Electrochemical Cells, Electrode Potentials, Nernst Equation, and Fuel Cells

Electrochemical Cells

Electrochemical cells are devices that convert chemical energy into electrical energy or vice versa. They are broadly classified into two types:

• Galvanic Cells (Voltaic Cells): These cells convert chemical energy from spontaneous redox reactions into electrical energy. They are used in batteries and fuel cells.
• Electrolytic Cells: These cells use electrical energy to drive non-spontaneous redox reactions. They are used in electrolysis, electroplating, and refining of metals.

Galvanic Cells

A typical galvanic cell consists of two half-cells, each containing an electrode immersed in an electrolyte solution. A salt bridge connects the two half-cells, allowing ion migration to maintain electrical neutrality. The two electrodes are connected externally by a wire, allowing electron flow.

Let's consider the Daniell cell as a classic example, which uses zinc and copper electrodes.

• Left Half-cell: A zinc electrode (anode) is immersed in a zinc sulfate (ZnSO₄) solution. Zinc metal is oxidized, losing electrons to form zinc ions (Zn²⁺). This is the oxidation half-reaction:

  Zn(s) → Zn²⁺(aq) + 2e^-

• Right Half-cell: A copper electrode (cathode) is immersed in a copper sulfate (CuSO₄) solution. Copper ions (Cu²⁺) from the solution gain electrons and are reduced to copper metal, depositing on the electrode. This is the reduction half-reaction:

  Cu²⁺(aq) + 2e^- → Cu(s)

• Overall Reaction: The sum of the half-reactions gives the overall cell reaction:

  Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

• Electron Flow: Electrons flow from the anode (zinc) to the cathode (copper) through the external circuit.
• Ion Flow: In the salt bridge, anions move towards the anode compartment to neutralize the excess positive charge from Zn²⁺ formation, and cations move towards the cathode compartment to neutralize the excess negative charge from the depletion of Cu²⁺.

Electrolytic Cells

In an electrolytic cell, an external power source (like a battery) is used to force a non-spontaneous reaction to occur. The electrode connected to the positive terminal of the power source is the anode (where oxidation occurs), and the electrode connected to the negative terminal is the cathode (where reduction occurs).

Example: Electrolysis of molten NaCl.

• Anode (+ve): Chloride ions are oxidized to chlorine gas:

  2Cl^-(l) → Cl₂(g) + 2e^-

• Cathode (-ve): Sodium ions are reduced to molten sodium metal:

  Na⁺(l) + e^- → Na(l)

• Overall Reaction:

  2Na⁺(l) + 2Cl^-(l) → 2Na(l) + Cl₂(g)

Electrode Potentials

When a metal is placed in an electrolyte solution containing its own ions, it tends to either lose electrons (oxidation) or gain electrons (reduction). This tendency is quantified by the electrode potential.

Every half-cell has an associated electrode potential. The potential difference between the two electrodes in a galvanic cell drives the electron flow.

• Oxidation Potential: The tendency of a metal to get oxidized (lose electrons).
• Reduction Potential: The tendency of a metal ion to get reduced (gain electrons).

By convention, the electrode potential refers to the reduction potential. The standard electrode potential (E°) is measured under standard conditions: 298 K (25 °C), 1 atm pressure for gases, and 1 M concentration for solutions.

The standard hydrogen electrode (SHE) is used as a reference electrode. It consists of a platinum electrode in contact with 1 M H⁺ ions and H₂ gas at 1 atm pressure. Its standard reduction potential is defined as 0 V.

The standard electrode potential of any electrode can be determined by coupling it with the SHE. If the measured cell potential is positive, the unknown electrode has a higher reduction potential than SHE. If it's negative, it has a lower reduction potential.

Cell Potential (EMF)

The cell potential, or electromotive force (EMF), is the difference in potential between the two electrodes of a galvanic cell. It's calculated as:

E_cell = E_cathode - E_anode

where E_cathode and E_anode are the standard reduction potentials of the cathode and anode, respectively.

Nernst Equation

The Nernst equation relates the electrode potential of a cell to the concentrations of the reactants and products involved in the redox reaction. It allows us to calculate the cell potential under non-standard conditions (i.e., when concentrations are not 1 M or pressures are not 1 atm).

Consider a general electrode reaction:

aMⁿ⁺(aq) + ne^- → bM(s)

The electrode potential (E) at non-standard conditions is given by:

E = E° - (RT / nF) * ln([M]/[Mⁿ⁺]^a)

Where:

• E = Electrode potential at non-standard conditions
• E° = Standard electrode potential
• R = Gas constant (8.314 J K^-1 mol^-1)
• T = Temperature in Kelvin
• n = Number of moles of electrons transferred
• F = Faraday constant (96485 C mol^-1)
• [M] = Activity (or concentration) of the metal M (usually taken as 1 for solids)
• [Mⁿ⁺] = Activity (or concentration) of the metal ion Mⁿ⁺

At 298 K (25 °C), the equation can be simplified using common logarithms:

E = E° - (0.0591 / n) * log([M]/[Mⁿ⁺]^a)

For a cell reaction involving a redox reaction:

aA + ne^- → bB

The cell potential (E_cell) is given by:

E_cell = E°_cell - (RT / nF) * ln(Q)

Or at 298 K:

E_cell = E°_cell - (0.0591 / n) * log(Q)

Where Q is the reaction quotient:

Q = ([Products]^coefficients) / ([Reactants]^coefficients)

For a cell reaction like:

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Q = [Zn²⁺] / [Cu²⁺]

And the Nernst equation for the cell becomes:

E_cell = E°_cell - (0.0591 / n) * log([Zn²⁺] / [Cu²⁺])

Relationship between E°_cell and Equilibrium Constant (K)

At equilibrium, the cell potential (E_cell) is zero, and the reaction quotient (Q) is equal to the equilibrium constant (K). Substituting these into the Nernst equation at 298 K:

0 = E°_cell - (0.0591 / n) * log(K)

E°_cell = (0.0591 / n) * log(K)

This equation shows that a positive standard cell potential corresponds to an equilibrium constant greater than 1, indicating a spontaneous reaction that favors product formation.

Fuel Cells

Fuel cells are a type of galvanic cell that converts the chemical energy of a fuel (like hydrogen, methane, or methanol) and an oxidant (like oxygen) directly into electrical energy through a redox reaction. Unlike batteries, fuel cells do not store energy; they produce energy as long as fuel and oxidant are supplied.

Hydrogen-Oxygen Fuel Cell

This is the most common and efficient type of fuel cell. It uses hydrogen as the fuel and oxygen as the oxidant.

• Anode (-ve): Hydrogen gas is oxidized.

  2H₂(g) + 2OH^-(aq) → 4H₂O(l) + 4e^- (in alkaline medium)

  H₂(g) → 2H⁺(aq) + 2e^- (in acidic medium)

• Cathode (+ve): Oxygen gas is reduced.

  O₂(g) + 2H₂O(l) + 4e^- → 4OH^-(aq) (in alkaline medium)

  O₂(g) + 4H⁺(aq) + 4e^- → 2H₂O(l) (in acidic medium)

• Overall Reaction: The net reaction, regardless of the medium, is the formation of water.

  2H₂(g) + O₂(g) → 2H₂O(l)

• Electrolyte: Can be acidic (e.g., H₃PO₄), alkaline (e.g., KOH), or molten carbonate.
• Electrodes: Typically made of porous materials like platinum or nickel, which act as catalysts.

The overall reaction is essentially the combustion of hydrogen, but it occurs electrochemically, producing electricity directly and water as the main byproduct. This makes them environmentally friendly.

Advantages of Fuel Cells

• High efficiency compared to internal combustion engines.
• Low or zero emissions (especially hydrogen fuel cells).
• Quiet operation.
• Scalability for various power needs.

Disadvantages of Fuel Cells

• High cost of catalysts (like platinum).
• Challenges in fuel storage and infrastructure (especially for hydrogen).
• Durability and maintenance issues.

Applications

• Galvanic Cells: Batteries (dry cells, lead-acid batteries, lithium-ion batteries), power sources for portable electronics.
• Electrolytic Cells: Electroplating of metals, extraction and refining of reactive metals (e.g., Al, Na), production of chemicals (e.g., Cl₂, NaOH).
• Fuel Cells: Power generation for vehicles (buses, cars), backup power systems, portable power sources, space exploration (Apollo missions).

Example Problem (Nernst Equation)

Calculate the EMF of the cell at 25 °C:

Zn(s) | Zn²⁺(0.01 M) || Cu²⁺(1.0 M) | Cu(s)

Given: E°_Zn²⁺/Zn = -0.76 V, E°_Cu²⁺/Cu = +0.34 V.

Solution:

1. Identify the anode and cathode: Since the standard reduction potential of Cu is higher than Zn, Cu will be the cathode and Zn will be the anode.

   Anode: Zn → Zn²⁺ + 2e^- (Oxidation)

   Cathode: Cu²⁺ + 2e^- → Cu (Reduction)

2. Calculate the standard cell potential (E°_cell):

   E°_cell = E°_cathode - E°_anode = E°_Cu²⁺/Cu - E°_Zn²⁺/Zn

   E°_cell = (+0.34 V) - (-0.76 V) = +1.10 V

3. Write the reaction quotient (Q):

   Q = [Zn²⁺] / [Cu²⁺] = (0.01 M) / (1.0 M) = 0.01

4. Apply the Nernst equation at 25 °C (n=2 for this reaction):

   E_cell = E°_cell - (0.0591 / n) * log(Q)

   E_cell = 1.10 V - (0.0591 / 2) * log(0.01)

   E_cell = 1.10 V - (0.02955) * (-2)

   E_cell = 1.10 V + 0.0591 V = 1.1591 V

The EMF of the cell under these non-standard conditions is approximately 1.16 V.