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Electrochemistry

Electrochemistry studies the interconversion of chemical and electrical energy through redox reactions. The PMDC MDCAT 2026 syllabus expects you to assign oxidation states, balance redox equations, distinguish galvanic from electrolytic cells, and use the standard hydrogen electrode (SHE) to compute cell potentials. Expect 2-3 MCQs per paper.

PMC Table of Specifications. This chapter covers four PMDC subtopics — balancing chemical reactions, oxidation/reduction definitions, redox reactions, and the standard hydrogen electrode. Most MCQs target either the SHE convention or oxidation-state assignment.

Balancing Chemical Reactions

Every chemical equation must obey the law of conservation of mass (atoms balanced) and, in ionic form, conservation of charge. For redox reactions the simplest, fool-proof method is the half-reaction (ion-electron) method.

Half-reaction method (acidic medium)

  1. Write separate oxidation and reduction half-reactions.
  2. Balance atoms other than O and H first.
  3. Balance O by adding H2O.
  4. Balance H by adding H+.
  5. Balance charge by adding electrons (e).
  6. Multiply each half-reaction so electrons cancel; add the half-reactions.

For basic medium, add the same number of OH to both sides as there are H+ and combine with the H+ to form H2O.

Common trap. Students often forget to verify charge balance after balancing atoms. A correctly balanced redox equation must balance both atoms and net charge on each side — check both before moving on.

Oxidation and Reduction

Modern definitions are based on electron transfer:

Oxidation
Loss of electrons; increase in oxidation number; addition of oxygen or removal of hydrogen.
Reduction
Gain of electrons; decrease in oxidation number; removal of oxygen or addition of hydrogen.
Oxidising agent
Species that causes oxidation (gets reduced itself). Examples: KMnO4, K2Cr2O7, Cl2, O3.
Reducing agent
Species that causes reduction (gets oxidised itself). Examples: H2, C, Na, Zn, SnCl2.

Rules for assigning oxidation numbers

Mnemonic: OIL RIG. Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons). Or: LEO the lion says GERLoss of Electrons = Oxidation, Gain of Electrons = Reduction.

Redox Reactions

A redox reaction is one in which electrons are transferred from a reducing agent to an oxidising agent. Oxidation and reduction always occur simultaneously: the electrons lost by one species must be gained by another.

Galvanic (voltaic) cell

Spontaneous redox reaction generates electrical energy. The Daniell cell is the prototype: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s). Zinc is oxidised at the anode (negative electrode), copper ions are reduced at the cathode (positive electrode). A salt bridge maintains electrical neutrality. E°cell = E°cathode − E°anode = +0.34 − (−0.76) = +1.10 V.

Electrolytic cell

Non-spontaneous redox driven by an external power source. Used in electroplating, electrorefining of copper, and the chlor-alkali process. Anode is positive (oxidation still occurs there); cathode is negative.

Galvanic (voltaic) vs Electrolytic cell
PropertyGalvanic / VoltaicElectrolytic
Energy conversionChemical → ElectricalElectrical → Chemical
ReactionSpontaneous (ΔG < 0; E°cell > 0)Non-spontaneous (ΔG > 0; needs input)
Anode (oxidation)Negative electrodePositive electrode
Cathode (reduction)Positive electrodeNegative electrode
External power sourceNone — cell is the sourceRequired (battery or DC supply)
Two electrolyte compartments?Yes — connected by salt bridgeSingle compartment with electrolyte
ExamplesDaniell cell, dry cell, lead-acid battery, fuel cellElectroplating, electrorefining of Cu, electrolysis of H2O / NaCl, chlor-alkali process
Common ruleOxidation always at anode, reduction always at cathode — only the polarity flips

Faraday's laws of electrolysis

Standard Hydrogen Electrode (SHE)

An absolute electrode potential cannot be measured — only differences are accessible. The standard hydrogen electrode is therefore defined as the reference, with E° = 0.00 V.

Construction

A platinum electrode coated with platinum black is dipped into a 1 M H+ solution at 25°C while H2 gas is bubbled at 1 atm pressure over it. The half-cell reaction is 2H+(aq, 1 M) + 2e ↔ H2(g, 1 atm).

Standard conditions

Using SHE to find standard electrode potentials

Couple any half-cell with SHE. The voltmeter reading is the E° of that half-cell. Sign convention: electrodes that reduce H+ better than H2 have E° > 0 (e.g. Cu2+/Cu = +0.34 V); electrodes that are oxidised more easily than H2 have E° < 0 (e.g. Zn2+/Zn = −0.76 V).

High-yield exam point. Memorise the three SHE conditions: 1 M H+, 1 atm H2, 25°C. Examiners frequently change one parameter and ask whether the cell potential is still 0.00 V (it is not).

Worked MCQs

Five MCQs that capture the high-yield testing patterns for electrochemistry. Read every explanation — the deeper concept lives there.

Q1. The oxidation number of manganese in KMnO4 is:

  • +2
  • +4
  • +6
  • +7

K = +1, each O = −2 (four oxygens = −8). Total must equal 0: +1 + Mn + (−8) = 0 → Mn = +7. This is also why KMnO4 is such a strong oxidising agent.

Q2. In the reaction Zn + Cu2+ → Zn2+ + Cu, which species is the reducing agent?

  • Zn
  • Cu2+
  • Zn2+
  • Cu

Zinc gives up two electrons (it is oxidised) and forces Cu2+ to gain electrons. The species that gets oxidised is, by definition, the reducing agent.

Q3. The standard hydrogen electrode (SHE) operates at:

  • 1 M H+, 0°C, 1 atm H2
  • 1 M H+, 25°C, 1 atm H2
  • 0.1 M H+, 25°C, 1 atm H2
  • 1 M H+, 25°C, 760 mm H2O

Standard conditions for SHE are 1 mol/dm3 H+, 1 atm H2(g) and 298 K (25°C). E° is defined as exactly 0.00 V under these conditions.

Q4. For the Daniell cell with E°(Cu2+/Cu) = +0.34 V and E°(Zn2+/Zn) = −0.76 V, the cell EMF is:

  • 0.42 V
  • −1.10 V
  • +1.10 V
  • +0.42 V

cell = E°cathode − E°anode = +0.34 − (−0.76) = +1.10 V. A positive value confirms the reaction is spontaneous — consistent with our experience that zinc displaces copper from copper sulphate.

Q5. One Faraday of charge is approximately equal to:

  • 9 650 C
  • 96 500 C
  • 965 000 C
  • 6.022 × 1023 C

1 F = NA × e = (6.022×1023)(1.602×10−19) ≈ 96 485 C, rounded to 96 500 C. It is the charge carried by one mole of electrons.

Quick Recap

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