Electrochemistry: common doubts, answered
The questions students ask most often about Electrochemistry, each with a short answer. For the full chapter, read the Electrochemistry notes.
Galvanic and electrolytic cells
Read this section in the notes →What is the difference between a galvanic cell and an electrolytic cell?
A galvanic cell uses a redox reaction that runs by itself to generate an electric current, as the Daniell cell and ordinary batteries do. An electrolytic cell works the other way: an external electricity supply forces a non-spontaneous reaction to happen, as in electroplating or getting sodium from molten NaCl. In both, oxidation takes place at the anode and reduction at the cathode.
Why is the anode negative in a galvanic cell but positive in an electrolytic cell?
Oxidation always happens at the anode, but its sign depends on the cell. In a galvanic cell the reaction itself releases electrons at the anode, so they accumulate there and it becomes the negative terminal. In an electrolytic cell an external source drags electrons away from the anode, so it must be joined to the positive terminal of the supply and is positive.
Why is a salt bridge needed in a galvanic cell?
The salt bridge, usually KCl or KNO₃ set in agar gel, joins the two half-cell solutions and lets ions travel between them. Without it, positive charge would build up near the anode as metal ions form and negative charge near the cathode as cations are removed, and the electron flow would stop almost at once. The bridge keeps both solutions neutral so current continues.
Electrode potentials and the SHE
Read this section in the notes →Why is the potential of the standard hydrogen electrode taken as zero?
A single electrode's potential cannot be measured alone, because any measurement needs a second electrode to complete the circuit. So a reference is agreed: platinum dipped in 1 M H⁺ with hydrogen gas at 1 bar is assigned 0 V at all temperatures. Pairing any other electrode with it gives that electrode's potential relative to hydrogen, which is what E° tables list.
How do you calculate the standard emf of a cell?
Use E°cell = E°cathode − E°anode, taking both values as standard reduction potentials straight from the table without changing any sign. The electrode with the higher reduction potential acts as the cathode. For the Daniell cell, E°cell = 0.34 − (−0.76) = 1.10 V. A positive result means the reaction as written is spontaneous.
The Nernst equation
Read this section in the notes →How do you write the Nernst equation for the Daniell cell?
For Zn + Cu²⁺ → Zn²⁺ + Cu at 298 K, E = E° − (0.059/2) log([Zn²⁺]/[Cu²⁺]). The reaction quotient is products over reactants, so raising [Cu²⁺] increases the cell potential and raising [Zn²⁺] lowers it. The solid metals do not appear in Q, and n = 2 because two electrons are transferred in the overall reaction.
Why does the cell potential become zero when a battery is fully discharged?
As a cell discharges, reactants are used up and products build up, so the reaction quotient grows and, by the Nernst equation, E falls. Once Q equals the equilibrium constant, the reaction has no further drive in either direction and E reaches zero. A dead battery is simply a cell whose reaction has come to equilibrium.
Cell potential, Gibbs energy and K
Read this section in the notes →Why don't we multiply E° when a half-equation is multiplied?
Electrode potential is an intensive property: it is energy per unit charge, so it does not depend on how many electrons are transferred. When a half-equation is doubled to balance electrons, E° stays the same; only n, and therefore the Gibbs energy ΔrG° = −nFE°, scales up. Multiplying E° along with the equation is one of the most common errors in this chapter.
How do you find the equilibrium constant from E°cell?
At 298 K use E°cell = (0.059/n) log K, which comes from combining ΔrG° = −nFE° with ΔrG° = −RT ln K. Even about one volt corresponds to an enormous K: for the Daniell cell, log K = 2 × 1.10/0.059, so K is around 10³⁷ and the reaction goes practically to completion. A negative E° would give K below 1.
Conductivity of solutions
Read this section in the notes →What is the difference between conductivity and molar conductivity?
Conductivity, κ, is the conductance of a block of solution of unit length and unit cross-section, measured in S m⁻¹ or S cm⁻¹. Molar conductivity, Λm = κ/c, describes the conducting ability of all the ions produced by one mole of electrolyte, in S cm² mol⁻¹ or S m² mol⁻¹. With κ in S cm⁻¹ and c in mol L⁻¹, use Λm = 1000κ/c.
What is the cell constant and how is it determined?
The cell constant G* = l/A is the gap between the electrodes divided by their area, and it connects measured resistance to conductivity through κ = G*/R. Since l and A are hard to measure accurately in a real cell, G* is found by filling the cell with a KCl solution of known conductivity and measuring its resistance. The same cell is then used for other solutions.
Molar conductivity and dilution
Read this section in the notes →Why does molar conductivity increase on dilution while conductivity decreases?
Conductivity falls because each unit volume of a diluted solution contains fewer ions to carry current. Molar conductivity counts all the ions from one mole of electrolyte, and dilution spreads that mole through a larger volume that is still entirely counted. Ions also hinder each other less and weak electrolytes ionise more, so Λm rises for both strong and weak electrolytes.
Kohlrausch's law
Read this section in the notes →How is the limiting molar conductivity of acetic acid found?
A weak electrolyte's molar conductivity shoots up at very low concentration, so its curve against √c cannot be extended to zero. Kohlrausch's law says the limiting value is the sum of independent contributions from each ion, so strong electrolyte values can be combined: Λ°(CH₃COOH) = Λ°(HCl) + Λ°(CH₃COONa) − Λ°(NaCl). The sodium and chloride parts cancel, leaving H⁺ plus acetate.
How do you find the degree of dissociation of a weak electrolyte from conductivity?
Divide the measured molar conductivity by its limiting value from Kohlrausch's law: α = Λm/Λ°m. The dissociation constant then follows from Ka = cα²/(1 − α). This works because, for a weak electrolyte, the rise in molar conductivity on dilution comes mainly from a larger fraction of molecules breaking into ions.
Electrolysis and Faraday's laws
Read this section in the notes →How much charge is needed to deposit one mole of a metal?
As many faradays as the charge on its ion, since one faraday, about 96500 C, is the charge of one mole of electrons. So Ag⁺ needs 1 F per mole, Cu²⁺ needs 2 F and Al³⁺ needs 3 F. In general, moles deposited = Q/(nF), with Q = current × time in seconds. Faraday's first law states that the mass deposited is proportional to the charge passed.
Products of electrolysis
Read this section in the notes →Why is hydrogen, not sodium, produced at the cathode when aqueous NaCl is electrolysed?
In water, Na⁺ ions compete with water for electrons at the cathode, and reducing water or H⁺ to hydrogen is far easier than reducing Na⁺, whose standard potential is −2.71 V. So hydrogen gas bubbles off and the solution near the cathode turns alkaline. Sodium metal can be obtained only by electrolysing molten NaCl, where there is no water to compete.
Why is chlorine released instead of oxygen at the anode when brine is electrolysed?
From standard potentials alone, oxidising water to oxygen (1.23 V) looks easier than oxidising chloride to chlorine (1.36 V). But oxygen forms slowly at the electrode and needs an extra voltage, its overpotential, to proceed at a useful rate. That raises the practical voltage for oxygen above the one for chlorine, so chlorine is the gas set free at the anode.
Batteries
Read this section in the notes →What is the difference between primary and secondary batteries?
In a primary battery the reaction runs only once, so it cannot be recharged after it is used up; the dry cell and the mercury cell are examples. A secondary battery can be restored by driving current backwards through it from a charger, which runs the reaction in reverse. In the lead storage battery, both plates turn into PbSO₄ on discharge and are restored on charging.
Why does a mercury cell give a constant voltage throughout its life?
The overall reaction in a mercury cell involves only solids and liquid mercury, with no dissolved ion whose concentration changes as the cell is used. Since the Nernst equation depends on the concentrations in Q, and none of them change, the cell potential stays at about 1.35 V until the reactants run out. This steady voltage suits small devices such as hearing aids and watches.
Fuel cells and corrosion
Read this section in the notes →Why are fuel cells more efficient than thermal power plants?
A fuel cell converts the chemical energy of a fuel such as hydrogen directly into electricity, without first burning it to make heat and then steam. Hydrogen-oxygen fuel cells reach about 70% efficiency, compared with about 40% for thermal plants, and their only product is water. They also run continuously as long as hydrogen and oxygen are supplied.
Why is rusting an electrochemical process and how does zinc protect iron?
On a wet iron surface, one region acts as an anode, where iron is oxidised to Fe²⁺, and another as a cathode, where oxygen is reduced with the help of H⁺ from water. The Fe²⁺ is oxidised further to hydrated iron(III) oxide, which is rust. A zinc coating protects iron because zinc, with the more negative electrode potential, is oxidised in place of the iron.
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