What is the role of the salt bridge?

It maintains electrical neutrality by allowing inert ion migration between the two half-cells, preventing charge buildup that would otherwise stop the cell.

Is E° dependent on the amount of substance?

No — standard electrode potentials are intensive. Multiplying a half-equation by a constant does not change E°, though it does change ΔG = −nFE°.

Why does the standard hydrogen electrode have E° = 0?

It is the chosen reference; all other potentials are measured relative to SHE. The choice is arbitrary but universal.

How does a fuel cell differ from a battery?

Fuel cells consume reactants supplied externally and continue to operate as long as fuel is fed; batteries contain a fixed amount of reactants and discharge once they are used up.

Why is rusting faster in salt water?

Dissolved ions increase electrolyte conductivity, allowing the electrochemical anode and cathode regions to communicate more efficiently, so the corrosion rate rises.

In a galvanic cell, oxidation occurs at the:

Anode (negative terminal). Mnemonic AOXRC: Anode-Oxidation, Reduction-Cathode. In galvanic cells the anode is the negative terminal; in electrolytic cells it is positive.

Standard EMF of a cell with E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V is:

+1.10 V. E°_cell = E°_cathode − E°_anode = 0.34 − (−0.76) = 1.10 V (Cu is the cathode because its reduction potential is higher).

How long must a 2.0 A current flow to deposit 0.5 mol of Cu from CuSO₄ solution?

48 240 s. Cu²⁺ + 2 e⁻ → Cu needs 2F per mole; 0.5 mol → 96 485 C; t = q/I = 96 485/2 ≈ 48 240 s.

The Nernst equation at 298 K can be written as:

E = E° − (0.0592/n) log Q. E = E° − (RT/nF) ln Q; at 298 K with the 2.303 conversion this becomes E = E° − (0.0592/n) log Q.

Galvanising iron with zinc protects against rust because:

Zn is more easily oxidised than Fe. Zn (E° = −0.76 V) is more easily oxidised than Fe (E° = −0.44 V), so it acts as a sacrificial anode.

Electrochemistry MDCAT 2026 — Notes, Key Concepts & MCQs (Chemistry)

Electrochemistry

Electrochemistry averages 2 MCQs per MDCAT paper — standard electrode potentials, the Nernst equation, and electrolysis stoichiometry recur most.

Electrochemistry is a Chemistry chapter on the official PMDC MDCAT 2026 syllabus, contributing roughly 2 MCQs to the 45-MCQ Chemistry section. Mastering the core concepts below typically secures the full chapter weightage.

Oxidation, reduction, and half-cells

Oxidation is loss of electrons (OIL); reduction is gain (RIG). A redox reaction is split into two half-reactions; in a galvanic (voltaic) cell, oxidation occurs at the anode (negative terminal) and reduction at the cathode (positive). The Daniell cell Zn | Zn²⁺ ‖ Cu²⁺ | Cu generates 1.10 V at standard conditions because Zn(s) → Zn²⁺ + 2 e⁻ (E° = +0.76 V as oxidation) couples with Cu²⁺ + 2 e⁻ → Cu(s) (E° = +0.34 V as reduction). FSc XII Chapter 9 and Atkins Chapter 6 use this as the canonical example.

Standard electrode potentials

By convention, the standard hydrogen electrode (SHE) is assigned E° = 0 V: 2 H⁺(1 M) + 2 e⁻ → H₂(1 atm) on Pt. All other E° values are tabulated as reduction potentials relative to SHE. A more positive E° means stronger oxidising power; F₂/F⁻ at +2.87 V is the strongest common oxidiser, Li⁺/Li at −3.04 V the strongest reducer when reversed. Cell EMF E°_cell = E°_cathode − E°_anode; if positive, the reaction is spontaneous as written.

The Nernst equation

At non-standard concentrations: E = E° − (RT/nF) ln Q. At 298 K with the conversion ln 10 = 2.303 and RT/F ≈ 0.0257 V: E = E° − (0.0592/n) log Q. For Cu²⁺ + 2 e⁻ → Cu with [Cu²⁺] = 0.01 M: E = 0.34 − (0.0592/2) log(1/0.01) = 0.34 − 0.0592 = 0.281 V. The Nernst equation underlies pH meters (a glass electrode whose potential varies linearly with pH) and concentration cells, where the same metal at two different ion concentrations generates a small EMF.

Faraday's laws of electrolysis

First law: mass deposited m ∝ charge q = It. Second law: equal charges deposit chemically equivalent masses across different cells. The constant is the Faraday F = 96 485 C/mol of electrons. To deposit 32 g (1 mol) of Cu from Cu²⁺ requires 2 mol electrons = 192 970 C. At a current of 5 A, this takes t = q/I = 192 970/5 ≈ 38 594 s ≈ 10.7 h. The MDCAT often combines stoichiometry with t and I to back out either the metal or the current.

Batteries, fuel cells, and corrosion

Primary cells (Leclanché Zn-MnO₂) discharge irreversibly. Secondary cells — lead-acid (Pb/PbO₂/H₂SO₄, 2.04 V), Ni-Cd, lithium-ion (LiCoO₂ cathode, graphite anode, ~3.7 V) — recharge by reversing current. Hydrogen-oxygen fuel cells produce 1.23 V from H₂ + ½O₂ → H₂O with water as the only product. Corrosion is electrochemical: iron rusts because Fe → Fe²⁺ + 2 e⁻ (anodic) and ½O₂ + H₂O + 2 e⁻ → 2 OH⁻ (cathodic) couple in moisture; protection methods include galvanising (sacrificial Zn, more easily oxidised) and cathodic protection.

Key Concepts

Worked MCQs

Frequently Asked Questions

How Electrochemistry Is Tested

MDCAT questions on Electrochemistry are a mix of recall (definitions, classifications), application (predict outcomes, interpret diagrams), and basic numerical/analytical reasoning. PMDC papers from 2020–2025 emphasized the concepts above; older UHS papers (2008–2019) tested them too, with slight variations in question framing.