Electrochemistry Notes for NEET: Class 12 NCERT Chemistry

Q: How many questions come from Electrochemistry in NEET each year?

Electrochemistry typically gives 3 to 5 questions per NEET paper. The most common topics are standard electrode potential and cell EMF calculation (1 to 2 questions), the Nernst equation (1 question), and Faraday's laws (1 question). Conductance and batteries also appear occasionally. This makes it one of the highest-yield chapters in Class 12 Chemistry.

Q: What is standard electrode potential and what does a positive or negative value mean?

The standard electrode potential (E°) is the voltage of a half-cell measured under standard conditions (1 M concentration, 1 bar pressure, 25°C) relative to the Standard Hydrogen Electrode (SHE), which is assigned E° = 0.00 V. A positive E° means the half-cell is a stronger oxidising agent than hydrogen; it tends to accept electrons (get reduced). A negative E° means it is a stronger reducing agent than hydrogen; it tends to lose electrons (get oxidised). The more positive the E°, the stronger the oxidising agent.

Q: How do you identify the anode and cathode in a galvanic cell versus an electrolytic cell?

In a galvanic cell (spontaneous), the anode is negative (oxidation occurs here; the electrode loses electrons) and the cathode is positive (reduction occurs here; the electrode gains electrons). In an electrolytic cell (non-spontaneous, driven by external power), the anode is connected to the positive terminal of the battery and the cathode to the negative terminal. The rule is: oxidation always occurs at the anode, and reduction always occurs at the cathode, in both types of cells.

Q: What is the Nernst equation and when do you use it?

The Nernst equation is E = E° - (0.0591/n) x log Q at 25°C, where E is the actual cell potential, E° is the standard cell potential, n is the number of electrons transferred, and Q is the reaction quotient (products/reactants using concentrations or pressures). You use it whenever concentrations differ from 1 M. A common NEET use is concentration cells, where both electrodes are the same metal but in solutions of different concentrations; E° = 0, so E = -(0.0591/n) x log Q. At equilibrium, E = 0 and Q = K, giving the relationship between E° and the equilibrium constant.

Q: What is the difference between conductance and conductivity?

Conductance (G) is the ability of a conductor to allow current to flow; it is the reciprocal of resistance (G = 1/R) and is measured in siemens (S). Conductivity (kappa, k) is the conductance of a 1 cm cube of solution; it is measured in S cm-1 or S m-1. Molar conductivity (lambda_m) is the conductance of a solution containing 1 mole of electrolyte, with electrodes 1 cm apart, measured in S cm2 mol-1. As you dilute a solution, conductivity decreases (fewer ions per cm3) but molar conductivity increases (more volume per mole, so ions move more freely). At infinite dilution, molar conductivity reaches its maximum limiting value.

Q: How do you apply Faraday's first and second laws in calculation problems?

Faraday's first law states that the mass of substance deposited is directly proportional to the charge passed: m = Z x I x t, where Z is the electrochemical equivalent (Z = M / (n x F)), I is current in amperes, t is time in seconds, M is molar mass, n is the number of electrons per ion, and F = 96500 C/mol (Faraday's constant). Practical formula: m = (M x I x t) / (n x 96500). Faraday's second law states that for the same charge, the masses of different substances deposited are proportional to their equivalent masses (M/n). So if you pass the same charge through two different electrolytic cells, the ratio of masses deposited equals the ratio of their equivalent masses.

Q: What is Kohlrausch's law and how do you apply it?

Kohlrausch's law states that at infinite dilution, the molar conductivity of an electrolyte equals the sum of the individual molar conductivities of its ions (lambda_m_infinity = v+ x lambda+_infinity + v- x lambda-_infinity, where v+ and v- are the number of cations and anions per formula unit). You use it to find the limiting molar conductivity of a weak electrolyte like acetic acid (CH3COOH), which cannot be measured directly (it would require infinite dilution in practice). You calculate it indirectly: lambda_m_infinity(CH3COOH) = lambda_m_infinity(HCl) + lambda_m_infinity(CH3COONa) - lambda_m_infinity(NaCl). Then you can find the degree of dissociation of the weak electrolyte: alpha = lambda_m / lambda_m_infinity.

Q: What is a fuel cell and how is it different from a battery?

A fuel cell is an electrochemical cell that converts chemical energy of a fuel (like hydrogen) into electrical energy continuously, as long as fuel and oxidant are supplied from outside. The reactants are not stored inside the cell. In a hydrogen-oxygen fuel cell, hydrogen is oxidised at the anode (H2 - 2e- gives 2H+) and oxygen is reduced at the cathode (O2 + 4H+ + 4e- gives 2H2O). The only product is water, making it environmentally clean. A battery, by contrast, contains a fixed amount of reactants sealed inside; once the reactants are used up, a primary battery is discarded or a secondary battery is recharged by reversing the reaction.

Introduction

Electrochemistry connects chemistry and electricity. It studies how chemical reactions produce electrical energy (galvanic cells) and how electrical energy drives chemical reactions (electrolysis). For NEET, this chapter typically gives 3 to 5 questions per year and is one of the highest-yield topics in Class 12 Chemistry.

The most tested areas are: cell EMF calculations (E°cell = E°cathode − E°anode), the Nernst equation, molar conductivity and Kohlrausch's law, and Faraday's laws. Batteries and corrosion appear as conceptual questions. You should know the standard electrode potential table and be comfortable with numerical problems on all four areas.

Galvanic Cells and Electrode Potential

The Daniell Cell

A galvanic cell (also called a voltaic cell) converts chemical energy into electrical energy through a spontaneous redox reaction. The classic example is the Daniell cell, which uses zinc and copper electrodes.

Anode (negative electrode):zinc rod dipped in ZnSO₄ solution. Oxidation occurs here: Zn(s) → Zn²⁺(aq) + 2e⁻. Zinc loses electrons and dissolves into solution.
Cathode (positive electrode): copper rod dipped in CuSO₄ solution. Reduction occurs here: Cu²⁺(aq) + 2e⁻→ Cu(s). Copper ions deposit as solid copper on the electrode.
Salt bridge: a U-shaped tube filled with concentrated KCl or KNO₃ in agar gel. It allows ions to flow between the two half-cells to maintain electrical neutrality, but prevents the solutions from mixing.
External circuit: electrons flow through the wire from the anode (Zn) to the cathode (Cu). This electron flow is the electrical current you measure.

The cell notation for the Daniell cell is written as:

Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)

The single vertical line (|) represents a phase boundary between the electrode and the electrolyte. The double line (||) represents the salt bridge. The anode is always written on the left, the cathode on the right.

Standard Electrode Potential (E°)

Each half-cell has a tendency to undergo reduction (gain of electrons). This tendency is measured as the standard electrode potential (E°), defined under standard conditions: 1 M concentration of ions, 1 bar pressure for gases, and 25°C.

All E° values are measured relative to the Standard Hydrogen Electrode (SHE), which is assigned E° = 0.00 V. The SHE consists of a platinum electrode in 1 M H⁺solution with H₂ gas bubbling at 1 bar.

Half-reaction (reduction)	E° (V)
F₂ + 2e⁻ → 2F⁻	+2.87
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O	+1.51
Cl₂ + 2e⁻ → 2Cl⁻	+1.36
O₂ + 4H⁺ + 4e⁻ → 2H₂O	+1.23
Fe³⁺ + e⁻ → Fe²⁺	+0.77
Cu²⁺ + 2e⁻ → Cu	+0.34
2H⁺ + 2e⁻ → H₂ (SHE reference)	0.00
Pb²⁺ + 2e⁻ → Pb	−0.13
Ni²⁺ + 2e⁻ → Ni	−0.25
Fe²⁺ + 2e⁻ → Fe	−0.44
Zn²⁺ + 2e⁻ → Zn	−0.76
Al³⁺ + 3e⁻ → Al	−1.66
Na⁺ + e⁻ → Na	−2.71
Li⁺ + e⁻ → Li	−3.04

Key rules: A more positive E° means a stronger oxidising agent (more easily reduced). A more negative E° means a stronger reducing agent (more easily oxidised). In a galvanic cell, the half-cell with the higher E° acts as the cathode (reduced) and the one with the lower E° acts as the anode (oxidised).

Cell EMF Calculation

The electromotive force (EMF) of a cell, also called the cell potential, is calculated as:

E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ}

For the Daniell cell: E°cell = E°(Cu²⁺/Cu) − E°(Zn²⁺/Zn) = +0.34 − (−0.76) = +1.10 V.

A positive E°cell means the cell reaction is spontaneous (the cell can do work). A negative E°cell means the reaction is non-spontaneous and requires external energy. This is directly related to Gibbs energy:

Δ G^{\circ} = - n F E_{ce l l}^{\circ}

where n is the number of electrons transferred and F = 96500 C/mol (Faraday's constant). When E°cell > 0, ΔG° < 0 (spontaneous).

Select two half-cells. The one with the higher E° acts as the cathode (reduction). The one with the lower E° acts as the anode (oxidation). E°cell = E°cathode − E°anode.

Cathode (reduction)

F₂/F⁻

+2.87 V

MnO₄⁻/Mn²⁺

+1.51 V

Cl₂/Cl⁻

+1.36 V

Cr₂O₇²⁻/Cr³⁺

+1.33 V

O₂/H₂O

+1.23 V

Fe³⁺/Fe²⁺

+0.77 V

Cu²⁺/Cu

+0.34 V

H⁺/H₂ (SHE)

+0.00 V

Pb²⁺/Pb

-0.13 V

Ni²⁺/Ni

-0.25 V

Fe²⁺/Fe

-0.44 V

Zn²⁺/Zn

-0.76 V

Al³⁺/Al

-1.66 V

Na⁺/Na

-2.71 V

Li⁺/Li

-3.04 V

Anode (oxidation)

F₂/F⁻

+2.87 V

MnO₄⁻/Mn²⁺

+1.51 V

Cl₂/Cl⁻

+1.36 V

Cr₂O₇²⁻/Cr³⁺

+1.33 V

O₂/H₂O

+1.23 V

Fe³⁺/Fe²⁺

+0.77 V

Cu²⁺/Cu

+0.34 V

H⁺/H₂ (SHE)

+0.00 V

Pb²⁺/Pb

-0.13 V

Ni²⁺/Ni

-0.25 V

Fe²⁺/Fe

-0.44 V

Zn²⁺/Zn

-0.76 V

Al³⁺/Al

-1.66 V

Na⁺/Na

-2.71 V

Li⁺/Li

-3.04 V

Cell notation

Zn|Zn²⁺||Cu²⁺|Cu

Cathode (reduction): Cu²⁺ + 2e⁻ → Cu

Anode (oxidation, reversed): Zn²⁺ + 2e⁻ ← Zn

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

Spontaneous (E°cell > 0)

ΔG° = −nFE°cell is negative. The cell does electrical work spontaneously.

NEET tip: higher E° = stronger oxidising agent, lower E° = stronger reducing agent

The metal with the more negative E° is always the anode in a galvanic cell. In NEET, the most tested pair is Zn/Cu (Daniell cell, E°cell = +1.10 V).

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Nernst Equation and Gibbs Energy

The Nernst Equation

The standard electrode potential E° applies only at standard conditions (all concentrations = 1 M, all gases at 1 bar). When concentrations differ, you use the Nernst equation:

E = E^{\circ} - \frac{R T}{n F} ln Q

At 25°C, substituting R = 8.314 J/mol K and F = 96500 C/mol and converting ln to log₁₀:

E = E^{\circ} - \frac{0.0591}{n} lo g Q

where Q is the reaction quotient (concentrations of products over reactants, each raised to their stoichiometric coefficients). This is the form used in most NEET numericals.

Example for the Daniell cell (Zn + Cu²⁺ → Zn²⁺ + Cu): Q = [Zn²⁺]/[Cu²⁺]. If [Zn²⁺] = 0.1 M and [Cu²⁺] = 0.01 M:

E = 1.10 − (0.0591/2) × log(0.1/0.01) = 1.10 − 0.02955 × 1 = 1.07 V.

Relationship between E°, ΔG°, and K

At equilibrium, Q = K and E = 0. Substituting E = 0 into the Nernst equation:

E^{\circ} = \frac{0.0591}{n} lo g K

Combined with ΔG° = −nFE°cell and ΔG° = −RT ln K:

E°cell	ΔG°	K	Reaction
Positive	Negative	K > 1	Spontaneous; products favoured
Zero	Zero	K = 1	At equilibrium
Negative	Positive	K < 1	Non-spontaneous; reactants favoured

Concentration Cells

A concentration cell uses the same electrode material but different ion concentrations in the two half-cells. Since both electrodes are identical, E° = 0. The EMF arises purely from the concentration difference:

E = - \frac{0.0591}{n} lo g \frac{[ ion ] _{an o d e}}{[ ion ] _{c a t h o d e}}

The electrode in the more dilute solution acts as the anode (oxidation lowers concentration further), and the more concentrated side acts as the cathode.

The Nernst equation: E = E° − (0.0591 / n) × log Q at 25°C. Use it to find the actual cell potential when concentrations are not 1 M, or to calculate Q when you know E.

Daniell cell (Q=10)

Ag/Ag+ conc. (Q=0.01)

Fe³⁺/Fe²⁺ (Q=0.1)

At equilibrium (E=0)

E°cell (standard EMF)

+1.100 V

n (electrons transferred)

log₁₀(Q) reaction quotient

+1.0 (Q = 10^1.0 ≈ 1.0e+1)

Q > 1: products accumulate, E decreases

Nernst equation calculation

E = +1.100 − (0.0591 / 2) × 1.0 = +1.100 − 0.0295 = +1.0705 V

Cell is spontaneous

Key relationships (NEET)

Q < K

Reaction proceeds forward, E > 0

Q = K

Equilibrium, E = 0

Q > K

Reaction proceeds reverse, E < 0

E° = (0.0591/n) log K

Standard EMF and equilibrium constant

Conductance and Kohlrausch's Law

Resistance and Conductance

The electrical resistance (R) of a solution depends on the length (l) and cross-sectional area (A) of the column of solution between the electrodes:

R = ρ \frac{l}{A}

where ρ (rho) is the resistivity (specific resistance). Conductance (G) is the reciprocal of resistance:

G = \frac{1}{R} (unit: siemens, S)

Conductivity (κ, kappa) is the conductance of a 1 cm cube of solution:

κ = G \times \frac{l}{A} (unit: S cm^{- 1})

The quantity l/A is called the cell constant (measured by calibrating the cell with a known KCl solution).

Molar Conductivity

Molar conductivity (Λ_m) is the conductance of a solution containing 1 mole of electrolyte, with electrodes 1 cm apart:

Λ_{m} = \frac{κ \times 1000}{M} (unit: S cm^{2} mol^{- 1})

where M is the molarity in mol/L and κ is in S/cm.

On dilution:

Conductivity (κ) decreases because there are fewer ions per unit volume.
Molar conductivity (Λ_m) increasesbecause the ions have more volume to move freely and interionic interactions decrease. At infinite dilution (c → 0), Λ_m reaches its maximum value Λ°_m.

Strong vs Weak Electrolytes on Dilution

Property	Strong electrolyte (e.g. KCl)	Weak electrolyte (e.g. CH₃COOH)
Degree of dissociation at dilution	Already 100%, does not change	Increases toward 100% on dilution
Λ_m vs concentration	Increases slowly; linear graph (Debye-Hückel-Onsager equation)	Increases rapidly; curve shoots up at low concentration
Can extrapolate Λ°_m?	Yes, by plotting Λ_m vs √c and extrapolating to c = 0	No (sharp rise makes extrapolation unreliable); use Kohlrausch's law

Kohlrausch's Law of Independent Migration of Ions

At infinite dilution, each ion migrates independently of all other ions. The limiting molar conductivity is the sum of contributions from individual ions:

Λ_{m}^{\circ} = ν_{+} λ_{+}^{\circ} + ν_{-} λ_{-}^{\circ}

where ν₊ and ν₋are the number of cations and anions per formula unit, and λ° values are the limiting molar conductivities of individual ions.

Application: For acetic acid (weak electrolyte, cannot extrapolate directly):

Λ°_m(CH₃COOH) = Λ°_m(HCl) + Λ°_m(CH₃COONa) − Λ°_m(NaCl)

This works because H⁺ comes from HCl, CH₃COO⁻ comes from CH₃COONa, and Na⁺ and Cl⁻ contributions cancel out.

Degree of dissociation: α = Λ_m / Λ°_m. For acetic acid at a given concentration, if Λ_m = 10 S cm² mol⁻¹and Λ°_m = 390.5 S cm² mol⁻¹, then α = 10/390.5 = 0.026 (about 2.6% dissociated).

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Electrolysis and Faraday's Laws

Electrolytic Cells

An electrolytic cell uses an external electrical source to drive a non-spontaneous chemical reaction. It is the reverse of a galvanic cell.

Anode (+):connected to the positive terminal of the battery. Oxidation occurs here. In electrolysis of CuSO₄ with copper electrodes, the copper anode dissolves: Cu → Cu²⁺ + 2e⁻.
Cathode (−):connected to the negative terminal. Reduction occurs here. Cu²⁺ + 2e⁻ → Cu deposits on the cathode.

During electrolysis, the metal that deposits is determined by the electrode potential: among the ions present, the one with the highest reduction potential is reduced first.

Products at Electrodes (Preferential Discharge)

Electrolyte	Cathode product	Anode product
Dilute H₂SO₄ (Pt electrodes)	H₂ gas	O₂ gas
CuSO₄ (Pt electrodes)	Cu metal	O₂ gas
CuSO₄ (Cu electrodes)	Cu metal	Cu dissolves (anode dissolves)
NaCl solution	H₂ gas (not Na, too negative)	Cl₂ gas
Molten NaCl (fused)	Na metal	Cl₂ gas

Faraday's First Law

The mass of substance deposited (or dissolved) at an electrode is directly proportional to the quantity of charge passed:

m = Z \times I \times t = Z \times Q

where Z is the electrochemical equivalent: Z = M / (n × F). Substituting:

m = \frac{M \times I \times t}{n \times F}

M = molar mass (g/mol), I = current (A), t = time (s), n = number of electrons per ion (n-factor), F = 96500 C/mol.

Faraday's Second Law

When the same charge passes through two different electrolytic cells connected in series, the masses of substances deposited are proportional to their equivalent masses (M/n):

\frac{m _{1}}{m _{2}} = \frac{M _{1} / n _{1}}{M _{2} / n _{2}}

Example: Same charge through AgNO₃ (M = 108, n = 1) and CuSO₄ (M = 63.5, n = 2): m(Ag)/m(Cu) = (108/1)/(63.5/2) = 108/31.75 ≈ 3.40.

Key numbers to memorise:1 Faraday = 96500 C. Passing 1 F deposits 1 equivalent (1 mole of monovalent ion, or ½ mole of divalent ion, etc.).

Faraday's first law: m = (M × I × t) / (n × F). Enter current, time, molar mass, and n-factor to find the mass deposited during electrolysis.

Cu²⁺ → Cu

Ag⁺ → Ag

Al³⁺ → Al

H₂ gas

O₂ gas

Current (I)

2 A

Time (t)

1800 s (0.50 hr)

Molar mass (M)

63.5 g/mol

n-factor (electrons per ion)

n = 2

Cu²⁺: n=2, Ag⁺: n=1, Al³⁺: n=3, Fe³⁺: n=3, Fe²⁺: n=2

Step 1: Total charge passed

Q = I × t = 2 × 1800

= 3,600 C

Step 2: Moles of electrons

n_e = Q / F = 3,600 / 96500

= 0.0373 mol

Step 3: Moles deposited (n-factor = 2)

moles = n_e / n = 0.0373 / 2

= 0.0187 mol

Step 4: Mass deposited (M = 63.5 g/mol)

m = moles × M = 0.0187 × 63.5

= 1.184 g

Faraday's second law (same charge through two cells)

If the same charge is passed through a CuSO₄ cell (M/n = 63.5/2 = 31.75) and an AgNO₃ cell (M/n = 108/1 = 108), the ratio of masses deposited = 31.75 : 108. NEET frequently tests this as a ratio problem.

Batteries and Commercial Applications

Primary Batteries (non-rechargeable)

Dry cell (Leclanché cell): the common 1.5 V battery.

Anode: zinc container (Zn → Zn²⁺ + 2e⁻)
Cathode: graphite (carbon) rod surrounded by MnO₂ paste
Electrolyte: paste of NH₄Cl and ZnCl₂
Cathode reaction: MnO₂ + NH₄⁺ + e⁻ → MnO(OH) + NH₃
EMF: ~1.5 V; drops with use as NH₃ builds up

Mercury cell: used in hearing aids. Zinc anode, mercury(II) oxide (HgO) as cathode material, KOH as electrolyte. Constant voltage (~1.35 V).

Secondary Batteries (rechargeable)

Lead-acid battery (car battery):

Anode: spongy lead (Pb)
Cathode: lead dioxide (PbO₂)
Electrolyte: 38% (by mass) H₂SO₄ solution
Discharge: both electrodes form PbSO₄; H₂SO₄ is consumed; density of acid falls
Charging: reverse reactions restore Pb and PbO₂; H₂SO₄ is regenerated
EMF: ~2 V per cell; a 12 V battery has 6 cells

Nickel-cadmium (Ni-Cd) battery: rechargeable, used in power tools. Anode: Cd, cathode: NiO(OH), electrolyte: KOH. EMF ~1.2 V.

Lithium-ion (Li-ion) battery: used in phones and laptops. Anode: graphite (stores Li as LiC₆), cathode: lithium metal oxide (e.g. LiCoO₂), electrolyte: organic solvent with Li salt. High energy density (~3.7 V per cell). The lithium ions move between electrodes; the electrodes do not dissolve.

Fuel Cells

A fuel cell converts the chemical energy of a fuel (supplied continuously from outside) directly into electrical energy. Unlike a battery, it does not run down as long as fuel is supplied.

Hydrogen-oxygen fuel cell:

Anode: H₂ + 2OH⁻ → 2H₂O + 2e⁻ (in alkaline medium) or H₂ → 2H⁺ + 2e⁻ (in acidic)
Cathode: O₂ + 2H₂O + 4e⁻ → 4OH⁻ (alkaline)
Net: 2H₂ + O₂ → 2H₂O (only product is water)
Efficiency ~70%, much higher than combustion engines (~40%)
Used in space shuttles and being developed for vehicles

Corrosion

Corrosion of iron (rusting) is an electrochemical process. When iron is exposed to moisture and oxygen, an electrochemical cell forms on the surface:

Anodic areas:Fe → Fe²⁺ + 2e⁻ (oxidation; iron dissolves)
Cathodic areas:O₂ + 2H₂O + 4e⁻ → 4OH⁻ (reduction of dissolved O₂)
Fe²⁺ and OH⁻ combine → Fe(OH)₂ → further oxidised to Fe₂O₃·xH₂O (rust)

Dissolved salts (e.g. NaCl) accelerate corrosion by increasing solution conductivity. Corrosion is prevented by: painting, galvanising (zinc coating), tinning, alloying (stainless steel), cathodic protection (connecting iron to a more active metal like Mg or Zn).

Worked NEET Problems

NEET-style problem · Cell EMF and Gibbs Energy

Question

The standard electrode potentials of Zn²⁺/Zn and Ag⁺/Ag are −0.76 V and +0.80 V respectively. Calculate (a) E°cell for Zn|Zn²⁺||Ag⁺|Ag, and (b) ΔG° if n = 2.

Solution

(a) In the cell notation, Zn is the anode (left) and Ag is the cathode (right). E°cell = E°cathode − E°anode = +0.80 − (−0.76) = +1.56 V.

(b) ΔG° = −nFE°cell = −2 × 96500 × 1.56 = −301,080 J/mol = −301.1 kJ/mol.

Since E°cell > 0, ΔG° < 0: the cell reaction is spontaneous.

NEET-style problem · Nernst Equation

Question

For the cell Cu|Cu²⁺(0.01 M)||Ag⁺(0.1 M)|Ag with E°cell = +0.46 V at 25°C, calculate the cell potential. (n = 2, use 0.0591/2 = 0.02955)

Solution

Cell reaction: Cu + 2Ag⁺ → Cu²⁺ + 2Ag. Q = [Cu²⁺]/[Ag⁺]² = 0.01/(0.1)² = 0.01/0.01 = 1.

E = E° − (0.0591/2) × log Q = 0.46 − 0.02955 × log(1) = 0.46 − 0 = 0.46 V.

Since log 1 = 0, concentrations cancel here. The cell potential equals the standard value.

NEET-style problem · Conductance and Kohlrausch's Law

Question

The conductivity of 0.02 M acetic acid solution is 9.0 × 10⁻⁵ S cm⁻¹. If Λ°_m(CH₃COOH) = 390.5 S cm² mol⁻¹, find the degree of dissociation (α) and K_a.

Solution

Step 1: Molar conductivity Λ_m = (κ × 1000) / M = (9.0 × 10⁻⁵ × 1000) / 0.02 = 4.5 S cm² mol⁻¹.

Step 2: α = Λ_m / Λ°_m= 4.5 / 390.5 = 0.01152 ≈ 1.15%.

Step 3: K_a= cα² / (1 − α) ≈ cα² (since α << 1). K_a = 0.02 × (0.01152)² = 0.02 × 1.328 × 10⁻⁴ = 2.66 × 10⁻⁶.

NEET-style problem · Faraday's First Law

Question

A current of 3 A is passed through a AgNO₃ solution for 30 minutes. What mass of silver is deposited? (M(Ag) = 108 g/mol, n = 1, F = 96500 C/mol)

Solution

Charge Q = I × t = 3 × 1800 = 5400 C.

m = (M × Q) / (n × F) = (108 × 5400) / (1 × 96500) = 583200 / 96500 = 6.04 g.

NEET-style problem · Equilibrium Constant from E°

Question

The standard EMF of a cell is +0.295 V at 25°C and involves 1 electron transfer. What is the equilibrium constant K?

Solution

E° = (0.0591/n) × log K → 0.295 = (0.0591/1) × log K.

log K = 0.295 / 0.0591 = 4.991 ≈ 5.

K = 10⁵ = 100,000. The reaction strongly favours products.

Summary Cheat Sheet

Formula / Concept	Details
E°cell = E°cathode − E°anode	Cathode has higher E°; anode has lower E°. Positive E°cell = spontaneous.
ΔG° = −nFE°cell	n = electrons transferred; F = 96500 C/mol
E° = (0.0591/n) log K	Connects standard EMF to equilibrium constant at 25°C
E = E° − (0.0591/n) log Q	Nernst equation at 25°C. Q = products/reactants.
κ = G × (l/A)	Conductivity from conductance and cell constant
Λ_m = κ × 1000 / M	Molar conductivity (S cm² mol⁻¹); M in mol/L, κ in S/cm
Λ°_m = ν₊λ°₊ + ν₋λ°₋	Kohlrausch's law: limiting molar conductivity from individual ions
α = Λ_m / Λ°_m	Degree of dissociation of weak electrolyte at a given concentration
m = (M × I × t) / (n × F)	Faraday's first law: mass deposited at electrode
m₁/m₂ = (M₁/n₁)/(M₂/n₂)	Faraday's second law: masses for same charge in series cells
1 Faraday = 96500 C = 1 mole of electrons	Deposits 1 equivalent of substance
Dry cell: Zn \| MnO₂ \| C, 1.5 V	Primary battery; non-rechargeable
Lead-acid: Pb \| H₂SO₄ \| PbO₂, ~2 V/cell	Secondary battery; rechargeable; used in cars
H₂-O₂ fuel cell: 2H₂ + O₂ → 2H₂O	Continuous fuel supply; only product is water

Frequently asked questions

How many questions come from Electrochemistry in NEET each year?

Electrochemistry typically gives 3 to 5 questions per NEET paper. The most common topics are standard electrode potential and cell EMF calculation (1 to 2 questions), the Nernst equation (1 question), and Faraday's laws (1 question). Conductance and batteries also appear occasionally. This makes it one of the highest-yield chapters in Class 12 Chemistry.

What is standard electrode potential and what does a positive or negative value mean?

The standard electrode potential (E°) is the voltage of a half-cell measured under standard conditions (1 M concentration, 1 bar pressure, 25°C) relative to the Standard Hydrogen Electrode (SHE), which is assigned E° = 0.00 V. A positive E° means the half-cell is a stronger oxidising agent than hydrogen; it tends to accept electrons (get reduced). A negative E° means it is a stronger reducing agent than hydrogen; it tends to lose electrons (get oxidised). The more positive the E°, the stronger the oxidising agent.

How do you identify the anode and cathode in a galvanic cell versus an electrolytic cell?

In a galvanic cell (spontaneous), the anode is negative (oxidation occurs here; the electrode loses electrons) and the cathode is positive (reduction occurs here; the electrode gains electrons). In an electrolytic cell (non-spontaneous, driven by external power), the anode is connected to the positive terminal of the battery and the cathode to the negative terminal. The rule is: oxidation always occurs at the anode, and reduction always occurs at the cathode, in both types of cells.

What is the Nernst equation and when do you use it?

The Nernst equation is E = E° - (0.0591/n) x log Q at 25°C, where E is the actual cell potential, E° is the standard cell potential, n is the number of electrons transferred, and Q is the reaction quotient (products/reactants using concentrations or pressures). You use it whenever concentrations differ from 1 M. A common NEET use is concentration cells, where both electrodes are the same metal but in solutions of different concentrations; E° = 0, so E = -(0.0591/n) x log Q. At equilibrium, E = 0 and Q = K, giving the relationship between E° and the equilibrium constant.

What is the difference between conductance and conductivity?

Conductance (G) is the ability of a conductor to allow current to flow; it is the reciprocal of resistance (G = 1/R) and is measured in siemens (S). Conductivity (kappa, k) is the conductance of a 1 cm cube of solution; it is measured in S cm-1 or S m-1. Molar conductivity (lambda_m) is the conductance of a solution containing 1 mole of electrolyte, with electrodes 1 cm apart, measured in S cm2 mol-1. As you dilute a solution, conductivity decreases (fewer ions per cm3) but molar conductivity increases (more volume per mole, so ions move more freely). At infinite dilution, molar conductivity reaches its maximum limiting value.

How do you apply Faraday's first and second laws in calculation problems?

Faraday's first law states that the mass of substance deposited is directly proportional to the charge passed: m = Z x I x t, where Z is the electrochemical equivalent (Z = M / (n x F)), I is current in amperes, t is time in seconds, M is molar mass, n is the number of electrons per ion, and F = 96500 C/mol (Faraday's constant). Practical formula: m = (M x I x t) / (n x 96500). Faraday's second law states that for the same charge, the masses of different substances deposited are proportional to their equivalent masses (M/n). So if you pass the same charge through two different electrolytic cells, the ratio of masses deposited equals the ratio of their equivalent masses.

What is Kohlrausch's law and how do you apply it?

Kohlrausch's law states that at infinite dilution, the molar conductivity of an electrolyte equals the sum of the individual molar conductivities of its ions (lambda_m_infinity = v+ x lambda+_infinity + v- x lambda-_infinity, where v+ and v- are the number of cations and anions per formula unit). You use it to find the limiting molar conductivity of a weak electrolyte like acetic acid (CH3COOH), which cannot be measured directly (it would require infinite dilution in practice). You calculate it indirectly: lambda_m_infinity(CH3COOH) = lambda_m_infinity(HCl) + lambda_m_infinity(CH3COONa) - lambda_m_infinity(NaCl). Then you can find the degree of dissociation of the weak electrolyte: alpha = lambda_m / lambda_m_infinity.

What is a fuel cell and how is it different from a battery?

A fuel cell is an electrochemical cell that converts chemical energy of a fuel (like hydrogen) into electrical energy continuously, as long as fuel and oxidant are supplied from outside. The reactants are not stored inside the cell. In a hydrogen-oxygen fuel cell, hydrogen is oxidised at the anode (H2 - 2e- gives 2H+) and oxygen is reduced at the cathode (O2 + 4H+ + 4e- gives 2H2O). The only product is water, making it environmentally clean. A battery, by contrast, contains a fixed amount of reactants sealed inside; once the reactants are used up, a primary battery is discarded or a secondary battery is recharged by reversing the reaction.

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Solutions

Chemical Kinetics

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ElectrochemistryNEET Chemistry · Class 12 · NCERT Chapter 2

Introduction

Galvanic Cells and Electrode Potential

The Daniell Cell

Standard Electrode Potential (E°)

Cell EMF Calculation

Nernst Equation and Gibbs Energy

The Nernst Equation

Relationship between E°, ΔG°, and K

Concentration Cells

Conductance and Kohlrausch's Law

Resistance and Conductance

Molar Conductivity

Strong vs Weak Electrolytes on Dilution

Kohlrausch's Law of Independent Migration of Ions

Electrolysis and Faraday's Laws

Electrolytic Cells

Products at Electrodes (Preferential Discharge)

Faraday's First Law

Faraday's Second Law

Batteries and Commercial Applications

Primary Batteries (non-rechargeable)

Secondary Batteries (rechargeable)

Fuel Cells

Corrosion

Worked NEET Problems

Summary Cheat Sheet

Frequently asked questions

How many questions come from Electrochemistry in NEET each year?

What is standard electrode potential and what does a positive or negative value mean?

How do you identify the anode and cathode in a galvanic cell versus an electrolytic cell?

What is the Nernst equation and when do you use it?

What is the difference between conductance and conductivity?

How do you apply Faraday's first and second laws in calculation problems?

What is Kohlrausch's law and how do you apply it?

What is a fuel cell and how is it different from a battery?

Continue with the next chapter notes

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