From the battery in your phone to the rust on an old gate — the chemistry of electrons moving through circuits and ions moving through solutions. With interactive Nernst calculator, Faraday's laws calculator, animated cell diagrams, and 20+ real-life examples.
Before diving in, here's the entire chapter on one page. This is your mental map — how cells, EMF, Nernst equation, conductance, and electrolysis all connect.
Electrochemistry studies the interconversion of chemical and electrical energy. Galvanic cells (spontaneous) generate EMF; electrolytic cells (non-spontaneous) consume electricity. Nernst equation links EMF to concentration. Conductance and Kohlrausch's law describe ion mobility. Faraday's laws quantify electrolysis. Applications: batteries, corrosion prevention, electroplating, fuel cells.
Electrochemical cell: A device that converts chemical energy of a redox reaction into electrical energy (galvanic cell) or vice versa (electrolytic cell).
Galvanic (Voltaic) cell: Produces electricity from a spontaneous redox reaction.
Electrolytic cell: Uses electricity to drive a non-spontaneous reaction.
In both cases, oxidation happens at anode and reduction happens at cathode.
Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)
Electrons flow from Zn to Cu through external circuit.
Salt bridge (KCl): Connects the two solutions. Allows ion flow (K⁺ to cathode, Cl⁻ to anode) to maintain electrical neutrality. Prevents mixing of solutions.
Anode | Anode solution || Cathode solution | Cathode
Single vertical line | = phase boundary
Double vertical line || = salt bridge
For Daniell cell: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)
Left = anode (oxidation), right = cathode (reduction)
Standard electrode potential: The potential difference between an electrode and the Standard Hydrogen Electrode (SHE) when the concentration of ions is 1 M, pressure is 1 bar, at 298 K.
SHE: Pt | H₂ (1 bar) | H⁺ (1 M). Assigned E° = 0.00 V by convention.
Half-cell with reference: E°_cell = E°_cathode − E°_anode
For Daniell cell: E°_cell = E°(Cu²⁺/Cu) − E°(Zn²⁺/Zn) = 0.34 − (−0.76) = 1.10 V
Galvanic cell = spontaneous redox → electricity. Anode = oxidation (Zn), cathode = reduction (Cu). Cell notation: anode | solution || solution | cathode. E°_cell = E°_cathode − E°_anode. Electrochemical series ranks elements by reducing/oxidizing power.
The standard cell potential E°_cell applies at 1 M concentration. For real cells, concentration affects EMF. The Nernst equation quantifies this.
For a general cell reaction: aA + bB → cC + dD
E_cell = E°_cell − (RT/nF) ln Q
Where Q = [C]^c [D]^d / [A]^a [B]^b (reaction quotient)
At 298 K (25°C), using base-10 log:
E_cell = E°_cell − (0.0591/n) log Q
Where n = number of electrons transferred in the balanced equation.
Reaction: Zn + Cu²⁺ → Zn²⁺ + Cu (n = 2)
Q = [Zn²⁺] / [Cu²⁺]
E_cell = E°_cell − (0.0591/2) log([Zn²⁺]/[Cu²⁺])
E°_cell = 1.10 V
Enter concentrations and temperature — calculate actual cell potential.
Think of a river flowing downhill. E°_cell is the "standard drop." If the "bottom" (product side) is crowded (high concentration), it's harder for water to flow → smaller drop. If "bottom" is empty (low product concentration), flow is easier → larger drop. Nernst equation tells you how the "crowding" affects the flow (voltage).
Mistake: "0.0591 is always divided by 2."
Correction:
0.0591 is divided by n, the number of electrons
transferred in the specific reaction. For Li-ion battery, n = 1
(per Li⁺). For Al³⁺/Al, n = 3. Always check the balanced equation!
Nernst: E = E° − (0.0591/n) log Q at 298 K. Q = products/reactants. Higher product concentration → lower E. At equilibrium E = 0, so K relates to E°_cell. Essential for concentration cells, pH meters, and understanding real battery behaviour.
ΔG = −nFE
ΔG° = −nFE°_cell
Where:
ΔG° = −RT ln K
Combining: −nFE°_cell = −RT ln K → E°_cell = (RT/nF) ln K
At 298 K: E°_cell = (0.0591/n) log K
| ΔG | E_cell | K | Meaning |
|---|---|---|---|
| ΔG < 0 | E > 0 | K > 1 | Spontaneous — galvanic cell works |
| ΔG = 0 | E = 0 | K = 1 | At equilibrium — dead battery |
| ΔG > 0 | E < 0 | K < 1 | Non-spontaneous — needs external voltage |
Maximum electrical work = −ΔG (at constant T, P)
W_max = nFE_cell (joules)
For Daniell cell with 1 mol Zn reacting: W = 2 × 96485 × 1.10 = 2.12 × 10⁵ J/mol ≈ 212 kJ/mol
Faraday's constant: F = N_A × e = 6.022×10²³ × 1.602×10⁻¹⁹ = 96,485 C/mol
ΔG = −nFE. Spontaneous redox → ΔG < 0 → E > 0. At equilibrium, E = 0, K from E°_cell = (0.0591/n) log K. Maximum work = nFE. Links thermodynamics to electrochemistry.
Resistance (R): Opposition to flow of current. Units: Ω (ohm)
Conductance (G): Reciprocal of resistance. G = 1/R. Units: S (siemens) or Ω⁻¹
Resistivity (ρ): R = ρ l/A. Units: Ω·cm
Conductivity (κ): κ = 1/ρ = G · (l/A). Units: S/cm
Cell constant: l/A (cm⁻¹)
Molar conductivity (Λ_m): Conducting power of 1 mole of electrolyte in solution
Λ_m = κ × 1000 / M
Where κ in S/cm, M in mol/L → Λ_m in S·cm²/mol
Enter conductivity and molarity — calculate molar conductivity and degree of dissociation.
At infinite dilution, the molar conductivity of an electrolyte is the sum of the contributions of its individual ions.
Λ°_m(A_xB_y) = x·λ°(A^y+) + y·λ°(B^x−)
Where λ° is the limiting molar conductivity of each ion.
Example: Λ°_m(NaCl) = λ°(Na⁺) + λ°(Cl⁻) = 50.1 + 76.3 = 126.4 S·cm²/mol
Example: Λ°_m(CH₃COOH) = Λ°_m(CH₃COONa) + Λ°_m(HCl) − Λ°_m(NaCl)
Conductivity κ = G(l/A). Molar conductivity Λ_m = 1000κ/M. For strong electrolytes, Λ_m vs √c is linear (Kohlrausch). Λ°_m = sum of ion contributions. Applications: degree of dissociation, K_a, solubility of salts, TDS meters.
Kohlrausch's Law: At infinite dilution, each ion makes a definite contribution to the total molar conductivity of an electrolyte, independent of the nature of the other ion.
Λ°_m = ν_+ λ°_+ + ν_− λ°_−
Where ν_+ and ν_− are numbers of cations and anions per formula unit.
1. Λ°_m of weak electrolyte
Λ°_m(CH₃COOH) = Λ°_m(CH₃COONa) + Λ°_m(HCl) − Λ°_m(NaCl)
because: CH₃COO⁻ + Na⁺ + H⁺ + Cl⁻ − Na⁺ − Cl⁻ = CH₃COO⁻ + H⁺
2. Degree of dissociation (α)
α = Λ_m / Λ°_m
For weak electrolyte (partial dissociation), α tells fraction dissociated.
3. Dissociation constant (K_a)
K_a = C·α² / (1 − α)
For dilute weak electrolyte (α << 1): K_a ≈ C·α²
4. Solubility of sparingly soluble salts
s = κ(saturated solution) × 1000 / Λ°_m
Where s is solubility in mol/L.
Weak electrolytes have very low Λ_m even at moderate concentrations, and their Λ_m vs √c curve is very steep at low concentrations. It never becomes linear, so no extrapolation to c=0 is possible. That's why Kohlrausch's law is essential — we build Λ°_m from strong electrolyte data.
Kohlrausch's law: Λ°_m = sum of independent ion contributions. Enables calculation of Λ°_m for weak electrolytes, degree of dissociation α = Λ_m/Λ°_m, K_a = Cα²/(1−α), and solubility of sparingly soluble salts.
| Feature | Galvanic Cell | Electrolytic Cell |
|---|---|---|
| Energy conversion | Chemical → Electrical | Electrical → Chemical |
| Reaction | Spontaneous (ΔG < 0) | Non-spontaneous (ΔG > 0) |
| Anode | Negative (−) | Positive (+) |
| Cathode | Positive (+) | Negative (−) |
| Electron flow | Anode → Cathode (external) | Anode → Cathode (external) |
| Example | Daniell cell | Electrolysis of water |
The mass of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed.
m = Z·I·t
Where:
Also: m = (M·I·t)/(n·F) = (M·Q)/(n·F)
When the same quantity of electricity is passed through different electrolytes, the masses deposited are proportional to their equivalent weights.
m₁/m₂ = (M₁/n₁) / (M₂/n₂)
Where M₁/n₁ and M₂/n₂ are equivalent masses.
1 Faraday = 96,485 C/mol = charge on 1 mole of electrons
Charge on 1 electron = 1.602 × 10⁻¹⁹ C
1 F = N_A × e = 6.022×10²³ × 1.602×10⁻¹⁹ = 96,485 C/mol
So passing 1 F of charge deposits 1 mole of monovalent ions (like Ag⁺ → Ag), 0.5 mol of divalent ions (like Cu²⁺ → Cu), etc.
Enter current, time, and ion details — calculate mass deposited.
Electrolysis of molten NaCl:
Electrolysis of aqueous NaCl (brine):
Electrolysis of water (acidified):
Electrolysis uses electricity to drive non-spontaneous reactions. Faraday's 1st law: m = MIt/nF. 2nd law: masses ∝ equivalent weights. 1 F = 96,485 C = 1 mol electrons. Applications: metal extraction (Al), chlor-alkali, electroplating, H₂ production.
Primary batteries: Cannot be recharged. Reaction is irreversible.
Secondary batteries: Can be recharged. Reaction is reversible.
Fuel cells: Continuously fed with fuel; produce electricity as long as fuel is supplied.
1. Dry cell (Leclanché cell):
2. Mercury cell:
1. Lead-acid battery (car battery):
2. Nickel-Cadmium (Ni-Cd) battery:
3. Lithium-ion battery (modern):
H₂-O₂ fuel cell (Bacon cell):
Other fuel cells: Methanol fuel cells (DMFC), solid oxide fuel cells (SOFC), phosphoric acid fuel cells (PAFC).
| Battery | EMF/cell | Rechargeable? | Application |
|---|---|---|---|
| Dry cell (Leclanché) | 1.5 V | No | Flashlights, toys |
| Mercury cell | 1.35 V | No | Watches, pacemakers |
| Lead-acid | 2.0 V | Yes | Cars, UPS, inverters |
| Ni-Cd | 1.2 V | Yes | Power tools (older) |
| Li-ion | 3.7 V | Yes | Phones, EVs, laptops |
| H₂-O₂ fuel cell | ~1.2 V (loaded) | Continuous | Spacecraft, prototypes |
The H₂-O₂ fuel cell was invented by William Grove in 1839, but it took 130 years to be practical. NASA used it in the Apollo missions — the fuel cells provided electricity AND drinking water for astronauts!
Batteries: primary (irreversible — dry cell, mercury) vs secondary (reversible — lead-acid, Ni-Cd, Li-ion) vs fuel cells (continuous fuel supply — H₂-O₂). Lead-acid: 2 V/cell. Li-ion: 3.7 V/cell, best energy density. Fuel cells: clean, high efficiency.
Corrosion: Gradual deterioration of metals due to chemical or electrochemical reaction with the environment.
Rusting of iron: The most common example. Iron is oxidized to hydrated iron(III) oxide (rust).
Electrochemical mechanism of rusting:
That's why iron rusts faster near the sea (salt spray) and in humid climates.
1. Galvanization: Coating iron with zinc. Zn has more negative E° (−0.76 V) than Fe (−0.44 V), so Zn acts as a sacrificial anode and corrodes first, protecting iron. Used in galvanized steel, dustbins, roofing sheets.
2. Cathodic protection: Connect iron to a more reactive metal (Mg, Zn, Al) — the reactive metal oxidizes instead of iron. Used for underground pipelines, ship hulls.
3. Painting / coating: Physical barrier prevents contact with O₂ and H₂O. Used for cars, gates, buildings.
4. Alloying: Stainless steel (Fe + Cr + Ni) resists rust because Cr forms a protective Cr₂O₃ layer.
5. Electroplating: Deposit a thin layer of a noble metal (Cr, Ni, Sn) on iron. Used in tin cans (tin-plated steel), chrome plating.
6. Oiling / greasing: Prevents air and moisture from reaching surface.
Corrosion costs the world economy an estimated $2.5 trillion per year — about 3-4% of global GDP. Rust alone damages infrastructure, transport, and machinery globally.
Corrosion = electrochemical degradation. Rusting of iron = Fe → Fe²⁺ at anode, O₂ reduction at cathode, forming Fe₂O₃·xH₂O. Prevention: galvanization (Zn sacrificial), cathodic protection (Mg/Zn anodes), painting, alloying (stainless steel), electroplating.
"Please Send Cats Monkeys And Zebras In Large Hairy Cages So Gold Prefers Costly Arguments"
K, Na, Ca, Mg, Al, Zn, Fe, Pb, H, Cu, Hg, Ag, Au
(K is most reactive, Au is least)
"AN OX and RED CAT"
ANode → OXidation
REDuction → CAThode
Always true — for both galvanic and electrolytic cells!
"Galvanic: Anode is negative; Electrolytic: Anode is positive"
Memory trick: In a galvanic cell, electrons flow FROM anode TO cathode externally, so anode is the source (negative terminal). In electrolysis, external power pushes electrons INTO anode, making it positive.
"0.0591 over n" at 298 K — remember 0.059 ≈ RT×ln(10)/F
At any temperature: E = E° − (2.303RT/nF) log Q
At 298 K: 2.303RT/F = 0.0591
First law: "m = MIt/nF" — say as "M-I-t over n-F"
Second law: "Masses to equivalent weights"
1 F = 96,485 C/mol ≈ 96,500
"Primary is Permanent (one-way); Secondary is Switchable (rechargeable); Fuel is Fresh (continuous)"
Zn = −0.76, Cu = +0.34, Ag = +0.80
"Zebra at −0.76, Cup +0.34, Silver +0.80"
Daniell cell E°: 0.34 − (−0.76) = 1.10 V — classic value!
"O₂ + H₂O + Salt = Rust"
Remove any one → no rusting. That's the basis of all prevention methods.
Use "AN OX, RED CAT" for anode/cathode, "0.0591/n" for Nernst equation, "m = MIt/nF" for Faraday. Practice electrochemical series recall for E°_cell calculations.
Q1. Calculate the EMF of a Zn-Cu cell at 298 K when [Zn²⁺] = 0.1 M and [Cu²⁺] = 0.01 M. E°_cell = 1.10 V.
E = E° − (0.0591/n) log Q, n = 2
Q = [Zn²⁺]/[Cu²⁺] = 0.1/0.01 = 10
E = 1.10 − (0.0591/2) × log(10)
E = 1.10 − 0.02955 × 1 = 1.070 V
Q2. Calculate ΔG° for the Daniell cell. (E°_cell = 1.10 V, n = 2, F = 96485 C/mol)
ΔG° = −nFE°
ΔG° = −2 × 96485 × 1.10
ΔG° = −212,267 J/mol ≈ −212.3 kJ/mol
Negative ΔG° → spontaneous. Maximum electrical work = 212 kJ/mol Zn.
Q3. Calculate the equilibrium constant for the reaction Cu + 2Ag⁺ → Cu²⁺ + 2Ag. E°_cell = 0.46 V.
log K = (n × E°_cell) / 0.0591
log K = (2 × 0.46) / 0.0591 = 0.92/0.0591 = 15.57
K = 10^15.57 = 3.7 × 10¹⁵
Very large K → reaction goes almost to completion
Q4. Calculate the molar conductivity of a 0.1 M solution with conductivity 0.012 S/cm.
Λ_m = κ × 1000 / M
Λ_m = 0.012 × 1000 / 0.1
Λ_m = 12 / 0.1 = 120 S·cm²/mol
Q5. A current of 5 A is passed through CuSO₄ solution for 30 min. Find mass of Cu deposited. (M_Cu = 63.5, n = 2)
t = 30 × 60 = 1800 s
Q = I × t = 5 × 1800 = 9000 C
n_e = Q/F = 9000/96485 = 0.0933 mol e⁻
For Cu²⁺ + 2e⁻ → Cu, moles Cu = n_e/2 = 0.0466 mol
m = 0.0466 × 63.5 = 2.96 g
Q6. Λ°_m for NaCl, HCl, CH₃COONa are 126.4, 426.2, 91.0 S·cm²/mol. Calculate Λ°_m for CH₃COOH.
Λ°_m(CH₃COOH) = λ°(CH₃COO⁻) + λ°(H⁺)
Λ°(CH₃COOH) = Λ°(CH₃COONa) + Λ°(HCl) − Λ°(NaCl)
RHS: (CH₃COO⁻ + Na⁺) + (H⁺ + Cl⁻) − (Na⁺ + Cl⁻) = CH₃COO⁻ + H⁺ ✓
Λ° = 91.0 + 426.2 − 126.4 = 390.8 S·cm²/mol
Now we can use α = Λ_m/Λ°_m for acetic acid at any concentration, then K_a = Cα²/(1−α).
Q7. The conductivity of 0.001 M CH₃COOH is 4.95 × 10⁻⁵ S/cm. Λ°_m = 390.5. Find degree of dissociation and K_a.
Λ_m = κ × 1000 / M = 4.95×10⁻⁵ × 1000 / 0.001
Λ_m = 49.5 S·cm²/mol
α = Λ_m/Λ°_m = 49.5/390.5 = 0.1268
K_a = Cα²/(1−α) = 0.001 × (0.1268)² / (1−0.1268)
= 0.001 × 0.01608 / 0.8732
= 1.84 × 10⁻⁵
Literature K_a for acetic acid = 1.8 × 10⁻⁵ ✓
Only ~12.7% of acetic acid molecules are ionized at 0.001 M — weak acid.
Selected exercise questions from NCERT Class 12 Chemistry Chapter 2. Click each to reveal the solution.
Molar conductivity (Λ_m): Conducting power of 1 mole of electrolyte in solution. Λ_m = κ × 1000 / M (S·cm²/mol).
Strong electrolytes: Λ_m decreases slightly with concentration. Λ_m vs √c is linear — Kohlrausch's law: Λ_m = Λ°_m − A√c. Extrapolation to c=0 gives Λ°_m.
Weak electrolytes: Λ_m drops sharply with concentration (because fewer ions are formed at higher concentration). Cannot extrapolate to Λ°_m — must use Kohlrausch's law from strong electrolyte data.
Solution: Cr is oxidized (anode) → 2Cr → 2Cr³⁺ + 6e⁻
Cd²⁺ is reduced (cathode) → 3Cd²⁺ + 6e⁻ → 3Cd
E°_cell = E°_cathode − E°_anode = −0.40 − (−0.74) = +0.34 V
n = 6 (per balanced equation)
ΔG° = −nFE° = −6 × 96485 × 0.34 = −196,829 J = −196.8 kJ/mol
Solution: Conductivity κ is the conductance of a unit-volume of solution. It depends on the number of ions per unit volume. When we dilute a solution, the number of ions per unit volume decreases (same total ions in more volume). Since fewer ions are available per unit length between electrodes, κ decreases.
However, molar conductivity Λ_m increases with dilution, because each mole has more "space" and can contribute fully (less ion-ion interference).
Solution: Λ°_m(HCOOH) = 349.6 + 54.6 = 404.2 S·cm²/mol
Degree of dissociation: α = Λ_m/Λ° = 46.1/404.2 = 0.114
K_a = Cα²/(1−α) = 0.025 × (0.114)² / (1−0.114)
= 0.025 × 0.0130 / 0.886 = 3.67 × 10⁻⁴
Solution: Each Cr₂O₇²⁻ accepts 6 electrons to become 2Cr³⁺.
1 mol Cr₂O₇²⁻ requires 6 mol electrons.
Q = nF = 6 × 96485 = 578,910 C ≈ 5.79 × 10⁵ C
Solution: For hydrogen electrode: H⁺ + e⁻ → ½H₂, E° = 0
E = E° − (0.0591/1) log(1/[H⁺])
E = 0 − 0.0591 × pH (because −log[H⁺] = pH)
E = −0.0591 × 10 = −0.591 V
Solution:
At cathode: Possible reductions — Cu²⁺ → Cu (E° = +0.34) vs 2H₂O → H₂ (E° = −0.83). Cu²⁺ wins (higher E°). Deposit of copper at cathode.
At anode: Possible oxidations — 2Cl⁻ → Cl₂ (E° = +1.36) vs 2H₂O → O₂ (E° = +1.23). Oxygen has lower oxidation potential, but overpotential of Cl₂ is low, so Cl₂ is actually produced preferentially in concentrated solution. In dilute solution, O₂ may form.
Products: Cu at cathode + Cl₂ at anode (with concentrated CuCl₂)
Fuel cell: A galvanic device that converts chemical energy of a continuously supplied fuel into electricity. Unlike batteries, it doesn't run down as long as fuel is supplied.
H₂-O₂ fuel cell:
Anode: 2H₂ + 4OH⁻ → 4H₂O + 4e⁻
Cathode: O₂ + 2H₂O + 4e⁻ → 4OH⁻
Overall: 2H₂ + O₂ → 2H₂O
Efficiency: ~70% (much higher than heat engines at 30-40%)
Byproduct: Only water (pollution-free)
Applications: Spacecraft (Apollo, Gemini, Space Shuttle — provide both electricity and drinking water), electric vehicles (Toyota Mirai, Honda Clarity)
Solution: Rusting occurs via electrochemical mechanism:
Anode (small area): Fe(s) → Fe²⁺(aq) + 2e⁻
Cathode (larger area, in presence of O₂ + moisture): O₂ + 4H⁺ + 4e⁻ → 2H₂O (acidic) OR O₂ + 2H₂O + 4e⁻ → 4OH⁻ (neutral)
Further oxidation: Fe²⁺ + ½O₂ + H₂O → Fe(OH)₂ → Fe(OH)₃ → Fe₂O₃·xH₂O (rust)
Rust is hydrated iron(III) oxide. Requires both oxygen and moisture. Salt accelerates rusting by making the solution more conducting. This mechanism is why zinc (more reactive) can protect iron — galvanization!
Solution: Area A = πr² = π(0.5)² = 0.785 cm²
Resistivity: ρ = R × A/l = 5.55×10³ × 0.785 / 50 = 87.1 Ω·cm
Conductivity: κ = 1/ρ = 1/87.1 = 0.01148 S/cm
Molar conductivity: Λ_m = κ × 1000 / M = 0.01148 × 1000 / 0.05 = 229.6 S·cm²/mol
| Faraday constant F | 96,485 C/mol | Charge of 1 mol electrons |
| R (gas constant) | 8.314 J/(mol·K) | or 0.0821 L·atm/(mol·K) |
| 1 F | 1 mol e⁻ | Deposits 1 mol monovalent ions |
| E°(Li⁺/Li) | −3.04 V | Most negative (strongest reducing) |
| E°(Zn²⁺/Zn) | −0.76 V | Anode in Daniell cell |
| E°(Fe²⁺/Fe) | −0.44 V | Rusting anode |
| E°(H⁺/H₂) | 0.00 V | SHE reference |
| E°(Cu²⁺/Cu) | +0.34 V | Cathode in Daniell cell |
| E°(Ag⁺/Ag) | +0.80 V | Common reference electrode |
| E°(Au³⁺/Au) | +1.50 V | Very positive (noble metal) |
| 2.303 RT/F at 298 K | 0.0591 V | Nernst constant |
| 1 S | 1 Ω⁻¹ | Siemens (conductance unit) |
| E, E° | Cell potential, standard cell potential |
| E°_cell | E°_cathode − E°_anode |
| n | Number of electrons transferred |
| F | Faraday constant (96,485 C/mol) |
| Q | Reaction quotient |
| K | Equilibrium constant |
| ΔG | Gibbs free energy change |
| R | Gas constant |
| G | Conductance (S or Ω⁻¹) |
| κ | Conductivity (S/cm) |
| Λ_m | Molar conductivity (S·cm²/mol) |
| Λ°_m | Limiting molar conductivity |
| λ° | Limiting ionic molar conductivity |
| α | Degree of dissociation |
| m | Mass deposited (g) |
| I | Current (A) |
| t | Time (s) |
| Z | Electrochemical equivalent (g/C) |
(A) Both true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
The formula is shown on every card for quick revision before exams.