrium constant?
Ans: The relationship between Gibbs energy \((\Delta {{\rm{G}}^0})\) and equilibrium constant \({\rm{(K)}}\) is as follows:
\(\Delta {{\rm{G}}^0} =  – {\rm{RT}}\ln {\rm{K}}\)

Q.2. What is the relationship between \({\rm{K}}\) and \({{\rm{E}}^0}_{{\rm{cell}}}\)?
Ans: The relationship between \({\rm{K}}\) and \({{\rm{E}}^0}_{{\rm{cell}}}\) is given as-
\( – {\rm{RT}}\ln {\rm{K}}\,{\rm{ = }}\,{\rm{ – nF}}{{\rm{E}}^0}_{{\rm{cell}}}\)

Q.3. What is the difference between \({{\rm{E}}_{{\rm{cell}}}}{\mkern 1mu}\) and \({{\rm{E}}^0}_{{\rm{cell}}}\)?
Ans: \({{\rm{E}}_{{\rm{cell}}}}\) means electrode potential of a cell. \({{\rm{E}}^0}_{{\rm{cell}}}\) means standard electrode potential which is the electrode potential measured at 1 atmosphere pressure, 1 molar solution at \(25\,^\circ {\rm{C}}\)

Q.4. Can \({{\rm{E}}^0}_{{\rm{cell}}}\) and \(\Delta {{\rm{G}}^0}\)  become equal to zero?
Ans: If a reaction is proceeding in a forward or backward direction \({{\rm{E}}^0}_{{\rm{cell}}}\) and \(\Delta {{\rm{G}}^0}\) cannot be zero. However, if \({{\rm{E}}^0}_{{\rm{cell}}}\) and \(\Delta {{\rm{G}}^0}\) are zero then there is no reaction taking place.

Q.5. Is a reaction spontaneous when Delta \({\rm{G}}\) is \({\rm{0}}\)?
Ans: If \(\Delta {\rm{G}}\,{\rm{ < }}\,{\rm{0}}\) the process occurs spontaneously. If \(\Delta {\rm{G}}\,{\rm{ = }}\,{\rm{0}}\)  the system is at equilibrium. If \(\Delta {\rm{G}}\,{\rm{ > }}\,{\rm{0}}\) the process is non-spontaneous, however it occurs spontaneously in the reverse direction.

Study Applications of Equilibrium Constant Here

We hope this article has helped you in understanding the concept of the relationship between Electrode Potential, Gibb’s Energy, and Equilibrium Constant. If you have any questions related to this post, reach us through the comment box below and we will get back to you as soon as possible.