Engineering Chemistry. Module 1, electrochemistry and corrosion science.

Electrochemical cells and the Daniell cell

Electrochemistry

Electrochemistry is the branch of chemistry that studies the interconversion of chemical energy and electrical energy. It covers both spontaneous chemical reactions that generate an electric current and non-spontaneous chemical reactions driven by an externally applied electric current.

Electrochemical device

An electrochemical device (or cell) consists of two electronic conductors (electrodes) immersed in one or more ionic conductors (electrolytes).

In broad scientific usage, electrochemical cells are divided into two categories:

1. Galvanic (or voltaic) cells
2. Electrolytic cells

(Note: In many engineering syllabi, the term electrochemical cell is used specifically as a synonym for a galvanic cell).

Galvanic cell vs electrolytic cell

Watch out. Never define the anode or cathode by its electrical sign ($+$ or $-$). Always define electrodes by the chemical reaction occurring at their surface: oxidation always occurs at the anode, and reduction always occurs at the cathode, regardless of whether the cell is galvanic or electrolytic. The electrical polarity flips between cell types because the anode is the source of electrons in a galvanic cell (making it negative), whereas an external power supply pulls electrons away from the anode in an electrolytic cell (making it positive).

Oxidation, reduction and electrolytes

Before studying electrochemical cells, review three foundational concepts:

Oxidation

Oxidation is the loss of electrons by a chemical species (leading to an increase in oxidation state).

Reduction

Reduction is the gain of electrons by a chemical species (leading to a decrease in oxidation state).

Ions and electrolytes

Ions are electrically charged atoms or groups of atoms (cations are positively charged; anions are negatively charged). Electrolytes are substances that produce mobile ions when dissolved in a polar solvent or melted, enabling ionic conduction.

Daniell cell construction and operation

The Daniell cell is a classic galvanic cell that converts the chemical energy of the spontaneous zinc–copper redox reaction into electrical energy.

Construction

1. Anode half-cell: A solid zinc rod ($\text{Zn}$) is immersed in an aqueous solution of zinc sulfate ($\text{ZnSO}_4$).
2. Cathode half-cell: A solid copper rod ($\text{Cu}$) is immersed in an aqueous solution of copper sulfate ($\text{CuSO}_4$).
3. Internal connection: The two electrolyte solutions are connected by an inverted U-tube called a salt bridge.
4. External circuit: The zinc and copper electrodes are connected externally through metallic wires to a voltmeter or electrical load.

Electrode reactions and standard cell potential

In the zinc half-cell, metallic zinc has a stronger tendency to lose electrons than copper. Zinc atoms undergo oxidation at the anode, releasing $\text{Zn}^{2+}$ ions into the solution and leaving electrons behind on the zinc rod (making it the negative terminal):

The released electrons travel through the external wire to the copper rod. In the copper half-cell, $\text{Cu}^{2+}$ ions from the $\text{CuSO}_4$ solution accept these electrons at the surface of the copper electrode and deposit as metallic copper (making the copper rod the positive cathode):

Adding the two half-reactions gives the net spontaneous cell reaction:

Under standard conditions (unit ion activities, often approximated as $1\text{ M}$ solutions, at $298\text{ K}$), the standard reduction potentials are $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\text{ V}$ and $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\text{ V}$. The standard cell electromotive force (EMF) is:

Conventional cell notation

By IUPAC convention, a galvanic cell is represented in a single line by placing the anode half-cell on the left and the cathode half-cell on the right. A single vertical line ($\mid$) represents a phase boundary between an electrode and its electrolyte, and a double vertical line ($\parallel$) represents a salt bridge:

In ionic shorthand, this is written as:

Direction of electron flow vs conventional current

• Electron flow: Electrons flow through the external metallic wire from the zinc anode ($-$) to the copper cathode ($+$).
• Conventional electric current: By definition, conventional current flows in the direction opposite to electron movement—that is, from the copper cathode ($+$) to the zinc anode ($-$) in the external circuit.

Watch out. Electrons never swim through the aqueous electrolyte or the salt bridge. Current in the external wire is carried entirely by electrons, whereas current inside the half-cells and salt bridge is carried entirely by the migration of ions.

Salt bridge composition and functions

A salt bridge is an inverted U-shaped glass tube filled with a concentrated solution of an inert electrolyte—such as potassium chloride ($\text{KCl}$), potassium nitrate ($\text{KNO}_3$), or ammonium nitrate ($\text{NH}_4\text{NO}_3$)—gelled with agar-agar or gelatine to prevent mechanical mixing of the liquids.

An electrolyte is chosen such that its cation and anion have nearly equal ionic mobilities (for example, $\text{K}^+$ and $\text{Cl}^-$ move at almost the same speed under an electric field) and do not react chemically with any species in the half-cells.

How ion migration maintains neutrality

As the Daniell cell operates, $\text{Zn}^{2+}$ ions accumulate in the anode compartment, creating a net positive charge. Simultaneously, $\text{Cu}^{2+}$ ions are consumed in the cathode compartment, leaving behind an excess of negatively charged sulfate ions ($\text{SO}_4^{2-}$). If unchecked, this charge build-up would immediately halt electron flow. To prevent this, anions (such as $\text{Cl}^-$) migrate from the salt bridge into the anode compartment, while cations (such as $\text{K}^+$) migrate into the cathode compartment.

Four essential functions of a salt bridge

1. Completes the electrical circuit: It provides an internal ionic pathway between the two half-cells so charge can circulate continuously.
2. Maintains electrical neutrality: By supplying anions to the anode half-cell and cations to the cathode half-cell, it prevents charge accumulation in either solution.
3. Prevents mechanical mixing and interfering diffusion: The gel limits bulk mixing and slows interdiffusion while still allowing ions to move; it is not a perfectly selective membrane.
4. Minimizes liquid junction potential: When two different electrolyte solutions come into direct contact, unequal diffusion rates of cations and anions across the boundary create an unwanted potential difference called the liquid junction potential. Because the salt bridge ions ($\text{K}^+$ and $\text{Cl}^-$) have nearly identical transport velocities, the junction potential is reduced; in practice it need not be exactly zero.

Cell reversibility

Reversible cell

A reversible cell operates at near-equilibrium conditions and satisfies three strict criteria when connected to an opposing external EMF ($E_{\text{ext}}$):

1. When $E_{\text{ext}} = E_{\text{cell}}$: If the opposing external voltage exactly equals the cell EMF, no current flows and the net cell reaction is zero, although microscopic forward and reverse processes can continue.
2. When $E_{\text{ext}} < E_{\text{cell}}$: If the opposing voltage is infinitesimally smaller than the cell EMF, an infinitesimal current flows from the cell, driving the spontaneous forward reaction ($\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$).
3. When $E_{\text{ext}} > E_{\text{cell}}$: If the opposing voltage is infinitesimally greater than the cell EMF, current flows in the opposite direction and the exact reverse chemical reaction occurs ($\text{Zn}^{2+} + \text{Cu} \rightarrow \text{Zn} + \text{Cu}$).

The Daniell cell is a standard example of a reversible cell when operated with very small currents.

Irreversible cell

An irreversible cell does not satisfy these conditions; applying an external voltage slightly greater than its EMF does not restore the original reactants, often because side reactions (such as gas evolution) occur at the electrodes. A common example is a primary zinc–carbon dry cell (or placing zinc and copper strips in dilute $\text{H}_2\text{SO}_4$, where $\text{H}_2$ gas escapes irreversibly).

Electrolysis of molten aluminium oxide

To contrast an electrolytic cell with a galvanic cell, consider the electrolysis of molten aluminium oxide ($\text{Al}_2\text{O}_3$). When $\text{Al}_2\text{O}_3$ is melted, it dissociates into mobile aluminium cations ($\text{Al}^{3+}$) and oxide anions ($\text{O}^{2-}$).

In the simplified textbook model shown above, two inert electrodes are dipped into a single bath of molten $\text{Al}_2\text{O}_3$ and connected to a direct-current (DC) power supply:

• At the negative cathode: The DC source pumps electrons into the cathode, making it negatively charged. Positively charged $\text{Al}^{3+}$ cations migrate to this negative electrode, gain electrons, and undergo reduction to form molten aluminium metal:

• At the positive anode: The DC source draws electrons away from the anode, making it positively charged. Negatively charged $\text{O}^{2-}$ anions migrate to this positive electrode, lose electrons, and undergo oxidation to evolve oxygen gas:

Distinction from industrial extraction (Hall–Héroult process)

The diagram above represents an idealized inert-electrode model useful for understanding electrode polarity and ion migration. In actual industrial aluminium smelting, pure $\text{Al}_2\text{O}_3$ (melting point $\approx 2072\ ^\circ\text{C}$) is dissolved in molten cryolite ($\text{Na}_3\text{AlF}_6$) to lower the operating temperature to about $950\ ^\circ\text{C}$, and consumable carbon (graphite) anodes are used instead of inert electrodes. At those high temperatures, the oxygen produced at the anode reacts with the carbon anode to form carbon monoxide and carbon dioxide ($\text{C} + \text{O}_2 \rightarrow \text{CO}_2$), requiring periodic replacement of the anode blocks.

Self-check questions

1. During the operation of a Daniell cell, which way do $\text{Cl}^-$ ions in a $\text{KCl}$ salt bridge migrate, and why? Answer: $\text{Cl}^-$ ions migrate into the anode ($\text{Zn}/\text{ZnSO}_4$) compartment. As zinc oxidizes to $\text{Zn}^{2+}$ ions, positive charge builds up in the anode solution; negatively charged $\text{Cl}^-$ ions enter from the salt bridge to maintain electrical neutrality.

2. Why is the anode negative in a Daniell cell but positive during the electrolysis of molten $\text{Al}_2\text{O}_3$? Answer: In both cells, the anode is the site of oxidation. In the spontaneous Daniell cell, $\text{Zn}$ atoms oxidize and leave electrons behind on the anode, giving it a negative charge. In the non-spontaneous $\text{Al}_2\text{O}_3$ electrolytic cell, the anode is connected to the positive terminal of an external DC power supply, which actively pulls electrons away from the electrode so $\text{O}^{2-}$ ions can be oxidized.

3. What happens to the reaction in a Daniell cell (standard $E_{\text{cell}} = 1.10\text{ V}$) if an external opposing voltage of $1.15\text{ V}$ is applied across its terminals? Answer: In the ideal reversible limit, $E_{\text{ext}} > E_{\text{cell}}$, so the cell functions as an electrolytic cell and the net reaction reverses: copper oxidizes at the copper electrode ($\text{Cu} \rightarrow \text{Cu}^{2+} + 2\text{e}^-$) and zinc ions reduce and plate onto the zinc electrode ($\text{Zn}^{2+} + 2\text{e}^- \rightarrow \text{Zn}$). Practical cells also require allowance for resistance, overpotential and competing reactions.

Module study map

Electrode potential and the electrochemical series

Every electrochemical cell relies on the charge separation that occurs where an electronic conductor meets an ionic solution. Understanding how a single electrode develops a potential, how that potential is standardized against a reference, and how tabulated values are organized into the electrochemical series allows engineers to predict cell voltages, material compatibility, and redox spontaneity before building a system.

Origin of single electrode potential and the electrical double layer

When a metal strip $\mathrm{M}$ is immersed in an aqueous solution containing its own ions $\mathrm{M}^{n+}$, two opposing processes compete at the solid–liquid interface:

1. Oxidation (dissolution tendency): Metal atoms leave the crystal lattice as hydrated cations, leaving electrons behind on the metal surface:

   $$\mathrm{M(s) \longrightarrow M^{n+}(aq) + n\,e^-}$$

2. Reduction (deposition tendency): Metal cations from the solution collide with the electrode, accept electrons, and deposit as neutral metal atoms:

   $$\mathrm{M^{n+}(aq) + n\,e^- \longrightarrow M(s)}$$

As charge is transferred, the interfacial electric field changes and opposes further net transfer. In an ideal reversible half-cell at open circuit, oxidation and reduction approach equal rates. A charged metal surface and nearby counter-ions form an electrical double layer. The simplified zinc/copper drawing below illustrates this idea; the actual surface charge depends on the electrode and solution conditions. A positive or negative potential quoted relative to the SHE does not, by itself, specify the absolute sign of the metal surface charge.

Figure: Charge separation and formation of the Helmholtz electrical double layer at zinc and copper electrode interfaces.

Helmholtz electrical double layer

The Helmholtz electrical double layer is an interfacial structure established by electrostatic attraction and thermal motion, consisting of a plane of charge on the metal surface facing a layer of oppositely charged ions in the solution. The resulting electric field opposes further net charge transfer, and interfacial equilibrium can be established in a reversible half-cell:

Single electrode potential

Single electrode potential ($E$) is the potential difference established across this equilibrium double layer between the metal and the bulk electrolyte. Its magnitude depends on:

• The chemical nature of the metal and its ions.
• The activity (effective concentration) of the ions in solution.
• The temperature of the system.
• The pressure, if a gaseous species participates in the electrode reaction.

Standard electrode potential and the IUPAC reduction convention

It is physically impossible to measure the absolute potential of an isolated half-cell because placing a second measuring probe into the electrolyte creates a second metal–solution interface with its own unknown potential. Voltmeters can only measure a potential difference between two electrodes. Therefore, every single electrode potential is reported as a relative value measured against a universally agreed reference: the Standard Hydrogen Electrode (SHE), which is assigned a potential of exactly $0.00\text{ V}$ at all temperatures.

Standard electrode potential

Standard electrode potential ($E^\circ$) is the potential of a half-cell measured relative to the SHE when all participating species are in their standard states:

• Solute species at unit activity ($a = 1$), which is routinely approximated in introductory calculations as a molar concentration of $1\text{ M}$.
• Gases at a partial pressure of $1\text{ bar}$ ($10^5\text{ Pa}$, very close to $1\text{ atm}$).
• Pure solids and pure liquids in their most stable physical forms.
• A specified temperature, conventionally $298\text{ K}$ ($25\,^\circ\text{C}$) when numerical tables are quoted.

By IUPAC convention, all standard electrode potentials are tabulated as standard reduction potentials ($E^\circ_{\text{red}}$). If a half-reaction is reversed to represent oxidation, the magnitude of its potential remains the same, but its algebraic sign is inverted:

When two half-cells are combined into an electrochemical cell, the standard cell electromotive force ($E^\circ_{\text{cell}}$) is calculated directly from the standard reduction potentials of the two electrodes:

Here, both $E^\circ_{\text{cathode}}$ and $E^\circ_{\text{anode}}$ must be substituted as reduction potentials without flipping their signs beforehand.

Watch out. Electrode potential ($E^\circ$) is an intensive property—it represents energy per unit charge ($1\text{ V} = 1\text{ J C}^{-1}$). Never multiply $E^\circ$ by stoichiometric coefficients when balancing an overall cell reaction. For example, in the reaction $\mathrm{Zn(s) + 2\,Ag^+(aq) \longrightarrow Zn^{2+}(aq) + 2\,Ag(s)}$, the silver half-reaction is doubled to balance electrons, but $E^\circ_{\mathrm{Ag^+/Ag}}$ remains $+0.80\text{ V}$, giving $E^\circ_{\text{cell}} = +0.80 - (-0.76) = +1.56\text{ V}$.

Electrochemical series

The electrochemical series is the arrangement of metallic and non-metallic redox couples in increasing order of their standard reduction potentials ($E^\circ$ at $298\text{ K}$).

In this arrangement, species with large negative $E^\circ$ values resist reduction; their reduced forms (such as $\mathrm{Li(s)}$ and $\mathrm{Mg(s)}$) lose electrons easily and act as strong reducing agents. Conversely, species with large positive $E^\circ$ values have a strong tendency to gain electrons; their oxidized ions (such as $\mathrm{Ag^+(aq)}$ and $\mathrm{Cu^{2+}(aq)}$) act as strong oxidizing agents.

Applications of the electrochemical series

1. Predicting anode, cathode, and standard cell EMF: When two half-cells are coupled in a galvanic cell, the couple with the higher (more positive) $E^\circ$ undergoes reduction and acts as the cathode, while the couple with the lower (more negative) $E^\circ$ undergoes oxidation and acts as the anode. Using $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ ensures a positive EMF for the spontaneous pairing.

2. Predicting spontaneity of redox reactions: A proposed redox reaction is thermodynamically spontaneous under standard conditions if and only if its calculated $E^\circ_{\text{cell}}$ is positive (which corresponds to a negative standard Gibbs free energy change, $\Delta G^\circ = -nFE^\circ_{\text{cell}} < 0$). If $E^\circ_{\text{cell}} < 0$, the forward reaction is non-spontaneous, and the reverse reaction is favoured.

3. Predicting metal and hydrogen displacement reactions: A metal with a lower (more negative) $E^\circ$ can thermodynamically favour reduction of ions whose metal couple has a higher $E^\circ$. In aqueous solutions, competing reactions with water and passivation must also be considered. For instance, dipping zinc ($E^\circ = -0.76\text{ V}$) into $\mathrm{CuSO_4}$ ($E^\circ = +0.34\text{ V}$) spontaneously deposits copper:

   $$\mathrm{Zn(s) + Cu^{2+}(aq) \longrightarrow Zn^{2+}(aq) + Cu(s)} \qquad (E^\circ_{\text{cell}} = +1.10\text{ V})$$

   Similarly, metals with negative standard reduction potentials ($E^\circ < 0.00\text{ V}$, such as $\mathrm{Mg}$, $\mathrm{Zn}$, and $\mathrm{Fe}$) can reduce $\mathrm{H^+}$ ions from dilute non-oxidizing acids (like dilute $\mathrm{HCl}$ or $\mathrm{H_2SO_4}$) to liberate $\mathrm{H_2}$ gas, whereas metals with positive potentials ($E^\circ > 0.00\text{ V}$, such as $\mathrm{Cu}$ and $\mathrm{Ag}$) cannot.

4. Predicting preferential discharge during electrolysis: When an aqueous solution containing multiple cations or anions is electrolysed, the species that requires the least electrical work is discharged first (ignoring overpotential effects). At the cathode, the cation with the highest (most positive) reduction potential is reduced first. At the anode, the species with the lowest reduction potential is oxidized first. For example, in an acidified aqueous mixture of $\mathrm{NaCl}$ and $\mathrm{KI}$:

   • Competing reductions at the cathode: $\mathrm{H^+}$ ($0.00\text{ V}$), $\mathrm{Na^+}$ ($-2.71\text{ V}$), and $\mathrm{K^+}$ ($-2.93\text{ V}$). Because $\mathrm{H^+}$ has the highest reduction potential, hydrogen gas is evolved: $\mathrm{2\,H^+(aq) + 2\,e^- \longrightarrow H_2(g)}$.
   • Competing oxidations at the anode: $\mathrm{I_2/I^-}$ ($+0.54\text{ V}$) and $\mathrm{Cl_2/Cl^-}$ ($+1.36\text{ V}$). Because iodide has the lower reduction potential, it is oxidized preferentially to iodine: $\mathrm{2\,I^-(aq) \longrightarrow I_2 + 2\,e^-}$.

Watch out. The electrochemical series is strictly a thermodynamic tool for standard aqueous conditions ($a=1$). It predicts whether a reaction is energetically feasible, not how fast it occurs (kinetics) or whether surface films interfere. For example, aluminium ($E^\circ = -1.66\text{ V}$) is thermodynamically much more reactive than zinc, yet it resists corrosion in air and neutral water because a dense, adherent $\mathrm{Al_2O_3}$ passive film blocks ion transport.

Self-check questions

1. Why is it impossible to measure the absolute potential of a single $\mathrm{Cu^{2+}/Cu}$ half-cell with a voltmeter? Answer: A voltmeter requires two electrical contacts to complete a circuit. Touching the electrolyte with a second metallic probe introduces a second metal–solution interface with its own interfacial potential, so the instrument always measures the difference between two electrode potentials.
2. Given $E^\circ_{\mathrm{Fe^{2+}/Fe}} = -0.44\text{ V}$ and $E^\circ_{\mathrm{Cu^{2+}/Cu}} = +0.34\text{ V}$, can you safely store an aqueous $\mathrm{CuSO_4}$ solution in an unprotected iron container? Answer: No. Because iron has a lower reduction potential than copper, iron will spontaneously oxidize and dissolve while displacing copper metal: $\mathrm{Fe(s) + Cu^{2+}(aq) \longrightarrow Fe^{2+}(aq) + Cu(s)}$ ($E^\circ_{\text{cell}} = +0.34 - (-0.44) = +0.78\text{ V} > 0$), corroding the container.
3. What is the standard oxidation potential of the $\mathrm{Mg^{2+}/Mg}$ electrode if its standard reduction potential is $-2.37\text{ V}$? Answer: By convention, $E^\circ_{\text{ox}} = -E^\circ_{\text{red}} = -(-2.37\text{ V}) = +2.37\text{ V}$.

The Nernst equation: derivation and applications

Standard electrode potentials ($E^\circ$) are measured under strict reference conditions: unit activity (approximated as $1\text{ mol L}^{-1}$ concentration for solutes and $1\text{ bar}$ standard pressure for gases; older texts sometimes approximate this as $1\text{ atm}$) at $298\text{ K}$ ($25^\circ\text{C}$). Practical electrochemical systems—ranging from discharging batteries to corroding pipelines and biological sensors—rarely operate at unit concentration. Derived by Walther Nernst, the Nernst equation connects chemical thermodynamics to electrochemistry, allowing us to calculate half-cell reduction potentials and full-cell electromotive force (EMF) at any electrolyte concentration, gas pressure, and temperature.

Thermodynamic foundation and derivation

In a reversible electrochemical cell (such as the Daniell cell, where an opposing voltage equal to the cell EMF makes the net current zero, and an infinitesimal increase reverses the ideal reaction), the maximum electrical work obtainable at constant temperature and pressure equals the decrease in Gibbs free energy. If $n$ moles of electrons are transferred in the balanced reaction against a cell potential $E$ (in volts, $\text{V} = \text{J C}^{-1}$), and $F$ is the Faraday constant (the magnitude of charge per mole of electrons, $F = N_A e \approx 96485\text{ C mol}^{-1}$), the electrical work leads directly to:

Here, $\Delta G$ and $\Delta G^\circ$ are the reaction Gibbs free energy change and standard Gibbs free energy change (in $\text{J mol}^{-1}$), and $E^\circ$ is the standard potential (in $\text{V}$).

Consider a general reversible electrochemical reaction:

From chemical thermodynamics, the van 't Hoff reaction isotherm relates $\Delta G$ under non-standard conditions to $\Delta G^\circ$:

where $R = 8.314\text{ J mol}^{-1}\text{ K}^{-1}$ is the universal gas constant, $T$ is the absolute temperature in kelvin ($\text{K}$), and $Q$ is the dimensionless thermodynamic reaction quotient defined in terms of activities ($a$):

For dilute electrolyte solutions, ionic activities are approximated by molar concentrations ($[\text{X}]$ in $\text{mol L}^{-1}$). For gases, activities are replaced by relative partial pressures ($p / p^\circ$, where $p^\circ$ is standard pressure). Pure solids and pure liquids have unit activity ($a = 1$) and are omitted from $Q$.

Watch out. Never include free electrons ($[\text{e}^-]^n$) inside the reaction quotient $Q$, and do not replace $Q$ with the equilibrium constant $K$ during a general derivation. $Q$ describes the instantaneous composition of the cell at any moment, whereas $K$ applies only when the cell has reached chemical equilibrium.

Substituting $\Delta G = -nFE$ and $\Delta G^\circ = -nFE^\circ$ into the reaction isotherm yields:

Dividing every term by $-nF$ gives the fundamental Nernst equation:

To convert from natural logarithms ($\ln$) to base-10 logarithms ($\log_{10}$, written simply as $\log$), we use the relation $\ln Q = 2.3026 \log Q \approx 2.303 \log Q$:

At the commonly tabulated temperature $T = 298\text{ K}$ ($25^\circ\text{C}$), the constant pre-factor evaluates to:

Substituting this value gives the practical working form of the Nernst equation at $298\text{ K}$:

Single-electrode potentials

By convention, all single-electrode potentials are written as reduction potentials.

Metal/metal-ion half-cell

For the reduction of a metal cation to the pure solid metal, $\text{M}^{n+}_{(\text{aq})} + n\text{e}^- \rightleftharpoons \text{M}_{(\text{s})}$, the solid metal has unit activity ($a_{\text{M}} = 1$). Therefore, in the dilute-solution approximation, $Q = 1/[\text{M}^{n+}]$. Substituting this into the Nernst equation gives:

At $298\text{ K}$, using $\log (1/[\text{M}^{n+}]) = -\log [\text{M}^{n+}]$:

For example, for a zinc electrode ($\text{Zn}^{2+} + 2\text{e}^- \rightleftharpoons \text{Zn}_{(\text{s})}$, $n=2$) or a copper electrode ($\text{Cu}^{2+} + 2\text{e}^- \rightleftharpoons \text{Cu}_{(\text{s})}$, $n=2$) at $298\text{ K}$:

Notice that the reduction potential varies logarithmically (not linearly) with the molar concentration of $[\text{M}^{n+}]$. A tenfold increase in $[\text{M}^{n+}]$ shifts $E$ positively by $0.0591/n\text{ V}$. At high concentrations, ion–ion interactions become significant and this dilute-concentration formula hits a physical limit, requiring true activities ($a = \gamma [\text{M}^{n+}]/c^\circ$, with $c^\circ = 1\text{ mol L}^{-1}$) instead of raw molarities.

Redox half-cell

When an inert electrode (such as platinum) is immersed in a solution containing two oxidation states of the same ion, both species are aqueous solutes and both appear in $Q$:

• Iron(III)/iron(II) couple: $\text{Fe}^{3+}_{(\text{aq})} + \text{e}^- \rightleftharpoons \text{Fe}^{2+}_{(\text{aq})}$ ($n = 1$, $Q = [\text{Fe}^{2+}]/[\text{Fe}^{3+}]$)

• Tin(IV)/tin(II) couple: $\text{Sn}^{4+}_{(\text{aq})} + 2\text{e}^- \rightleftharpoons \text{Sn}^{2+}_{(\text{aq})}$ ($n = 2$, $Q = [\text{Sn}^{2+}]/[\text{Sn}^{4+}]$)

Complete electrochemical cells

For a complete galvanic cell, $E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}$ and $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$. We can write the cell Nernst equation either by subtracting the two half-cell Nernst equations or directly from the balanced overall cell reaction.

1. Daniell cell: $\text{Zn}_{(\text{s})} + \text{Cu}^{2+}_{(\text{aq})} \rightleftharpoons \text{Zn}^{2+}_{(\text{aq})} + \text{Cu}_{(\text{s})}$ Here $n = 2$. Omitting the pure solids $\text{Zn}_{(\text{s})}$ and $\text{Cu}_{(\text{s})}$, the reaction quotient is $Q = [\text{Zn}^{2+}]/[\text{Cu}^{2+}]$. At $298\text{ K}$:

2. Copper–silver cell:

   • Anode (oxidation): $\text{Cu}_{(\text{s})} \rightarrow \text{Cu}^{2+}_{(\text{aq})} + 2\text{e}^-$
   • Cathode (reduction): $2\text{Ag}^+_{(\text{aq})} + 2\text{e}^- \rightarrow 2\text{Ag}_{(\text{s})}$
   • Overall reaction ($n = 2$): $\text{Cu}_{(\text{s})} + 2\text{Ag}^+_{(\text{aq})} \rightleftharpoons \text{Cu}^{2+}_{(\text{aq})} + 2\text{Ag}_{(\text{s})}$

   Because two moles of $\text{Ag}^+$ participate per two moles of electrons transferred, $[\text{Ag}^+]$ is squared in $Q = [\text{Cu}^{2+}] / [\text{Ag}^+]^2$. At $298\text{ K}$:

3. Aluminium–iron(II) cell:

   • Anode: $2\text{Al}_{(\text{s})} \rightarrow 2\text{Al}^{3+}_{(\text{aq})} + 6\text{e}^-$
   • Cathode: $3\text{Fe}^{2+}_{(\text{aq})} + 6\text{e}^- \rightarrow 3\text{Fe}_{(\text{s})}$
   • Overall reaction ($n = 6$): $2\text{Al}_{(\text{s})} + 3\text{Fe}^{2+}_{(\text{aq})} \rightleftharpoons 2\text{Al}^{3+}_{(\text{aq})} + 3\text{Fe}_{(\text{s})}$

   Here $n = 6$ and $Q = [\text{Al}^{3+}]^2 / [\text{Fe}^{2+}]^3$, giving at $298\text{ K}$:

pH dependence of gas electrodes

Electrode reactions that consume or produce $\text{H}^+$ ions exhibit potentials that depend directly on solution $\text{pH}$ ($\text{pH} = -\log a_{\text{H}^+} \approx -\log [\text{H}^+]$).

Hydrogen electrode

The reduction half-reaction is $2\text{H}^+_{(\text{aq})} + 2\text{e}^- \rightleftharpoons \text{H}_{2(\text{g})}$ ($n = 2$, $E^\circ_{\text{H}^+/\text{H}_2} = 0.000\text{ V}$). Writing $Q = (p_{\text{H}_2}/p^\circ) / [\text{H}^+]^2$, at standard hydrogen gas pressure ($p_{\text{H}_2} = p^\circ$) and $298\text{ K}$:

Oxygen electrode

In acidic or neutral aqueous media, oxygen reduction proceeds via:

Here $n = 4$ and $E^\circ_{\text{O}_2/\text{H}_2\text{O}} = +1.229\text{ V}$ (often rounded to $1.23\text{ V}$). Since liquid water has unit activity, at standard oxygen pressure ($p_{\text{O}_2} = p^\circ$) the reaction quotient is $Q = 1 / [\text{H}^+]^4$. At $298\text{ K}$:

Chemical equilibrium and applications

As a galvanic cell discharges, reactant ions are consumed and product ions accumulate. Consequently, $Q$ increases and the cell voltage $E_{\text{cell}}$ steadily drops. Eventually, the forward and reverse reaction rates become equal and the system reaches chemical equilibrium. At equilibrium, no net current flows ($\Delta G = 0$), so $E_{\text{cell}} = 0$ and the reaction quotient equals the equilibrium constant ($Q = K$).

Watch out. At chemical equilibrium, the instantaneous cell potential is zero ($E_{\text{cell}} = 0$), the standard cell potential need not be zero. $E^\circ_{\text{cell}}$ is a fixed thermodynamic constant characteristic of the reaction at unit activity.

Setting $E_{\text{cell}} = 0$ and $Q = K$ in the Nernst equation yields:

Equivalently, since $\Delta G^\circ = -nFE^\circ_{\text{cell}}$, this reproduces the fundamental thermodynamic relation $\Delta G^\circ = -RT \ln K$. At $298\text{ K}$:

Beyond calculating equilibrium constants, the Nernst equation is routinely applied to:

1. Determine non-standard cell EMF: Calculate reversible potentials at specified activities, temperature and pressure. Actual voltage under load also includes resistance and kinetic losses.
2. Find unknown ionic concentrations: Measure $E_{\text{cell}}$ potentiometrically against a reference electrode to quantify trace metal ions (potentiometric titration and ion-selective analysis).
3. Measure solution $\text{pH}$: Convert the voltage of a $\text{pH}$-sensitive electrode (such as a hydrogen or glass electrode) directly into $\text{pH}$.
4. Determine ionic valency ($n$): Measure the slope of an $E$ versus $\log [\text{M}^{n+}]$ plot ($0.0591/n\text{ V}$ per decade at $298\text{ K}$) to find the number of electrons transferred.

Self-check questions

1. Question: How does the reduction potential of a $\text{Cu}^{2+}/\text{Cu}$ electrode at $298\text{ K}$ change if the $\text{Cu}^{2+}$ concentration is diluted to one-hundredth ($0.01$ times) of its initial value? Answer: Since $n = 2$ and $E = E^\circ + (0.0591/2) \log [\text{Cu}^{2+}]$, multiplying $[\text{Cu}^{2+}]$ by $10^{-2}$ changes the potential by $(0.0591/2) \log(10^{-2}) = -0.0591\text{ V}$ (it decreases by $59.1\text{ mV}$).

2. Question: Write the reaction quotient $Q$ and identify $n$ for the cell reaction $\text{Zn}_{(\text{s})} + 2\text{H}^+_{(\text{aq})} \rightleftharpoons \text{Zn}^{2+}_{(\text{aq})} + \text{H}_{2(\text{g})}$ at $p_{\text{H}_2} = 1\text{ bar}$. Answer: Two electrons are transferred ($n = 2$). Omitting solid zinc and setting $p_{\text{H}_2}/p^\circ = 1$, the reaction quotient is $Q = [\text{Zn}^{2+}] / [\text{H}^+]^2$.

3. Question: Why does $E^\circ_{\text{cell}}$ remain non-zero when an ideal Daniell cell reaches chemical equilibrium ($E_{\text{cell}} = 0$)? Answer: $E^\circ_{\text{cell}}$ is defined specifically at standard state ($[\text{Zn}^{2+}] = [\text{Cu}^{2+}] = 1\text{ M}$), whereas a cell at chemical equilibrium has a composition where $[\text{Zn}^{2+}]/[\text{Cu}^{2+}] = K$ and $E_{\text{cell}} = 0$.

Worked problems: electrode and cell potentials

Solving electrochemical numerical problems requires a consistent convention for electrode potentials, reaction stoichiometry, and the reaction quotient. In accordance with IUPAC standards, every half-cell potential in this section is treated as a reduction potential ($E$ or $E^\circ$). This section consolidates the core problem types encountered in engineering chemistry—single-electrode potentials at standard and non-standard temperatures, complete cell electromotive force (EMF), unknown ion concentrations, unknown standard potentials, and equilibrium constants—into a systematic worked reference.

Problem-solving framework and conventions

Before substituting numbers into the Nernst equation, define all physical quantities clearly:

• $E$ and $E^\circ$: Non-standard and standard reduction potentials (in volts, $\text{V}$).
• $E_{\text{cell}}$ and $E^\circ_{\text{cell}}$: Non-standard and standard cell potentials ($\text{V}$), where $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$.
• $R$: Universal gas constant, $8.314\text{ J mol}^{-1}\text{ K}^{-1}$.
• $T$: Absolute temperature in kelvin ($\text{K}$), where $T(\text{K}) = t(^\circ\text{C}) + 273.15$ (often approximated as $+273$ in routine problems).
• $F$: Faraday constant, $96485\text{ C mol}^{-1}$.
• $n$: Number of moles of electrons transferred in the balanced half-reaction or overall cell reaction.
• $Q$: Reaction quotient, constructed with product activities (approximated here by molar concentrations, $[\text{M}^{n+}]$) in the numerator and reactant concentrations in the denominator, each raised to its stoichiometric coefficient. Pure solids and pure liquids have unit activity ($a = 1$) and are omitted from $Q$.

At any temperature $T$, the Nernst equation for a general reduction or cell process is:

At $T = 298\text{ K}$ ($25\,^\circ\text{C}$), the factor $\frac{2.303RT}{F}$ is approximately $0.0591\text{ V}$. For a single metal/metal-ion reduction $\text{M}^{n+}(aq) + ne^- \rightarrow \text{M}(s)$, $Q = \frac{1}{[\text{M}^{n+}]}$, which gives the convenient working forms:

Follow these four steps for every calculation:

1. Identify given data and convert Celsius temperatures to kelvin.
2. Write the balanced reduction half-reaction or overall cell reaction to determine $n$ unambiguously.
3. Construct the reaction quotient $Q$ (strictly $\frac{[\text{products}]^{\nu_p}}{[\text{reactants}]^{\nu_r}}$ for the reaction as written).
4. Substitute values and interpret the physical meaning of the sign and magnitude.

Watch out. A frequent mistake when using the minus-sign Nernst equation ($E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n}\log_{10} Q$) is placing reactant (cathode) ions in the numerator of $Q$. If you write $\frac{[\text{cathode ions}]}{[\text{anode ions}]}$, you must use a plus sign before the logarithm. To avoid sign reversals, always write $Q = \frac{[\text{products}]}{[\text{reactants}]}$ with a minus sign.

Single-electrode potentials