Class 12Chemistry · Physical ChemistryFull chapter

Electrochemistry

The whole chapter in one place — read it, then test yourself. Clear notes, key equations, a practice quiz, and worked NCERT solutions & PYQs.

Electrochemical Cells and Electrode Potential

Quick answer A galvanic cell converts the free energy of a spontaneous redox reaction into electrical energy using two electrodes connected externally and internally by a salt bridge.

An electrochemical cell converts chemical energy into electrical energy (galvanic/voltaic cell) or electrical energy into chemical energy (electrolytic cell). In a galvanic cell, a spontaneous redox reaction is split into two half-reactions that occur at physically separated electrodes dipped in their own electrolyte solutions, connected by a metallic wire externally and a salt bridge internally. The salt bridge (a U-tube containing an inert electrolyte like KNO3 or KCl in agar-agar gel) completes the internal circuit and maintains electrical neutrality in both half-cells by allowing ion migration, without mixing the two solutions.

By convention, oxidation occurs at the anode (negative electrode in a galvanic cell) and reduction occurs at the cathode (positive electrode in a galvanic cell). Electrons flow through the external circuit from anode to cathode, while conventional current flows from cathode to anode. A cell is represented in shorthand notation with the anode written on the left and the cathode on the right, a single vertical line for a phase boundary, and a double vertical line for the salt bridge. For the Daniell cell, the reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) is written as:

Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)

The potential difference between the two electrodes of a galvanic cell is called the cell potential or electromotive force (emf), measured in volts. Since the absolute potential of a single electrode cannot be measured, all electrode potentials are measured relative to a reference electrode — the Standard Hydrogen Electrode (SHE), whose standard reduction potential is arbitrarily fixed at exactly 0.00 V at all temperatures. The SHE consists of platinum foil coated with platinum black, dipped in 1 M H+ solution with H2 gas at 1 bar bubbled over it. By measuring the emf of a cell formed with SHE and any other electrode under standard conditions (1 M concentration, 1 bar pressure, 298 K), the standard electrode potential (E°) of that electrode is obtained. Arranging elements in order of increasing standard reduction potential gives the electrochemical series, which predicts relative oxidising/reducing strength, the feasibility of a redox reaction, and displacement reactions.

Worked example: Calculate the standard emf of the Daniell cell, given E°(Cu2+/Cu) = +0.34 V and E°(Zn2+/Zn) = −0.76 V.

Here Cu2+/Cu is the cathode (higher, more positive reduction potential — reduction favoured) and Zn2+/Zn is the anode (lower reduction potential — oxidation favoured).

cell = E°cathode − E°anode = 0.34 − (−0.76) = +1.10 V

Since E°cell is positive, the reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) is spontaneous as written, which matches the observed behaviour of the Daniell cell.

Standard cell potential E°cell = E°cathode − E°anode V · Cathode is where reduction occurs (higher E°); anode is where oxidation occurs (lower E°).
SHE reference potential E°(H⁺/H₂) = 0.00 V Arbitrary universal reference used to assign standard electrode potentials to all other electrodes.
Remember
  • Galvanic cells convert spontaneous redox chemical energy into electrical energy; oxidation always occurs at the anode, reduction at the cathode.
  • The salt bridge maintains electrical neutrality by ion migration and completes the internal circuit without mixing the two electrolyte solutions.
  • Standard Hydrogen Electrode (SHE) is the universal reference with E° = 0.00 V, used to measure standard electrode potentials of all other electrodes.
  • Cell notation places the anode (oxidation, left) and cathode (reduction, right) with || representing the salt bridge.
  • E°cell = E°cathode − E°anode; a positive E°cell indicates a spontaneous cell reaction under standard conditions.
  • The electrochemical series (arrangement of E° values) predicts relative oxidising and reducing power and feasibility of redox/displacement reactions.

Nernst Equation and Electrochemical Thermodynamics

Quick answer The Nernst equation gives the cell potential under non-standard concentrations and links electrochemistry to Gibbs energy and the equilibrium constant.

Standard electrode potentials apply only when all species are at 1 M concentration (or 1 bar for gases) at 298 K. For any other concentration, the actual electrode/cell potential is given by the Nernst equation. For a general electrode reaction Mn+(aq) + ne → M(s), the electrode potential is:

E(Mn+/M) = E°(Mn+/M) − (RT/nF) ln [1/[Mn+]]

For a full cell reaction, the Nernst equation is written in terms of the reaction quotient Q:

Ecell = E°cell − (RT/nF) ln Q

Substituting R = 8.314 J K−1 mol−1, F = 96500 C mol−1, T = 298 K, and converting ln to log10 (factor 2.303), this simplifies to the commonly used form:

Ecell = E°cell − (0.0591/n) log Q (at 298 K)

At equilibrium, the cell can do no further work, so Ecell = 0 and Q becomes the equilibrium constant K. Substituting gives an important relation between the standard cell potential and the equilibrium constant of the reaction:

log K = nE°cell/0.0591

Electrochemistry is also linked directly to thermodynamics: the maximum electrical work obtainable from a cell equals the decrease in Gibbs energy of the reaction, giving:

ΔG = −nFEcell and, under standard conditions, ΔG° = −nFE°cell

Worked example 1 (Nernst equation): Find Ecell at 298 K for Zn(s) | Zn2+(0.1 M) || Cu2+(1.0 M) | Cu(s), given E°cell = 1.10 V.

The cell reaction is Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s), with n = 2 and Q = [Zn2+]/[Cu2+] = 0.1/1.0 = 0.1, so log Q = −1.

Ecell = 1.10 − (0.0591/2) × (−1) = 1.10 + 0.02955 = 1.13 V

Worked example 2 (K and ΔG° from E°cell): For the same Daniell cell (n = 2, E°cell = 1.10 V):

log K = (2 × 1.10)/0.0591 = 2.20/0.0591 ≈ 37.2, so K ≈ 1.7 × 1037, showing the reaction goes essentially to completion.

ΔG° = −nFE°cell = −(2)(96500 C mol−1)(1.10 V) = −212300 J mol−1 = −212.3 kJ mol−1, confirming the reaction is highly spontaneous.

Nernst equation (general) Ecell = E°cell − (RT/nF) ln Q V · R = 8.314 J K⁻¹ mol⁻¹, F = 96500 C mol⁻¹, n = moles of electrons transferred, Q = reaction quotient.
Nernst equation at 298 K Ecell = E°cell − (0.0591/n) log Q V · Practical form obtained by substituting T = 298 K and converting to log₁₀.
Gibbs energy–EMF relation ΔG = −nFEcell ; ΔG° = −nFE°cell J mol⁻¹ · Electrical work done by the cell equals the decrease in Gibbs free energy of the reaction.
Equilibrium constant from E°cell log K = nE°cell / 0.0591 Valid at 298 K; derived by setting Ecell = 0 and Q = K in the Nernst equation.
Remember
  • The Nernst equation, Ecell = E°cell − (0.0591/n) log Q at 298 K, gives cell potential at any concentration, not just standard conditions.
  • As the cell reaction proceeds, concentrations change, Q changes, and Ecell keeps decreasing until it becomes zero at equilibrium.
  • At equilibrium, Ecell = 0 and Q = K, giving log K = nE°cell/0.0591 — this links electrochemistry directly to chemical equilibrium.
  • ΔG = −nFEcell and ΔG° = −nFE°cell connect cell potential to the thermodynamic spontaneity and maximum useful work of the reaction.
  • A large positive E°cell corresponds to a very large K and a large negative ΔG°, meaning the reaction proceeds nearly to completion.

Conductance of Electrolytic Solutions

Quick answer Conductivity and molar conductivity describe how well an electrolyte solution conducts electricity, and Kohlrausch's law lets us predict the behaviour of weak electrolytes at infinite dilution.

Electrolytic solutions conduct electricity through the movement of ions. The conductivity (κ) of a solution is the reciprocal of its resistivity and is measured using a conductivity cell with two platinum electrodes of area A separated by distance l. The ratio l/A is called the cell constant (G*). Since resistance R is measured experimentally using a Wheatstone-bridge-based conductivity meter:

κ = G*/R = (l/A) × (1/R)

Conductivity depends on the number of ions per unit volume, so a more useful quantity for comparing electrolytes is molar conductivity (Λm), defined as the conductivity contributed by one mole of the electrolyte:

Λm = κ × 1000 / C (C in mol L−1, Λm in S cm2 mol−1)

On dilution, κ decreases (fewer ions per unit volume), but Λm always increases because the increase in volume containing one mole of electrolyte more than compensates. For strong electrolytes (fully dissociated), Λm increases only slightly with dilution and can be extrapolated linearly to zero concentration to obtain the limiting molar conductivity Λ°m (molar conductivity at infinite dilution). For weak electrolytes (partially dissociated, e.g. acetic acid), Λm rises sharply near infinite dilution because dissociation increases, and Λ°m cannot be obtained by extrapolation — it is instead calculated using Kohlrausch's law of independent migration of ions, which states that at infinite dilution each ion contributes to the total molar conductivity independently of the other ion present:

Λ°m = ν+λ°+ + νλ°

where λ°+ and λ° are the limiting molar ionic conductivities of the cation and anion, and ν+, ν are the number of cations and anions per formula unit. This lets us calculate Λ°m for a weak electrolyte from the Λ°m values of strong electrolytes containing the same ions (e.g. Λ°m(CH3COOH) = Λ°m(HCl) + Λ°m(CH3COONa) − Λ°m(NaCl)). Kohlrausch's law also allows calculation of the degree of dissociation (α) of a weak electrolyte at any concentration, α = Λm/Λ°m, and hence its dissociation constant via the Ostwald dilution law.

Worked example: The conductivity of 1.028 × 10−3 mol L−1 acetic acid is 4.95 × 10−5 S cm−1. Given Λ°m(CH3COOH) = 390.5 S cm2 mol−1, find α and Ka.

Λm = κ × 1000/C = (4.95 × 10−5 × 1000)/(1.028 × 10−3) = 48.15 S cm2 mol−1

α = Λm/Λ°m = 48.15/390.5 = 0.1233 (12.33%)

Using the Ostwald dilution law, Ka = Cα2/(1−α) = (1.028 × 10−3 × 0.12332)/(1−0.1233) = (1.028 × 10−3 × 0.01520)/0.8767 = 1.78 × 10−5 mol L−1, close to the accepted value of acetic acid's dissociation constant.

Conductivity from resistance κ = G*/R = (l/A)/R S cm⁻¹ · G* = cell constant = l/A, obtained by calibrating the cell with a solution of known κ.
Molar conductivity Λm = κ × 1000 / C S cm² mol⁻¹ · C is molar concentration in mol L⁻¹ (mol dm⁻³).
Kohlrausch's law Λm° = ν₊λ°₊ + ν₋λ°₋ S cm² mol⁻¹ · Limiting molar conductivity is the sum of independent ionic contributions at infinite dilution.
Degree of dissociation α = Λm / Λm° Valid for weak electrolytes at concentration C.
Ostwald's dilution law Ka = Cα² / (1 − α) mol L⁻¹ · Relates dissociation constant of a weak electrolyte to concentration and degree of dissociation.
Remember
  • Conductivity (κ) is measured using a conductivity cell; κ = cell constant / resistance, and it decreases with dilution as ion concentration falls.
  • Molar conductivity Λm = κ×1000/C always increases on dilution because it accounts for the growing volume containing one mole of electrolyte.
  • For strong electrolytes, Λm° is found by linear extrapolation of Λm vs √C to zero concentration; for weak electrolytes this extrapolation fails.
  • Kohlrausch's law (Λm° = ν+λ°+ + ν−λ°−) gives Λm° of weak electrolytes indirectly, from strong electrolytes sharing common ions.
  • Degree of dissociation of a weak electrolyte at concentration C is α = Λm/Λm°, and the Ostwald dilution law gives Ka = Cα²/(1−α).
  • Applications of conductance measurements include determining Λm°, α, Ka of weak acids, and solubility/Ksp of sparingly soluble salts.

Electrolytic Cells and Faraday's Laws of Electrolysis

Quick answer In an electrolytic cell, an external electrical source drives a non-spontaneous redox reaction, and Faraday's laws quantitatively relate the charge passed to the amount of substance deposited or liberated.

Unlike a galvanic cell, an electrolytic cell uses an external source of electrical energy (like a battery) to force a non-spontaneous redox reaction to occur — this process is called electrolysis. In electrolysis, the electrode connected to the positive terminal of the external battery is the anode (oxidation still occurs here), and the electrode connected to the negative terminal is the cathode (reduction still occurs here) — note that in an electrolytic cell the anode is positive and the cathode is negative, the reverse polarity compared to a galvanic cell, even though oxidation is always at the anode and reduction is always at the cathode in both cases. During electrolysis of molten salts or aqueous solutions, cations migrate to the cathode and are reduced, while anions migrate to the anode and are oxidised; in aqueous solutions, competing reduction/oxidation of water itself may occur depending on relative electrode potentials and overpotential effects.

Faraday's laws of electrolysis quantify the relationship between the charge passed and the amount of chemical change:

First law: The mass (m) of a substance deposited or liberated at an electrode is directly proportional to the quantity of charge (Q) passed through the electrolyte: m ∝ Q, i.e. m = ZQ = ZIt, where Z is the electrochemical equivalent of the substance and I is the current in amperes passed for time t seconds, so that Q = It (in coulombs).

Second law: When the same quantity of charge is passed through different electrolytes connected in series, the masses of different substances deposited/liberated at their respective electrodes are proportional to their chemical equivalent weights (M/n, where M is molar mass and n is the number of electrons involved per ion/atom in the electrode reaction).

Combining both laws, the mass deposited is given by:

m = (M × I × t)/(n × F)

where F is the Faraday constant (≈ 96500 C mol−1), the charge carried by one mole of electrons.

Worked example: Calculate the mass of copper deposited at the cathode when a current of 2 A is passed through a CuSO4 solution for 1 hour (Cu = 63.5 g mol−1, n = 2 since Cu2+ + 2e → Cu).

Charge passed, Q = It = 2 A × 3600 s = 7200 C

m = (M × Q)/(n × F) = (63.5 × 7200)/(2 × 96500) = 457200/193000 = 2.37 g of copper

Charge passed Q = I × t C · I is current in amperes, t is time in seconds.
Faraday's first law m = Z × Q = Z × I × t g · Z = electrochemical equivalent (mass deposited per coulomb).
Electrochemical equivalent Z = M / (n × F) g C⁻¹ · M = molar mass, n = number of electrons transferred per ion/atom.
Combined Faraday's law m = (M × I × t) / (n × F) g · F ≈ 96500 C mol⁻¹; gives mass deposited/liberated at an electrode.
Remember
  • Electrolysis uses external electrical energy to drive a non-spontaneous redox reaction inside an electrolytic cell.
  • In an electrolytic cell the anode is the positive terminal and cathode is the negative terminal (opposite polarity to a galvanic cell), but oxidation is always at the anode and reduction always at the cathode.
  • Faraday's first law: mass deposited/liberated is directly proportional to the quantity of charge passed, m = ZIt.
  • Faraday's second law: for the same charge, masses of substances deposited are proportional to their equivalent weight (M/n).
  • Combined form m = MIt/(nF) allows calculation of mass deposited, current required, or time needed given the other quantities.
  • The Faraday constant F ≈ 96500 C mol⁻¹ is the charge carried by one mole of electrons and links moles of electrons to coulombs.

Batteries and Fuel Cells

Quick answer Batteries are practical arrangements of galvanic cells classified as primary (non-rechargeable) or secondary (rechargeable), while fuel cells generate electricity continuously by consuming fuel and oxidant supplied from outside.

A battery is a practical, self-contained source of electrical energy consisting of one or more galvanic cells connected in series. Batteries are broadly of two types.

Primary batteries cannot be recharged — once the reactants are consumed, the cell becomes dead and is discarded. The common dry cell (Leclanché cell) has a zinc container as anode and a carbon rod surrounded by manganese dioxide and powdered carbon as cathode, with a moist paste of NH4Cl and ZnCl2 as electrolyte; it gives about 1.5 V and is used in torches, transistor radios, and clocks. The mercury cell uses a zinc-mercury amalgam anode and a paste of HgO and carbon as cathode, in a KOH-ZnO electrolyte paste; it gives a very steady voltage of about 1.35 V throughout its life (since the overall cell reaction does not involve ions whose concentration changes), making it suitable for watches and hearing aids.

Secondary batteries can be recharged by passing a current in the reverse direction, regenerating the original reactants (essentially running the discharge reaction backward as an electrolytic process). The lead storage battery, widely used in vehicles, has a lead anode and a grid of lead packed with lead dioxide (PbO2) as cathode, both dipped in dilute sulphuric acid. On discharge:

Anode: Pb(s) + SO42−(aq) → PbSO4(s) + 2e

Cathode: PbO2(s) + SO42−(aq) + 4H+(aq) + 2e → PbSO4(s) + 2H2O(l)

On recharging by an external source, these reactions are reversed, regenerating Pb, PbO2, and H2SO4. The nickel-cadmium (Ni-Cd) cell is another rechargeable battery, more expensive but longer-lasting than the lead storage battery, useful where a lightweight rechargeable source is needed.

A fuel cell is a galvanic cell that generates electricity continuously as long as fuel and oxidant are supplied from outside, unlike batteries where reactants are stored inside. The best-developed example is the hydrogen-oxygen fuel cell, used in spacecraft, where H2 and O2 gases are bubbled through porous carbon electrodes containing catalysts (Pt or Pd) into concentrated aqueous KOH electrolyte:

Cathode (reduction): O2(g) + 2H2O(l) + 4e → 4OH(aq)

Anode (oxidation): 2H2(g) + 4OH(aq) → 4H2O(l) + 4e

Overall: 2H2(g) + O2(g) → 2H2O(l)

Fuel cells are highly efficient and pollution-free (the only product is water), and are actively being developed as clean alternative power sources for vehicles and stationary power generation.

Lead storage battery (discharge, anode) Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻ Lead is oxidised at the anode during discharge.
Lead storage battery (discharge, cathode) PbO₂(s) + SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l) Lead dioxide is reduced at the cathode during discharge.
H₂–O₂ fuel cell (overall) 2H₂(g) + O₂(g) → 2H₂O(l) Net reaction of the hydrogen-oxygen fuel cell used in spacecraft; only by-product is water.
Remember
  • Primary batteries (e.g. dry cell, mercury cell) cannot be recharged; the electrode reactions are not reversible once reactants are consumed.
  • Dry cell (Leclanché cell) gives ~1.5 V using Zn anode and MnO2/carbon cathode; mercury cell gives a very steady ~1.35 V.
  • Secondary batteries (e.g. lead storage battery, Ni-Cd cell) can be recharged by passing current in the reverse direction, regenerating original reactants.
  • Lead storage battery: Pb anode, PbO2 cathode, dilute H2SO4 electrolyte; both electrodes form PbSO4 on discharge.
  • Fuel cells (e.g. H2-O2 fuel cell) generate electricity continuously from externally supplied fuel and oxidant, are highly efficient, and produce only water as by-product.
  • Unlike batteries, fuel cell reactants are not stored inside the cell — they are fed continuously, so the cell does not 'run down' as long as supply continues.

Corrosion

Quick answer Corrosion is a spontaneous electrochemical process in which a metal is slowly oxidised by atmospheric oxygen and moisture, most commonly seen as the rusting of iron.

Corrosion is the slow, spontaneous degradation of metals by the action of atmospheric gases, moisture, and other substances in their environment, converting the metal into an oxide, sulphide, carbonate, or similar compound. It is fundamentally an electrochemical phenomenon: it involves oxidation of the metal at one region (behaving as anode) and reduction of atmospheric oxygen at another region (behaving as cathode) of the same metal surface, with the moisture film on the metal acting as the electrolyte connecting these micro-electrochemical cells.

The most familiar example is the rusting of iron. At the anodic region of the iron surface, iron is oxidised:

Fe(s) → Fe2+(aq) + 2e

The electrons released flow through the metal to a nearby cathodic region, where dissolved oxygen (in the presence of H+ ions, from atmospheric CO2 dissolving in the moisture film) is reduced:

O2(g) + 4H+(aq) + 4e → 2H2O(l)

The Fe2+ formed is further oxidised by dissolved oxygen to Fe3+, which combines with water to form hydrated iron(III) oxide, Fe2O3·xH2O, commonly known as rust. Corrosion causes enormous economic loss by damaging buildings, bridges, ships, and all objects made of iron, and can also weaken structures leading to safety hazards.

Corrosion can be minimised by several practical methods:

  • Barrier protection: covering the metal surface with paint, oil, grease, or a coating so that moisture and oxygen cannot reach it.
  • Galvanisation: coating iron with a layer of a more reactive metal like zinc; even if the coating is scratched, zinc (being more easily oxidised) corrodes preferentially, protecting the iron underneath.
  • Cathodic (sacrificial) protection: connecting the iron structure to a more active metal (like magnesium or zinc) which acts as a sacrificial anode and is oxidised in preference to iron, commonly used to protect ships' hulls and underground pipelines.
  • Alloying: mixing iron with other metals (e.g. chromium and nickel) to form stainless steel, which resists corrosion much better than pure iron.

Corrosion prevention is essential in engineering and construction, and understanding its electrochemical basis (formation of tiny anodic and cathodic regions on the same metal surface) is key to designing effective protective strategies.

Anodic reaction (rusting) Fe(s) → Fe²⁺(aq) + 2e⁻ Iron is oxidised at the anodic spot on the metal surface.
Cathodic reaction (rusting) O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l) Dissolved oxygen is reduced at the cathodic spot, consuming the electrons released by iron oxidation.
Remember
  • Corrosion is a spontaneous electrochemical process where a metal is oxidised by atmospheric oxygen and moisture over time.
  • In rusting of iron, one region of the metal surface acts as anode (Fe → Fe²⁺ + 2e⁻) and another as cathode (O2 reduction), with the moisture film as electrolyte.
  • Fe²⁺ is further oxidised to Fe³⁺, which forms hydrated iron(III) oxide, Fe2O3·xH2O, known as rust.
  • Prevention methods include barrier protection (paint/grease/coating), galvanisation with zinc, cathodic/sacrificial protection with a more reactive metal, and alloying (e.g. stainless steel).
  • Corrosion causes major economic loss and structural safety concerns, making prevention an important practical application of electrochemistry.

Key equations

Every formula in this chapter, in one place — screenshot it before your exam.

E°cell = E°cathode − E°anode
Standard cell potentialV
E°(H⁺/H₂) = 0.00 V
SHE reference potential
Ecell = E°cell − (RT/nF) ln Q
Nernst equation (general)V
Ecell = E°cell − (0.0591/n) log Q
Nernst equation at 298 KV
ΔG = −nFEcell ; ΔG° = −nFE°cell
Gibbs energy–EMF relationJ mol⁻¹
log K = nE°cell / 0.0591
Equilibrium constant from E°cell
κ = G*/R = (l/A)/R
Conductivity from resistanceS cm⁻¹
Λm = κ × 1000 / C
Molar conductivityS cm² mol⁻¹
Λm° = ν₊λ°₊ + ν₋λ°₋
Kohlrausch's lawS cm² mol⁻¹
α = Λm / Λm°
Degree of dissociation
Ka = Cα² / (1 − α)
Ostwald's dilution lawmol L⁻¹
Q = I × t
Charge passedC
m = Z × Q = Z × I × t
Faraday's first lawg
Z = M / (n × F)
Electrochemical equivalentg C⁻¹
m = (M × I × t) / (n × F)
Combined Faraday's lawg
Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻
Lead storage battery (discharge, anode)
PbO₂(s) + SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l)
Lead storage battery (discharge, cathode)
2H₂(g) + O₂(g) → 2H₂O(l)
H₂–O₂ fuel cell (overall)
Fe(s) → Fe²⁺(aq) + 2e⁻
Anodic reaction (rusting)
O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l)
Cathodic reaction (rusting)

Test yourself

Tap an answer to check it instantly — you'll see why it's right, and what to revise if it isn't.

0 correct · 0/12 answered
Q1 Electrochemical cells easy

In a galvanic cell, oxidation takes place at the:

Q2 Electrode potential easy

The standard reduction potential of the Standard Hydrogen Electrode (SHE) is taken as:

Q3 Electrochemical cells easy

In the Daniell cell Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s), the negative terminal (anode) is:

Q4 Conductance easy

The SI-derived unit commonly used for molar conductivity is:

Q5 Conductance medium

On dilution, the molar conductivity of a weak electrolyte:

Q6 Faraday's laws medium

How many Faradays of charge are required to completely reduce 1 mole of MnO₄⁻ to Mn²⁺?

Q7 Nernst equation medium

For the cell Zn(s) | Zn²⁺(1 M) || Cu²⁺(0.01 M) | Cu(s) at 298 K with E°cell = 1.10 V, the cell potential Ecell is approximately:

Q8 Kohlrausch's law medium

Given Λm°(HCl) = 425.9, Λm°(CH₃COONa) = 91.0, and Λm°(NaCl) = 126.4 S cm² mol⁻¹, the value of Λm°(CH₃COOH) by Kohlrausch's law is closest to:

Q9 Nernst equation hard

For a redox reaction with n = 2 and E°cell = 0.0591 V at 298 K, the equilibrium constant K is closest to:

Q10 Faraday's laws hard

A current of 2 A is passed through a solution of AgNO₃ until 5.4 g of Ag (atomic mass 108) is deposited. The approximate time required is:

Q11 Fuel cells hard

In an alkaline H₂-O₂ fuel cell, the correct cathode (reduction) half-reaction is:

Q12 Corrosion hard

During the rusting of iron, which statement correctly describes the electrochemical mechanism?

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 Depict the galvanic cell in which the reaction Mg(s) + 2Ag⁺(aq) → Mg²⁺(aq) + 2Ag(s) takes place. Further show: (i) which electrode is negative and which is positive, (ii) the carriers of current in the cell, (iii) individual reactions at each electrode. Also calculate the standard emf given E°(Ag⁺/Ag) = +0.80 V and E°(Mg²⁺/Mg) = −2.37 V.Galvanic cells and cell notation

In this reaction, Mg is oxidised (loses electrons) and Ag⁺ is reduced (gains electrons), so Mg is the anode and Ag is the cathode.

Cell representation: Mg(s) | Mg2+(aq) || Ag+(aq) | Ag(s)

(i) Mg electrode (anode) is the negative terminal; Ag electrode (cathode) is the positive terminal.

(ii) Inside the cell, current is carried by the migration of ions (Mg2+ moving away from the anode, Ag+ moving toward the cathode, and ions in the salt bridge maintaining neutrality); in the external circuit, current is carried by electrons flowing from Mg (anode) to Ag (cathode).

(iii) Electrode reactions:

Anode (oxidation): Mg(s) → Mg2+(aq) + 2e

Cathode (reduction): 2Ag+(aq) + 2e → 2Ag(s)

Standard emf: E°cell = E°cathode − E°anode = E°(Ag+/Ag) − E°(Mg2+/Mg) = 0.80 − (−2.37) = 3.17 V

2 Calculate the standard cell potential and ΔrG° for the reaction 2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd(s), given E°(Cr³⁺/Cr) = −0.74 V and E°(Cd²⁺/Cd) = −0.40 V.Standard cell potential and Gibbs energy

Here Cd2+ is reduced (cathode) and Cr is oxidised (anode). The number of electrons transferred, n, is found from balancing: 2Cr → 2Cr3+ + 6e and 3Cd2+ + 6e → 3Cd, so n = 6.

cell = E°cathode − E°anode = E°(Cd2+/Cd) − E°(Cr3+/Cr) = −0.40 − (−0.74) = +0.34 V

Since E°cell is positive, the reaction is spontaneous as written.

ΔrG° = −nFE°cell = −(6)(96500 C mol−1)(0.34 V) = −196860 J mol−1 = −196.86 kJ mol−1

3 Write the Nernst equation and calculate the emf of the cell: Mg(s) | Mg²⁺(0.001 M) || Cu²⁺(0.0001 M) | Cu(s), given E°(Cu²⁺/Cu) = +0.34 V and E°(Mg²⁺/Mg) = −2.37 V.Nernst equation

Cell reaction: Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s), with n = 2.

cell = E°(Cu2+/Cu) − E°(Mg2+/Mg) = 0.34 − (−2.37) = 2.71 V

Nernst equation: Ecell = E°cell − (0.0591/n) log ([Mg2+]/[Cu2+])

Q = [Mg2+]/[Cu2+] = 0.001/0.0001 = 10, so log Q = 1

Ecell = 2.71 − (0.0591/2)(1) = 2.71 − 0.0296 = 2.68 V

4 The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm⁻¹. Calculate its molar conductivity.Molar conductivity calculation

Given κ = 0.0248 S cm−1, C = 0.20 mol L−1.

Λm = κ × 1000/C = (0.0248 × 1000)/0.20 = 24.8/0.20 = 124 S cm2 mol−1

5 How many coulombs of charge are required to reduce 1 mole of Cr₂O₇²⁻ ions to Cr³⁺ ions?Faraday's laws of electrolysis

The reduction half-reaction is: Cr2O72− + 14H+ + 6e → 2Cr3+ + 7H2O

So 1 mole of Cr2O72− requires 6 moles of electrons, i.e. 6 Faradays of charge.

Charge required = 6 × F = 6 × 96500 C mol−1 = 5.79 × 105 C

6 Suggest two ways by which the corrosion (rusting) of an iron object such as a ship's hull or an underground pipeline can be prevented.Corrosion prevention

1. Barrier protection / galvanisation: Coating the iron surface with paint, grease, or a layer of a more reactive metal such as zinc (galvanisation) prevents moisture and oxygen from reaching the iron surface. Even if the zinc coating is scratched, zinc being more reactive than iron gets oxidised preferentially, protecting the iron underneath.

2. Cathodic (sacrificial) protection: The iron structure (e.g. ship's hull or pipeline) is connected by a wire to a block of a more reactive metal such as magnesium or zinc. This sacrificial anode gets oxidised in preference to iron, because it has a lower (more negative) reduction potential, thereby protecting the iron from corrosion until the sacrificial metal itself is consumed and needs replacement.

Previous-year board questions 4

Q1 Calculate the emf of the cell in which the following reaction takes place: Ni(s) + 2Ag⁺(0.002 M) → Ni²⁺(0.160 M) + 2Ag(s). Given E°cell = 1.05 V. 2023 3 marks

Here n = 2 (two electrons transferred per formula unit of reaction).

Reaction quotient: Q = [Ni2+]/[Ag+]2 = 0.160/(0.002)2 = 0.160/0.000004 = 40000

log Q = log(4 × 104) = 4.602

Using the Nernst equation: Ecell = E°cell − (0.0591/n) log Q

Ecell = 1.05 − (0.0591/2)(4.602) = 1.05 − (0.02955)(4.602) = 1.05 − 0.136 = 0.914 V

Q2 The molar conductivity of 0.025 mol L⁻¹ methanoic acid (HCOOH) solution is 46.1 S cm² mol⁻¹. Given λ°(H⁺) = 349.6 S cm² mol⁻¹ and λ°(HCOO⁻) = 54.6 S cm² mol⁻¹, calculate the degree of dissociation and the dissociation constant of methanoic acid. 2022 4 marks

By Kohlrausch's law: Λ°m(HCOOH) = λ°(H+) + λ°(HCOO) = 349.6 + 54.6 = 404.2 S cm2 mol−1

Degree of dissociation: α = Λm/Λ°m = 46.1/404.2 = 0.1140 (11.40%)

Using the Ostwald dilution law: Ka = Cα2/(1−α)

Ka = (0.025 × (0.114)2)/(1 − 0.114) = (0.025 × 0.012996)/0.886 = 0.0003249/0.886 = 3.67 × 10−4 mol L−1

Q3 Explain why the conductivity of an electrolytic solution decreases on dilution, while its molar conductivity increases. 2023 2 marks

Conductivity (κ) depends on the number of ions present per unit volume of solution. On dilution, the concentration of ions per unit volume decreases, so fewer charge carriers are present in a given volume, and κ decreases.

Molar conductivity (Λm), however, is the conductivity contributed by one mole of the electrolyte, given by Λm = κ × 1000/C. On dilution, C decreases faster (or, equivalently, the volume containing one mole of electrolyte increases) than κ decreases. Additionally, for weak electrolytes, dilution increases the degree of dissociation, releasing more ions per mole. Both effects together cause Λm to increase on dilution, even though κ itself decreases.

Q4 State Faraday's laws of electrolysis. A solution of Ni(NO₃)₂ is electrolysed for 20 minutes with a current of 5 A. What mass of nickel will be deposited at the cathode? (Ni = 58.7 g mol⁻¹) 2022 5 marks

Faraday's first law: The mass of a substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electric charge (Q) passed through the electrolyte: m = ZQ = ZIt, where Z is the electrochemical equivalent.

Faraday's second law: When the same quantity of charge is passed through different electrolytes connected in series, the masses of substances deposited/liberated are proportional to their chemical equivalent weights (M/n).

Numerical part:

Ni2+ + 2e → Ni, so n = 2.

Charge passed: Q = It = 5 A × (20 × 60) s = 5 × 1200 = 6000 C

Mass deposited: m = (M × Q)/(n × F) = (58.7 × 6000)/(2 × 96500) = 352200/193000 = 1.825 g of nickel

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