Every chemistry question lost on this paper, rebuilt in full. Each one carries what was given, what was asked, the concept behind it, the formula, the steps written out, a table justifying the right option and ruling out each of the others, the fastest route through, and an animated figure wherever seeing the thing settles the answer.
16Questions lost
14Attempted, wrong
2Left blank
78Marks at stake
The shape of this paper
Sixteen questions: fourteen attempted and missed, only two left blank. That is an attempt
rate of 88 per cent — the hesitation is gone. On NEET marking the cost is 56 marks not
gained plus 14 in negatives.
Six of the fourteen come down to getting n wrong — the number of electrons in
the half reaction. Another four turn on the direction of a relationship: which of two
quantities is larger, or which way a trend runs.
All sixteen come from one chapter, and fourteen of them were attempted. The accuracy is the problem, not the willingness — and a striking number turn on the single number n, the count of electrons transferred.
Q 137Attempted · wrongMolar conductivity · two statements
Statement I (S-I): The limiting molar conductivity of KCl (strong electrolyte) is higher compared to that of CH₃COOH (weak electrolyte). Statement II (S-II): Molar conductivity decreases with decrease in concentration of electrolyte.
Both S-I & S-II are true.
S-I is false and S-II is true
Her answerS-I is true and S-II is false
CorrectBoth S-I and S-II are false
Given
KCl is a strong electrolyte; CH₃COOH is weak.
Two statements about limiting molar conductivity and about dilution.
Asked
The truth of each statement.
Concept to use
The word limiting changes everything in S-I. At infinite dilution every electrolyte is fully ionised, so the comparison is no longer between strong and weak — it is between the ions themselves. Acetic acid gives H⁺, the most mobile ion of all, so its Λ° (390.7) actually exceeds that of KCl (149.9). And S-II has the dilution trend backwards.
DiagramAnimatedMolar conductivity rises on dilution — and the weak acid wins at infinity.
Formula to useΛ°(CH₃COOH) = 390.7 > Λ°(KCl) = 149.9 S cm² mol⁻¹
Baby steps
S-I. At infinite dilution both are completely ionised. Acetic acid releases H⁺, which has exceptional mobility, so its limiting molar conductivity is the higher of the two. FALSE.
S-II. On dilution the ions get further apart and interfere with each other less, so molar conductivity increases, not decreases. FALSE.
Both statements are false.
Answer
Both S-I and S-II are false
Why this option and not the others
Option
Verdict
Reason
Both true
rule out
Neither is. S-I inverts the limiting values and S-II inverts the dilution trend.
S-I false, S-II true
rule out
S-I is indeed false, but molar conductivity rises on dilution.
S-I true, S-II false
rule out
S-II is correctly rejected, but S-I fails too — the word limiting is what breaks it.
Both false
keep
The weak acid has the higher Λ°, and Λ rises on dilution.
Shortcut
Two facts to lock in. Molar conductivity always RISES on dilution, for strong and weak alike — only the shape of the rise differs. And at infinite dilution the strong/weak distinction disappears, so limiting values are decided by which ions are present, not by electrolyte strength. H⁺ and OH⁻ are the two fastest ions there are.
Where it went wrong
S-II was correctly rejected, so the dilution trend was known. S-I slipped because “strong electrolyte” makes a higher conductivity feel obvious — and at any finite concentration it would be. But the statement says limiting, which means infinite dilution, where acetic acid is fully ionised and its H⁺ ions win. One word changes the answer, which is the same read-the-adjective failure seen elsewhere in these papers.
Q 142Attempted · wrongΔG° from cell potential
The E°cell for the reaction 2Fe³⁺ + Fe → 3Fe²⁺ is 1.21 V at 298 K. ΔG° of this process is
Her answer−116.76 kJ
−350.29 kJ
−175.14 kJ
Correct−233.53 kJ
Given
2Fe³⁺ + Fe → 3Fe²⁺, E°cell = 1.21 V.
F = 96500 C mol⁻¹.
Asked
ΔG° for the process.
Concept to use
ΔG° = −nFE°, and the only thing to work out is n, the number of electrons transferred. Balance the half reactions: Fe → Fe²⁺ + 2e⁻ releases two electrons, and 2Fe³⁺ + 2e⁻ → 2Fe²⁺ accepts them. So n = 2.
DiagramAnimatedΔG° = −nFE°, with n from the electrons transferred.
Formula to useΔG° = −nFE°
Baby steps
Oxidation: Fe → Fe²⁺ + 2e⁻.
Reduction: 2Fe³⁺ + 2e⁻ → 2Fe²⁺.
Electrons balance at n = 2.
ΔG° = −2 × 96500 × 1.21 = −233,530 J.
= −233.53 kJ.
Answer
−233.53 kJ
Why this option and not the others
Option
Verdict
Reason
−116.76 kJ
rule out
Exactly half the answer — this is n = 1. But iron goes from 0 to +2, which is two electrons.
−350.29 kJ
rule out
This is n = 3, taken from the coefficient of Fe²⁺ in the products rather than from the electrons transferred.
−175.14 kJ
rule out
n = 1.5, which is not physically meaningful here.
−233.53 kJ
keep
n = 2, from Fe → Fe²⁺ + 2e⁻.
Shortcut
Get n from the oxidation-state change, not from the stoichiometric coefficients. Iron metal goes 0 → +2, so two electrons leave — n = 2. The coefficient 3 on Fe²⁺ and the coefficient 2 on Fe³⁺ are both offered as distractors, and both give wrong answers.
Where it went wrong
−116.76 is exactly half the correct value, so n was taken as 1. The tempting reading is that each Fe³⁺ gains only one electron — which is true per ion, but there are two of them, and the balanced reaction transfers two electrons in total. Counting from the metal being oxidised (Fe → Fe²⁺) gives n directly and avoids the ambiguity.
Q 143Attempted · wrongElectrochemical equivalent
Electro chemical equivalent is highest for
CorrectSodium
Hydrogen
Her answerAluminium
Magnesium
Given
Four elements: Na (23, n = 1), H (1, n = 1), Al (27, n = 3), Mg (24, n = 2).
Asked
Which has the highest electrochemical equivalent.
Concept to use
The electrochemical equivalent Z is the mass deposited per coulomb: Z = M/(nF). Since F is a constant, ranking Z is just ranking M/n — the equivalent mass. A large atomic mass helps, but a large valency hurts, and here aluminium's n = 3 is what sinks it.
DiagramAnimatedElectrochemical equivalent is M/n, not M.
Formula to useZ = M / (nF) → rank by M/n
Baby steps
Sodium: 23/1 = 23.
Hydrogen: 1/1 = 1.
Aluminium: 27/3 = 9.
Magnesium: 24/2 = 12.
Highest M/n is sodium, despite aluminium having the larger atomic mass.
Answer
Sodium
Why this option and not the others
Option
Verdict
Reason
Sodium
keep
M/n = 23, the largest of the four.
Hydrogen
rule out
M/n = 1 — the smallest, because its atomic mass is tiny.
Aluminium
rule out
The largest atomic mass at 27, but n = 3 divides it down to 9. This is the trap: mass alone is not the criterion.
Magnesium
rule out
M/n = 12, second highest but still below sodium.
Shortcut
Ignore F entirely and just compute M/n for each option in your head — four small divisions. The question is really testing whether you remember to divide by the valency at all, and aluminium is placed there precisely to catch anyone who ranks by atomic mass.
Where it went wrong
Aluminium has the biggest atomic mass, so it looks like the obvious answer — but electrochemical equivalent is mass per coulomb, and depositing one aluminium atom costs three electrons rather than one. Dividing by n is the whole point of the quantity. Sodium's valency of 1 more than compensates for its slightly lower mass.
Q 147Attempted · wrongWhen Kohlrausch's law applies
Kohlrausch's law is applicable (A) for concentrated solution (B) at infinite dilution (C) for concentrated as well as dilute solutions
A and C only
CorrectB only
B and C only
Her answerA, B, and C
Given
Three claims about where Kohlrausch's law holds.
Asked
Which are correct.
Concept to use
Kohlrausch's law says the limiting molar conductivity of an electrolyte is the sum of independent contributions from its ions. That independence only holds when the ions are so far apart that they do not interact — which is precisely the condition of infinite dilution. At any real concentration the ions attract one another and the simple sum fails.
DiagramAnimatedIndependent migration only holds at infinite dilution.
Formula to useΛ°ₘ = ν₊λ°₊ + ν₋λ°₋ — valid at infinite dilution only
Baby steps
The law treats each ion's contribution as independent of the others.
That is only true when interionic attractions vanish, i.e. at infinite dilution.
In a concentrated solution the ions crowd each other, so (A) and (C) both fail.
Only B is correct.
Answer
B only
Why this option and not the others
Option
Verdict
Reason
A and C only
rule out
Both refer to concentrated solutions, where interionic forces break the independence the law assumes.
B only
keep
Infinite dilution is exactly the condition under which ion contributions become independent.
B and C only
rule out
B is right, but C extends the law to concentrated solutions where it does not hold.
A, B, and C
rule out
Only B holds. Note that A and C contradict each other in spirit with B, so an option containing all three cannot be right.
Shortcut
The word “limiting” is built into the law's own name — it is the law of independent migration of ions at infinite dilution. Any option extending it to concentrated solutions is wrong by definition. And structurally: when options A and C both describe concentrated solutions and B describes the opposite, an “all of these” option is self-contradictory.
Where it went wrong
“A, B and C” accepts everything, including two claims that contradict the condition in B. In a set like this, check whether the options are mutually consistent before choosing the generous one — a law cannot hold both only at infinite dilution and also at all concentrations. This is the third “all of these” over-acceptance across these papers.
Q 153Attempted · wrongUnits of molar conductivity
Conductance is directly proportional to area of the vessel and the concentration of solution in it and is inversely proportional to the length of the vessel, then the unit of the constant of proportionality is
Her answerS m mol⁻¹
CorrectS m² mol⁻¹
S⁻² m² mol
S² m² mol⁻²
Given
G ∝ A × c / l, so G = K(Ac/l).
G in siemens (S), A in m², l in m, c in mol m⁻³.
Asked
The units of the proportionality constant K.
Concept to use
This is a units question dressed as a conductance question. Rearrange for K and substitute the SI units of everything else. The constant here is in fact the molar conductivity, whose units are S m² mol⁻¹ — and the squared metre is what the question is really testing.
DiagramAnimatedMolar conductivity is always S m² mol⁻¹.
Formula to useK = G l / (A c)
Baby steps
Rearrange: K = Gl/(Ac).
Units: S × m ÷ (m² × mol m⁻³).
The denominator is mol m⁻¹, so K has units S × m ÷ (mol m⁻¹).
= S × m × m ÷ mol = S m² mol⁻¹.
Answer
S m² mol⁻¹
Why this option and not the others
Option
Verdict
Reason
S m mol⁻¹
rule out
One power of metre short. It comes from forgetting that concentration carries m⁻³, which supplies the extra metre.
S m² mol⁻¹
keep
The standard unit of molar conductivity, and what Gl/(Ac) reduces to.
S⁻² m² mol
rule out
Siemens cannot appear with a negative power here — G is in the numerator.
S² m² mol⁻²
rule out
Squaring both S and mol has no basis in the rearrangement.
Shortcut
Recognise the quantity rather than deriving it: the combination Gl/(Ac) is molar conductivity, and Λₘ is always in S m² mol⁻¹ (or S cm² mol⁻¹). The metre is squared — that single detail separates the correct answer from the distractor. Remember it as “area units per mole”.
Where it went wrong
S m mol⁻¹ is one power of metre short, which happens when concentration is treated as dimensionless or as mol per litre without converting. In SI, concentration is mol m⁻³, and that m⁻³ in the denominator flips up to supply the second metre. Writing every unit out explicitly before cancelling — rather than tracking them mentally — is what catches this.
Q 154Attempted · wrongMolar and equivalent conductivity
The expression showing the relationship between equivalent conductivity and molar conductivity is (Z = n-factor of the electrolyte)
Correctλₘ = Z × λₑₚ
λₑₚ = Z × λₘ
Her answerλₘ = λₑₚ / Z
λₘ = λₑₚ²
Given
Z is the n-factor of the electrolyte.
λₘ is molar conductivity; λₑₚ is equivalent conductivity.
Asked
The correct relationship.
Concept to use
One mole contains Z equivalents. Molar conductivity is measured per mole, equivalent conductivity per equivalent — so the molar value counts Z times as much material and must be the larger of the two: λₘ = Z × λₑₚ.
DiagramAnimatedOne mole holds Z equivalents, so the molar value is the larger.
Formula to use1 mole = Z equivalents → λₘ = Z × λₑₚ
Baby steps
Equivalent conductivity is defined per equivalent: λₑₚ = κ/(normality).
Molar conductivity is per mole: λₘ = κ/(molarity).
Normality = Z × molarity, so the equivalent value is the smaller one.
Therefore λₘ = Z × λₑₚ.
Check with H₂SO₄ (Z = 2): its molar conductivity is twice its equivalent conductivity, which is right — a mole of H₂SO₄ supplies two equivalents of H⁺.
Answer
λₘ = Z × λₑₚ
Why this option and not the others
Option
Verdict
Reason
λₘ = Z×λₑₚ
keep
A mole contains Z equivalents, so the molar value is Z times the equivalent value.
λₑₚ = Z×λₘ
rule out
Inverted — this would make the equivalent value the larger one.
λₘ = λₑₚ/Z
rule out
Also inverted. Dividing would make the molar value smaller than the equivalent value.
λₘ = λₑₚ²
rule out
Squaring is dimensionally impossible — both sides must have the same units.
Shortcut
Reason from the physical meaning: a mole is bigger than an equivalent (there are Z equivalents in one mole), so molar conductivity must be the bigger number. Whichever option makes λₘ larger is the right one, and that settles it without recalling any formula.
Where it went wrong
The relation was inverted — dividing where it should multiply. The safeguard is the size check above: since one mole contains several equivalents, the per-mole quantity has to be the larger. Testing on a concrete case such as H₂SO₄, where Z = 2, makes the direction obvious in a few seconds.
Q 161Attempted · wrongKohlrausch's law · applying the stoichiometry
Molar ionic conductance of Ca²⁺ and PO₄³⁻ at infinite dilution are 'X' mho cm² mol⁻¹ and 'Y' mho cm² mol⁻¹ respectively. Molar conductance of calcium phosphate at infinite dilution is … mho cm² mol⁻¹.
(X + Y)
Correct3X + 2Y
6(X + Y)
Her answer(3X + 2Y)/6
Given
λ°(Ca²⁺) = X, λ°(PO₄³⁻) = Y.
Calcium phosphate is Ca₃(PO₄)₂.
Asked
The limiting molar conductance of calcium phosphate.
Concept to use
Kohlrausch's law adds up the ionic contributions, each multiplied by how many of that ion one formula unit produces. Calcium phosphate is Ca₃(PO₄)₂, giving 3 calcium ions and 2 phosphate ions — so the coefficients are 3 and 2, and they multiply rather than divide.
DiagramAnimatedThe subscripts in the formula are the coefficients.
Formula to useΛ°ₘ = ν₊λ°₊ + ν₋λ°₋
Baby steps
Write the formula correctly: calcium phosphate is Ca₃(PO₄)₂.
Dissociation: Ca₃(PO₄)₂ → 3Ca²⁺ + 2PO₄³⁻.
Apply Kohlrausch: Λ° = 3λ°(Ca²⁺) + 2λ°(PO₄³⁻).
= 3X + 2Y.
Answer
3X + 2Y
Why this option and not the others
Option
Verdict
Reason
(X + Y)
rule out
Ignores the stoichiometry entirely — correct only for a 1:1 salt like NaCl.
3X + 2Y
keep
Three calcium ions and two phosphate ions per formula unit.
6(X + Y)
rule out
Multiplies both by 6, as though the total ion count applied to each.
(3X + 2Y)/6
rule out
The right numerator divided by the total number of ions. Kohlrausch's law is a sum, not an average — there is nothing to divide by.
Shortcut
Write the chemical formula first, then read the subscripts straight into the equation. Ca₃(PO₄)₂ gives 3 and 2 — the subscripts are the coefficients. And remember that molar conductivity is already a per-mole quantity, so there is never any dividing by the ion count.
Where it went wrong
The numerator was right, so the stoichiometry was read correctly — the error was dividing by 6 at the end. That extra step looks like an averaging instinct, as though the answer should be per ion. But Λ°ₘ is defined per mole of electrolyte, and one mole of Ca₃(PO₄)₂ genuinely does supply five ions' worth of conduction. Nothing is averaged.
During the discharge of lead storage battery 1 F of electricity is produced. Weight of lead sulphate formed at anode is (Atomic weight of lead is 208 u)
Correct152 g
608 g
Her answer103 g
228 g
Given
Lead storage battery on discharge; 1 F of electricity passed.
Anode reaction: Pb + SO₄²⁻ → PbSO₄ + 2e⁻.
Atomic weight of Pb = 208, so M(PbSO₄) = 208 + 32 + 64 = 304.
Asked
The mass of PbSO₄ formed at the anode.
Concept to use
The anode half reaction releases two electrons per lead atom, so one Faraday — one mole of electrons — produces only half a mole of lead sulphate. Getting n = 2 from the half equation is the whole question.
DiagramAnimatedThe anode half equation gives n = 2, so 1 F makes half a mole.
Formula to usemoles of PbSO₄ = (charge in F) / n , with n = 2
Baby steps
Anode: Pb + SO₄²⁻ → PbSO₄ + 2e⁻.
1 F = 1 mole of electrons, and 2 moles of electrons are needed per mole of PbSO₄.
So moles formed = 1/2 = 0.5 mol.
M(PbSO₄) = 208 + 32 + (4 × 16) = 304 g mol⁻¹.
Mass = 0.5 × 304 = 152 g.
Answer
152 g
Why this option and not the others
Option
Verdict
Reason
152 g
keep
0.5 mol of PbSO₄, because two electrons are released per lead atom.
608 g
rule out
Two moles — four times too much. It would need 4 F.
103 g
rule out
Not obtainable from these numbers; it looks like a molar mass slip, perhaps using 206 for Pb and then halving twice.
228 g
rule out
0.75 mol, which no whole-number electron count produces.
Shortcut
Always write the half equation before touching the arithmetic — it hands you n directly. Here n = 2, so 1 F gives half a mole. And note the symmetry worth knowing: PbSO₄ forms at both electrodes during discharge, so the total is 304 g, but the question asks only about the anode.
Where it went wrong
103 g does not follow from any consistent route, which suggests the molar mass rather than the electron count went wrong — PbSO₄ is 304, not 206 or 208. Building the molar mass explicitly (208 for Pb, 32 for S, 64 for four oxygens) before dividing removes the guesswork, and the half equation gives the factor of two.
Q 164Attempted · wrongNernst equation · swapping the concentrations
The electro-chemical cell Zn|ZnSO₄(0.01 M) || CuSO₄(1.0 M)|Cu has emf E₁. When the concentration of ZnSO₄ is changed to 1.0 M and that of CuSO₄ to 0.01 M, the emf changes to E₂. The relationship between E₁ and E₂ is:
E₁ = E₂
E₁ < E₂
CorrectE₁ > E₂
Her answer2E₁ = E₂
Given
Cell 1: [Zn²⁺] = 0.01 M, [Cu²⁺] = 1.0 M.
Cell 2: [Zn²⁺] = 1.0 M, [Cu²⁺] = 0.01 M.
Cell reaction: Zn + Cu²⁺ → Zn²⁺ + Cu, n = 2.
Asked
Whether E₁ is greater than, less than or equal to E₂.
Concept to use
The Nernst equation says E falls as the reaction quotient Q rises. Here Q = [Zn²⁺]/[Cu²⁺] — product over reactant. Cell 1 has a small Q (0.01), cell 2 has a large Q (100). More product and less reactant means the reaction has less driving force, so E₂ must be smaller.
DiagramAnimatedMore reactant and less product means a bigger push forward.
Formula to useE = E° − (0.0591/n) log ([Zn²⁺]/[Cu²⁺])
Baby steps
Cell 1: Q = 0.01/1.0 = 0.01, so log Q = −2 and E₁ = E° + 0.0591.
Cell 2: Q = 1.0/0.01 = 100, so log Q = +2 and E₂ = E° − 0.0591.
The difference is 0.118 V, with cell 1 higher.
So E₁ > E₂.
Answer
E₁ > E₂
Why this option and not the others
Option
Verdict
Reason
E₁ = E₂
rule out
The concentrations are not symmetric in their effect — one is a product and one is a reactant.
E₁ < E₂
rule out
The reverse. Cell 2 has more product and less reactant, so it has less driving force.
E₁ > E₂
keep
Small Q gives the higher emf; cell 1 has Q = 0.01 against cell 2's Q = 100.
2E₁ = E₂
rule out
There is no doubling relationship here — the concentration effect is logarithmic and additive, not multiplicative.
Shortcut
Skip the Nernst equation and reason with Le Chatelier: more reactant and less product means a stronger push forward, so a higher emf. Cell 1 has plenty of Cu²⁺ (the reactant) and little Zn²⁺ (the product), so it wins. That gets the answer in one sentence and cannot be inverted by an algebra slip.
Where it went wrong
“2E₁ = E₂” treats the concentration change as scaling the emf, but the Nernst term is logarithmic and additive — it shifts E by a fixed amount, here 0.0591 V, rather than multiplying it. Emf values are typically around a volt while Nernst corrections are hundredths of a volt, so a doubling is never plausible. Noticing the scale of the correction is a useful sanity check.
Q 166Attempted · wrongElectrolysis · what forms at the anode
Which of the following upon electrolysis liberate O₂ gas as major product at anode? A) AgNO₃ (aq) (inert electrodes) B) CuSO₄ (aq) (copper electrodes) C) Conc. HCl (Pt electrodes) D) Dil. H₂SO₄ (Pt electrodes)
CorrectA and D only
Her answerA, B, D only
B, C only
C only
Given
Four electrolysis set-ups, with the electrode material specified in each.
Asked
Which liberate O₂ at the anode.
Concept to use
At the anode, whichever species is easiest to oxidise reacts. Two things can pre-empt water: an easily oxidised anion such as Cl⁻, or an active electrode that dissolves instead. Oxygen appears only when the electrode is inert and the anion is hard to oxidise.
DiagramAnimatedRead the electrode material before the solution.
Formula to useinert electrode + hard-to-oxidise anion (NO₃⁻, SO₄²⁻) → O₂ from water
Baby steps
A. AgNO₃ with inert electrodes. NO₃⁻ is very hard to oxidise, so water goes instead: O₂. ✓
B. CuSO₄ with COPPER electrodes. The copper anode itself dissolves — an active electrode. No oxygen at all. ✗
C. Conc. HCl with Pt. Chloride is plentiful and easily oxidised, so Cl₂ comes off, not O₂. ✗
D. Dil. H₂SO₄ with Pt. SO₄²⁻ is hard to oxidise, so water goes: O₂. ✓
So A and D only.
Answer
A and D only
Why this option and not the others
Option
Verdict
Reason
A and D only
keep
Both have inert electrodes and anions that resist oxidation, so water is oxidised to O₂.
A, B, D only
rule out
B has copper electrodes — the anode dissolves rather than releasing gas. The electrode material is stated in the question precisely to signal this.
B, C only
rule out
Neither gives oxygen: B gives Cu²⁺ and C gives Cl₂.
C only
rule out
Concentrated HCl gives chlorine, not oxygen.
Shortcut
Read the electrode material first. If it says copper, silver or any active metal, the anode dissolves and nothing is liberated — that eliminates the option before you think about the solution at all. Then, among inert electrodes, only halides beat water; nitrate and sulphate do not.
Where it went wrong
Option B was included, and its electrode material is stated explicitly as copper. An active electrode changes the whole anode reaction: instead of oxidising something in solution, the metal itself goes into solution. The bracketed electrode material in each option is not decoration — it is the deciding information, and it is easy to read past when scanning the chemical formulas.
Q 167Attempted · wrongCost of electrolysis
The cost of electricity to liberate 24 grams of oxygen is Rs 3000. Cost of electricity to deposit 24 grams of magnesium will be (Assuming current efficiency is 100%)
CorrectRs 2000
Rs 1500
Her answerRs 4300
Rs 2500
Given
24 g of O₂ costs Rs 3000 to liberate.
M(O₂) = 32, M(Mg) = 24.
O₂ needs 4 F per mole; Mg needs 2 F per mole.
Asked
The cost of depositing 24 g of magnesium.
Concept to use
Cost is proportional to charge, so convert each mass into Faradays and compare. The two 24s in the question are a deliberate distraction — they are equal masses of substances with different molar masses and different electron requirements.
DiagramAnimatedFind the cost of one Faraday, then count Faradays.
Formula to useF required = (mass/M) × n ; cost ∝ F
Baby steps
24 g O₂ = 24/32 = 0.75 mol. Each mole needs 4 F, so 0.75 × 4 = 3 F.
3 F costs Rs 3000, so 1 F costs Rs 1000.
24 g Mg = 24/24 = 1 mol. Mg²⁺ + 2e⁻ → Mg needs 2 F.
Cost = 2 × 1000 = Rs 2000.
Answer
Rs 2000
Why this option and not the others
Option
Verdict
Reason
Rs 2000
keep
2 F at Rs 1000 per Faraday.
Rs 1500
rule out
1.5 F — would need magnesium to be monovalent.
Rs 4300
rule out
Not obtainable from any consistent electron count; it looks like a proportion taken directly from the masses.
Rs 2500
rule out
2.5 F, which no whole-number valency produces.
Shortcut
Find the cost of one Faraday first and everything else is one multiplication. Here 3 F for Rs 3000 makes it Rs 1000 per Faraday, a number so clean it confirms the working. Then just count the Faradays for the second substance.
Worth memorising: O₂ needs 4 F per mole (2H₂O → O₂ + 4H⁺ + 4e⁻), which is the fact this question hinges on.
Where it went wrong
Rs 4300 does not correspond to a whole number of Faradays, which suggests the masses were compared directly rather than converted to charge. The two 24s look invitingly similar, but 24 g of O₂ is 0.75 mol needing 3 F while 24 g of Mg is 1 mol needing 2 F. Converting mass → moles → Faradays every time removes the temptation to compare grams.
Q 168Attempted · wrongSame charge, two different products
The weight of silver (at. wt. = 108) displaced by a quantity of electricity which displaces 5600 mL of O₂ at STP will be
5.4 g
Her answer10.8 g
54.0 g
Correct108.0 g
Given
5600 mL of O₂ at STP = 5.6 L.
At STP, 22.4 L = 1 mol.
Ag⁺ + e⁻ → Ag needs 1 F per mole; O₂ needs 4 F per mole.
Asked
The mass of silver deposited by the same quantity of electricity.
Concept to use
The charge is the bridge between the two substances. Convert the oxygen volume to moles, then to Faradays, then use those Faradays for silver. Because silver is monovalent, one Faraday deposits exactly one mole, which makes the final step trivial.
DiagramAnimated5600 mL at STP is a quarter mole, and O₂ costs 4 F per mole.
Formula to usemoles = V/22.4 ; F = moles × n ; mass = (F/n) × M
Baby steps
Moles of O₂ = 5.6/22.4 = 0.25 mol.
Charge needed = 0.25 × 4 = 1 F.
For silver, n = 1, so 1 F deposits 1 mole of Ag.
Mass = 1 × 108 = 108 g.
Answer
108.0 g
Why this option and not the others
Option
Verdict
Reason
5.4 g
rule out
0.05 mol — twenty times too little.
10.8 g
rule out
0.1 mol. This would follow from taking 5600 mL as 0.25 F rather than 1 F, i.e. forgetting that oxygen needs four electrons.
54.0 g
rule out
0.5 mol, which would need 0.5 F.
108.0 g
keep
1 F, and silver's equivalent mass equals its atomic mass because n = 1.
Shortcut
Two numbers make this a ten-second question. 5600 mL at STP is exactly 0.25 mol (a quarter of 22.4 L), and oxygen costs 4 F per mole — so the charge is exactly 1 F. And 1 F of silver is exactly 108 g, because n = 1. The numbers in these questions are always chosen to land on whole Faradays.
Where it went wrong
10.8 g is one tenth of the answer, which points at the oxygen side rather than the silver side — the factor of 4 electrons per O₂ molecule is the most likely omission. Writing the half equation 2H₂O → O₂ + 4H⁺ + 4e⁻ explicitly before converting is what supplies that 4, and it is the same fact that decided Q167 on this paper.
Q 169Left blankEquilibrium constant from E°
The equilibrium constant of the reaction: Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s); E° = 0.46 V at 298 K is
2.0 × 10¹⁰
4.0 × 10¹⁰
Correct4.0 × 10¹⁵
2.4 × 10¹⁰
Given
Cu + 2Ag⁺ → Cu²⁺ + 2Ag, E° = 0.46 V.
n = 2 (copper goes 0 → +2).
At 298 K, 0.0591 V is the Nernst constant.
Asked
The equilibrium constant K.
Concept to use
At equilibrium the cell has run down, so E = 0 and the Nernst equation collapses to a direct link between E° and K. The result to remember is log K = nE°/0.0591 at 298 K.
DiagramAnimatedlog K = nE°/0.0591 — the exponent alone picks the answer.
Formula to uselog K = nE° / 0.0591
Baby steps
n = 2, because copper is oxidised from 0 to +2.
log K = (2 × 0.46)/0.0591 = 0.92/0.0591.
= 15.57.
K = 1015.57 = 100.57 × 10¹⁵ ≈ 3.7 × 10¹⁵.
Nearest option: 4.0 × 10¹⁵.
Answer
4.0 × 10¹⁵
Why this option and not the others
Option
Verdict
Reason
2.0 × 10¹⁰
rule out
Would need log K = 10.3, i.e. E° about 0.30 V.
4.0 × 10¹⁰
rule out
log K = 10.6 — the right digits but five powers of ten short.
4.0 × 10¹⁵
keep
log K = 15.57, so K ≈ 3.7 × 10¹⁵.
2.4 × 10¹⁰
rule out
Again five orders of magnitude too small.
Shortcut
The exponent alone picks the answer here: log K = nE°/0.0591 = 0.92/0.0591 ≈ 15.6, so K is of order 10¹⁵. Three of the four options are 10¹⁰, so you never need the leading digit at all. Estimating the power of ten first is the fastest route through any log-based option list.
Q 170Left blankHydrogen electrode in pure water
The pressure of H₂ required to make the potential of H₂ electrode zero in pure water at 298 K
10⁻¹⁰ atm
10⁻⁴ atm
Correct10⁻¹⁴ atm
10⁻¹² atm
Given
Pure water at 298 K, so pH = 7 and [H⁺] = 10⁻⁷ M.
Hydrogen electrode: 2H⁺ + 2e⁻ → H₂, E° = 0.
Asked
The H₂ pressure that makes the electrode potential zero.
Concept to use
Apply the Nernst equation to the hydrogen electrode and set E = 0. The reaction quotient is P(H₂)/[H⁺]², and with [H⁺] = 10⁻⁷ the denominator is already 10⁻¹⁴ — so the pressure must be extraordinarily small to balance it.
DiagramAnimatedSetting E = 0 gives log P = −2 × pH.
Formula to useE = E° − (0.0591/2) log (P / [H⁺]²)
Baby steps
E° = 0 for the hydrogen electrode by definition.
E = −(0.0591/2) log (P / (10⁻⁷)²) = −(0.0591/2)(log P + 14).
Set E = 0: log P + 14 = 0.
log P = −14, so P = 10⁻¹⁴ atm.
Answer
10⁻¹⁴ atm
Why this option and not the others
Option
Verdict
Reason
10⁻¹⁰ atm
rule out
Would correspond to [H⁺] = 10⁻⁵, i.e. pH 5, not pure water.
10⁻⁴ atm
rule out
Would need pH 2.
10⁻¹⁴ atm
keep
log P = −2 × pH = −14 for pure water.
10⁻¹² atm
rule out
Would correspond to pH 6.
Shortcut
There is a clean shortcut: setting E = 0 gives log P = −2 × pH. For pure water pH = 7, so log P = −14 immediately. And a sanity check on the sign — pure water is a very poor source of H⁺, so a vanishingly small hydrogen pressure is needed to hold the potential at zero. Any positive power of ten would be nonsense.
Q 172Attempted · wrongDaniell cell with an applied voltage
For the cell Zn|ZnSO₄(1 M) || CuSO₄(1 M)|Cu with E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V, identify the incorrect statement:
If E₋ₓₜ = 1.1 V, no flow of e⁻ or current occurs
If E₋ₓₜ < 1.1 V, Zn dissolves at anode and Cu deposits at cathode
Her answerIf E₋ₓₜ > 1.1 V, electrons flow from Cu to Zn
CorrectIf E₋ₓₜ > 1.1 V, Zn dissolves at Zn electrode and Cu deposits at Cu electrode
Given
E°cell = 0.34 − (−0.76) = 1.10 V.
An external emf E₋ₓₜ is applied in opposition.
Asked
Which statement is incorrect.
Concept to use
Three regimes, decided by comparing the applied voltage with the cell's own 1.1 V. Below 1.1 V the cell works normally as a galvanic cell. At 1.1 V everything balances and nothing flows. Above 1.1 V the cell is driven backwards and becomes electrolytic — every reaction reverses.
DiagramAnimatedThree regimes, decided by comparing with 1.1 V.
Formula to useE₋ₓₜ < 1.1 V: galvanic · = 1.1 V: no current · > 1.1 V: electrolytic (reversed)
Baby steps
Option 1. At exactly 1.1 V the two emfs cancel and no current flows. Correct statement.
Option 2. Below 1.1 V the cell runs normally: Zn dissolves at the anode, Cu deposits at the cathode. Correct statement.
Option 3. Above 1.1 V the current reverses, so electrons are pushed from Cu towards Zn. Correct statement.
Option 4. Above 1.1 V the reactions reverse: Zn is deposited and Cu dissolves. Saying Zn dissolves and Cu deposits describes the galvanic behaviour, which no longer applies. INCORRECT — this is the answer.
Answer
“If E₋ₓₜ > 1.1 V, Zn dissolves at Zn electrode and Cu deposits at Cu electrode”
Why this option and not the others
Option
Verdict
Reason
= 1.1 V, no current
rule out
True — the applied emf exactly opposes the cell emf.
< 1.1 V, Zn dissolves, Cu deposits
rule out
True — normal galvanic operation.
> 1.1 V, electrons flow Cu to Zn
rule out
True — the current has reversed, so the electron flow reverses too.
> 1.1 V, Zn dissolves, Cu deposits
keep
This is the answer. Above 1.1 V everything reverses, so Zn is deposited and Cu dissolves.
Shortcut
Sort the four options by which regime they describe, then check consistency. Two options here describe E₋ₓₜ > 1.1 V and they say different things about the same situation — so one of those two must be the incorrect one. That halves the work before any electrochemistry. Then recall that above 1.1 V the cell is being forced backwards.
Where it went wrong
The statement chosen — electrons flowing from Cu to Zn above 1.1 V — is actually correct, because the current has reversed. In a “which is incorrect” question it is easy to pick the option that sounds counter-intuitive rather than the one that is actually false. The reliable method is the structural one: find the two options describing the same conditions and decide between them, since they cannot both be right.
Q 180Attempted · wrongNickel–cadmium cell
During the discharge of Nickel-Cadmium battery, the change in oxidation state at positive electrode will be
+4 to +3
Correct+3 to +2
Her answer0 to +2
+2 to 0
Given
Ni–Cd cell on discharge.
Negative electrode: Cd → Cd(OH)₂.
Positive electrode: NiO(OH) → Ni(OH)₂.
Asked
The oxidation state change at the positive electrode.
Concept to use
The positive electrode is the cathode on discharge, so it is the one being reduced. In NiO(OH) nickel is +3; in Ni(OH)₂ it is +2. The cadmium electrode is the negative one, and that is where 0 → +2 happens.
DiagramAnimatedNickel at the positive electrode, cadmium at the negative.
Formula to usepositive (cathode): NiO(OH) + H₂O + e⁻ → Ni(OH)₂ + OH⁻
Baby steps
Work out Ni in NiO(OH): oxygen is −2 and the OH group is −1, so Ni must be +3.
Work out Ni in Ni(OH)₂: two OH groups at −1 each, so Ni is +2.
The positive electrode is reduced on discharge, so nickel goes +3 → +2.
The 0 → +2 change belongs to cadmium, at the negative electrode.
Answer
+3 to +2
Why this option and not the others
Option
Verdict
Reason
+4 to +3
rule out
Nickel is +3 in NiO(OH), not +4.
+3 to +2
keep
NiO(OH) → Ni(OH)₂ at the positive electrode.
0 to +2
rule out
This is the cadmium change, and cadmium is the NEGATIVE electrode. Right change, wrong electrode.
+2 to 0
rule out
The reverse of the cadmium change — what happens on charging, not discharging.
Shortcut
Two things to settle before answering. Which electrode? Positive means cathode, means reduction on discharge. And which metal sits there? Nickel at the positive, cadmium at the negative — the name of the cell lists them in that order, negative first, which is a useful mnemonic.
Where it went wrong
0 → +2 is a real change in this cell, but it belongs to cadmium at the negative electrode. Only pure elements start at 0, and nickel in NiO(OH) is already at +3 — so a 0 starting point could never describe the positive electrode. Checking which species actually sits at the named electrode, before computing any oxidation states, is what separates the two.
What the sixteen have in common
Reading the paper as a whole
The number n decides six questions
Q
Where n came in
What went wrong
142
ΔG° = −nFE°
n taken as 1; Fe → Fe²⁺ gives n = 2, so the answer halved.
143
Z = M/nF
Ranked by atomic mass; dividing by valency makes sodium beat aluminium.
163
Pb → PbSO₄ + 2e⁻
1 F gives half a mole, so 152 g.
167
O₂ needs 4 F, Mg needs 2 F
Masses compared directly instead of converting to Faradays.
168
O₂ needs 4 F, Ag needs 1 F
The factor of 4 for oxygen was dropped.
169
log K = nE°/0.0591
(left blank — n = 2 and one division.)
Write the half equation before anything else. It hands you n directly, and n is the
one number that appears in almost every formula in this chapter. Two facts worth committing:
O₂ costs 4 F per mole, and Ag costs 1 F per mole.
Four questions where a relationship was inverted
Q137 — molar conductivity rises on dilution, and at infinite dilution
the weak acid actually has the higher value.
Q154 — a mole contains Z equivalents, so the molar value is the larger
one: λₘ = Z × λₑₚ.
Q161 — Kohlrausch's law is a sum, so the coefficients multiply. Nothing is
divided by the ion count.
Q164 — more reactant and less product means a bigger emf, so E₁ >
E₂.
Each of these has a one-sentence physical check that beats the algebra. For Q154: a mole
is bigger than an equivalent, so the per-mole number must be bigger. For Q164: more
reactant means a stronger push forward.
Two questions where the electrode material was the answer
Q166 asks which electrolyses liberate O₂ at the anode. Option B specifies
copper electrodes — an active anode dissolves instead of releasing gas, so no oxygen
at all. The bracketed electrode material in each option is the deciding information, not
decoration.
Q172 asks for the incorrect statement about a Daniell cell under an applied voltage.
Two options describe the same condition (E₋ₓₜ > 1.1 V) and say different
things — so one of those two must be the answer. Above 1.1 V everything reverses.