9.4 Electrochemistry: Galvanic Cells, Standard Potentials, and the Nernst Equation

Electrochemistry: Galvanic Cells, Standard Potentials, and the Nernst Equation

Electrochemistry investigates the interconversion of chemical energy and electrical energy through oxidation-reduction (redox) reactions occurring at phase interfaces. Electrochemistry appears in both Physical Chemistry (equilibrium electrochemistry) and Analytical Chemistry (electroanalytical methods), demanding quantitative proficiency in galvanic cell notation, Nernst equation calculations, battery chemistries, Faraday's electrolysis laws, and electrochemical corrosion mechanisms.

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1. Electrochemical Cells and IUPAC Cell Notation

Electrochemical cells are classified into two fundamental thermodynamic categories:

Universal Mnemonic:

• AN OX: Anode is the site of Oxidation.
• RED CAT: Reduction occurs at the Cathode.

IUPAC Cell Notation Rules

By convention, the anode is written on the left and the cathode on the right:

1. A single vertical line ($|$) represents a phase boundary (e.g., solid electrode in contact with aqueous solution).
2. A double vertical line ($||$) represents a salt bridge or porous partition, which eliminates the liquid junction potential.
3. Solute concentrations or gas partial pressures are explicitly indicated in parentheses.
4. If an electrode reaction involves only dissolved species, an inert electronic conductor (such as $Pt(s)$ or graphite) is included at the terminal.

• Example (The Classic Daniell Cell): $Zn(s) \mid Zn^{2+}(aq, 1.0\text{ M}) \parallel Cu^{2+}(aq, 1.0\text{ M}) \mid Cu(s)$
  • Anode reaction (left): $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$
  • Cathode reaction (right): $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$
  • Overall cell reaction: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$

Function of the Salt Bridge

The salt bridge contains an inert, non-interfering concentrated electrolyte (typically $KCl$, $KNO_3$, or $NH_4NO_3$) immobilized in agar gel. It serves two functions:

1. Completes the electrical circuit by allowing ions to migrate (anions toward the anode, cations toward the cathode) to preserve electroneutrality in the half-cells.
2. Minimizes the liquid junction potential (potential difference arising at the boundary between two electrolyte solutions of differing mobility) by choosing salts whose cation and anion mobilities are virtually identical ($u_{K^+} \approx u_{Cl^-}$).

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2. Electromotive Force (EMF), Free Energy, and Standard Potentials

The maximum reversible non-expansion (electrical) work that an electrochemical cell can perform at constant temperature and pressure is equal to the change in Gibbs free energy:

$w_{elec, max} = \Delta G$

Because electrical work is the product of total charge transported ($q = nF$) and cell potential difference ($E_{cell}$):

$w_{elec} = -q E_{cell} = -n F E_{cell}$

$\Delta G = -n F E_{cell}$

Under standard state conditions ($1\text{ bar}$, $1\text{ M}$ activities, $25^\circ\text{C}$):

$\Delta G^\circ = -n F E_{cell}^\circ$

where:

• $n$ = stoichiometric number of moles of electrons transferred in the balanced cell reaction.
• $F$ = Faraday's constant $= 96485.3\text{ C/mol } e^-$ (charge of one mole of electrons).
• $E_{cell}$ = cell electromotive force in volts ($1\text{ V} = 1\text{ J/C}$).

The Standard Hydrogen Electrode (SHE)

Standard reduction potentials ($E^\circ$) are defined relative to the Standard Hydrogen Electrode (SHE), which is assigned a potential of exactly $0.000\text{ V}$ at all temperatures:

$2H^+(aq, a=1) + 2e^- \rightleftharpoons H_2(g, 1\text{ bar, } Pt) \quad E^\circ = 0.000\text{ V}$

The standard cell potential is computed by subtracting the anode standard reduction potential from that of the cathode:

$E_{cell}^\circ = E_{cathode}^\circ - E_{anode}^\circ$

Exam Rule: Both $E_{cathode}^\circ$ and $E_{anode}^\circ$ are tabulated as reduction potentials. Never change the algebraic sign of $E_{anode}^\circ$ when using this subtraction formula, and never multiply $E^\circ$ by stoichiometric coefficients because potential is an intensive thermodynamic property.

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3. The Nernst Equation and Chemical Equilibrium

Substituting $\Delta G = -nFE_{cell}$ and $\Delta G^\circ = -nFE_{cell}^\circ$ into the fundamental free energy equation $\Delta G = \Delta G^\circ + RT \ln Q$:

$-nFE_{cell} = -nFE_{cell}^\circ + RT \ln Q$

Dividing both sides by $-nF$ produces the Nernst Equation:

$E_{cell} = E_{cell}^\circ - \frac{RT}{nF} \ln Q$

At $T = 298.15\text{ K}$ ($25.0^\circ\text{C}$), converting from natural to common logarithms using $\ln(10) \approx 2.302585$:

$\frac{RT}{F} \ln(10) = \frac{(8.31446\text{ J}/(\text{mol}\cdot\text{K}))(298.15\text{ K})}{96485.3\text{ C/mol}} \times 2.302585 = 0.05916\text{ V} \approx 0.0592\text{ V}$

$E_{cell} = E_{cell}^\circ - \frac{0.0592}{n} \log_{10} Q$

Relationship Between $E_{cell}^\circ$ and the Equilibrium Constant ($K$)

When a galvanic cell completely discharges, it reaches dynamic chemical equilibrium:

1. $E_{cell} = 0$ (the cell is dead; zero electrical driving force).
2. $Q = K$ (the reaction quotient equals the equilibrium constant).

Substituting these into the Nernst equation:

$0 = E_{cell}^\circ - \frac{0.0592}{n} \log_{10} K \implies E_{cell}^\circ = \frac{0.0592}{n} \log_{10} K$

Solving for $\log_{10} K$:

$\log_{10} K = \frac{n E_{cell}^\circ}{0.0592} \iff K = 10^{\frac{n E_{cell}^\circ}{0.0592}}$

Concentration Cells

A concentration cell is an electrochemical cell composed of two half-cells with identical chemical electrodes and electrolytes, differing only in solute concentration (or gas pressure):

$M(s) \mid M^{z+}(aq, c_{dilute}) \parallel M^{z+}(aq, c_{conc}) \mid M(s)$

Because the two half-cells are chemically identical, $E_{cell}^\circ = E_{cathode}^\circ - E_{anode}^\circ = 0.000\text{ V}$.

The cell potential is driven purely by the entropy of dilution:

$E_{cell} = 0 - \frac{0.0592}{n} \log_{10} \left( \frac{c_{dilute}}{c_{conc}} \right) = +\frac{0.0592}{n} \log_{10} \left( \frac{c_{conc}}{c_{dilute}} \right) > 0$

Concentration cells form the physical operating basis for analytical potentiometric ion-selective electrodes and the glass membrane pH electrode, where cell potential varies linearly with $\text{pH}$ ($E = \text{constant} - 0.0592\cdot\text{pH}$).

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4. Commercial Batteries and Fuel Cells

Electrochemical power sources are classified as primary batteries, secondary batteries, or fuel cells:

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5. Electrolysis, Faraday's Laws, Overpotential, and Corrosion

Faraday's Laws of Electrolysis

1. First Law: The mass of substance altered at an electrode is directly proportional to the quantity of electric charge ($Q$) passed through the electrolyte.
2. Second Law: For a given quantity of electric charge, the mass of an elemental substance deposited is proportional to its chemical equivalent weight ($M / n$).

Combining both laws into the universal quantitative electrolysis equation:

$Q = I \cdot t = n \cdot F \cdot \left(\frac{m}{M}\right) \implies m = \frac{M \cdot I \cdot t}{n \cdot F}$

where $m$ is mass in grams, $M$ is molar mass (g/mol), $I$ is current in amperes, $t$ is time in seconds, $n$ is the number of electrons per atom/molecule, and $F = 96485\text{ C/mol}$.

Overpotential and Polarization

In practical electrolysis, the actual potential required to drive a reaction exceeds the thermodynamic reversible potential ($E_{rev}$):

$V_{applied} = E_{cathode} - E_{anode} + |\eta_{cathode}| + |\eta_{anode}| + iR$

where $\eta$ represents overpotential (voltage above thermodynamic potential required to drive electron transfer kinetics at a finite rate) and $iR$ is the ohmic potential drop across the electrolyte solution.

• Activation Overpotential: Arises from the activation energy barrier of the charge-transfer step at the electrode surface (governed by the Butler-Volmer and Tafel equations: $\eta = a + b \log i$).
• Concentration Overpotential: Arises when rapid reaction depletes electroactive species in the diffusion layer near the electrode.
• Crucial Industrial Example: Hydrogen has an exceptionally high activation overpotential on mercury electrodes ($\eta_{H_2} > 1.0\text{ V}$). This kinetic inhibition prevents $H^+$ reduction and permits the industrial reduction of $Na^+$ in the chlor-alkali mercury cell to produce sodium amalgam and chlorine gas.

Corrosion and Sacrificial Cathodic Protection

Corrosion of iron (rusting) is an electrochemical process occurring on metal surfaces:

• Anodic site: $Fe(s) \rightarrow Fe^{2+}(aq) + 2e^- \quad (E^\circ = -0.44\text{ V})$
• Cathodic site: $O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l) \quad (E^\circ = +1.23\text{ V})$
• Subsequent oxidation and precipitation: $4Fe^{2+} + O_2 + (4+2x)H_2O \rightarrow 2[Fe_2O_3\cdot xH_2O](s) + 8H^+$

Cathodic Protection:

1. Sacrificial Anode: Galvanically coupling the iron structure (pipeline, ship hull) to a metal with a more negative reduction potential, such as Zinc ($E^\circ = -0.76\text{ V}$) or Magnesium ($E^\circ = -2.37\text{ V}$). The sacrificial metal corrodes preferentially, supplying electrons to polarize iron as the cathode.
2. Impressed Current: Connecting the steel structure to the negative terminal of an external DC power supply.

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6. Worked Examination Calculations

Problem A: Cell Potential, Free Energy, and Equilibrium Constant

Task: A galvanic cell consists of a cadmium electrode in $Cd^{2+}(aq)$ and a copper electrode in $Cu^{2+}(aq)$: $Cd(s) \mid Cd^{2+}(aq) \parallel Cu^{2+}(aq) \mid Cu(s)$ Given standard potentials: $E^\circ(Cu^{2+}/Cu) = +0.340\text{ V}$, $E^\circ(Cd^{2+}/Cd) = -0.403\text{ V}$.

1. Write the spontaneous cell reaction and calculate $E_{cell}^\circ$.
2. Calculate $\Delta_r G^\circ$ at $25.0^\circ\text{C}$.
3. Calculate the thermodynamic equilibrium constant $K$ at $25.0^\circ\text{C}$.
4. Calculate $E_{cell}$ when $[Cd^{2+}] = 0.500\text{ M}$ and $[Cu^{2+}] = 0.00100\text{ M}$.

Solution: 1. Standard Cell Potential: Cathode: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s) \quad (E_{cat}^\circ = +0.340\text{ V})$ Anode: $Cd(s) \rightarrow Cd^{2+}(aq) + 2e^- \quad (E_{an}^\circ = -0.403\text{ V})$ $E_{cell}^\circ = E_{cat}^\circ - E_{an}^\circ = +0.340\text{ V} - (-0.403\text{ V}) = +0.743\text{ V}$ Net reaction: $Cd(s) + Cu^{2+}(aq) \rightarrow Cd^{2+}(aq) + Cu(s)$, with $n = 2$.

2. Standard Free Energy: $\Delta_r G^\circ = -n F E_{cell}^\circ = -(2\text{ mol})(96485\text{ C/mol})(0.743\text{ V}) = -143377\text{ J/mol} = -143.4\text{ kJ/mol}$

3. Equilibrium Constant: $\log_{10} K = \frac{n E_{cell}^\circ}{0.05916} = \frac{2 \times 0.743}{0.05916} = \frac{1.486}{0.05916} = 25.118$ $K = 10^{25.118} = 1.31 \times 10^{25}$

4. Non-Standard Potential (Nernst Equation): $Q = \frac{[Cd^{2+}]}{[Cu^{2+}]} = \frac{0.500}{0.00100} = 500.0$ $E_{cell} = E_{cell}^\circ - \frac{0.05916}{n} \log_{10} Q = 0.743 - \frac{0.05916}{2} \log_{10}(500.0)$ $\log_{10}(500) = 2.6990$ $E_{cell} = 0.743 - (0.02958)(2.6990) = 0.743 - 0.0798 = +0.6632\text{ V} \approx +0.663\text{ V}$

Problem B: Quantitative Electroplating via Faraday's Law

Task: An industrial plating bath deposits elemental chromium from an acidic aqueous bath containing chromium(III) sulfate, $Cr_2(SO_4)_3$. A steady current of $5.000\text{ A}$ is passed through the bath for exactly $45.00\text{ minutes}$. Calculate the mass of chromium metal deposited on the cathode workpiece.

• Data: $M(Cr) = 51.996\text{ g/mol}$; $F = 96485\text{ C/mol } e^-$.

Solution: Cathode half-reaction: $Cr^{3+}(aq) + 3e^- \rightarrow Cr(s) \implies n = 3$. Total time in seconds: $t = 45.00\text{ min} \times 60\text{ s/min} = 2700.0\text{ s}$ Total electrical charge: $Q = I \cdot t = (5.000\text{ A})(2700.0\text{ s}) = 13500.0\text{ C}$ Using Faraday's equation: $m = \frac{M \cdot Q}{n \cdot F} = \frac{(51.996\text{ g/mol})(13500.0\text{ C})}{(3)(96485\text{ C/mol})} = \frac{701946}{289455} = 2.4251\text{ g}$

Exactly $2.425\text{ g}$ of chromium is electroplated onto the cathode.