JEE Chemistry Electrochemistry MCQs Set 07

Answer: (c) dissociate into ions even at high concentration

Explanation:
1. A strong electrolyte is almost completely ionised in solution at all concentrations.
2. Solubility and the ability to conduct alone do not define a strong electrolyte.

Teacher's Note:
1. Strength of an electrolyte is about the degree of ionisation, not about solubility.
2. Weak electrolytes ionise to a large extent only at high dilution.

Answer: (b) free ions

Explanation:
1. On melting, the ionic lattice breaks and Na⁺ and Cl⁻ ions become free to move.
2. These mobile ions carry the current.

Teacher's Note:
1. Ionic compounds conduct only when the ions are free to move.
2. Solid NaCl does not conduct.

Answer: (c) ions

Explanation:
1. In an electrolyte, current is carried by positive and negative ions moving towards the electrodes.
2. Electrons move only in the external wire and metallic conductors.

Teacher's Note:
1. Metallic conduction is by electrons, electrolytic conduction is by ions.
2. Chemical changes occur at the electrodes in electrolytic conduction.

Answer: (d) 10,000 c.c.

Explanation:
1. An N/10 solution has 0.1 equivalent in 1000 cc.
2. So 1 equivalent is present in 10,000 cc.

Teacher's Note:
1. V (in cc) = 1000/N.
2. A more dilute solution has a larger V.

Answer: (d) 1.0 N

Explanation:
1. Specific conductance is the conductance of 1 cm³ of solution, which depends on the number of ions per cm³.
2. The most concentrated solution, 1.0 N, has the most ions per cm³.

Teacher's Note:
1. Specific conductance rises with concentration for these strong electrolyte solutions.
2. Equivalent conductance, however, falls.

Answer: (b) the resistance decreases with increasing temperature

Explanation:
1. In an electrolyte, heating lowers the viscosity and increases ion mobility, so resistance falls.
2. In a metal, vibration of atoms increases with temperature and resistance rises.

Teacher's Note:
1. Opposite temperature behaviour is a standard difference between the two types.
2. Both types obey Ohm's law.

Answer: (a) Specific conductance increases on dilution

Explanation:
1. On dilution the number of ions per cm³ decreases, so specific conductance decreases.
2. Equivalent and molar conductance increase because of lower inter-ionic attraction (and greater ionisation for weak electrolytes). So statement (a) is wrong.

Teacher's Note:
1. Specific conductance depends on ions per unit volume.
2. Molar quantities refer to a fixed amount of electrolyte, so they behave differently.

Answer: (a) 126

Explanation:
1. By Kohlrausch's law Λ°(NaCl) = Λ°(CH₃COONa) + Λ°(HCl) − Λ°(CH₃COOH).
2. = 91 + 426 − 391 = 126 mho cm² eq⁻¹.

Teacher's Note:
1. Choose the combination in which the unwanted ions cancel.
2. Add the salts containing the required ions and subtract the other.

Answer: (c) 128 S cm² mol⁻¹

Explanation:
1. Λ°(NaBr) = Λ°(NaCl) + Λ°(KBr) − Λ°(KCl).
2. = 126 + 152 − 150 = 128 S cm² mol⁻¹.

Teacher's Note:
1. Write the ions: Na⁺ + Br⁻ = (Na⁺ + Cl⁻) + (K⁺ + Br⁻) − (K⁺ + Cl⁻).
2. The K⁺ and Cl⁻ contributions cancel.

Answer: (a) Λ°(NaCl)

Explanation:
1. Λ°(HOAc) = Λ°(NaOAc) + Λ°(HCl) − Λ°(NaCl).
2. So the molar conductivity of NaCl at infinite dilution is needed.

Teacher's Note:
1. The Na⁺ ions of NaOAc must be cancelled using NaCl.
2. This also removes the extra Cl⁻ from HCl.

Answer: (d) 4 × 10⁻⁶

Explanation:
1. Solubility S = κ × 1000 / Λm = 3.06 × 10⁻⁶ × 1000 / 1.53 = 2 × 10⁻³ mol/L.
2. Ksp = S² = 4 × 10⁻⁶.

Teacher's Note:
1. A saturated solution of a sparingly soluble salt is treated as infinitely dilute.
2. For a 1:1 salt Ksp = S².

Answer: (b) from Cu to Ag

Explanation:
1. Cu is oxidised at the anode, which is the left electrode, and Ag⁺ is reduced at the cathode.
2. Electrons flow from the anode (Cu) to the cathode (Ag) through the external wire.

Teacher's Note:
1. Electrons always flow from anode to cathode in the outer circuit.
2. The metal with the lower reduction potential is the anode.

Answer: (b) from Cu to Zn outside the cell

Explanation:
1. Electrons flow from Zn to Cu in the external wire.
2. Conventional current is opposite to electron flow, so it goes from Cu to Zn outside the cell.

Teacher's Note:
1. Current direction is opposite to the direction of electron flow.
2. Inside the cell the current goes from Zn to Cu through the electrolyte.

Answer: (d) Zn²⁺(aq.) + 2e⁻ → Zn(s)

Explanation:
1. A standard reduction potential refers to the reduction of the aqueous ion to the metal.
2. Hence Zn²⁺(aq) + 2e⁻ → Zn(s) defines it. The other equations use wrong states or are oxidations.

Teacher's Note:
1. Reduction potential is for gain of electrons by the ion in solution.
2. Check the states of matter in each equation.

Answer: (b) negative

Explanation:
1. E°cell = E°cathode − E°anode = E°(Zn²⁺/Zn) − E°(Cu²⁺/Cu) = −0.76 − 0.34 = −1.10 V.
2. The emf is negative, so the cell reaction as written is not spontaneous.

Teacher's Note:
1. Copper has a higher reduction potential than zinc.
2. The spontaneous cell would have zinc on the left.

Answer: (a) Ag⁺ → Ag − e⁻

Explanation:
1. Anodic reactions are oxidations (loss of electrons).
2. Ag⁺ → Ag − e⁻ is the same as Ag⁺ + e⁻ → Ag, which is a reduction, so it is not an anodic reaction.

Teacher's Note:
1. Rewrite each equation with electrons on the correct side.
2. Oxidation = loss of electrons.

Answer: (a) Oxidation occurs at the anode in both galvanic and electrolytic cells

Explanation:
1. By definition the anode is the electrode where oxidation takes place in any cell.
2. Only the sign of the anode differs (positive in electrolytic, negative in galvanic).

Teacher's Note:
1. Anode always means oxidation, cathode always means reduction.
2. Only the polarity changes between the two types of cell.

Answer: (d) Na⁺

Explanation:
1. Zn displaces metals that are below it in the electrochemical series.
2. Na is above Zn (E° = −2.71 V against −0.76 V), so Zn cannot displace Na⁺.

Teacher's Note:
1. A metal displaces only less reactive metals from solution.
2. Highly reactive metals like Na react with water instead.

Answer: (b) ZYX

Explanation:
1. Carbon cannot reduce ZO, so Z is the most active metal.
2. Y removes oxygen from XO, so Y is more active than X. Hence the order is Z > Y > X.

Teacher's Note:
1. A more active metal holds oxygen more strongly.
2. Use each piece of evidence to build the order.

Answer: (c) more of Pb and Fe²⁺ ions

Explanation:
1. Fe is above Pb in the electrochemical series, so Fe displaces Pb²⁺ from solution.
2. The reaction Fe + Pb²⁺ → Fe²⁺ + Pb gives more Pb and Fe²⁺.

Teacher's Note:
1. The more reactive metal goes into solution as ions.
2. The reverse reaction does not occur.

Answer: (c) Cl₂

Explanation:
1. Cl₂ has a higher reduction potential (1.36 V) than Br₂ (1.09 V), so Cl₂ oxidises Br⁻ to Br₂.
2. I₂ is a weaker oxidising agent than Br₂ and the others cannot oxidise Br⁻.

Teacher's Note:
1. A stronger oxidising agent oxidises the anion of a weaker one.
2. Halogen displacement follows the order F₂ > Cl₂ > Br₂ > I₂.

Answer: (c) II and IV

Explanation:
1. A metal displaces another when its reduction potential is lower.
2. Fe (−0.44 V) can displace Cu and Ag, so II and IV are correct. Cu cannot displace Fe, and Ag cannot displace Cu.

Teacher's Note:
1. Order by E°: Fe < Cu < Ag.
2. Each metal displaces those that lie after it in the series.

Answer: (d) B > C > A

Explanation:
1. Reducing power increases as the standard reduction potential decreases.
2. B (−2.50) is the strongest, then C (0.50), then A (0.68), so B > C > A.

Teacher's Note:
1. Lower E° means a stronger reducing agent.
2. Rank from the most negative to the most positive.

Answer: (b) Zn + 2Ag⁺ → 2Ag + Zn²⁺

Explanation:
1. Cell voltage = oxidation potential of the species oxidised + reduction potential of the species reduced.
2. For Zn and Ag⁺: 0.76 + 0.80 = 1.56 V, which is larger than 1.10 V (a), 0.34 V (c) and −0.25 V (d).

Teacher's Note:
1. Choose the pair with the highest oxidation potential for the reductant and the highest reduction potential for the oxidant.
2. Reduction potential is the negative of oxidation potential.

Answer: (b) E°(W⁻/W₂) < E°(Y⁻/Y₂) < E°(X⁻/X₂) < E°(Z⁻/Z₂)

Explanation:
1. Y₂ oxidises X⁻, so Y₂ is a stronger oxidising agent than X₂. Y₂ does not oxidise W⁻, so W₂ is stronger than Y₂, and X₂ oxidises Z⁻, so X₂ is stronger than Z₂.
2. Oxidising strength is W₂ > Y₂ > X₂ > Z₂, so the oxidation potentials follow the reverse order, W⁻ < Y⁻ < X⁻ < Z⁻.

Teacher's Note:
1. A stronger oxidising agent has a higher reduction potential and a lower oxidation potential.
2. Build the order step by step from the three observations.

Answer: (c) z + salt of x → x + salt of z

Explanation:
1. x displaces both y and z, and y displaces z, so the order of activity is x > y > z.
2. z is the least active, so it cannot displace x from its salt, and statement (c) is incorrect.

Teacher's Note:
1. Write the activity order first.
2. A metal displaces only less active metals.

Answer: (c) D

Explanation:
1. A metal displaces another metal that has a higher (less negative) reduction potential.
2. Only D (−0.402 V) is lower than A (−0.250 V), so D displaces A.

Teacher's Note:
1. Lower reduction potential means a more reactive metal.
2. B and C are less reactive than A.

Answer: (d) AgNO₃

Explanation:
1. Cu displaces only metals with a higher reduction potential.
2. Only Ag⁺ is reduced by Cu, giving blue Cu²⁺ ions in solution.

Teacher's Note:
1. Blue colour is typical of hydrated Cu²⁺.
2. Zn, Mg and K are more reactive than Cu and are not displaced.