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Two chapters, seventeen questions

Every chemistry question saved from this paper, grouped by chapter and sub-topic. Each carries what was given, what was asked, the concept and formula behind it, the steps in full, and the fastest route through.

17
Questions lost
2
Attempted, wrong
14
Left blank
2
Chapters involved

Solutions 6 · Electrochemistry 11. On NEET marking these seventeen were worth 68 marks. One question (Q161) had its option list cut off in the screenshot.

The headline here is the skip rate: fourteen blank against two attempted — the highest in any paper reviewed so far. And the blanks are not the hard questions: Q147 is one division, Q163 is substituting zero into a printed equation, and Q145 needs no arithmetic at all.

Chapters in this paper  —  red = wrong · amber = blank · green = correct

Solutions

6 questions · 6 blank

Q137Q139Q163Q168Q170Q178

Electrochemistry

11 questions · 2 wrong · 8 blank · 1 uncaptured

Q136Q138Q140Q141Q145Q147Q148Q161Q162Q165Q166

Chapter

Solutions

Raoult's law, colligative properties and Henry's law  ·  6 questions · 6 blank

Vapour pressure and Raoult's law

3 questions · 3 blank
Q137 Molecular formula from vapour pressure lowering Not attempted

The vapour pressure of a solution having 2 g of solute X (molar atomic mass = 32 g/mol) in 100 g of CS2 (vapour pressure = 854 torr) is 848.95 torr. The molecular formula of the solute is

  •  X2
  •  X4
  • KEYX8
  •  X6
Given
  • 2 g of solute X dissolved in 100 g CS2.
  • Atomic mass of X = 32 g/mol (so X is sulfur).
  • P° = 854 torr, Psolution = 848.95 torr.
  • Molar mass of CS2 = 76 g/mol.
Asked
The molecular formula (how many X atoms per molecule).
Concept to use
Relative lowering of vapour pressure gives the molar mass of the molecule as it exists in solution, which need not match the atomic mass. Dividing the measured molar mass by the atomic mass tells you how many atoms are joined together — the atomicity. Sulfur famously exists as S8 rings, and this experiment is the classic way of showing it.
Formula to use
(P° − P)/P° = nsolute/nsolvent  →  Msolute = (w2 × M1 × P°) / (w1 × ΔP)
Baby steps
  1. Relative lowering: ΔP/P° = (854 − 848.95)/854 = 5.05/854 = 0.005913.
  2. Moles of solvent: 100/76 = 1.316 mol of CS2.
  3. Moles of solute: n2 = 0.005913 × 1.316 ≈ 0.00783 mol.
  4. Molar mass of the solute: M = 2 g / 0.00783 mol ≈ 256 g/mol.
  5. Atoms per molecule: 256 / 32 = 8.
  6. Formula: X8 — sulfur as S8.
Answer
X8
Shortcut
Recognise the chemistry and skip most of the arithmetic. Atomic mass 32 and a solvent of CS2 point straight at sulfur, and dissolved sulfur is S8. The calculation is then only a confirmation — useful if you have time, but the answer is guessable in seconds.
The reason this works is that colligative properties count particles, not atoms. The same principle in reverse is how association and dissociation are detected: an abnormally high molar mass means the solute has clumped together, an abnormally low one means it has split apart.
Q163 Reading a vapour pressure equation Not attempted

Vapour pressure of a solution of heptane and octane is given by the equation Psol(mm of Hg) = 32 + 63x, where ‘x’ is the mole fraction of heptane. Vapour pressure of pure octane will be

read the equation at the two ends of the x-axis3295x = 0x = 1no heptaneall heptanemole fraction of heptane, xP (mm Hg)x = 0 → pure octaneP = 32 mm Hgx = 1 → pure heptaneP = 32 + 63 = 95the intercept is the other component; the slope is the difference, not a pressure
  • KEY32 mm Hg
  •  95 mm Hg
  •  47.5 mm Hg
  •  63 mm Hg
Given
  • Psol = 32 + 63x, with x the mole fraction of heptane.
Asked
Vapour pressure of pure octane.
Concept to use
For an ideal binary solution, P = P°B + (P°A − P°B)xA. So when the equation is written in terms of one component’s mole fraction, the constant term is the pure vapour pressure of the other component, and the slope is the difference between the two. Pure octane means no heptane, so x = 0.
Formula to use
P = P°octane + (P°heptane − P°octane)xheptane
Baby steps
  1. Pure octane means the mole fraction of heptane is x = 0.
  2. Substitute: P = 32 + 63(0) = 32 mm Hg.
  3. Cross-check the other end: x = 1 gives P = 32 + 63 = 95 mm Hg, which must be pure heptane.
  4. So the intercept 32 is octane and 95 is heptane, consistent with heptane being the more volatile (lighter) of the two.
Answer
32 mm Hg
Shortcut
Match the equation to the standard form by eye: the intercept belongs to the component whose mole fraction is not written, and intercept + slope gives the other one. Reading it that way answers both possible versions of this question at once.
63 is offered as a trap because it is the visible number attached to x, but a slope is a difference of pressures (95 − 32), not a pressure. And 47.5 is the average of 32 and 63 — a number with no physical meaning here.
Q168 Mass of solute to reduce vapour pressure to 80% Not attempted

What is the mass of non-volatile solute (molar mass 40 g/mol) which should be dissolved in 114 g of octane (MW = 114) to reduce its vapour pressure to 80 %?

which form of the relative-lowering formula is intendedNCERT dilute-solution form(P° − P)/P° = n2 / n10.2 = n2 / 1n2 = 0.2 molmass = 0.2 × 40 = 8 gexact mole-fraction form(P° − P)/P° = n2 / (n1 + n2)0.2 = n2 / (1 + n2)n2 = 0.25 molmass = 10 g8 g is on the option list, 10 g is notwhen both forms are defensible, the option list tells you which approximation the paper wants
  •  4 gm
  • KEY8 gm
  •  1.0 gm
  •  15 gm
Given
  • Solvent: 114 g octane, M = 114 g/mol → exactly 1 mol.
  • Solute: non-volatile, M = 40 g/mol.
  • Vapour pressure reduced to 80 % of the pure value.
Asked
Mass of solute required.
Concept to use
“Reduced to 80 %” means P/P° = 0.8, so the relative lowering is 0.2. NCERT’s working formula for dilute solutions writes that relative lowering as n2/n1 — moles of solute over moles of solvent. The solvent quantity is chosen to be exactly one mole, which makes the arithmetic trivial and signals which formula is intended.
Formula to use
(P° − P)/P° = n2/n1  (dilute-solution form used in NCERT)
Baby steps
  1. Moles of octane: 114 g ÷ 114 g/mol = 1 mol.
  2. Relative lowering: (P° − P)/P° = 1 − 0.8 = 0.2.
  3. Apply the formula: 0.2 = n2/1 → n2 = 0.2 mol.
  4. Mass = 0.2 × 40 = 8 g.
Answer
8 gm
Shortcut
The solvent mass being numerically equal to its molar mass is a deliberate gift — it makes n1 = 1, so the relative lowering is the number of moles of solute. Multiply by the molar mass and you are done in two lines.
Worth knowing: the exact treatment uses mole fraction, 0.2 = n2/(1 + n2), which gives n2 = 0.25 mol and a mass of 10 g. Both are defensible, but only 8 g appears among the options — so the paper wants the dilute approximation. When two forms disagree, the option list tells you which one is intended.

Colligative properties

2 questions · 2 blank
Q139 What happens during freezing point depression Not attempted

During a freezing point depression experiment, which of the following statements best describes the system?
A) The vapour pressure of the solution is lower than that of the pure solvent
B) The vapour pressure of the solution is higher than that of the pure solvent
C) Only solute particles freeze at the freezing point
D) Only solvent particles freeze at the freezing point

the solution curve lies below the solvent curve everywherevapourpressuretemperaturepure solventsolutionTf solventTf solutionΔTfA — solution vapour pressure is lower ✓D — only solvent particles freeze ✓the solute stays in the liquid — that is why the freezing point drops
  •  A only
  •  B and C only
  •  A and C only
  • KEYA and D only
Given
  • Four statements about a solution during freezing point depression.
Asked
Which pair correctly describes the system.
Concept to use
Both correct statements follow from one idea: the solute stays in the liquid and gets in the way. Non-volatile solute particles occupy part of the surface, so fewer solvent molecules escape and the vapour pressure falls. And when freezing begins it is the solvent that crystallises out in pure form, leaving the solute behind in the remaining liquid — which is exactly why the freezing point drops in the first place.
Formula to use
Freezing point = temperature where solid and liquid vapour pressures are equal; lowering the solution’s vapour pressure shifts that crossing to a lower T
Baby steps
  1. A. Adding a non-volatile solute always lowers the vapour pressure — this is Raoult’s law. Correct.
  2. B. Directly contradicts A, so it must be wrong. Incorrect.
  3. C. The solute does not freeze out; it remains dissolved in the shrinking liquid. Incorrect.
  4. D. Pure solvent crystals separate first — this is the basis of freeze purification. Correct.
  5. So A and D.
Answer
A and D only
Shortcut
A and B are direct opposites, so exactly one is true — and Raoult’s law says it is A. That single observation eliminates half the options before you look at C or D.
The fact that only solvent freezes is the reason for the depression, not a side effect. Because the solid separating is pure solvent, the remaining solution keeps getting more concentrated as freezing proceeds — which is why a solution freezes over a range of temperatures rather than at a single sharp point.
Q170 Molal depression constant from latent heat Not attempted

Calculate the molal depression constant of a solvent, which freezes at 15°C, the latent heat of fusion is 180.7 J g−1

  • KEY3.81 K molal−1
  •  0.381 K molal−1
  •  1.90 K molal−1
  •  0.19 K molal−1
Given
  • Freezing point of the solvent Tf = 15°C = 288 K.
  • Latent heat of fusion ΔHfus = 180.7 J g−1 (per gram, not per mole).
  • R = 8.314 J K−1 mol−1.
Asked
The molal depression constant Kf.
Concept to use
The theoretical expression for Kf is R Tf² M1 / (1000 ΔHfus) with ΔH per mole. Here the latent heat is given per gram, so the molar mass M1 cancels against it and the formula simplifies to R Tf² / (1000 × Lf). Recognising that cancellation is the whole trick — otherwise you would be stuck without a molar mass.
Formula to use
Kf = R Tf² / (1000 × Lf)  — with Lf in J g−1
Baby steps
  1. Convert the temperature: 15°C = 288 K. Using 15 directly is fatal here, since the formula needs T² in kelvin.
  2. Tf² = 288² = 82 944.
  3. Numerator: R Tf² = 8.314 × 82 944 ≈ 689 600.
  4. Denominator: 1000 × 180.7 = 180 700.
  5. Kf = 689 600 / 180 700 ≈ 3.81 K molal−1.
Answer
3.81 K molal−1
Shortcut
Check the units of what you are given before choosing a formula. Latent heat per gram means the molar mass has already been divided out, so use Kf = RT²/1000Lf. If it had been given per mole you would need M1 as well — and its absence from the question is the clue.
The distractors 0.381 and 0.19 are factor-of-ten and factor-of-twenty slips from the 1000 in the denominator. That 1000 is there to convert kilograms to grams in the definition of molality — keep it attached to the latent heat term so it cannot be dropped.

Henry's law

1 question · 1 blank
Q178 Henry's law with a gas mixture Not attempted

Henry’s law constant, KH for N2 gas at 289 K is 105 atm. The mole fraction of N2 in air is 0.6. The number of moles N2 from air dissolved in 10 moles of water at 5 atm air Pressure is (Assume nN2 << nH2O)

  • KEY3 × 10−4
  •  4 × 10−5
  •  5 × 10−4
  •  6 × 10−6
Given
  • KH = 105 atm for N2 at 289 K.
  • Mole fraction of N2 in air = 0.6; total air pressure = 5 atm.
  • 10 moles of water.
Asked
Moles of N2 dissolved.
Concept to use
Henry’s law uses the partial pressure of the individual gas, not the total pressure of the mixture. So the first step is always Dalton’s law: multiply the total pressure by the gas’s mole fraction. The law then gives the mole fraction dissolved, and the stated assumption lets you convert that to moles by multiplying by the moles of water.
Formula to use
pN2 = xair × Ptotal  |  xdissolved = p/KH  |  n = x × nwater
Baby steps
  1. Partial pressure of N2: p = 0.6 × 5 = 3 atm.
  2. Henry’s law: xN2 = p/KH = 3/105 = 3 × 10−5.
  3. Since nN2 << nwater, the mole fraction is essentially nN2/nwater.
  4. nN2 = 3 × 10−5 × 10 = 3 × 10−4 mol.
Answer
3 × 10−4
Shortcut
Three multiplications in a fixed order — fraction × pressure, divide by KH, times moles of solvent. Written as one line it is n = (0.6 × 5 / 105) × 10, and every Henry’s law question with a gas mixture has this same shape.
4 × 10−5 is what you get by stopping at the mole fraction and forgetting the last multiplication; using the total 5 atm instead of the partial 3 atm gives 5 × 10−4. Both distractors correspond to skipping exactly one of the three steps, so name each step as you do it.

Chapter

Electrochemistry

Electrode potentials, the Nernst equation, conductance and electrolysis  ·  11 questions · 2 wrong · 8 blank · 1 uncaptured

Electrode potentials and cell reactions

4 questions · 1 wrong · 3 blank
Q136 Identifying the strongest reducing agent Marked wrong

Based on the data given below: E°Cl2/Cl− = 1.36 V, E°MnO4−/Mn2+ = 1.51 V, E°Cr3+/Cr = −0.74 V. The strongest reducing agent is:

more negative E° = better reducing agentMnO4− / Mn2++1.51 VCl2 / Cl−+1.36 VCr3+ / Cr−0.74 VE° fallshigh E° → wants electronsstrong oxidising agentlow E° → gives electrons awaystrong reducing agentlowest E° here is Cr³⁺/Cr → Cr is the reductantand the reducing agent is the reduced form — the metal Cr, not the ion
  • MARKEDMn2+
  • KEYCr
  •  MnO4−
  •  Cl−
Given
  • Three standard reduction potentials: +1.36 V, +1.51 V and −0.74 V.
Asked
The strongest reducing agent.
Concept to use
Two decisions, and the question is built so that getting only one right still loses the mark. Which couple? A reducing agent gives electrons away, so it belongs to the couple with the most negative reduction potential — that is Cr3+/Cr at −0.74 V. Which side of that couple? The species that donates electrons is the reduced form, written on the right of the couple — the metal Cr, not the ion Cr3+.
Formula to use
Most negative E° → best reducing agent  |  the reducing agent is the reduced (right-hand) species of the couple
Baby steps
  1. Rank the potentials: MnO4−/Mn2+ (+1.51) > Cl2/Cl− (+1.36) > Cr3+/Cr (−0.74).
  2. A high positive E° means the couple is eager to accept electrons — so MnO4− is the strongest oxidising agent, not reducing.
  3. The most negative E° means the couple is most willing to give electrons away: that is the chromium couple.
  4. Within that couple, the electron donor is the reduced form: Cr metal. Cr3+ is the oxidised form and cannot donate further.
  5. Answer: Cr.
Answer
Cr
Shortcut
Ask two questions in order: which couple has the lowest E°? and then which side of it is already reduced? The answer is always the species written after the slash. Applying the same routine to the highest E° gives the strongest oxidising agent — here MnO4−.
Where it went wrong
Mn2+ is the reduced form of a couple, so it satisfies the second test — but its couple has the highest E°, which makes it a very poor electron donor. Passing one filter and failing the other is exactly what this option is designed to catch. Check the potential first, and only then pick the side.
Q138 When is a displacement reaction spontaneous Not attempted

The standard reduction potentials for Zn2+/Zn, Ni2+/Ni and Fe2+/Fe are −0.76 V, −0.23 V and −0.44 V respectively. The reaction X + Y2+ → X2+ + Y will be spontaneous when

more negative E° = better reducing agentMnO4− / Mn2++1.51 VCl2 / Cl−+1.36 VCr3+ / Cr−0.74 VE° fallshigh E° → wants electronsstrong oxidising agentlow E° → gives electrons awaystrong reducing agentlowest E° here is Cr³⁺/Cr → Cr is the reductantand the reducing agent is the reduced form — the metal Cr, not the ion
  •  X = Ni, Y = Fe
  •  X = Ni, Y = Zn
  • KEYX = Zn, Y = Ni
  •  X = Fe, Y = Zn
Given
  • E°: Zn2+/Zn = −0.76 V, Ni2+/Ni = −0.23 V, Fe2+/Fe = −0.44 V.
Asked
Which pairing makes the reaction spontaneous.
Concept to use
In the reaction as written, X is oxidised (it loses electrons and becomes X2+) and Y2+ is reduced. Spontaneity needs E°cell > 0, and E°cell = E°cathode − E°anode = E°Y − E°X. So the requirement is simply that Y has the higher (less negative) potential — the metal being displaced must be the nobler one.
Formula to use
E°cell = E°Y couple − E°X couple > 0  →  E°Y > E°X
Baby steps
  1. Order the metals by E°: Zn (−0.76) < Fe (−0.44) < Ni (−0.23).
  2. The condition is E°Y > E°X: the more reactive metal X displaces the less reactive Y from solution.
  3. X = Ni, Y = Fe: −0.44 − (−0.23) = −0.21 V. Negative. ✗
  4. X = Ni, Y = Zn: −0.76 − (−0.23) = −0.53 V. Negative. ✗
  5. X = Zn, Y = Ni: −0.23 − (−0.76) = +0.53 V. Positive. ✓
  6. X = Fe, Y = Zn: −0.76 − (−0.44) = −0.32 V. Negative. ✗
Answer
X = Zn, Y = Ni
Shortcut
Skip the subtractions entirely. X must be the metal with the lower E° of the pair, so scan the options and keep only the one where the first-named metal sits lower in the list. Zn is the lowest of the three, so any option with X = Zn is the candidate — and there is exactly one.
This is the reactivity series in disguise: a more reactive metal displaces a less reactive one from its salt solution. Zinc displacing nickel is spontaneous; nickel displacing zinc is not — which is why zinc, not nickel, is used for galvanising iron.
Q145 Does E° change when the equation is halved Not attempted

E° for 2AgCl(s) + 2e → 2Ag(s) + 2Cl− is +0.22 V. Then the E° value for AgCl(s) + e → Ag(s) + Cl− is

ΔG scales with the equation, E° does not2AgCl + 2e⁻ → 2Ag + 2Cl⁻n = 2E° = +0.22 VΔG° = −2F(0.22)halvethe equationAgCl + e⁻ → Ag + Cl⁻n = 1E° = +0.22 VΔG° = −1F(0.22) — halvedE° is unchanged: +0.22 Velectrode potential is an intensive property — like temperature, not like mass
  •  0.11 V
  • KEY+0.22 V
  •  −0.22 V
  •  −0.44 V
Given
  • E° = +0.22 V for the reaction written with 2 electrons.
  • The same reaction is rewritten with 1 electron.
Asked
E° for the halved equation.
Concept to use
Electrode potential is an intensive property — it does not depend on how much material is involved, in the same way that temperature does not. What does scale is the free energy: ΔG° = −nFE°. When you halve the equation, n halves and ΔG° halves with it, but E° is the ratio of the two and stays exactly the same.
Formula to use
ΔG° = −nFE°  →  E° = −ΔG°/nF  — both numerator and denominator halve, so E° is unchanged
Baby steps
  1. For the given equation, n = 2: ΔG° = −2F(0.22).
  2. Halving the equation halves the free energy change: ΔG° becomes −F(0.22), and now n = 1.
  3. Compute the new potential: E° = −ΔG°/nF = F(0.22)/(1 × F) = +0.22 V.
  4. The sign is untouched too, because the direction of the reaction has not been reversed — only its scale.
Answer
+0.22 V
Shortcut
Sort every electrochemical quantity into intensive or extensive once and for all. Intensive (unchanged by scaling): E°, temperature, concentration. Extensive (scales with the equation): ΔG°, ΔH°, n. The only operation that changes E° is reversing the reaction, which flips its sign.
0.11 V is the answer you get by treating E° as if it scaled — the single most common error in this topic. A useful physical check: potential is measured in volts, i.e. joules per coulomb. It is already a per-unit quantity, so scaling the equation cannot change it.
Q165 Overall reaction from a cell representation Not attempted

What is the overall cell reaction for the voltaic cell Fe / Fe2+ ‖ Fe3+/Fe2+ / Pt ?

anode on the left, cathode on the right — alwaysLEFT = ANODEoxidation happens heremetal | its ione.g. Zn | Zn²⁺RIGHT = CATHODEreduction happens hereion | metale.g. Cu²⁺ | Cu‖salt bridgeZn | Zn²⁺ ‖ Cu²⁺ | Cu    for   Cu²⁺ + Zn → Cu + Zn²⁺a gas electrode needs an inert Pt contact, written on the outside: Pt, Cl2 | Cl⁻
  •  2Fe3+ ⇌ 3Fe2+
  •  Fe3+ + Fe2+ ⇌ Fe3+ + Fe2+
  •  3Fe2+ ⇌ 2Fe3+ + Fe
  • KEY2Fe3+ + Fe ⇌ 3Fe2+
Given
  • Cell: Fe | Fe2+ ‖ Fe3+/Fe2+ | Pt.
Asked
The overall cell reaction.
Concept to use
Read the representation left to right. The left half-cell is always the anode, so Fe is oxidised to Fe2+, releasing two electrons. The right half-cell is the cathode, where Fe3+ is reduced to Fe2+, accepting one electron each. Balance the electrons, then add.
Formula to use
Anode: Fe → Fe2+ + 2e−  |  Cathode: Fe3+ + e− → Fe2+
Baby steps
  1. Anode (left): Fe → Fe2+ + 2e−.
  2. Cathode (right): Fe3+ + e− → Fe2+. The Pt is only an inert contact, since both species there are ions in solution.
  3. Balance electrons: multiply the cathode half by 2 → 2Fe3+ + 2e− → 2Fe2+.
  4. Add the two halves and cancel the electrons: Fe + 2Fe3+ → Fe2+ + 2Fe2+.
  5. Collect: 2Fe3+ + Fe ⇌ 3Fe2+.
Answer
2Fe3+ + Fe ⇌ 3Fe2+
Shortcut
Check the atom count as a filter. The left electrode is solid Fe, so metallic iron must be consumed — it has to appear on the left of the overall equation. Only one option has Fe as a reactant, and the option with Fe as a product is simply this reaction reversed.
The Pt in the representation signals an inert electrode, used whenever both members of a couple are dissolved species (Fe3+/Fe2+) or a gas is involved. It takes no part in the chemistry and never appears in the overall equation.

Cell representation

1 question · 1 uncaptured
Q161 Matching cell reactions to cell representations Options not captured

Match column I (cell reaction) with column II (cell representation).

Column I — reactionColumn II — representation
P. F2 + 2Cl− → 2F− + Cl21. Pt, Cl2 | Cl− ‖ F− | F2, Pt
Q. Cl2 + Zn → 2Cl− + Zn2+2. Zn | Zn2+ ‖ Cl− | Cl2 | Pt
R. Zn2+ + Mg → Zn + Mg2+3. Zn | Zn2+ ‖ Mg2+ | Mg
S. Cu2+ + Zn → Cu + Zn2+4. Zn | Zn2+ ‖ Cu2+ | Cu
5. Mg | Mg2+ ‖ Zn2+ | Zn
The option list was cut off in the screenshot, so the derived matching is given below.

anode on the left, cathode on the right — alwaysLEFT = ANODEoxidation happens heremetal | its ione.g. Zn | Zn²⁺RIGHT = CATHODEreduction happens hereion | metale.g. Cu²⁺ | Cu‖salt bridgeZn | Zn²⁺ ‖ Cu²⁺ | Cu    for   Cu²⁺ + Zn → Cu + Zn²⁺a gas electrode needs an inert Pt contact, written on the outside: Pt, Cl2 | Cl⁻
  • KEYP–1, Q–2, R–5, S–4  (derived — options not captured)
Given
  • Four cell reactions and five candidate cell representations — one is a decoy.
Asked
The correct matching.
Concept to use
The convention is fixed and mechanical: anode on the left, cathode on the right, salt bridge in the middle. So for each reaction, identify which species is oxidised (that becomes the left half-cell, written metal | ion) and which is reduced (right half-cell, written ion | metal). A gas electrode needs an inert platinum contact, written on the outer edge.
Formula to use
Anode (oxidation) | its ion  ‖  ion | cathode (reduction)
Baby steps
  1. P. Cl− → Cl2 is oxidation (left, needs Pt); F2 → F− is reduction (right, needs Pt). Gives Pt, Cl2 | Cl− ‖ F− | F2, Pt → 1.
  2. Q. Zn → Zn2+ is oxidation (left); Cl2 → Cl− is reduction (right, with Pt). Gives Zn | Zn2+ ‖ Cl− | Cl2 | Pt → 2.
  3. R. Mg → Mg2+ is oxidation, so Mg is on the left; Zn2+ → Zn is reduction, on the right. Gives Mg | Mg2+ ‖ Zn2+ | Zn → 5. (Entry 3 is this cell written backwards — the decoy.)
  4. S. Zn → Zn2+ oxidation on the left; Cu2+ → Cu reduction on the right → 4.
  5. Matching: P–1, Q–2, R–5, S–4.
Answer
P–1, Q–2, R–5, S–4
Shortcut
For each reaction, find the species being consumed as a free element — that is the anode metal, and it goes on the far left. In R the free element consumed is Mg, so Mg starts the representation, which distinguishes 5 from the reversed decoy 3 in one glance.
Entries 3 and 5 are the same cell written in opposite directions, and only one can be right. Whenever a matching list contains a reversed pair, that pair is where the question is decided — check it first. The options were not captured in the screenshot, so this matching is derived from the convention rather than read off a key.

Nernst equation

2 questions · 2 blank
Q148 Nernst equation on dilution Not attempted

A nickel electrode is immersed in an aqueous nickel(II) sulfate solution. If the concentration of Ni2+ ions is reduced to one-tenth of its initial value by dilution, how does the reduction potential of the Ni2+/Ni electrode change?

  •  decreases by 60 mV
  •  increases by 30 mV
  • KEYdecreases by 30 mV
  •  decreases by 60 V
Given
  • Ni2+/Ni electrode, n = 2.
  • [Ni2+] reduced to one tenth of its original value.
Asked
The change in reduction potential.
Concept to use
The Nernst equation for a metal-ion electrode is E = E° + (0.059/n) log[Mn+]. Diluting the solution lowers the ion concentration, which makes the log term more negative and so lowers the reduction potential — the electrode becomes less inclined to accept electrons when there are fewer ions to reduce. The factor n = 2 halves the size of the shift.
Formula to use
E = E° + (0.059/n) log[Ni2+]  →  ΔE = (0.059/n) log(1/10)
Baby steps
  1. n = 2 for Ni2+ + 2e− → Ni.
  2. Concentration falls to one tenth, so log(new/old) = log(0.1) = −1.
  3. ΔE = (0.059/2) × (−1) = −0.0295 V.
  4. That is a decrease of about 30 mV.
  5. Direction check: fewer Ni2+ ions means less tendency to be reduced, so the reduction potential must fall. ✓
Answer
decreases by 30 m volt
Shortcut
For a tenfold change in concentration the shift is always 59/n millivolts — 59 mV for a one-electron couple, ~30 mV for a two-electron couple, ~20 mV for three. Memorising that one number turns every dilution question into a division and a sign.
The sign is where marks are lost. Diluting the ion always decreases the reduction potential; concentrating it increases it. And note the last option says 60 V, not mV — a unit trap sitting quietly at the end of the list.
Q166 Concentration ratio from a cell potential Not attempted

In the cell Mg(s) / Mg2+(C1) ‖ Ni2+(C2) / Ni(s), Ecell − E°cell = 0.059 V. The ratio C1/C2 at 298 K will be

  •  2
  •  100
  • KEY10−2
  •  11
Given
  • Cell: Mg | Mg2+(C1) ‖ Ni2+(C2) | Ni.
  • Ecell − E°cell = 0.059 V at 298 K.
Asked
The ratio C1/C2.
Concept to use
The Nernst equation gives Ecell = E°cell − (0.059/n) log Q, so the quantity given is simply −(0.059/n) log Q. The two things to get right are n = 2 (both metals are divalent) and the form of Q: products over reactants, which for this cell means [Mg2+]/[Ni2+] = C1/C2.
Formula to use
Ecell − E°cell = −(0.059/n) log(C1/C2), with n = 2
Baby steps
  1. Overall reaction: Mg + Ni2+ → Mg2+ + Ni, so Q = C1/C2 and n = 2.
  2. Substitute: 0.059 = −(0.059/2) log(C1/C2).
  3. Divide both sides by 0.059: 1 = −(1/2) log(C1/C2).
  4. log(C1/C2) = −2.
  5. C1/C2 = 10−2.
Answer
10−2
Shortcut
The 0.059 on the right cancels the 0.059 in the Nernst term, leaving log Q = −n. With n = 2 that is log Q = −2 straight away. Papers choose the given value to make exactly this cancellation happen, so look for it before reaching for a calculator.
The sign decides between 10−2 and 100, and both are on the list. Ecell being above E°cell means Q is less than 1 — fewer products than reactants — so C1 must be the smaller concentration. Sanity-checking the direction takes a second and rules out the wrong one of the pair.

Conductance and Kohlrausch's law

3 questions · 1 wrong · 2 blank
Q140 Which ion has the highest molar conductivity Not attempted

Which one of the following has the highest limiting molar conductivity at 298 K?

  •  Ca2+
  •  K+
  •  Na+
  • KEYH+
Given
  • Four cations: Ca2+, K+, Na+, H+.
Asked
Which has the highest limiting molar conductivity.
Concept to use
The proton is a special case. Every other ion has to physically push through the solvent, and its speed is limited by size and hydration. H+ does not move that way at all: it travels by the Grotthuss mechanism, hopping along a chain of hydrogen-bonded water molecules so that no single proton has to cross the distance. That gives it a conductivity several times larger than any other cation.
Formula to use
λ°: H+ (350) >> Ca2+ (119) > K+ (73) > Na+ (50) S cm² mol−1
Baby steps
  1. Among the ordinary cations, the smaller the hydrated size the faster the ion: K+ beats Na+, because the smaller bare Na+ attracts a bigger water shell and so moves more slowly.
  2. Ca2+ carries two charges, so it contributes more per mole than either alkali metal despite being heavily hydrated.
  3. But H+ conducts by proton hopping rather than by bodily movement, which is far faster than any migration.
  4. Its limiting molar conductivity, about 350 S cm² mol−1, is roughly three times that of Ca2+ and seven times that of Na+.
  5. Answer: H+.
Answer
H+
Shortcut
If H+ or OH− appears in a conductivity comparison, it wins — H+ first, OH− second. No other ion comes close, so the rest of the list need not be ranked at all.
The counter-intuitive part is worth holding on to: Na+ is smaller than K+ yet conducts worse, because a smaller bare ion holds a larger hydration shell and the shell is what actually moves. Ionic mobility follows hydrated size, not bare size.
Q141 pH of a weak base from conductivity data Not attempted

On the basis of following data, select the correct pH value for 10−2 M NH4OH(aq) solution at 25°C. Λm(NH4OH) of 10−2 M solution = 10.2 S cm² mol−1; Λ°m(NH4Cl) = 130, Λ°m(NaOH) = 320, Λ°m(NaCl) = 110 S cm² mol−1

build the weak electrolyte from three strong ones+Λ° NH4Cl130+Λ° NaOH320−Λ° NaCl110Λ° NH4OH = 340 S cm² mol⁻¹α = Λm / Λ°m= 10.2 / 340 = 0.03[OH⁻] = cα = 10⁻² × 0.03= 3 × 10⁻⁴ MpOH = 3.52 → pH = 10.48the Na⁺ and Cl⁻ contributions cancel, leaving exactly NH4OH
  • KEY10.48
  •  3.52
  •  2
  •  12
Given
  • Λm of the 10−2 M NH4OH solution = 10.2 S cm² mol−1.
  • Λ°m: NH4Cl = 130, NaOH = 320, NaCl = 110.
  • Concentration c = 10−2 M.
Asked
The pH of the solution.
Concept to use
NH4OH is a weak electrolyte, so its limiting conductivity cannot be measured directly by extrapolation — it must be assembled from strong electrolytes using Kohlrausch’s law of independent migration. The combination NH4Cl + NaOH − NaCl leaves exactly NH4+ + OH−, since the Na+ and Cl− contributions cancel. Then the degree of dissociation follows from the ratio of the two conductivities.
Formula to use
Λ°(NH4OH) = Λ°(NH4Cl) + Λ°(NaOH) − Λ°(NaCl)  |  α = Λm/Λ°m
Baby steps
  1. Build the weak base: Λ° = 130 + 320 − 110 = 340 S cm² mol−1. (Check the cancellation: NH4++Cl− plus Na++OH− minus Na++Cl− leaves NH4++OH−. ✓)
  2. Degree of dissociation: α = 10.2 / 340 = 0.03.
  3. [OH−] = cα = 10−2 × 0.03 = 3 × 10−4 M.
  4. pOH = −log(3 × 10−4) = 4 − 0.477 = 3.52.
  5. pH = 14 − 3.52 = 10.48.
Answer
10.48
Shortcut
Check the direction before computing anything: NH4OH is a base, so its pH must be above 7. That immediately removes 3.52 and 2, leaving only 10.48 and 12 — and 12 would require a far higher dissociation than a weak base at 10−2 M can manage.
3.52 is the pOH, placed among the options to catch anyone who stops one line early. Whenever a question about a base ends in a logarithm, write “pOH” next to the number before converting — the label is what prevents the slip.
Q162 Types of electrical conduction Marked wrong

Match the following: 1) Graphite, 2) CuO, 3) HCl, 4) Ag  with  A) Semiconductor, B) Insulator, C) Electronic conductor, D) Electrolytic conductor

  •  1-C, 2-B, 3-D, 4-A
  •  1-A, 2-B, 3-D, 4-C
  • MARKED1-B, 2-A, 3-D, 4-C
  • KEY1-C, 2-A, 3-D, 4-C
Given
  • Four substances and four conduction categories.
Asked
The correct matching.
Concept to use
Three categories to keep apart. Electronic conductors carry current by free electrons — metals, and graphite, which has delocalised electrons in its layers. Electrolytic conductors carry current by moving ions — HCl in solution. Semiconductors conduct weakly and conduct better when heated — CuO is a standard example. Nothing on this list is an insulator, which is why category B goes unused.
Formula to use
Free electrons → electronic  |  free ions → electrolytic  |  small band gap → semiconductor
Baby steps
  1. Graphite conducts through delocalised π electrons within its sheets — a non-metal that is nevertheless an electronic conductor → C.
  2. CuO is a metal oxide with a small band gap; its conductivity rises with temperature → semiconductor, A.
  3. HCl ionises in water to H+ and Cl−, which carry the current → electrolytic conductor, D.
  4. Ag is a metal, the archetypal electronic conductor → C.
  5. Both graphite and silver are electronic conductors, so C is used twice and B is not used at all: 1-C, 2-A, 3-D, 4-C.
Answer
1-C, 2-A, 3-D, 4-C
Shortcut
Notice that the four items do not map one-to-one onto the four categories. Graphite and silver both conduct by electrons, so one category is repeated and one is unused — and the only option showing C twice is the answer. Counting the repeats settles it before you check any individual pairing.
Where it went wrong
Graphite was matched to insulator, probably because it is a non-metal. But graphite is the standard exception: it is used for electrodes and brushes precisely because it conducts. The deeper issue is the assumption that a matching question must pair items one-to-one — here it does not, and that structural clue was the fastest route to the answer.

Faraday's laws of electrolysis

1 question · 1 blank
Q147 Current from a rate of gas evolution Not attempted

In the electrolysis of acidulated water, hydrogen is produced at a rate of 1.12 cm³ s−1 at STP. The current flowing through the electrolytic cell is:

  • KEY9.65 A
  •  19.3 A
  •  0.193 A
  •  965 A
Given
  • Rate of H2 production = 1.12 cm³ s−1 at STP.
  • Molar volume at STP = 22400 cm³ mol−1; F = 96500 C mol−1.
Asked
The current.
Concept to use
Faraday’s first law connects charge to moles of product, and the bridge between them is the number of electrons per molecule. For hydrogen the half-reaction is 2H+ + 2e− → H2, so every mole of H2 costs two faradays. Since a rate is given, the answer comes out directly as charge per second, which is current.
Formula to use
moles/s = (volume/s)/22400  |  I = (moles/s) × n × F, with n = 2 for H2
Baby steps
  1. Convert volume rate to mole rate: 1.12 / 22400 = 5 × 10−5 mol s−1.
  2. Electrons required: n = 2 per molecule of H2.
  3. Charge per second: I = 5×10−5 × 2 × 96500.
  4. 5×10−5 × 96500 = 4.825, and doubling gives 9.65 A.
Answer
9.65 A
Shortcut
1.12 cm³ is not an accident — it is one twenty-thousandth of 22400, so the mole rate is 5×10−5 exactly. Spotting that 22.4 L and its decimal relatives have been planted in the numbers lets you skip the division and go straight to the multiplication.
19.3 A is what you get by using n = 4 (the value for O2, which needs four electrons per molecule) and 0.193 A comes from a decimal slip. In any electrolysis question, write the half-reaction down first — n is where these questions are decided, and it differs for each gas.

What the seventeen have in common

Two chapters, and the striking figure is the split: two attempted, fourteen left blank, plus one whose options were cut off. This is the highest skip rate in any paper reviewed so far.

1 · The blanks are overwhelmingly short questions
Q147 is one division and one multiplication (1.12/22400 × 2 × 96500 = 9.65 A). Q163 is substituting x = 0 into an equation printed in the question. Q145 needs no arithmetic at all — E° simply does not change when an equation is halved. Q166 is a single Nernst line where the 0.059 cancels. Between them, four questions worth 16 marks need perhaps three minutes. Fix: a timed set of twenty one-line electrochemistry problems, answers only — the aim is to make starting automatic rather than to teach anything new.

2 · Both wrong answers came from applying half a rule — Q136, Q162
Q136 needs two decisions: which couple (lowest E°) and which side of it (the reduced form). Mn2+ passes the second test and fails the first. Q162 assumed a matching question must pair one-to-one, when in fact graphite and silver are both electronic conductors and one category goes unused. Fix: when a rule has two parts, write both down before choosing — and count the categories in a matching question before assuming a one-to-one map.

3 · Several blanks are decided by a direction check alone
Q141 asks the pH of a base, so the answer must exceed 7 — two options die immediately, and the remaining trap is that 3.52 is the pOH. Q166 turns on whether C1/C2 is 10−2 or 100, settled by noticing Ecell exceeds E°cell. Q148 turns on whether dilution raises or lowers the reduction potential. None of these requires a full calculation to eliminate most of the list.

4 · One question has a genuine ambiguity worth knowing about — Q168
The exact mole-fraction form gives 10 g; NCERT's dilute-solution approximation gives 8 g. Only 8 g is on the option list, so the paper wants the approximation. That is not a trick — recognising which form a paper expects is a real skill, and the option list is the evidence.

Seven anchors that cover both chapters

The correct option is marked KEY and the option selected in the test is marked MARKED; questions with no marked option were left unattempted. Every numerical answer here was checked computationally before being written in.