The six that finish the chapterConcepts · Animated figures · Formula sheet · 20 worked PYQs

The equivalent conductance of a strong electrolyte increases on dilution. This is mainly because of:

• 1. an increase in the ionic mobility of the ions
• 2. an increase in the degree of dissociation
• 3. a rise in the number of ions per unit volume
• 4. a rise in the viscosity of the solvent

Given

A strong electrolyte, being diluted

Asked

The physical reason for the rise, not the arithmetic

Concept

A strong electrolyte is already fully dissociated at any concentration. Whatever changes on dilution, it is not the degree of dissociation.

Formula

Λ_m = Λ°_m − A√c — the gentle linear rise characteristic of strong electrolytes

Baby steps

1. Ask what is already true at high concentration: for NaCl or KCl, α is essentially 1 throughout.
2. So option 1 is ruled out — it describes weak electrolytes.
3. What does change is crowding. At high concentration each ion is surrounded by a cloud of oppositely charged ions that drags on it as it moves.
4. Dilution thins that cloud, interionic attraction falls, and each ion travels faster under the same field.
5. Faster ions per mole means higher conductance per mole.

Answer

Option 1 — increased ionic mobility

Why not others

Option 2 is the correct explanation for a weak electrolyte, and is the most-picked wrong answer. Option 3 is backwards — dilution lowers the ion count per unit volume, which is exactly why κ falls. Option 4 is unrelated; the solvent is barely changed.

Shortcut

Strong electrolyte → mobility explanation. Weak electrolyte → dissociation explanation. One line, two cases, no working.

An aqueous solution of which of the following is the best conductor of electric current?

• 1. Ammonia, NH₃
• 2. Fructose, C₆H₁₂O₆
• 3. Hydrochloric acid, HCl
• 4. Acetic acid, C₂H₄O₂

Given

Four solutes: one strong acid, one weak acid, one weak base, one non-electrolyte

Asked

Which conducts best in water

Concept

Conduction in solution needs mobile ions. More ions means better conduction, so the strongest electrolyte wins.

Formula

No formula — classify each solute as strong, weak, or non-electrolyte

Baby steps

1. Fructose is a sugar. It dissolves but does not ionise at all — a non-electrolyte, essentially zero conductance.
2. Ammonia is a weak base, only slightly ionised to NH₄⁺ and OH⁻.
3. Acetic acid is a weak acid, only about 1–5% ionised at ordinary concentrations.
4. HCl is a strong acid, completely ionised to H⁺ and Cl⁻.
5. Complete ionisation gives the most ions, and H⁺ additionally has the highest ionic conductivity of any ion (349.6 S cm² mol⁻¹).

Answer

Option 3 — hydrochloric acid

Why not others

Acetic acid is the tempting distractor because it is an acid and does conduct — but NCERT's own table shows 0.1 M HCl at 3.91 S m⁻¹ against 0.1 M acetic acid at just 0.047 S m⁻¹, about eighty times worse.

Shortcut

Sort into three bins: non-electrolyte, weak, strong. Pick the strong one. If two are strong, pick the one containing H⁺ or OH⁻.

The equivalent conductances of Ba²⁺ and Cl⁻ are 127 and 76 ohm⁻¹ cm² eq⁻¹ respectively. What is the equivalent conductance of BaCl₂ at infinite dilution?

• 1. 330 ohm⁻¹ cm² eq⁻¹
• 2. 279 ohm⁻¹ cm² eq⁻¹
• 3. 101.5 ohm⁻¹ cm² eq⁻¹
• 4. 203 ohm⁻¹ cm² eq⁻¹

Given

λ_eq(Ba²⁺) = 127, λ_eq(Cl⁻) = 76, both already in equivalent units

Asked

Λ_eq(BaCl₂), not Λ_m

Concept

The whole point of equivalent conductance is that the charge has already been divided out. Once values are per equivalent, you simply add them — no stoichiometric multiplier.

Formula

Λ_eq = λ_eq(cation) + λ_eq(anion)

Baby steps

1. Check the units on the given data: eq⁻¹, so these are equivalent conductances.
2. Add them directly: 127 + 76
3. Λ_eq(BaCl₂) = 203 ohm⁻¹ cm² eq⁻¹
4. Cross-check by converting: Λ_m = 2 × 203 = 406, which also equals λ_m(Ba²⁺) + 2λ_m(Cl⁻) = 254 + 152 = 406 ✓

Answer

Option 4 — 203 ohm⁻¹ cm² eq⁻¹

Why not others

Option 2 (279) applies the molar rule 127 + 2(76) to data that is already per equivalent — the trap. Option 1 doubles everything. The units on the given numbers are the whole clue, which is why reading them first matters.

Shortcut

See eq⁻¹ in the data, just add. See mol⁻¹, apply the ν multipliers. The unit tells you which rule to use.

A conductivity cell filled with 0.1 mol L⁻¹ KCl has resistance 100 Ω. The same cell filled with 0.02 mol L⁻¹ KCl has resistance 520 Ω. Given that κ for the 0.1 M KCl is 1.29 S m⁻¹, find the conductivity and molar conductivity of the 0.02 M solution.

• 1. 0.248 S m⁻¹ and 12.4 S cm² mol⁻¹
• 2. 0.248 S m⁻¹ and 124 S cm² mol⁻¹
• 3. 2.48 S m⁻¹ and 124 S cm² mol⁻¹
• 4. 0.248 S m⁻¹ and 1240 S cm² mol⁻¹

Given

Standard: κ = 1.29 S m⁻¹, R = 100 Ω. Unknown: R = 520 Ω, M = 0.02 mol L⁻¹

Asked

κ and Λ_m of the dilute solution

Concept

The classic two-step: calibrate the cell with the known solution, then use that constant on the unknown. The cell constant is the shared link.

Formula

G* = κR then κ = G*/R then Λ_m = κ/c

Baby steps

1. G* = 1.29 S m⁻¹ × 100 Ω = 129 m⁻¹ (equivalently 1.29 cm⁻¹)
2. κ(0.02 M) = G*/R = 129 / 520 = 0.248 S m⁻¹
3. Convert the concentration to SI: 0.02 mol L⁻¹ × 1000 = 20 mol m⁻³
4. Λ_m = 0.248 / 20 = 0.0124 S m² mol⁻¹
5. Convert: × 10⁴ → 124 S cm² mol⁻¹

Answer

Option 2 — κ = 0.248 S m⁻¹, Λ_m = 124 S cm² mol⁻¹

Why not others

Options 1 and 4 are the two directions of the 10⁴ conversion slip between S m² mol⁻¹ and S cm² mol⁻¹. Option 3 misplaces a decimal in κ. Every wrong option here has the right chemistry and the wrong unit handling — which is exactly what the question is for.

Shortcut

Do the whole thing in centimetres instead: G* = 1.29 cm⁻¹, κ = 1.29/520 = 2.48 × 10⁻³ S cm⁻¹, Λ_m = 2.48 × 10⁻³ × 1000/0.02 = 124. No power-of-ten conversion at all.

Which of these relations for the emf of an electrochemical cell is correct?
(i) oxidation potential of anode − reduction potential of cathode
(ii) oxidation potential of anode + reduction potential of cathode
(iii) reduction potential of anode + reduction potential of cathode
(iv) oxidation potential of anode − oxidation potential of cathode

• 1. (i) only
• 2. (iii) only
• 3. (ii) and (iv)
• 4. (i) and (iii)

Given

Four candidate expressions for cell emf

Asked

Which are correct — note this may be more than one

Concept

Start from the one relation you trust, E°_cell = E°_cathode − E°_anode in reduction potentials, and rewrite each candidate into that form.

Formula

E_ox = −E_red for the same couple

Baby steps

1. Baseline: emf = E_red(cathode) − E_red(anode)
2. Since E_ox(anode) = −E_red(anode), substitute: emf = E_red(cathode) + E_ox(anode). That is statement (ii) ✓
3. Now rewrite the cathode term too: E_red(cathode) = −E_ox(cathode), giving emf = E_ox(anode) − E_ox(cathode). That is statement (iv) ✓
4. (i) mixes an oxidation term with a reduction term in a subtraction — double-counting the sign flip. ✗
5. (iii) adds two reduction potentials, which would give the wrong value for any real cell. ✗

Answer

Option 3 — (ii) and (iv)

Why not others

Options containing (i) or (iii) fail the substitution test. Test any candidate relation against the Daniell cell: E_ox(Zn) = +0.76, E_red(Cu) = +0.34, and (ii) gives 1.10 V correctly while (iii) would give −0.42 V.

Shortcut

A correct expression must mix the two types when adding, or match types when subtracting. Add oxidation to reduction; subtract like from like. Any expression that adds like types or subtracts unlike types is wrong.

For the cell A | A⁺(x M) ‖ B⁺(y M) | B the measured emf is +0.20 V. What is the cell reaction?

• 1. A + B⁺ → A⁺ + B
• 2. A⁺ + B → A + B⁺
• 3. A + B → A⁺ + B⁺
• 4. A⁺ + B⁺ → A + B

Given

Cell notation with A on the left and B on the right, emf positive

Asked

The spontaneous reaction taking place

Concept

The notation itself declares the roles. Left is anode (oxidation), right is cathode (reduction). A positive emf confirms the cell is written the right way round.

Formula

left = anode = oxidation  ;  right = cathode = reduction

Baby steps

1. A is on the left, so A is oxidised: A → A⁺ + e⁻
2. B is on the right, so B⁺ is reduced: B⁺ + e⁻ → B
3. Add the two half-reactions; the electrons cancel.
4. A + B⁺ → A⁺ + B
5. emf is +0.20 V, positive, so this is indeed spontaneous as written — no need to reverse anything.

Answer

Option 1 — A + B⁺ → A⁺ + B

Why not others

Option 2 is the reverse, which would be correct only if the emf had come out negative. Options 3 and 4 are not redox reactions at all — one oxidises both species, the other reduces both, and neither balances electrons.

Shortcut

Read the notation left to right: the leftmost solid goes into solution, the rightmost ion comes out of it. The concentrations x and y are supplied only as distraction.

A cell reaction is spontaneous when:

• 1. Δ_rG° is positive
• 2. Δ_rG° is negative
• 3. E°_cell is negative
• 4. E°_red is positive for both electrodes

Given

Four statements about signs

Asked

The condition for spontaneity

Concept

Spontaneity is defined thermodynamically by Gibbs energy. The cell potential is a re-expression of the same thing with the sign inverted.

Formula

Δ_rG° = −nFE°_cell

Baby steps

1. A process runs on its own when it lowers the Gibbs energy: Δ_rG° < 0
2. Since n and F are always positive, a negative Δ_rG° requires a positive E°_cell.
3. So options 1 and 2 both describe non-spontaneous cells.
4. Option 4 says nothing useful — what matters is the difference between the two electrode potentials, not their individual signs. Two positive electrodes still make a working cell if one is larger.
5. Chain: Δ_rG° < 0 ⟺ E°_cell > 0 ⟺ K > 1

Answer

Option 2 — Δ_rG° is negative

Why not others

Option 4 is the thoughtful-looking trap. The Cu/Ag cell has both electrodes at positive E° (+0.34 and +0.80) and works perfectly well, giving 0.46 V.

Shortcut

Memorise the triple equivalence as one object: negative ΔG, positive E, K greater than one. Any question about spontaneity is asking you to point at one of these three.

What pressure of H₂ would make the potential of a hydrogen electrode zero in pure water at 298 K?

• 1. 10⁻⁷ atm
• 2. 1 atm
• 3. 10⁻¹⁰ atm
• 4. 10⁻¹⁴ atm

Given

Pure water at 298 K, so pH = 7 and [H⁺] = 10⁻⁷ M. Required: E = 0

Asked

The hydrogen pressure that achieves it

Concept

The SHE is zero only when both conditions hold — 1 M H⁺ and 1 bar H₂. Spoil one and you can compensate with the other.

Formula

E = E° − 0.059 log[ (p_H₂)^½ / [H⁺] ]

Baby steps

1. Set E = 0 and E° = 0, so the log term itself must vanish.
2. log[ (p)^½/[H⁺] ] = 0, which means (p)^½/[H⁺] = 1
3. So (p)^½ = [H⁺] = 10⁻⁷
4. Square both sides: p = (10⁻⁷)² = 10⁻¹⁴ atm

Answer

Option 4 — 10⁻¹⁴ atm

Why not others

Option 1 (10⁻⁷) is the answer if you forget to square — the exponent ½ on hydrogen is the entire question. Option 2 assumes standard conditions, which pure water is not.

Shortcut

Whenever the half-reaction is written H⁺ + e⁻ → ½H₂, the pressure carries a power of ½, so any pressure you solve for gets squared. Recognising that turns this into a one-line question.

On electrolysing dilute sulphuric acid with platinum electrodes, what is obtained at the anode?

• 1. O₂ gas
• 2. H₂ gas
• 3. SO₂ gas
• 4. S₂O₈²⁻ ions

Given

Dilute H₂SO₄, inert Pt electrodes

Asked

The anode product only

Concept

At the anode two oxidations compete: water going to oxygen, or sulphate going to peroxodisulphate. The lower E° wins.

Formula

2H₂O → O₂ + 4H⁺ + 4e⁻ (1.23 V) versus 2SO₄²⁻ → S₂O₈²⁻ + 2e⁻ (1.96 V)

Baby steps

1. List the anode candidates: water and sulphate ion.
2. Compare their potentials: 1.23 V for water, 1.96 V for sulphate.
3. At the anode the species with the lower E° is oxidised, so water wins.
4. Water is oxidised to oxygen gas.
5. The word "dilute" confirms this — plenty of water available, little sulphate competition.

Answer

Option 1 — O₂ gas

Why not others

Option 4 is what happens with concentrated H₂SO₄, where sulphate is abundant enough to be oxidised instead. Option 2 is the cathode product, not the anode's. Sulphur is already in its highest oxidation state in sulphate, so SO₂ cannot form by oxidation.

Shortcut

Dilute sulphuric acid behaves as electrolysis of water: H₂ at the cathode, O₂ at the anode, in a 2:1 volume ratio. Concentrated is the only case that changes.

During electrolysis of aqueous NaCl, a platinum cathode liberates H₂ while a mercury cathode gives sodium amalgam. Why?

• 1. Mercury is a better conductor than platinum
• 2. Hydrogen has a high overpotential on mercury
• 3. Sodium has a higher reduction potential than hydrogen
• 4. Mercury forms a complex with chloride

Given

Same solution, two different cathode materials, two different products

Asked

The reason for the switch

Concept

Since the solution is unchanged, thermodynamics cannot explain the difference. Only a kinetic factor tied to the electrode surface can — and that factor is overpotential.

Formula

Effective discharge potential = E° + overpotential, which depends on the electrode material

Baby steps

1. On platinum: H⁺ (0.00 V) beats Na⁺ (−2.71 V) comfortably, so hydrogen is released.
2. On mercury, hydrogen evolution is very sluggish and needs a large extra voltage before it will proceed.
3. That overpotential pushes hydrogen's effective discharge potential far more negative.
4. Sodium, which suffers no such penalty, is now the easier option and is discharged instead.
5. The sodium produced dissolves in the mercury to form an amalgam — the basis of the Castner–Kellner process for making pure NaOH.

Answer

Option 2 — high hydrogen overpotential on mercury

Why not others

Option 3 is factually wrong: sodium's reduction potential is far lower. Option 1 confuses conductivity with electrode kinetics — both metals conduct perfectly well.

Shortcut

If the electrolyte is the same and only the electrode changes, the answer is always overpotential. It is the only variable left.

For ⅔Al₂O₃ → 4⁄3Al + O₂ at 500 °C, Δ_rG = +960 kJ mol⁻¹. What is the least potential difference needed to reduce aluminium oxide electrolytically at that temperature?

• 1. 4.5 V
• 2. 5.0 V
• 3. 1.25 V
• 4. 2.5 V

Given

Δ_rG = +960 kJ mol⁻¹ of O₂ produced, F = 96500 C mol⁻¹

Asked

Minimum applied voltage

Concept

The reaction is non-spontaneous, so you must supply at least as much electrical work as the Gibbs energy demands. The threshold voltage is exactly ΔG/nF.

Formula

Δ_rG = −nFE → E_min = Δ_rG / nF

Baby steps

1. Find n from the equation as written. One O₂ is produced, and O₂ formation from O²⁻ releases 4 electrons, so n = 4.
2. Cross-check with aluminium: 4⁄3 mol Al × 3 electrons each = 4 electrons ✓
3. Convert ΔG to joules: 960 kJ = 960000 J
4. E = 960000 / (4 × 96500) = 960000 / 386000
5. E = 2.487 V, so at least about 2.5 V must be applied.

Answer

Option 4 — about 2.5 V

Why not others

Option 2 (5.0 V) comes from n = 2. Option 3 from n = 8. Getting n right is the whole question — and the safest route is to count electrons on the oxygen side, since the aluminium coefficient is a fraction.

Shortcut

Whenever a reaction is written per mole of O₂, n = 4. That single fact solves the entire family of Al₂O₃ decomposition questions, which have appeared in both AIPMT 2003 and 2012.

How long must a 3 A current flow through molten sodium chloride to produce 0.10 mol of chlorine gas?

• 1. 55 minutes
• 2. 0.895 hours
• 3. 1.79 hours
• 4. 3.58 hours

Given

I = 3 A, n(Cl₂) = 0.10 mol, F = 96500 C mol⁻¹

Asked

Time, in hours

Concept

Faraday's law run backwards. The only subtlety is that chlorine is diatomic, so each molecule needs two electrons.

Formula

2Cl⁻ → Cl₂ + 2e⁻  ;  Q = nFz  ;  t = Q/I

Baby steps

1. Write the anode reaction: 2Cl⁻ → Cl₂ + 2e⁻. So 2 mol of electrons per mol of Cl₂.
2. Moles of electrons needed = 0.10 × 2 = 0.20 mol
3. Q = 0.20 × 96500 = 19300 C
4. t = Q/I = 19300/3 = 6433 s
5. Convert: 6433/3600 = 1.787 h, about 1.79 hours

Answer

Option 3 — about 1.79 hours

Why not others

Option 2 (0.895 h) is exactly half — the result of forgetting that Cl₂ requires two electrons and treating it as one. Option 4 doubles again. The diatomic gas is where this question is won or lost.

Shortcut

For any diatomic gas released at an electrode — Cl₂, H₂, O₂ — count electrons per molecule, not per atom. H₂ and Cl₂ take 2; O₂ takes 4.

A button cell used in watches works by Zn(s) + Ag₂O(s) + H₂O(l) ⇌ 2Ag(s) + Zn²⁺(aq) + 2OH⁻(aq). Given Zn²⁺ + 2e⁻ → Zn with E° = −0.76 V and Ag₂O + H₂O + 2e⁻ → 2Ag + 2OH⁻ with E° = +0.34 V, what is the cell potential?

• 1. 0.42 V
• 2. 1.10 V
• 3. −1.10 V
• 4. 0.34 V

Given

E°(Zn²⁺/Zn) = −0.76 V, E°(Ag₂O/Ag) = +0.34 V

Asked

E°_cell

Concept

Read the overall equation to see which species is oxidised, then subtract. Zinc appears as a solid on the left and as Zn²⁺ on the right, so zinc is oxidised — it is the anode.

Formula

E°_cell = E°_cathode − E°_anode

Baby steps

1. Zn(s) → Zn²⁺: oxidation, so the zinc couple is the anode.
2. Ag₂O → Ag: silver goes from +1 to 0, a reduction, so the silver couple is the cathode.
3. E°_cell = 0.34 − (−0.76)
4. = 0.34 + 0.76 = 1.10 V

Answer

Option 2 — 1.10 V

Why not others

Option 1 (0.42 V) drops the sign and computes 0.34 − 0.76. Option 3 assigns the roles backwards. Note the coincidence that this equals the Daniell cell's 1.10 V — different chemistry, same number, so do not let familiarity substitute for working.

Shortcut

Subtracting a negative adds. Any cell pairing a negative anode with a positive cathode gives a potential larger than either value, so if your answer is smaller than the cathode figure, you have dropped a sign.

Which statement about the lead storage battery during recharging is correct?

• 1. PbSO₄ forms on both plates and the acid is consumed
• 2. Pb is converted to PbO₂ on both plates
• 3. Water is consumed and the electrolyte becomes more dilute
• 4. PbSO₄ on the anode becomes Pb and on the cathode becomes PbO₂, regenerating H₂SO₄

Given

Lead storage battery: Pb anode, PbO₂ cathode, 38% H₂SO₄

Asked

What happens on charging, not on discharging

Concept

Charging is simply the discharge reaction driven backwards by an external supply. Write the discharge equation, then read it right to left.

Formula

Pb + PbO₂ + 2H₂SO₄ ⇌ 2PbSO₄ + 2H₂O

Baby steps

1. Discharge: Pb + SO₄²⁻ → PbSO₄ + 2e⁻ at the anode; PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O at the cathode.
2. So on discharge both plates end up coated in PbSO₄ and acid is used up.
3. Charging reverses every arrow: 2PbSO₄ + 2H₂O → Pb + PbO₂ + 2H₂SO₄
4. The sulphate on the plate that was the anode returns to Pb; the sulphate on the plate that was the cathode returns to PbO₂.
5. Water is consumed and sulphuric acid regenerated, so the electrolyte becomes more concentrated and denser.

Answer

Option 4 — PbSO₄ reverts to Pb and PbO₂, acid regenerated

Why not others

Option 1 describes discharging, not charging — the most common confusion. Option 3 has the water and acid the wrong way round: water is consumed on charging, produced on discharging.

Shortcut

Memorise only the discharge equation and the fact that charging is its reverse. Then any charging question is answered by reading your own equation backwards.

Why does the potential of a mercury cell stay at about 1.35 V throughout its working life?

• 1. Because no ion in solution changes concentration during the reaction
• 2. Because the electrolyte is a paste rather than a liquid
• 3. Because mercury is a liquid metal with constant density
• 4. Because the cell is recharged continuously by ambient heat

Given

Mercury cell: Zn–Hg amalgam anode, HgO cathode, KOH/ZnO paste electrolyte

Asked

The reason for the constant voltage

Concept

Voltage drifts during use because Nernst's log term responds to changing concentrations. Remove the concentration dependence and the voltage cannot drift.

Formula

Zn(Hg) + HgO(s) → ZnO(s) + Hg(l) — every species is a solid, a liquid or an amalgam

Baby steps

1. Write the overall reaction and label the states: Zn amalgam, HgO solid, ZnO solid, Hg liquid.
2. None of them is a dissolved ion.
3. In the Nernst equation, solids and pure liquids are taken as unity.
4. So Q = 1, log Q = 0, and E stays permanently at E°.
5. Contrast the dry cell, where Zn²⁺ and NH₃ concentrations do change — and whose voltage duly falls in use.

Answer

Option 1 — no dissolved ion changes concentration

Why not others

Option 2 is true of the cell but is not the reason; plenty of paste-electrolyte cells drift. Option 4 is not a thing. This exact reasoning appears in both NCERT and CBSE board papers, worded almost identically.

Shortcut

Constant voltage means no concentration term. Scan the overall equation for anything marked (aq): if there is none, the voltage is flat.

Which of the following is not an example of corrosion?

• 1. Rusting of an iron object
• 2. Tarnishing of silver
• 3. Production of hydrogen by electrolysis of water
• 4. Green coating on copper and bronze

Given

Four processes, three of which involve a metal surface changing

Asked

The odd one out

Concept

Corrosion is the spontaneous oxidation of a metal surface by its environment. A process driven by an external power supply is electrolysis, not corrosion.

Formula

Test: is a metal being oxidised by its surroundings, without any applied voltage?

Baby steps

1. Rusting — iron oxidised by air and water, spontaneous. Corrosion ✓ (NCERT names it explicitly)
2. Tarnishing of silver — silver reacting with sulphur compounds in air. Corrosion ✓
3. Green coating on copper and bronze — basic copper carbonate forming in damp air. Corrosion ✓
4. Electrolysis of water — requires an external DC supply, and no metal is being consumed. Not corrosion ✗

Answer

Option 3 — electrolysis of water

Why not others

Options 1, 2 and 4 are the three examples NCERT lists in the opening sentence of the corrosion section, so they are the safest possible distractors to eliminate.

Shortcut

Corrosion is something that happens to a metal by itself. If a battery or power supply is involved, or if no metal is present, it is not corrosion.

Zinc can be coated onto iron to make galvanised iron, but iron cannot be coated onto zinc for the same purpose. Why?

• 1. Zinc has a lower melting point than iron
• 2. Iron is denser than zinc and will not adhere
• 3. Zinc forms a soluble oxide while iron does not
• 4. Zinc has a more negative electrode potential, so it is oxidised in preference to iron

Given

E°(Zn²⁺/Zn) = −0.76 V, E°(Fe²⁺/Fe) = −0.44 V

Asked

Why the protection only works one way round

Concept

Cathodic protection works because the coating metal is the easier one to oxidise. It is a sacrificial arrangement, not merely a physical barrier.

Formula

More negative E° → oxidised preferentially → acts as the anode

Baby steps

1. Compare the two: zinc at −0.76 V is more negative than iron at −0.44 V.
2. So in any zinc–iron couple, zinc is the anode and corrodes first.
3. That protects the iron even where the coating is scratched, because the exposed iron becomes the cathode.
4. Reverse the roles: with iron coating zinc, iron is now the less negative metal and would be the cathode — but the zinc underneath is what you were trying to protect. The arrangement is useless.
5. Compare tinning: tin at −0.14 V is less negative than iron, so a scratched tin can actually rusts faster than bare iron.

Answer

Option 4 — zinc's more negative electrode potential

Why not others

Options 1 and 3 are physical properties with no bearing on which metal oxidises. Option 3 is chemically wrong — zinc oxide is not soluble in the sense implied.

Shortcut

Sacrificial protection always runs from more negative E° to less negative. Mg protects Zn, Zn protects Fe, Fe protects nothing above it. Tin does not protect iron at all once scratched.

In the rusting of iron, the anodic and cathodic reactions have E° values of −0.44 V and +1.23 V. What is the overall E°_cell, and what is the immediate product at the anodic spot?

• 1. 0.79 V and Fe³⁺
• 2. 1.67 V and Fe³⁺
• 3. 1.67 V and Fe²⁺
• 4. 0.79 V and Fe₂O₃·xH₂O

Given

Anode: Fe → Fe²⁺ + 2e⁻, E° = −0.44 V. Cathode: O₂ + 4H⁺ + 4e⁻ → 2H₂O, E° = +1.23 V

Asked

Two things — the cell potential, and the oxidation state formed first

Concept

Rusting happens in two stages. The electrochemical step gives iron(II); atmospheric oxidation to iron(III) comes afterwards, outside the cell.

Formula

E°_cell = E°_cathode − E°_anode

Baby steps

1. E°_cell = 1.23 − (−0.44) = 1.67 V
2. Read the anodic half-reaction: Fe → Fe²⁺ + 2e⁻. The product is iron(II).
3. Only later does air oxidise it further: 2Fe²⁺ + 2H₂O + ½O₂ → Fe₂O₃ + 4H⁺
4. Rust itself is hydrated ferric oxide, Fe₂O₃·xH₂O — the end product, not the immediate one.

Answer

Option 3 — 1.67 V, and Fe²⁺ forms first

Why not others

Option 2 has the right voltage but jumps to iron(III) too early. Option 1 drops the sign in the subtraction. Option 4 names the final rust rather than the anodic product.

Shortcut

Two numbers to hold: 1.67 V for the cell, and Fe(II) before Fe(III). Almost every corrosion question tests one or the other.

A device that converts the energy of combustion of fuels such as hydrogen and methane directly into electrical energy is called a:

• 1. fuel cell
• 2. dynamo
• 3. electrolytic cell
• 4. secondary battery

Given

A definition, with the key words "combustion of fuels" and "directly into electrical energy"

Asked

The name of the device

Concept

The word "directly" is doing all the work. A power station also turns combustion into electricity, but through heat and a turbine — indirectly, and at about 40% efficiency.

Formula

2H₂ + O₂ → 2H₂O, run as a galvanic cell with continuous reactant supply

Baby steps

1. A dynamo converts mechanical energy, not chemical, so it is out.
2. An electrolytic cell runs the other way — electrical energy into chemical change.
3. A secondary battery stores a fixed quantity of reactants and must be recharged; it is not fed fuel.
4. A fuel cell is a galvanic cell with reactants supplied continuously and products removed continuously.
5. Efficiency about 70%, against roughly 40% for a thermal plant.

Answer

Option 1 — fuel cell

Why not others

Option 4 is the near-miss. Both are galvanic, but a battery holds a finite charge while a fuel cell runs indefinitely as long as fuel arrives. That distinction is the definition.

Shortcut

Fuel supplied continuously plus energy converted directly equals fuel cell. Both halves of that phrase appear in NCERT's own sentence.

Standard free energies of formation at 298 K are −237.2, −394.4 and −8.2 kJ mol⁻¹ for H₂O(l), CO₂(g) and pentane(g). Find E°_cell for a pentane–oxygen fuel cell.

• 1. +0.0968 V
• 2. +1.0968 V
• 3. +2.0968 V
• 4. +4.0968 V

Given

Δ_fG° values as above; F = 96500 C mol⁻¹; elements in their standard states have Δ_fG° = 0

Asked

E°_cell for the combustion run as a fuel cell

Concept

Three moves: balance the combustion, get Δ_rG° from formation values, then convert to a voltage. The hard part is n.

Formula

Δ_rG° = ΣΔ_fG°(products) − ΣΔ_fG°(reactants) then E° = −Δ_rG°/nF

Baby steps

1. Balance: C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O
2. Products: 5(−394.4) + 6(−237.2) = −1972.0 − 1423.2 = −3395.2 kJ
3. Reactants: −8.2 + 8(0) = −8.2 kJ
4. Δ_rG° = −3395.2 − (−8.2) = −3387.0 kJ mol⁻¹
5. Find n: each O₂ accepts 4 electrons, and there are 8 of them, so n = 32
6. E° = 3387000 / (32 × 96500) = 3387000 / 3088000 = 1.0968 V

Answer

Option 2 — +1.0968 V

Why not others

Every wrong option here corresponds to a wrong n: 32 is the only value that gives an answer near 1 V, which is where real fuel cells actually operate. If your voltage comes out at 4 V or 0.1 V for a combustion cell, recheck the electron count.

Shortcut

n = 4 × (moles of O₂ in the balanced equation). That works for every combustion fuel cell — hydrogen, methane, methanol or pentane — and it is the only step in this question that is genuinely hard.