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Find the emf of the half-cell Pt(s) | H₂(g, 2 atm) | HCl(aq, 0.02 M), given E°(H₂/H⁺) = 0 V, 2.303RT/F = 0.059 and log 2 = 0.3010.

• 1. +0.109 V
• 2. −0.100 V
• 3. −0.109 V
• 4. −0.089 V

Given

p(H₂) = 2 atm, [H⁺] = 0.02 M (HCl is strong, so fully ionised), E° = 0, n = 1

Asked

The electrode potential of this hydrogen half-cell

Concept

A non-standard hydrogen electrode. Both the pressure and the acid concentration are off standard, so both enter Q.

Formula

H⁺ + e⁻ → ½H₂  ;  E = E° − 0.059 log[ (p_H₂)^½ / [H⁺] ]

Baby steps

1. Write the half-reaction as a reduction: H⁺ + e⁻ → ½H₂, so n = 1 and H₂ carries the power ½.
2. Build Q. Product H₂ on top with exponent ½; reactant H⁺ underneath. Q = (2)^½ / 0.02
3. Take logs separately: log(2^½) = ½ × 0.3010 = 0.1505
4. log(1/0.02) = −log(0.02) = 1.6990
5. log Q = 0.1505 + 1.6990 = 1.8495
6. E = 0 − 0.059 × 1.8495 = −0.1091 V

Answer

Option 3 — E = −0.109 V

Why not others

Option 2 (−0.100 V) is what you get by ignoring the 2 atm pressure entirely — this is the trap the question was built around. Option 1 flips the sign, which happens if you invert Q. Option 4 comes from halving the log 2 term twice.

Shortcut

For a hydrogen electrode at 1 bar the whole thing collapses to E = −0.059 pH. Here pH = 1.70, giving −0.100 V, and the extra pressure adds a further −0.0089 V. Getting to −0.109 V that way takes ten seconds.

A hydrogen gas electrode is set up by dipping platinum wire into HCl of pH 10 and passing hydrogen at 1 atm over it. What is the oxidation potential of this electrode?

• 1. −0.59 V
• 2. +0.59 V
• 3. +0.118 V
• 4. 0.00 V

Given

pH = 10, p(H₂) = 1 atm, E° = 0

Asked

Oxidation potential — note the word, this is not the reduction potential

Concept

At 1 bar the hydrogen electrode's reduction potential is fixed by pH alone. Oxidation potential is its negative.

Formula

E_red = −0.059 pH  ;  E_ox = −E_red

Baby steps

1. Since p(H₂) = 1, Q reduces to 1/[H⁺], so E = −0.059 log(1/[H⁺]) = −0.059 pH.
2. E_reduction = −0.059 × 10 = −0.59 V
3. The question asks for oxidation potential, so reverse the sign.
4. E_oxidation = +0.59 V

Answer

Option 2 — +0.59 V

Why not others

Option 1 is the reduction potential — the correct number attached to the wrong quantity, and the option most students pick. Option 4 assumes standard conditions, ignoring that pH 10 is nowhere near 1 M H⁺.

Shortcut

Memorise the pair: reduction potential = −0.059 × pH, oxidation potential = +0.059 × pH. Then read the question's noun carefully and pick one.

In the cell Zn | ZnSO₄(0.01 M) ‖ CuSO₄(1.0 M) | Cu the emf is E₁. The concentrations are then swapped — ZnSO₄ to 1.0 M and CuSO₄ to 0.01 M — and the emf becomes E₂. How do E₁ and E₂ compare?

• 1. E₁ < E₂
• 2. E₁ = E₂
• 3. E₂ = 0, E₁ ≠ 0
• 4. E₁ > E₂

Given

Case 1: [Zn²⁺] = 0.01, [Cu²⁺] = 1.0. Case 2: [Zn²⁺] = 1.0, [Cu²⁺] = 0.01. n = 2

Asked

A comparison, not a number

Concept

Zn²⁺ is the product and Cu²⁺ the reactant. Less product and more reactant means a stronger push, so a higher emf.

Formula

E = E° − (0.059/2) log([Zn²⁺]/[Cu²⁺])

Baby steps

1. Case 1: Q = 0.01/1.0 = 10⁻², log Q = −2. Correction = −(0.0295)(−2) = +0.059. So E₁ = E° + 0.059
2. Case 2: Q = 1.0/0.01 = 10², log Q = +2. Correction = −(0.0295)(+2) = −0.059. So E₂ = E° − 0.059
3. E₁ − E₂ = 0.118 V, a definite positive number.

Answer

Option 4 — E₁ > E₂, by 0.118 V

Why not others

Option 1 is what you get from writing Q upside down as [Cu²⁺]/[Zn²⁺]. Option 2 tempts anyone who thinks E° is the whole answer and concentration is decoration.

Shortcut

No arithmetic needed. Case 1 has plenty of the reactant and almost none of the product — the best possible situation for the forward reaction. So E₁ is the larger. Answer in five seconds.

Find the emf at 298 K of the cell in which Ni(s) + 2Ag⁺(0.001 M) → Ni²⁺(0.001 M) + 2Ag(s) takes place. Take E°_cell = 1.05 V and 2.303RT/F = 0.059.

• 1. 0.96 V
• 2. 1.14 V
• 3. 1.05 V
• 4. 0.91 V

Given

[Ni²⁺] = 0.001 M, [Ag⁺] = 0.001 M, E°_cell = 1.05 V

Asked

E_cell under these non-standard concentrations

Concept

Two electrons move overall even though silver's own half-reaction moves one. Silver's coefficient of 2 becomes a squared term in Q.

Formula

E = E° − (0.059/2) log([Ni²⁺]/[Ag⁺]²)

Baby steps

1. n = 2, read off the balanced equation (2 Ag⁺ each take one electron).
2. Q = [Ni²⁺]/[Ag⁺]² = 10⁻³ / (10⁻³)² = 10⁻³/10⁻⁶ = 10³
3. log Q = 3
4. E = 1.05 − (0.059/2)(3) = 1.05 − 0.0885 = 0.9615 V

Answer

Option 1 — E ≈ 0.96 V

Why not others

Option 2 flips the sign of the correction. Option 3 ignores concentrations. Option 4 comes from using n = 1 and forgetting the square on [Ag⁺]. Note that some reprints of this paper show E° as 10.5 V; the NCERT original this question is lifted from (Intext 2.5) uses 1.05 V.

Shortcut

When both concentrations are equal and one is squared, Q collapses to 1/c. Here 1/10⁻³ = 10³ immediately.

A centimolar KCl solution has conductivity 0.0210 ohm⁻¹ cm⁻¹ at 25 °C, and the cell containing it has resistance 60 ohm. What is the cell constant?

• 1. 0.42 cm⁻¹
• 2. 1.26 cm⁻¹
• 3. 2.10 cm⁻¹
• 4. 0.35 cm⁻¹

Given

κ = 0.0210 S cm⁻¹, R = 60 Ω. ("Centimolar" = 0.01 M, and here it is a red herring.)

Asked

Cell constant G*

Concept

The cell constant is the geometry factor l/A. It is found by measuring a solution whose κ is already known.

Formula

G* = l/A = κ × R

Baby steps

1. Both quantities are already in centimetre units, so no conversion is needed.
2. G* = 0.0210 × 60
3. G* = 1.26 cm⁻¹

Answer

Option 2 — 1.26 cm⁻¹

Why not others

Options 2 and 4 come from dividing instead of multiplying. The 0.01 M is deliberately supplied to tempt you into computing Λ_m, which was never asked for.

Shortcut

Cell constant is the only quantity in this chapter that is a product of κ and R. Everything else involving κ and R is a quotient. If you multiply, you are computing G*.

Find the molar conductivity of a 0.5 mol dm⁻³ AgNO₃ solution whose electrolytic conductivity at 298 K is 5.76 × 10⁻³ S cm⁻¹.

• 1. 0.086 S cm² mol⁻¹
• 2. 28.8 S cm² mol⁻¹
• 3. 2.88 S cm² mol⁻¹
• 4. 11.52 S cm² mol⁻¹

Given

κ = 5.76 × 10⁻³ S cm⁻¹, c = 0.5 mol dm⁻³ (dm³ = litre, so this is 0.5 M)

Asked

Λ_m in S cm² mol⁻¹

Concept

κ is per unit volume; Λ_m is per mole. Dividing by molarity converts one to the other, with a factor of 1000 because a litre is 1000 cm³.

Formula

Λ_m = κ × 1000 / M

Baby steps

1. Recognise dm⁻³ as L⁻¹, so M = 0.5 mol L⁻¹.
2. κ is already in S cm⁻¹, so the 1000-form applies directly.
3. Λ_m = (5.76 × 10⁻³ × 1000) / 0.5
4. Numerator = 5.76. Divide by 0.5 → 11.52
5. Λ_m = 11.52 S cm² mol⁻¹

Answer

Option 4 — 11.52 S cm² mol⁻¹

Why not others

Option 1 is κ/M with the 1000 dropped — the single most common error in this chapter. Option 3 divides by 2 as if AgNO₃ carried an n-factor of 2; molar conductivity never uses n-factors, only equivalent conductivity does.

Shortcut

Dividing by 0.5 is the same as doubling. 5.76 × 10⁻³ × 1000 = 5.76, doubled is 11.52. No calculator.

Molar conductance rises on dilution according to Λ_m = Λ°_m − A√c. Which of these statements are true?
(A) The equation applies to both strong and weak electrolytes.
(B) The value of A depends on the nature of the solvent.
(C) A is the same for BaCl₂ and MgSO₄.
(D) A is the same for BaCl₂ and Mg(OH)₂.

• 1. (B) and (D) only
• 2. (A) and (B) only
• 3. (A) and (C) only
• 4. (C) and (D) only

Given

The Debye–Hückel–Onsager form of the strong-electrolyte relation

Asked

Which two statements survive

Concept

A depends on the solvent, the temperature and the charge type of the electrolyte — nothing else.

Formula

Λ_m = Λ°_m − A√c  (strong electrolytes only)

Baby steps

1. (A) — False. A weak electrolyte's Λ_m curve is steep and non-linear near zero concentration; the straight-line law does not describe it.
2. (B) — True. NCERT states directly that A depends on the solvent and temperature for a given electrolyte type.
3. (C) — Classify: BaCl₂ gives Ba²⁺ and 2Cl⁻, so 2-1. MgSO₄ gives Mg²⁺ and SO₄²⁻, so 2-2. Different types, different A. False.
4. (D) — Mg(OH)₂ gives Mg²⁺ and 2OH⁻, so 2-1, matching BaCl₂. Same type, same A. True.

Answer

Option 1 — (B) and (D) only

Why not others

Any option containing (A) fails on the strong-electrolyte restriction. Any option containing (C) fails because a 2-1 and a 2-2 electrolyte are not the same charge type, even though both contain a doubly charged ion.

Shortcut

Read the charge type off the formula as (charge on cation)-(charge on anion). BaCl₂ → 2-1. MgSO₄ → 2-2. Mg(OH)₂ → 2-1. Match the pairs and the statement answers itself.

For the strong electrolyte XY, a plot of Λ_m against √c has slope −90.0 S cm² mol⁻³ᐟ² L¹ᐟ² at 298 K. At 0.01 M, Λ_m = 145.0 S cm² mol⁻¹. Given λ°(X⁺) = 74.0 S cm² mol⁻¹, what is λ°(Y⁻)?

• 1. 71.0 S cm² mol⁻¹
• 2. 62.0 S cm² mol⁻¹
• 3. 80.0 S cm² mol⁻¹
• 4. 145.0 S cm² mol⁻¹

Given

−A = −90.0 so A = 90.0; at c = 0.01 M, Λ_m = 145.0; λ°(X⁺) = 74.0

Asked

The limiting molar conductivity of the anion alone

Concept

Two steps stitched together: first recover Λ°_m from the straight-line law, then split it with Kohlrausch's law.

Formula

Λ°_m = Λ_m + A√c  then  Λ°_m = λ°₊ + λ°₋

Baby steps

1. √c = √0.01 = 0.1
2. A√c = 90.0 × 0.1 = 9.0
3. Rearrange the line: Λ°_m = Λ_m + A√c = 145.0 + 9.0 = 154.0 S cm² mol⁻¹
4. XY is a 1-1 electrolyte, so both ν values are 1: Λ°_m = λ°(X⁺) + λ°(Y⁻)
5. λ°(Y⁻) = 154.0 − 74.0 = 80.0 S cm² mol⁻¹

Answer

Option 3 — 80.0 S cm² mol⁻¹

Why not others

Option 1 (71.0) comes from subtracting A√c instead of adding it — the sign in the original equation is minus, so recovering Λ°_m requires a plus. Option 4 forgets the Kohlrausch split entirely.

Shortcut

√0.01 = 0.1 is worth spotting instantly; question-setters choose concentrations whose square roots are clean. 0.0001 → 0.01, 0.04 → 0.2, 0.09 → 0.3.

At infinite dilution the molar conductances of NaCl, HCl and CH₃COONa are 126.45, 426.16 and 91.0 S cm² mol⁻¹. What is the molar conductance of CH₃COOH at infinite dilution?

• 1. 698.28 S cm² mol⁻¹
• 2. 540.61 S cm² mol⁻¹
• 3. 201.28 S cm² mol⁻¹
• 4. 390.71 S cm² mol⁻¹

Given

Λ°_m(NaCl) = 126.45, Λ°_m(HCl) = 426.16, Λ°_m(CH₃COONa) = 91.0

Asked

Λ°_m(CH₃COOH), which cannot be measured directly

Concept

Acetic acid is weak, so its Λ_m curve cannot be extrapolated to zero concentration. Kohlrausch's law lets you assemble it from three strong electrolytes instead.

Formula

Λ°_m(HAc) = Λ°_m(HCl) + Λ°_m(NaAc) − Λ°_m(NaCl)

Baby steps

1. Write what you want in ions: H⁺ + CH₃COO⁻
2. Find a combination that produces exactly those. HCl gives H⁺ + Cl⁻. NaAc gives Na⁺ + Ac⁻. Adding them gives H⁺ + Cl⁻ + Na⁺ + Ac⁻.
3. Subtract NaCl (Na⁺ + Cl⁻) to cancel the two unwanted ions. Left with H⁺ + Ac⁻ ✓
4. Apply the same operations to the numbers: 426.16 + 91.0 − 126.45
5. = 517.16 − 126.45 = 390.71 S cm² mol⁻¹

Answer

Option 4 — 390.71 S cm² mol⁻¹

Why not others

Option 1 adds all three. Option 2 subtracts the wrong member (426.16 + 126.45 − 91.0 style errors land near here). Always verify by cancelling ions, never by remembering which sign goes where.

Shortcut

Λ°_m of acetic acid is 390.5 in the NCERT data set and 390.71 here. If a question about acetic acid produces something near 390, you have almost certainly done it right.

The molar conductivity of 0.007 M acetic acid is 20 S cm² mol⁻¹. Find its dissociation constant, given λ°(H⁺) = 350 and λ°(CH₃COO⁻) = 50 S cm² mol⁻¹.

• 1. 1.75 × 10⁻⁵
• 2. 1.75 × 10⁻⁴
• 3. 3.50 × 10⁻⁵
• 4. 2.50 × 10⁻³

Given

c = 0.007 M, Λ_m = 20, λ°(H⁺) = 350, λ°(Ac⁻) = 50

Asked

K_a

Concept

Three moves: Kohlrausch to get Λ°_m, ratio to get α, Ostwald to get K_a.

Formula

Λ°_m = λ°₊ + λ°₋  ;  α = Λ_m/Λ°_m  ;  K_a = cα²/(1−α)

Baby steps

1. Λ°_m = 350 + 50 = 400 S cm² mol⁻¹
2. α = 20/400 = 0.05, that is 5% dissociated
3. α² = 0.0025
4. cα² = 0.007 × 0.0025 = 1.75 × 10⁻⁵
5. Since α is only 0.05, the (1 − α) denominator is 0.95 and barely matters. Exact value = 1.75 × 10⁻⁵/0.95 = 1.84 × 10⁻⁵; the keyed option is the approximation.

Answer

Option 1 — K_a ≈ 1.75 × 10⁻⁵ (exact 1.84 × 10⁻⁵)

Why not others

Option 2 is a decimal slip. Option 3 uses cα instead of cα². Option 4 forgets to square α altogether. Note the real acetic acid K_a is 1.8 × 10⁻⁵, so anything far from that order of magnitude is wrong on sight.

Shortcut

When α comes out under 0.05, drop the (1 − α) and use K_a = cα². NEET options are almost always spaced widely enough that the approximation still picks the right one.

Copper sulphate solution is electrolysed for 10 minutes at 1.5 A. What mass of copper is deposited at the cathode? (M(Cu) = 63 g mol⁻¹, 1 F = 96487 C mol⁻¹)

• 1. 0.5876 g
• 2. 0.1469 g
• 3. 0.2938 g
• 4. 0.3938 g

Given

I = 1.5 A, t = 10 min, M = 63, Cu²⁺ + 2e⁻ → Cu so n = 2

Asked

Mass of copper deposited

Concept

Charge counts electrons; electrons are converted to moles of metal through the half-reaction's n.

Formula

m = M I t / (n F)

Baby steps

1. Convert time: t = 10 × 60 = 600 s
2. Q = I t = 1.5 × 600 = 900 C
3. Moles of electrons = 900 / 96487 = 9.328 × 10⁻³
4. Cu²⁺ needs 2 electrons each, so moles of Cu = 9.328 × 10⁻³ / 2 = 4.664 × 10⁻³
5. m = 4.664 × 10⁻³ × 63 = 0.2938 g

Answer

Option 3 — 0.2938 g

Why not others

Option 1 is exactly double — the result of using n = 1 and forgetting that Cu²⁺ takes two electrons. Option 2 is half, from using n = 4. This is NCERT Example 2.10 reprinted almost unchanged.

Shortcut

Equivalent mass of Cu is 63/2 = 31.5. Then m = 31.5 × 900/96487 ≈ 0.294 g in one step. Learning the equivalent mass shortcut cuts every Faraday question to a single division.

What mass of copper is deposited when 9.6487 A flows for 100 seconds through a voltameter containing copper sulphate solution? (M(Cu) = 63 g mol⁻¹, 1 F = 96487 C)

• 1. 0.630 g
• 2. 0.315 g
• 3. 0.158 g
• 4. 3.150 g

Given

I = 9.6487 A, t = 100 s, M = 63, n = 2

Asked

Mass of copper deposited

Concept

Identical to Q11 — the numbers have been chosen so the charge is a clean fraction of a faraday.

Formula

m = M I t / (n F)

Baby steps

1. Q = 9.6487 × 100 = 964.87 C
2. Notice: 964.87 / 96487 = 0.01 exactly. That is 0.01 mol of electrons.
3. Moles of Cu = 0.01/2 = 0.005
4. m = 0.005 × 63 = 0.315 g

Answer

Option 2 — 0.315 g

Why not others

Option 1 doubles by dropping n. Option 4 is a factor-of-ten slip in the charge. The setter built the number 9.6487 so the faraday cancels cleanly — treat that as a hint that you have the method right.

Shortcut

Whenever the current has 96487 hiding in it, divide first. Seeing "0.01 mol of electrons" straight away turns this into mental arithmetic.

How many faradays are needed to produce 20 g of calcium from molten CaCl₂? (Atomic mass of Ca = 40 g mol⁻¹)

• 1. 1 F
• 2. 2 F
• 3. 0.5 F
• 4. 4 F

Given

Mass = 20 g, M(Ca) = 40, Ca²⁺ + 2e⁻ → Ca so n = 2

Asked

Charge in faradays, not coulombs

Concept

Faradays are just moles of electrons. Convert mass to moles of metal, then multiply by n.

Formula

faradays = (mass / M) × n

Baby steps

1. Moles of Ca = 20/40 = 0.5 mol
2. Each Ca²⁺ needs 2 electrons.
3. Moles of electrons = 0.5 × 2 = 1.0
4. 1.0 mol of electrons = 1 F

Answer

Option 1 — 1 F

Why not others

Option 3 forgets to multiply by n. Option 2 uses a full mole of Ca instead of half. The question deliberately gives half a mole so that the n = 2 factor cancels it back to a whole number — do not read that clean answer as evidence you can skip a step.

Shortcut

Faradays = mass ÷ equivalent mass. Equivalent mass of Ca = 40/2 = 20, and 20/20 = 1 F.

Match each conversion with the number of faradays it requires.
A. 1 mol H₂O → O₂  ·  B. 1 mol MnO₄⁻ → Mn²⁺  ·  C. 1.5 mol Ca from molten CaCl₂  ·  D. 1 mol FeO → Fe₂O₃
I. 3F  ·  II. 2F  ·  III. 1F  ·  IV. 5F

• 1. A-III, B-IV, C-I, D-II
• 2. A-II, B-I, C-IV, D-III
• 3. A-II, B-IV, C-I, D-III
• 4. A-IV, B-II, C-III, D-I

Given

Four conversions, two oxidations and two reductions

Asked

Electrons per mole for each

Concept

Only oxidation-state change matters. Balance each half-reaction and count.

Formula

F required = (moles) × (change in oxidation number per formula unit)

Baby steps

1. A. Oxygen goes from −2 in H₂O to 0 in O₂. That is 2 electrons lost per oxygen atom, and 1 mol H₂O holds 1 mol of oxygen atoms → 2F (II)
2. B. Mn goes from +7 in MnO₄⁻ to +2 in Mn²⁺. That is a gain of 5 electrons → 5F (IV)
3. C. Ca²⁺ + 2e⁻ → Ca needs 2F per mole; 1.5 mol needs 1.5 × 2 = 3F (I)
4. D. Fe goes from +2 in FeO to +3 in Fe₂O₃, losing 1 electron per iron. 1 mol FeO has 1 mol Fe → 1F (III)

Answer

Option 3 — A-II, B-IV, C-I, D-III

Why not others

The trap in D is assuming Fe₂O₃ means two irons and therefore 2F. It does not — you are told 1 mol of FeO, which supplies only one iron. Read what is being consumed, not what is being formed.

Shortcut

Work out only the ones you are certain of, then eliminate. B is unambiguously 5F, and only two options place B at IV. Between those two, A is clearly 2F because it is the only value left that makes sense for oxygen.

The standard cell potential for Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) is 1.1 V. Find the standard Gibbs energy change. (F = 96487 C mol⁻¹)

• 1. −106.14 kJ mol⁻¹
• 2. −212.27 kJ mol⁻¹
• 3. +212.27 kJ mol⁻¹
• 4. −21227 kJ mol⁻¹

Given

E°_cell = 1.1 V, F = 96487, n = 2 (Zn → Zn²⁺ loses two electrons)

Asked

Δ_rG° in kJ mol⁻¹

Concept

Electrical work at maximum efficiency equals the drop in Gibbs energy.

Formula

Δ_rG° = −nFE°_cell

Baby steps

1. Identify n from the balanced equation: Zn → Zn²⁺ + 2e⁻, so n = 2
2. Δ_rG° = −2 × 96487 × 1.1
3. 2 × 96487 = 192974; × 1.1 = 212271.4
4. Δ_rG° = −212271 J mol⁻¹
5. Divide by 1000 → −212.27 kJ mol⁻¹

Answer

Option 2 — −212.27 kJ mol⁻¹

Why not others

Option 4 is the joules answer with the kJ label attached — the unit trap this question exists to set. Option 1 uses n = 1. Option 3 drops the minus sign, which would make a spontaneous reaction look non-spontaneous.

Shortcut

nF ≈ 2 × 96500 = 193000 ≈ 193 kJ per volt. Multiply by 1.1 to get 212 kJ, then attach the minus. NCERT works this exact example (2.3).

For 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(aq), E°_cell = 0.24 V at 298 K. Find the standard Gibbs energy change. (F = 96500 C mol⁻¹)

• 1. −23.16 kJ mol⁻¹
• 2. +46.32 kJ mol⁻¹
• 3. −92.64 kJ mol⁻¹
• 4. −46.32 kJ mol⁻¹

Given

E°_cell = 0.24 V, F = 96500

Asked

Δ_rG°

Concept

The only real work here is finding n. Fe³⁺ + e⁻ → Fe²⁺ is one electron, but the equation as written has two Fe³⁺.

Formula

Δ_rG° = −nFE°_cell

Baby steps

1. Split into half-reactions: 2Fe³⁺ + 2e⁻ → 2Fe²⁺ and 2I⁻ → I₂ + 2e⁻
2. Electrons transferred per equation as written: n = 2
3. Δ_rG° = −2 × 96500 × 0.24
4. 2 × 96500 = 193000; × 0.24 = 46320
5. Δ_rG° = −46320 J mol⁻¹ = −46.32 kJ mol⁻¹

Answer

Option 4 — −46.32 kJ mol⁻¹

Why not others

Option 1 is the n = 1 answer, produced by looking only at the Fe³⁺/Fe²⁺ half-reaction and missing the coefficient of 2. Option 3 uses n = 4.

Shortcut

n is always readable from the anode half-reaction after balancing. Here 2I⁻ → I₂ releases 2 electrons, so n = 2 without any further thought about iron.

A cell involving one electron has E°_cell = 0.59 V at 298 K. What is the equilibrium constant of the cell reaction? (2.303RT/F = 0.059 V)

• 1. 1.0 × 10⁵
• 2. 1.0 × 10⁻¹⁰
• 3. 1.0 × 10¹⁰
• 4. 5.0 × 10⁹

Given

n = 1, E°_cell = 0.59 V

Asked

K for the cell reaction

Concept

At equilibrium the cell is flat — E = 0 and Q = K. Substituting those into the Nernst equation gives the relation directly.

Formula

E°_cell = (0.059/n) log K

Baby steps

1. Rearrange: log K = n E°_cell / 0.059
2. log K = (1 × 0.59)/0.059
3. log K = 10
4. K = 10¹⁰

Answer

Option 3 — K = 1.0 × 10¹⁰

Why not others

Option 2 inverts the ratio, giving log K = −10. Option 1 divides 0.59 by 0.118 as if n were 2, when the question states one electron explicitly.

Shortcut

For n = 1, every 0.059 V is one power of ten. 0.59 is ten lots of 0.059, so K = 10¹⁰ on sight. For n = 2, every 0.0295 V is one power of ten.

Find E° for Fe³⁺ + e⁻ → Fe²⁺ at 298 K, given E°(Fe³⁺/Fe) = −0.04 V and E°(Fe²⁺/Fe) = −0.44 V.

• 1. +0.76 V
• 2. −0.76 V
• 3. +0.40 V
• 4. −0.48 V

Given

Fe³⁺ + 3e⁻ → Fe, E° = −0.04 V (n = 3). Fe²⁺ + 2e⁻ → Fe, E° = −0.44 V (n = 2)

Asked

E° for the one-electron step between them

Concept

Electrode potentials cannot be added or subtracted when the electron counts differ. Gibbs energies can. This is the single most important rule in the topic.

Formula

ΔG° = −nFE°  ;  ΔG°(target) = ΔG°(1) − ΔG°(2)

Baby steps

1. Target = (Fe³⁺ → Fe) minus (Fe²⁺ → Fe), because subtracting removes the metal and leaves Fe³⁺ → Fe²⁺.
2. ΔG°₁ = −3F(−0.04) = +0.12F
3. ΔG°₂ = −2F(−0.44) = +0.88F
4. ΔG°(target) = 0.12F − 0.88F = −0.76F
5. The target moves 1 electron, so ΔG° = −1 × F × E° → −0.76F = −F E°
6. E° = +0.76 V

Answer

Option 1 — +0.76 V

Why not others

Option 3 (+0.40 V) is what you get by naively subtracting the potentials: −0.04 − (−0.44) = 0.40. That is the trap, and it is wrong because n differs between the two half-cells. Option 2 has the right magnitude and the wrong sign.

Shortcut

Use the weighted form: n₃E₃ = n₁E₁ − n₂E₂. Here 1 × E = 3(−0.04) − 2(−0.44) = −0.12 + 0.88 = 0.76. Cross-check: the real tabulated value for Fe³⁺/Fe²⁺ is +0.77 V, so 0.76 is clearly right.

Two half-cells are connected: Fe²⁺ + 2e⁻ → Fe with E° = −0.44 V, and Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O with E° = +1.33 V. What is the cell potential of the reaction that occurs?

• 1. +2.65 V
• 2. +0.89 V
• 3. +0.01 V
• 4. +1.77 V

Given

E°(Fe²⁺/Fe) = −0.44 V, E°(Cr₂O₇²⁻/Cr³⁺) = +1.33 V

Asked

E°_cell for the spontaneous combination

Concept

The couple with the higher reduction potential is reduced and becomes the cathode. The other is forced to run in reverse as the anode.

Formula

E°_cell = E°_cathode − E°_anode

Baby steps

1. Compare: +1.33 > −0.44, so dichromate is reduced. Cathode = Cr₂O₇²⁻ couple.
2. Iron is therefore oxidised: Fe → Fe²⁺ + 2e⁻. Anode = Fe²⁺/Fe couple.
3. E°_cell = 1.33 − (−0.44)
4. = 1.33 + 0.44 = 1.77 V
5. Positive, so the reaction is feasible as identified.

Answer

Option 4 — +1.77 V

Why not others

Option 2 (+0.89 V) is 1.33 − 0.44, from failing to carry the minus sign through the subtraction. Option 3 would result from flipping the anode's sign and then subtracting as well — the double-correction error. Note also that E°_cell is not multiplied by any coefficient when balancing electrons; potential is intensive.

Shortcut

Subtracting a negative always makes the answer bigger. If your cell potential comes out smaller than the larger of the two E° values, you have dropped a sign.

Given E° for Cu²⁺(aq) + e⁻ → Cu⁺(aq) is +0.15 V and for Cu⁺(aq) + e⁻ → Cu(s) is +0.50 V, what is E° for Cu²⁺/Cu?

• 1. +0.650 V
• 2. +0.325 V
• 3. +0.150 V
• 4. +0.175 V

Given

Step 1: Cu²⁺ → Cu⁺, n₁ = 1, E°₁ = 0.15 V. Step 2: Cu⁺ → Cu, n₂ = 1, E°₂ = 0.50 V

Asked

E° for the combined two-electron step Cu²⁺ → Cu

Concept

Same principle as Q18, but here the two steps are being added rather than subtracted. Add the Gibbs energies, not the potentials.

Formula

n₃E₃ = n₁E₁ + n₂E₂

Baby steps

1. The two steps chain together: Cu²⁺ → Cu⁺ → Cu, total 2 electrons, so n₃ = 2
2. ΔG total = ΔG₁ + ΔG₂ = −1F(0.15) + −1F(0.50) = −0.65F
3. For the combined step, ΔG = −2F E°₃
4. −2F E°₃ = −0.65F → E°₃ = 0.65/2
5. E°₃ = +0.325 V

Answer

Option 2 — +0.325 V

Why not others

Option 1 (+0.650 V) is the straight sum, which is exactly the error the question is designed to catch. Adding potentials is only valid when the electron counts already match, which they do not here (1 + 1 combining into 2).

Shortcut

When two one-electron steps combine into a two-electron step, the answer is simply the average of the two potentials. 0.15 and 0.50 average to 0.325. That shortcut only works for the 1+1→2 case, so verify the electron counts before using it.