CBSE Class 12 Chemistry Electrochemistry MCQs Set 10

Answer: (b) 20 ohm⁻¹ cm² eq⁻¹

Explanation:
1. κ = \( \frac{1}{R} \) × \( \frac{l}{A} \) = \( \frac{1}{40} \) × \( \frac{2}{5} \)
2. Λ_eq = κ × \( \frac{1000}{N} \) = \( \frac{1}{40} \) × \( \frac{2}{5} \) × \( \frac{1000}{0.5} \) = 20 ohm⁻¹ cm² eq⁻¹

Teacher's Note:
1. Normality (N) is used here instead of molarity, which is the only difference from the usual Λ_m formula – the geometry part (κ from R, l, A) is identical.

Answer: (c) 2Ag⁺ + Pb → Pb²⁺ + 2Ag

Explanation:
1. At cathode: Ag⁺ + e⁻ → Ag; E° = +0.80 V
2. At anode: Pb → Pb²⁺ + 2e⁻ (E°_Pb²⁺/Pb = −0.13 V)
3. E°_cell = E°_cathode − E°_anode = 0.80 − (−0.13) = 0.93 V
4. Hence, the reaction will be 2Ag⁺ + Pb → Pb²⁺ + 2Ag

Teacher's Note:
1. Ag⁺ has the higher reduction potential, so it is reduced; Pb, with the lower one, is oxidised – the reaction only runs in the direction that makes Ecell positive.

Answer: (c) Specific conductance = S m²

Explanation:
1. Specific conductance (conductivity, κ) has units of S m⁻¹, not S m², so option (c) is the incorrectly matched one.

Teacher's Note:
1. A quick way to remember: conductivity is per unit length (S m⁻¹), while molar/equivalent conductance are per mole or equivalent and carry an m² in the numerator.

Answer: (a) Conductivity decreases

Explanation:
1. Conductivity (specific conductance) decreases on dilution because the number of ions per unit volume decreases, even though molar conductivity increases.

Teacher's Note:
1. Don't mix up κ and Λ_m: κ falls on dilution, but Λ_m rises – they respond oppositely because Λ_m already accounts for volume.

Answer: (c) 2.004 × 10⁻² g L⁻¹

Explanation:
1. Solubility (mol L⁻¹) = \( \frac{\kappa\times1000}{\Lambda_m^{\circ}} \) = \( \frac{1.821\times10^{-5}\times1000}{130.26} \) = 13.97 × 10⁻⁵ mol L⁻¹
2. Solubility (g L⁻¹) = 13.97 × 10⁻⁵ × 143.5 = 2.004 × 10⁻² g L⁻¹ (AgCl = 108 + 35.5 = 143.5)

Teacher's Note:
1. For a sparingly soluble salt, κ of the saturated solution and Λ°_m together give molar solubility directly – a very common numerical pattern.

Answer: (b) +1.35 V

Explanation:
1. Pt + Cl₂ → Pt²⁺ + 2Cl⁻; E° = 0.15 V (reverse of the given cell reaction)
2. Pt²⁺ + 2e⁻ → Pt; E° = 1.20 V
3. Adding: Cl₂ + 2e⁻ → 2Cl⁻; E° = 0.15 + 1.20 = 1.35 V

Teacher's Note:
1. When you reverse a given cell reaction, flip the sign of its E° before combining it with another half-reaction.

Answer: (d) NO₃⁻ is reduced in preference to H⁺ ion

Explanation:
1. Due to the reduction of NO₃⁻ in preference to H⁺ ion, H⁺ ion is not reduced, so no H₂ gas is given off with HNO₃.

Teacher's Note:
1. HNO₃ is an oxidising acid: the NO₃&supminus; ion itself gets reduced (to NO or NO₂) in preference to H⁺, which is why it doesn't release hydrogen gas like other acids.

Answer: (b) For the cell, 2Ag⁺ + H₂(1 atm) → 2Ag + 2H⁺(1 M); E_cell = E°_cell − \( \frac{0.0591}{2} \) log \( \frac{[Ag^{+}]^{2}}{[H^{+}]^{2}} \)

Explanation:
1. For 2Ag⁺ + H₂(1 atm) → 2Ag + 2H⁺(1 M), the correct Nernst equation is E_cell = E°_cell − \( \frac{0.0591}{2} \) log \( \frac{[H^{+}]^{2}}{[Ag^{+}]^{2}} \) (products over reactants), so option (b) has the ratio inverted and is the incorrectly matched expression.

Teacher's Note:
1. Always write the log term as [products]/[reactants], each raised to its coefficient – inverting the ratio is the single most common transcription error in these questions.

Answer: (c) 2I⁻ + Br₂ → I₂ + 2Br⁻

Explanation:
1. 2I⁻ → I₂ + 2e⁻ (oxidation, lower E°); Br₂ + 2e⁻ → 2Br⁻ (reduction, higher E°)
2. Net reaction: 2I⁻ + Br₂ → I₂ + 2Br⁻

Teacher's Note:
1. Br2/Br&supminus; has the higher E°, so bromine is the oxidiser here and I&supminus; gets oxidised to I₂, not the other way round.

Answer: (d) unchanged

Explanation:
1. E°_cell is defined under standard conditions and does not depend on the actual concentrations used in the cell, so it remains unchanged.

Teacher's Note:
1. Changing concentration changes Ecell (through the Nernst equation), never E°cell, which is a fixed reference value.

Answer: (c) Λ°_NH4OH = \( \frac{\Lambda^{\circ}_{Ba(OH)_2} + 2\Lambda^{\circ}_{NH_4Cl} - \Lambda^{\circ}_{BaCl_2}}{2} \)

Explanation:
1. Λ°_Ba(OH)2 = Λ°_Ba²⁺ + 2Λ°_OH⁻; Λ°_BaCl2 = Λ°_Ba²⁺ + 2Λ°_Cl⁻; Λ°_NH4Cl = Λ°_NH4⁺ + Λ°_Cl⁻
2. Combining these three to isolate Λ°_NH4⁺ + Λ°_OH⁻ gives Λ°_NH4OH = \( \frac{\Lambda^{\circ}_{Ba(OH)_2} + 2\Lambda^{\circ}_{NH_4Cl} - \Lambda^{\circ}_{BaCl_2}}{2} \)

Teacher's Note:
1. This is the hardest Kohlrausch's-law combination in the chapter – work it out algebraically rather than memorising it, since the coefficients change with the salts chosen.

Answer: (a) 3.66%

Explanation:
1. α = \( \frac{\Lambda^{c}_{m}}{\Lambda^{\circ}_{m}} \) = \( \frac{14.3}{390.71} \) = 0.0366, i.e., 3.66%

Teacher's Note:
1. Equivalent conductivity at infinite dilution plays exactly the role that Λ°m does for molar conductivity – same formula, different units.

Answer: (c) For a cell reaction to occur spontaneously, the EMF of the cell should be negative.

Explanation:
1. E°_cell should be positive, not negative, for a spontaneous reaction, since ΔG° = −nFE°_cell. So statement (c) is the incorrect one.

Teacher's Note:
1. This is a classic sign-error trap – students often remember "spontaneous means negative" from ΔG and wrongly carry the negative sign over to E°cell.

Answer: (d) S₂O₈²⁻ is formed at the anode during electrolysis of CuSO₄ solution

Explanation:
1. Possible reactions at the anode during electrolysis of a CuSO₄ solution are: 2SO₄²⁻_(aq) → S₂O₈²⁻_(aq) + 2e⁻; E°_red = +2.07 V, and 2H₂O_(l) → O_2(g) + 4H⁺_(aq) + 4e⁻; E°_red = +1.23 V.
2. Comparing the reduction potential values, water (lower E°) is preferentially oxidised at the anode, giving oxygen gas, not S₂O₈²⁻. So statement (d) is the incorrect observation.

Teacher's Note:
1. S2O8²&supminus; forms at the anode only in concentrated H₂SO₄, where there's no water competing – in a CuSO₄ solution, water is oxidised to O₂ instead.

Answer: (b) I and II only

Explanation:
1. Salts that have polyvalent cations or anions are intermediate electrolytes (AlPO₄).
2. For a strong electrolyte Λ_m falls almost linearly with √C, for a weak electrolyte it falls steeply, and an intermediate electrolyte lies between the two. Λ_m never rises with √C, so graph III is wrong; graphs I and II are correct.

Teacher's Note:
1. Molar conductivity always decreases as concentration increases, so any Λ_m vs √C graph that rises is wrong straight away.
2. Salts with polyvalent ions, such as AlPO₄, behave as intermediate electrolytes.

Answer: (c) II and IV only

Explanation:
1. At cathode (reduction): 2Ag⁺_(aq) + 2e⁻ → 2Ag_(s)
2. At anode (oxidation): Mg_(s) → Mg²⁺_(aq) + 2e⁻
3. E_cell = E°_cell − \( \frac{RT}{2F} \) ln \( \frac{[Mg^{2+}]}{[Ag^{+}]^{2}} \) = E°_cell − \( \frac{0.059}{2} \) log \( \frac{[Mg^{2+}]}{[Ag^{+}]^{2}} \)
4. E°_cell = E°_cathode − E°_anode = E°_Ag⁺/Ag − E°_Mg²⁺/Mg
5. E_cell = E°_cell − \( \frac{0.059}{2} \) log \( \frac{(0.130)}{(0.0001)^{2}} \)
6. So II (0.059 used with ln) and IV (ratio inverted) are not correct.

Teacher's Note:
1. The factor 0.059 already includes the 2.303 that converts ln to log, so it must always go with log, never ln.
2. Statement III is correct: flipping the ratio inside the log changes the minus sign to a plus.

Answer: (c) III only

Explanation:
1. A more negative E° value means that the redox couple is a stronger reducing agent than the other one.
2. I: Fe (−0.44 V) would reduce Sn²⁺ (−0.14 V), so the vessel would dissolve; SnSO₄ cannot be stored in Fe.
3. II: Zn (−0.76 V) would reduce Fe²⁺ (−0.44 V); FeSO₄ cannot be stored in Zn.
4. III: Sn (−0.14 V) cannot reduce Cr³⁺ (−0.74 V); Cr₂(SO₄)₃ can be stored in Sn.
5. IV: Fe (−0.44 V) cannot reduce Zn²⁺ (−0.76 V), so ZnSO₄ can be stored in iron; IV is wrong.
6. Hence, only III is correct.

Teacher's Note:
1. A solution is safe in a metal vessel only if the vessel metal has a higher E° than the metal ion in solution.
2. If the vessel metal has the lower E°, it displaces the ion and the vessel corrodes.

Answer: (c) Zn_(s)|ZnSO_4(aq)||CuSO_4(aq)|Cu_(s); E_cell = E°_cell − \( \frac{2.303RT}{2F} \) log \( \frac{[Cu^{2+}]}{[Zn^{2+}]} \)

Explanation:
1. For the cell given in option (c), the correct Nernst equation is
2. E_cell = E°_cell − \( \frac{2.303RT}{2F} \) log \( \frac{[Zn^{2+}]}{[Cu^{2+}]} \)
3. Option (c) has the ratio inverted, so it is the incorrect equation.

Teacher's Note:
1. In the log term, the product ion (Zn²⁺, formed at the anode) goes on top and the reactant ion (Cu²⁺) goes below.
2. Options (a), (b) and (d) all follow this rule correctly.

Answer: (c) \( \Lambda = \Lambda_{o} \) as C → 0

Explanation:
1. At infinite dilution, concentration approaches zero and molar conductivity at infinite dilution is Λ_o. When C → 0, then Λ = Λ_o.

Teacher's Note:
1. Λ_o (limiting molar conductivity) is the intercept of the Λ_m vs √C straight line for a strong electrolyte; putting C = 0 in the equation gives it directly.

Answer: (c) 390.5 S cm² mol⁻¹

Explanation:
1. The limiting molar conductivity (Λ°_m) for strong and weak electrolyte can be determined by using Kohlrausch’s law which states that “the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.”
2. Λ°_CH3COONa = λ°_CH3COO⁻ + λ°_Na⁺
3. 91 S cm² mol⁻¹ = λ°_CH3COO⁻ + 50.1 S cm² mol⁻¹
4. ⇒ λ°_CH3COO⁻ = 40.9 S cm² mol⁻¹
5. For acetic acid, Λ°_CH3COOH = λ°_CH3COO⁻ + λ°_H⁺ = 40.9 + 349.6 = 390.5 S cm² mol⁻¹

Teacher's Note:
1. First pull out the acetate ion value from CH₃COONa, then add H⁺; the K⁺ and OH⁻ data are distractors here.

Answer: (c) Conductivity does not depend upon solvation of ions present in solution.

Explanation:
1. Conductivity depends upon the nature of the solute added, size of the ions produced, nature of solvent, viscosity of solvent and solvation of ions present in solution. Greater the solvation of ions of an electrolyte, lesser will be the electrical conductivity of the solution.

Teacher's Note:
1. A heavily solvated ion drags a shell of solvent molecules with it, so it moves slowly – that is why solvation lowers conductivity.

Answer: (b) 1 : 2 electrolyte

Explanation:
1. When 0.1 mole of CoCl₃(NH₃)₅ was reacted with excess of AgNO₃, we get 0.2 moles of AgCl. As 1 mol AgNO₃ can give only 1 mole AgCl, in the given reaction 1 mol Co(NH₃)₅Cl₃ will give 2 mol Cl⁻ ions, when 2Cl⁻ are not in coordinate sphere of complex. Thus, the formula for complex has to be [Co(NH₃)₅Cl]Cl₂.
2. [Co(NH₃)₅Cl]Cl₂ → [Co(NH₃)₅]²⁺ + 2Cl⁻
3. Therefore, the conductivity of the solution will be 1 : 2 electrolyte.

Teacher's Note:
1. Only the chloride ions outside the square bracket (ionisable) are precipitated by AgNO₃; count moles of AgCl per mole of complex to find them.

Answer: (b) Λ°_m[NH4Cl] + Λ°_m[NaOH] − Λ°_m[NaCl]

Explanation:
1. NH₄Cl ⇌ NH₄⁺ + Cl⁻ …(i)
2. NaCl ⇌ Na⁺ + Cl⁻ …(ii)
3. NaOH ⇌ Na⁺ + OH⁻ …(iii)
4. NH₄OH ⇌ NH₄⁺ + OH⁻ …(iv)
5. To get equation (iv): Λ°_m(NH4Cl) + Λ°_m(NaOH) − Λ°_m(NaCl) = Λ°_m(NH4OH)

Teacher's Note:
1. Check any such combination by cancelling ions: NH₄⁺ + Cl⁻ + Na⁺ + OH⁻ − Na⁺ − Cl⁻ leaves exactly NH₄⁺ + OH⁻.

Answer: (d) Cl⁻(aq) → ½Cl₂(g) + e⁻; E°_cell = 1.36 V

Explanation:
1. During electrolysis: NaCl → Na⁺ + Cl⁻; H₂O → H⁺ + OH⁻
2. Na⁺ + e⁻ → Na (E°_cell = −2.71 V); H⁺ + e⁻ → ½H₂ (E°_cell = 0.00 V)
3. At cathode, H₂O + e⁻ → ½H₂ + OH⁻
4. At anode, two reactions are possible where lower of E° is preferred: Cl⁻ → ½Cl₂ + e⁻ (E°_cell = 1.36 V); 2H₂O → O₂ + 4H⁺ + 4e⁻ (E°_cell = 1.23 V)
5. But due to slow oxidation process of O₂ it needs high voltage (overpotential). Thus, Cl₂ will form rather than O₂ as preference.

Teacher's Note:
1. Options (a) and (c) are reductions, so they can only happen at the cathode – you can eliminate them at once for an anode question.

Answer: (c) E_cell is an intensive property while Δ_rG of cell reaction is an extensive property.

Explanation:
1. E_cell is an intensive property and it does not depend upon number of particles but Δ_rG of the cell reaction is an extensive property because this depends upon number of particles.

Teacher's Note:
1. That is why doubling a cell equation doubles Δ_rG (through n) but leaves E_cell unchanged.

Answer: (c) E_external > E_cell

Explanation:
1. If an external opposite potential is applied on the electrochemical cell, the reaction continues to take place till the opposite voltage reaches the value 1.1 V. At this stage, no current flows through the cell and if there is any further increase in the external potential (E_external), then reaction starts functioning in opposite direction, i.e., an electrochemical cell behaves like an electrolytic cell.
2. E_external > E_cell

Teacher's Note:
1. Remember the three cases: E_ext < E_cell galvanic, E_ext = E_cell no current, E_ext > E_cell electrolytic.

Answer: (b)

Explanation:
1. E_Mg²⁺/Mg = E°_Mg²⁺/Mg + \( \frac{0.059}{2} \) log [Mg²⁺]
2. Compare this equation with the equation of straight line y = mx + c.
3. The graph of E_Mg²⁺/Mg vs log [Mg²⁺] is a straight line with a positive slope (0.059/2) and intercept E°_Mg²⁺/Mg.

Teacher's Note:
1. Rewrite −log(1/x) as +log x first; the sign of the slope then becomes obvious.
2. The intercept is not zero because E° itself is the value of E when log[Mg²⁺] = 0, i.e. [Mg²⁺] = 1 M.

Answer: (c) to complete the circuit so that current can flow

Explanation:
1. In an electrochemical cell, a salt bridge is used to keep solution electrically neutral and allow the flow of ions from one cell to another so that reaction can not stop, otherwise due to accumulation of ions on cathode and anode can stop reactions.

Teacher's Note:
1. The salt bridge carries current by ions, not electrons; electrons travel only through the external wire.

Answer: (a) Ag⁺(aq) + e⁻ → Ag(s)   E° = +0.80 V

Explanation:
1. Ag⁺(aq) + e⁻ → Ag(s); E° = +0.80 V. H⁺(aq) + e⁻ → ½H₂(g); E° = 0.00 V.
2. On the basis of their standard reduction potential (E°) values, cathode reaction is given by the one with higher E° values.
3. Thus, Ag⁺(aq) + e⁻ → Ag(s) reaction will be more feasible at cathode.

Teacher's Note:
1. At the cathode the species with the higher (more positive) reduction potential is reduced first; at the anode the one with the lower reduction potential is oxidised first.

Answer: (b) Cu(s) | Cu²⁺(aq) || Ag⁺(aq) | Ag(s)

Explanation:
1. Oxidation half reaction: Cu(s) → Cu²⁺(aq) + 2e⁻
2. Reduction half reaction: Ag⁺ + e⁻ → Ag(s)
3. Cu(s) | Cu²⁺(aq) || Ag⁺(aq) | Ag(s)   (Anode (oxidation) on the left, salt bridge, cathode (reduction) on the right)

Teacher's Note:
1. In cell notation the anode is written on the left as metal | ion, and the cathode on the right as ion | metal.

Answer: (a) Copper liberates hydrogen from acids.

Explanation:
1. Copper does not liberate hydrogen from acids because copper lies below hydrogen in electrochemical series. So, copper does not have sufficient electrode potential to liberate elemental hydrogen from compounds in which oxidation state of hydrogen is +1.

Teacher's Note:
1. E°(Cu²⁺/Cu) = +0.34 V is positive, so Cu cannot reduce H⁺ to H₂; only metals with negative E° do this with dilute acids.

Answer: (c) 2.651 V

Explanation:
1. E_cell = E°_cell − \( \frac{0.059}{n} \) log \( \frac{[Mg^{2+}]}{[Cu^{2+}]} \)
2. = 2.71 V − \( \frac{0.059}{2} \) log \( \frac{0.1}{0.001} \)
3. = 2.71 V − \( \frac{0.059}{2} \) log 10²
4. E_cell = 2.651 V

Teacher's Note:
1. Product ion over reactant ion: [Mg²⁺] goes on top. A quick check – here Q > 1, so E_cell must come out below E°_cell.

Answer: (c) BrO₄⁻ > IO₄⁻ > ClO₄⁻

Explanation:
1. Higher the reduction potential, higher is its tendency to get reduced. Hence, the order of oxidising power is: ClO₄⁻ < IO₄⁻ < BrO₄⁻

Teacher's Note:
1. Oxidising power simply follows the order of reduction potentials – just sort the E° values; read carefully whether the question asks increasing or decreasing order.

Answer: (b) Cr

Explanation:
1. The negative value of standard reduction potential for Cr³⁺ to Cr means that the redox couple is a stronger reducing agent. So Cr, the reduced form of this couple, is the strongest reducing agent.

Teacher's Note:
1. A reducing agent is the reduced form (right-hand species) of a couple; the lowest E° gives the strongest one. Cr³⁺ is a trap – it is the oxidised form.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Both statements are correct: molar conductivity falls as concentration rises, and this happens because increased interionic attraction at higher concentration slows ion migration – the reason correctly explains the assertion.

Teacher's Note:
1. This decrease is gentle and linear-ish for strong electrolytes (per the Kohlrausch plot), unlike the steep rise-on-dilution seen for weak electrolytes.

Answer: (c) A is true, but R is false.

Explanation:
1. Λ°_AB = λ°_A⁺ + λ°_B⁻ holds for any electrolyte at infinite dilution, so the assertion is correct.
2. Stating that Kohlrausch's law applies only to strong electrolytes is too restrictive – it is exactly what lets Λ°_m of weak electrolytes be found indirectly, so the reason as stated is incorrect, giving (c).

Teacher's Note:
1. Kohlrausch's law is actually most useful for weak electrolytes, since it's the only way to get their Λ°m (direct extrapolation doesn't work for them).

Answer: (c) A is true, but R is false.

Explanation:
1. The assertion is correct: equivalent (and molar) conductance falls as concentration rises.
2. At higher concentration, the mobility of ions decreases (due to stronger interionic attraction) – that is the real reason conductance falls, not a change in the number of ions available, so the stated reason is incorrect, giving (c).

Teacher's Note:
1. The number of ions per gram equivalent is fixed by definition; what changes with concentration is how fast those ions can move, which is the real driver here.

Answer: (c) A is true, but R is false.

Explanation:
1. The assertion is correct – this is the standard opposite-trend behaviour of κ and Λ on dilution.
2. On dilution, the total number of ions actually increases only slightly (for a weak electrolyte, as more dissociates) or stays the same (for a strong electrolyte) – not "considerably", so the reason as stated is incorrect, giving (c).

Teacher's Note:
1. Be wary of words like "considerably" in a reason statement – the direction can be right while the degree is overstated, which is enough to make the reason false.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. The ratio κ/G equals the cell constant, a fixed geometric quantity independent of concentration, so the assertion is correct.
2. The reason describes a true but separate fact about how κ and G individually vary with dilution; it does not explain why their ratio is concentration-independent, so it is not the correct explanation, giving (b).

Teacher's Note:
1. κ/G = cell constant by definition – the real explanation is that geometric constant, not the opposing trends of κ and G individually.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Both statements are correct, and the reason correctly explains why Kohlrausch's law is needed in the first place: since direct experimental measurement is impossible for weak electrolytes, the law provides an indirect route.

Teacher's Note:
1. This closes the loop with item 83 above – here the same two facts are paired the right way round, with the reason now correctly explaining the assertion.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Conductivity of an electrolyte decreases with decrease in concentration because of ions per unit volume decreases on dilution. Molar conductivity of electrolyte increases with decrease in concentration as total volume of solution of 1 mole electrolyte increases.

Teacher's Note:
1. Do not mix up the two quantities: on dilution κ falls (fewer ions per cm³) but Λ_m rises.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Weak electrolytes dissociate partially in concentrated solution. On dilution, their degree of dissociation increases hence, their Λ_m increases sharply.

Teacher's Note:
1. This sharp, non-linear rise is why the Λ_m vs √C curve of a weak electrolyte cannot be extrapolated to zero concentration.

Answer: (c) A is true, but R is false.

Explanation:
1. As the temperature of electrolytic solution is increased, the degree of ionisation increases, that ions carry charges. This results in the increase of electrical conductance of electrolytic solutions.

Teacher's Note:
1. Electrolytic conduction is by ions, not electrons; electronic (metallic) conduction actually decreases with rise in temperature.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Molar conductivity of an ionic solution depends on the temperature as well as on the concentration of the electrolytes in the solution. Therefore, on increase in temperature, molar conductivity increases whereas molar conductivity decreases on increase in concentration.

Teacher's Note:
1. Two true facts about the same quantity are not automatically cause and effect; ask whether R actually explains A.