1Nernst
For Zn|Zn²⁺(0.1 M)‖Cu²⁺(0.01 M)|Cu with E°cell = 1.10 V, Ecell at 298 K is:
A — E = 1.10 − (0.059/2)log(0.1/0.01) = 1.10 − 0.0295 = 1.0705 V
NEET 2027 · Electrochemistry · Priority 03 & chapter close-out
Priority 03 is a single topic worth about 3% of the chapter's questions, and it needs one careful read rather than a full pack. So this file gives Topic 13 the space it actually deserves and then spends the rest of its length on what the chapter genuinely still lacks: a clean, unseen, self-scoring test across all thirteen topics that reports where the marks were lost.
Everything here is background reading with one genuinely testable fact buried in it. Read the section once, memorise the temperature rule, and move on — do not spend practice time here.
Conductivity spans an almost unimaginable range — around thirty-four orders of magnitude between Teflon and silver. NCERT sorts materials into four bands by where they fall on that scale.
| Class | Conductivity / S m⁻¹ | Examples | Note |
|---|---|---|---|
| Conductors | 10³ and above | Ag 6.2×10³ · Cu 5.9×10³ · Au 4.5×10³ · Na 2.1×10³ · Fe 1.0×10³ | Metals and their alloys; also graphite and carbon black |
| Semiconductors | 10⁻⁷ to 10² | Ge 2.0 · Si 1.5×10⁻² · CuO 1×10⁻⁷ | Includes doped silicon and gallium arsenide |
| Insulators | 10⁻¹⁶ and below | Glass 1.0×10⁻¹⁶ · Teflon 1.0×10⁻¹⁸ | Ceramics too |
| Superconductors | infinite (zero resistivity) | Metals near 0–15 K; ceramics and mixed oxides up to 150 K | Defined by zero resistivity, not merely very high conductivity |
The two mechanisms differ in another way worth holding. When electrons pass through a metal, they enter at one end and leave at the other and the metal is chemically unchanged. When ions carry current through a solution, the ions are consumed and produced at the electrodes, so prolonged direct current genuinely changes the solution's composition — which is exactly why an AC source is used to measure conductivity.
In 1977 MacDiarmid, Heeger and Shirakawa found that polyacetylene — made by polymerising acetylene gas — takes on a metallic lustre and conducts electricity when exposed to iodine vapour. Several more followed: polyaniline, polypyrrole, polythiophene.
These are built almost entirely from carbon and hydrogen, with occasional nitrogen, oxygen or sulphur, which makes them far lighter than metals — useful for lightweight batteries. They also keep the mechanical flexibility of polymers, so transistors can be made that bend like a sheet of plastic. The three shared the Nobel Prize in Chemistry in 2000.
Burning fossil fuels releases carbon dioxide, which drives the greenhouse effect, warming the surface, melting polar ice and raising sea levels — NCERT names the Maldives as a nation facing total submergence. Hydrogen is offered as the alternative because its combustion produces only water.
The catch is that hydrogen is not a source of energy, only a carrier. It has to be made, and for the scheme to be worth anything it must be made by splitting water using solar energy rather than from fossil fuels. Both halves of the vision — producing hydrogen by electrolysis of water, and consuming it in a fuel cell — rest on electrochemical principles, which is why the box sits at the end of this chapter.
One line per topic. If any row here draws a blank, go back to that topic's pack before attempting the test below.
| # | Topic | Weight | The one thing to hold |
|---|---|---|---|
| 01 | Nernst equation | 18% | E = E° − (0.059/n) log Q · products on top · more product, lower voltage |
| 02 | Conductivity vs molar | 15% | Λm = κ×1000/M · κ falls on dilution, Λm rises |
| 04 | Faraday's laws | 13% | m = MIt/nF · n from the balanced half-reaction, never the group number |
| 03 | Kohlrausch | 11% | Λ°m = ν₊λ°₊ + ν₋λ°₋ · α = Λm/Λ°m · Ka = cα²/(1−α) |
| 05 | ΔG and K | 9% | ΔG° = −nFE° in joules · E° = (0.059/n) log K · E intensive, ΔG extensive |
| 06 | E° series | 7% | E°cell = cathode − anode · combine half-cells through ΔG, never by adding E° |
| 07 | Measurement | 6% | G* = κR to calibrate, κ = G*/R to measure · AC because DC would electrolyse |
| 08 | Cell notation | 5% | Anode left, cathode right · anode negative in galvanic, positive in electrolytic |
| 09 | Electrolysis products | 4% | Cathode takes higher E°, anode lower · overpotential gives Cl₂ from brine |
| 10 | Batteries | 4% | Pb/PbO₂/38% H₂SO₄ · mercury cell holds 1.35 V because no ion changes concentration |
| 11 | Corrosion | 3% | E°cell 1.67 V · Fe²⁺ forms first, Fe³⁺ later · Zn protects scratched iron, Sn does not |
| 12 | Fuel cells | 2% | Alkaline equations with OH⁻ · 70% against 40% · n = 4 × moles of O₂ |
| 13 | Materials | 3% | Electronic conductance falls with temperature, ionic rises |
Question count follows the chapter's own weighting, so the test spends its time where NEET does. Sit it in one go without notes. The timer starts when you press Start; 50 minutes matches NEET's real pace of roughly a minute a question with a little room to spare. Nothing is stored anywhere — take a screenshot of the breakdown before closing the page.
0 / 45 correct
| Topic | Score | Accuracy | Questions missed |
|---|
For Zn|Zn²⁺(0.1 M)‖Cu²⁺(0.01 M)|Cu with E°cell = 1.10 V, Ecell at 298 K is:
A — E = 1.10 − (0.059/2)log(0.1/0.01) = 1.10 − 0.0295 = 1.0705 V
The reduction potential of a hydrogen electrode at pH 5 and 1 bar H₂ is:
A — E = −0.059 × pH = −0.059 × 5 = −0.295 V
The emf of the concentration cell Cu|Cu²⁺(0.001 M)‖Cu²⁺(0.1 M)|Cu is:
D — E° = 0; E = (0.059/2)log(0.1/0.001) = 0.0295 × 2 = 0.059 V
In a Daniell cell, increasing [Cu²⁺] while holding [Zn²⁺] fixed will:
A — Cu²⁺ is a reactant; more reactant lowers Q and raises E.
For Ni|Ni²⁺(0.01 M)‖Cu²⁺(0.1 M)|Cu with E°cell = 0.59 V, Ecell is:
D — E = 0.59 − (0.059/2)log(0.01/0.1) = 0.59 + 0.0295 = 0.6195 V
Which species does NOT appear in the reaction quotient Q for Zn + Cu²⁺ → Zn²⁺ + Cu?
D — Solids and pure liquids are taken as unity in Q.
Mg(s) + 2Ag⁺(0.0001 M) → Mg²⁺(0.130 M) + 2Ag(s) has E°cell = 3.17 V. Ecell is:
B — log Q = log(0.130/10⁻⁸) = 7.11; E = 3.17 − 0.0295(7.11) = 2.96 V
For a two-electron cell, a tenfold change in one ion's concentration shifts Ecell by:
B — 0.059/n with n = 2 gives 0.0295 V per decade.
The conductivity of 0.20 M KCl at 298 K is 0.0248 S cm⁻¹. Its molar conductivity is:
C — Λm = κ × 1000/M = 0.0248 × 1000/0.20 = 124 S cm² mol⁻¹
A cell containing 0.001 M KCl (κ = 0.146 × 10⁻³ S cm⁻¹) has resistance 1500 Ω. Its cell constant is:
B — G* = κR = 0.146×10⁻³ × 1500 = 0.219 cm⁻¹
On diluting an electrolyte solution:
A — Fewer ions per unit volume lowers κ; the volume holding one mole grows faster, raising Λm.
1 S m² mol⁻¹ is equal to:
C — 1 m² = 10⁴ cm², so the numerical value is multiplied by 10⁴.
Which pair of electrolytes has the same value of the constant A in Λm = Λ°m − A√c?
D — A depends only on charge type. NaCl and KNO₃ are both 1-1.
A 0.05 M NaOH column of diameter 1 cm and length 50 cm has resistance 5.55 × 10³ Ω. Its molar conductivity is:
A — A = 0.785 cm²; ρ = RA/l = 87.135 Ω cm; κ = 0.01148 S cm⁻¹; Λm = 0.01148 × 1000/0.05 = 229.6
The relation Λm = Λ°m − A√c is valid for:
C — Weak electrolyte curves are steeply non-linear near zero concentration.
Given λ°(Ca²⁺) = 119.0 and λ°(Cl⁻) = 76.3 S cm² mol⁻¹, Λ°m(CaCl₂) is:
C — Λ°m = λ°(Ca²⁺) + 2λ°(Cl⁻) = 119.0 + 152.6 = 271.6
Given λ°(Mg²⁺) = 106.0 and λ°(SO₄²⁻) = 160.0 S cm² mol⁻¹, Λ°m(MgSO₄) is:
B — MgSO₄ is 1:1 in ions, so Λ°m = 106.0 + 160.0 = 266
If λ°(Al³⁺) = 189 and λ°(SO₄²⁻) = 160 S cm² mol⁻¹, Λ°m for Al₂(SO₄)₃ is:
D — Λ°m = 2(189) + 3(160) = 378 + 480 = 858
For 0.001028 M acetic acid, κ = 4.95 × 10⁻⁵ S cm⁻¹ and Λ°m = 390.5 S cm² mol⁻¹. Ka is about:
A — Λm = 48.15; α = 48.15/390.5 = 0.1233; Ka = cα²/(1−α) = 1.78 × 10⁻⁵
λ°(H⁺) is far larger than that of any other cation because:
B — Charge is relayed along a chain of hydrogen bonds rather than the ion pushing through the solvent.
The charge required to reduce 1 mol of Al³⁺ to Al is:
C — 3F = 3 × 96487 = 289461 C
Ni(NO₃)₂ solution is electrolysed with 5 A for 20 minutes. Mass of Ni deposited is (M = 58.7):
C — Q = 6000 C; mol e⁻ = 0.0622; mol Ni = 0.0311; m = 1.83 g
Charge needed to reduce 1 mol of Cr₂O₇²⁻ to Cr³⁺ is:
A — Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O, so 6F = 578922 C
How many electrons flow when 0.5 A passes through a metallic wire for 2 hours?
B — Q = 0.5 × 7200 = 3600 C; n = 3600/1.6021×10⁻¹⁹ = 2.25 × 10²²
Faradays required to produce 40.0 g of Al from molten Al₂O₃ (M = 27):
C — mol Al = 40/27 = 1.481; × 3 = 4.44 F
Cells of ZnSO₄, AgNO₃ and CuSO₄ are in series. 1.45 g Ag is deposited. Mass of Cu deposited is (Ag 108, Cu 63.5):
D — mol Ag = 0.01343 = mol e⁻; mol Cu = 0.006713; m = 0.426 g
For Zn|Zn²⁺‖Fe²⁺|Fe with E°cell = 0.32 V, ΔrG° is (F = 96487 C):
D — ΔG° = −nFE° = −2 × 96487 × 0.32 = −61752 J = −61.75 kJ mol⁻¹
If E°cell for a reaction is negative, then:
C — ΔG° = −nFE°, so negative E° gives positive ΔG° and K below 1.
For Zn + Cu²⁺ → Zn²⁺ + Cu with E°cell = 1.1 V, Kc at 298 K is about:
B — log Kc = 1.1 × 2/0.059 = 37.29, so Kc ≈ 2 × 10³⁷
If a cell reaction is multiplied by 2, then:
B — E is intensive; ΔG is extensive and scales with n.
Using standard electrode potentials, which of these can oxidise Fe²⁺ to Fe³⁺?
D — Only a couple with E° above +0.77 V (Fe³⁺/Fe²⁺) can oxidise Fe²⁺.
Given K⁺/K = −2.93, Ag⁺/Ag = 0.80, Hg²⁺/Hg = 0.79, Mg²⁺/Mg = −2.37, Cr³⁺/Cr = −0.74 V, the increasing order of reducing power is:
D — More negative E° means stronger reducing agent, so order follows increasing negativity.
Copper does not dissolve in dilute HCl but does dissolve in dilute HNO₃ because:
B — E°(Cu²⁺/Cu) = +0.34 V is above H⁺/H₂, so H⁺ cannot oxidise Cu; NO₃⁻ can.
A cell filled with a solution of κ = 1.29 S m⁻¹ has resistance 100 Ω. Its cell constant is:
D — G* = κR = 1.29 × 100 = 129 m⁻¹
Alternating current is used when measuring the resistance of an electrolyte because direct current would:
B — DC drives electrolysis, altering the very solution being measured.
If Λm(MgSO₄) = 266 S cm² mol⁻¹, its equivalent conductance is:
C — n-factor for MgSO₄ is 2, so Λeq = Λm/2 = 133
In a galvanic cell the anode is:
D — Electrons accumulate at the anode in a galvanic cell, making it negative.
The correct cell notation for Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s) is:
A — Anode on the left, cathode on the right, double bar for the salt bridge.
Electrolysis of aqueous AgNO₃ using silver electrodes gives:
C — A reactive anode is oxidised in preference to water — the basis of electroplating.
Electrolysis of aqueous CuCl₂ with platinum electrodes gives, at the anode:
A — Chloride discharges ahead of water because of oxygen overpotential.
In a lead storage battery, the anode and cathode are respectively:
A — Pb is oxidised at the anode; PbO₂ is reduced at the cathode. Both become PbSO₄ on discharge.
In a dry cell, the oxidation state of manganese changes from:
A — MnO₂ → MnO(OH), a one-electron reduction from +4 to +3.
Rust is best represented as:
C — Rust is hydrated ferric oxide; the water of hydration is part of the formula.
The approximate efficiencies of a fuel cell and a thermal power plant are:
A — Direct chemical-to-electrical conversion avoids the losses of the heat–turbine route.
With rising temperature, electronic conductance in metals and ionic conductance in solution respectively:
B — Metal lattices scatter electrons more when hot; warm solutions are less viscous so ions move faster.
The breakdown table is the point of this file. A raw score out of 45 says almost nothing; a topic with 2 out of 6 says exactly where the next study hour goes.