Electrochemistry Class 12: Notes in One Read

These Electrochemistry Class 12 notes cover the whole chapter in the order you meet it: cells, electrode potential, the Nernst equation, conductivity, electrolysis, batteries and corrosion. Along the way you will find key terms, solved examples, a short revision list, important questions with model answers, and the mistakes to avoid.

Electrochemical cells: galvanic and electrolytic

Electrochemistry studies two things: producing electricity from the energy of spontaneous chemical reactions, and using electrical energy to drive non-spontaneous reactions. It is used to produce chemicals like sodium hydroxide, chlorine and fluorine, and in batteries and fuel cells.

An electrochemical cell has two metallic electrodes dipped in electrolytic solutions. There are two types:

• Galvanic (voltaic) cell: converts the chemical energy of a spontaneous redox reaction into electrical energy. Example: the Daniell cell.
• Electrolytic cell: uses an external source of electricity to force a non-spontaneous reaction.

You can see the switch by applying an opposing external voltage (Eext) to a Daniell cell:

Galvanic cells and electrode potential

A galvanic cell is made of two half-cells, each a metal electrode in contact with its ions. In the Daniell cell, zinc in ZnSO4 is the oxidation half-cell and copper in CuSO4 is the reduction half-cell.

• Electrode potential: the potential difference at the electrode-electrolyte interface. When all concentrations are unity (1 mol/dm³), it is the standard electrode potential.
• Anode: where oxidation occurs; negative in a galvanic cell.
• Cathode: where reduction occurs; positive in a galvanic cell.
• Salt bridge: connects the two solutions and keeps them electrically neutral.
• Cell potential: Ecell = Eright − Eleft (reduction potentials of cathode minus anode).

In cell notation, the anode goes on the left and the cathode on the right, with a double vertical line for the salt bridge. Example: Cu(s) | Cu²⁺(aq) || Ag⁺(aq) | Ag(s), where copper is oxidised and silver ions are reduced.

Standard Hydrogen Electrode (SHE): the reference electrode, given a potential of exactly 0 V at all temperatures. It is platinum foil coated with platinum black, dipped in 1 M H⁺ solution, with pure hydrogen gas bubbled at 1 bar. Platinum is inert: it only provides a surface for the reaction and conducts electrons.

When an electrode is connected as the cathode with SHE as the anode, the measured cell potential is its standard reduction potential. Copper gives 0.34 V. Zinc is −0.76 V, which means H⁺ ions are more easily reduced than Zn²⁺, so zinc can reduce H⁺ to hydrogen gas.

The Nernst equation

Standard potentials assume unit concentration. The Nernst equation handles other concentrations. For Mⁿ⁺ + ne⁻ → M:

E = E° − (RT/nF) ln(1/Mⁿ⁺), which at 298 K becomes E = E° − (0.059/n) log(1/Mⁿ⁺).

For the Daniell cell: Ecell = E°cell − (0.059/2) log(Zn²⁺/Cu²⁺). So Ecell rises if Cu²⁺ increases or Zn²⁺ decreases. In general, Ecell = E°cell − (RT/nF) ln Q, where Q is the reaction quotient.

Solved example. For Mg | Mg²⁺(0.130 M) || Ag⁺(0.0001 M) | Ag, with E°cell = 3.17 V and n = 2:

Ecell = 3.17 − (0.059/2) log(0.130 / (0.0001)²) = 3.17 − 0.21 = 2.96 V.

Equilibrium constant and Gibbs energy

As a cell runs, concentrations change until Ecell = 0 and the reaction is at equilibrium. Then E°cell = (2.303 RT/nF) log Kc, so Kc can be found from E°cell.

The electrical work done equals the decrease in Gibbs energy: ΔrG = −nFEcell, and ΔrG° = −nFE°cell. A positive E°cell gives a negative ΔrG°, meaning the reaction is spontaneous.

Solved example. Daniell cell, E°cell = 1.1 V, n = 2: ΔrG° = −2 × 96487 C/mol × 1.1 V = −212271 J/mol = −212.27 kJ/mol.

Remember: ΔrG is extensive (depends on n in the balanced equation), while Ecell is intensive (does not change with the size of the system).

Conductance and conductivity

The cell constant G* = l/A, usually found using a solution of known conductivity such as KCl. Resistance is measured with AC, because DC changes the composition of the solution.

Conductivity depends on the nature of the electrolyte, ion size, solvent viscosity, concentration and temperature. It increases with temperature, because ions move faster and the solvent becomes less viscous.

Effect of dilution and Kohlrausch law

• Conductivity (κ) decreases on dilution, because there are fewer ions per unit volume.
• Molar conductivity (Λm) increases on dilution. For strong electrolytes it rises slowly, because inter-ionic attractions decrease. For weak electrolytes it rises steeply at low concentration, because the degree of dissociation increases.

For weak electrolytes, the limiting molar conductivity Λ°m cannot be found by extrapolation. Instead we use Kohlrausch law of independent migration of ions: Λ°m of an electrolyte is the sum of the contributions of its cations and anions, for example Λ°m(NaCl) = λ°(Na⁺) + λ°(Cl⁻).

For acetic acid: Λ°m(HAc) = Λ°m(HCl) + Λ°m(NaAc) − Λ°m(NaCl). Then the degree of dissociation is α = Λm / Λ°m.

Electrolysis and Faraday's laws

In an electrolytic cell, the cathode is negative (reduction) and the anode is positive (oxidation).

• First law: the mass deposited or liberated is proportional to the quantity of electricity passed, Q = It.
• Second law: for the same quantity of electricity, the masses liberated are proportional to their chemical equivalent weights.
• One mole of electrons carries about 96500 C, called one Faraday (F).

Products depend on electrode potentials and the electrodes used, and water may also react. In aqueous NaCl, hydrogen forms at the cathode instead of sodium, because H⁺ is more easily reduced than Na⁺.

Solved example. A current of 1.5 A for 10 minutes: Q = 1.5 A × 600 s = 900 C. Since Cu²⁺ + 2e⁻ → Cu, 2 Faradays deposit 63 g. Mass = (63 × 900) / (2 × 96487) = 0.2938 g of copper.

Batteries, fuel cells and corrosion

Primary batteries cannot be recharged. The mercury cell gives a constant 1.35 V, because its overall reaction involves no ions whose concentration changes. Secondary batteries can be recharged; in the lead storage battery, PbSO4 forms at both electrodes on discharge, and charging reverses this. Fuel cells turn the energy of combustion of fuels like hydrogen directly into electricity; the H2-O2 cell is about 70% efficient compared with 40% for thermal plants, and produces only water.

Corrosion: in rusting, iron is oxidised at an anodic spot (Fe → Fe²⁺ + 2e⁻) and oxygen is reduced at a cathodic spot in the presence of H⁺. Fe²⁺ is further oxidised to rust, Fe2O3·xH2O. Salt speeds it up. Prevention: paint or oil, galvanisation with zinc, or a sacrificial metal such as magnesium or zinc that corrodes instead of iron.

Remember this

• Galvanic: chemical → electrical. Electrolytic: electrical → chemical.
• Anode = oxidation, cathode = reduction, in both kinds of cell.
• Ecell = Eright − Eleft; anode on the left in cell notation.
• ΔrG° = −nFE°cell; positive E°cell means spontaneous.
• κ falls on dilution; Λm rises on dilution.

Important questions with model answers

1. What happens when 1.5 V is applied against a Daniell cell? 1.5 V is more than 1.1 V, so the cell works as an electrolytic cell: electrons flow from Cu to Zn, zinc deposits and copper dissolves.

2. Why is platinum used in the SHE? It is inert. It provides a surface for the hydrogen reaction and conducts electrons without taking part.

3. How does the EMF of a Daniell cell change if Zn²⁺ increases? In Ecell = E°cell − (0.059/2) log(Zn²⁺/Cu²⁺), the log term grows and is subtracted, so Ecell decreases.

4. Why does molar conductivity increase with dilution? The volume holding one mole of electrolyte increases. In strong electrolytes inter-ionic attractions fall; in weak electrolytes the degree of dissociation rises. Both increase Λm.

5. How much electricity is needed to reduce 1 mol of Al³⁺ to Al? Al³⁺ + 3e⁻ → Al needs 3 mol of electrons, that is 3 F = 3 × 96487 C = 289461 C.

6. Why does a mercury cell give a constant voltage? Its overall reaction involves no ions in solution whose concentration could change, so the potential stays at 1.35 V.

7. How does galvanisation protect iron even when scratched? Zinc is more reactive than iron (more negative reduction potential), so it acts as the anode and is oxidised in preference to iron.

Common mistakes to avoid

• Writing the cathode on the left in cell notation.
• Using oxidation potentials in Ecell = Eright − Eleft; both must be reduction potentials.
• Forgetting to square Ag⁺ when n = 2 and two silver ions react.
• Mixing up conductivity and molar conductivity trends on dilution.

To work through more Nernst equation and Faraday's law problems step by step, study this chapter with Joy.