CBSE Class 12 Chemistry Electrochemistry MCQs Set 11

Answer: (a) Copper will deposit at cathode and oxygen will be released at anode.

Explanation:
1. CuSO₄ ⇌ Cu²⁺ + SO₄²⁻; H₂O ⇌ H⁺ + OH⁻
2. At cathode: Cu²⁺ + 2e⁻ → Cu (E°_cell = 0.34 V); H⁺ + e⁻ → ½H₂ (E°_cell = 0.00 V)
3. This reaction where copper metal is deposited at cathode will take place due to higher reduction potential.
4. At anode: 2SO₄²⁻ → S₂O₈²⁻ + 2e⁻ (E° = 1.96 V); 2H₂O → O₂ + 4H⁺ + 4e⁻ (E° = 1.23 V)
5. The reaction with lower value of E° will be preferred at anode, hence O₂ is released at anode.

Teacher's Note:
1. With inert (Pt) electrodes the electrode itself never reacts, so look only at the ions and water in solution.

Answer: (a) Copper will deposit at cathode and copper will dissolve at anode.

Explanation:
1. Electrolysis of CuSO₄ can be represented by two half-cell reactions occurring at cathode and anode, respectively, as given below:
2. At cathode: Cu²⁺ + 2e⁻ → Cu(s)
3. At anode: Cu(s) → Cu²⁺ + 2e⁻
4. On dissolution of CuSO₄ in water it will dissociate as Cu²⁺, SO₄²⁻, H⁺, OH⁻.
5. At anode, SO₄²⁻ and OH⁻ are stable to exist on their own, so these do not oxidise. Thus, Cu on anode oxidises to form Cu²⁺ ion and dissolves the anode gradually. At cathode, Cu²⁺ ion is reduced rather than H⁺ because of higher potential of copper.
6. Thus, copper will deposit at cathode while copper will dissolve at anode.

Teacher's Note:
1. Compare with Q18: a reactive (Cu) anode is itself oxidised, so no O₂ is evolved – this is the principle of electrorefining of copper.

Answer: (b) \( \frac{G^{*}}{R} \)

Explanation:
1. κ = \( \frac{1}{R} \) · \( \frac{l}{A} \) or \( \frac{G^{*}}{R} \)

Teacher's Note:
1. Conductivity = conductance × cell constant; it is the conductance of a 1 cm cube of solution, not the molar conductivity.

Answer: (c) E_ext > E_cell

Explanation:
1. When an external potential greater than the cell emf is applied, the electrons are pushed in the opposite direction and the cell reaction is reversed.
2. The cell then works as an electrolytic cell.

Teacher's Note:
1. At E_ext = E_cell no current flows; only an opposing potential larger than E_cell reverses the reaction.

Answer: (b) Cell emf

Explanation:
1. The potential difference between the two electrodes measured when no current is drawn from the cell is called the emf of the cell.

Teacher's Note:
1. The words “no current is drawn” are the signal for emf; cell potential or voltage is usually measured while current flows.

Answer: (a) A + B⁺ → A⁺ + B

Explanation:
1. In the cell notation the left electrode is the anode (oxidation: A → A⁺ + e⁻) and the right electrode is the cathode (reduction: B⁺ + e⁻ → B).
2. Since the emf is positive, the cell reaction A + B⁺ → A⁺ + B is spontaneous as written.

Teacher's Note:
1. Write oxidation at the left electrode and reduction at the right one, then add; a positive emf confirms the direction.

Answer: (c) Pt(s) | H₂(g, 1 bar) | H⁺(aq., 1 M) || Cu²⁺(aq., 1M) | Cu

Explanation:
1. Standard electrode potential is measured against the standard hydrogen electrode (H₂ at 1 bar, [H⁺] = 1 M) with the copper ions also at unit concentration (1 M).

Teacher's Note:
1. Check every condition in the cell: gas pressure 1 bar and all ion concentrations 1 M – one wrong value makes it non-standard.

Answer: (d) Both (b) and (c)

Explanation:
1. A positive E° means Cu²⁺ is reduced more easily than H⁺, so Cu²⁺/Cu is a stronger oxidising agent than H⁺/H₂.
2. For the same reason Cu cannot reduce H⁺ ions, i.e., Cu cannot displace H₂ from acids.

Teacher's Note:
1. Metals with positive E° (Cu, Ag, Au) lie below hydrogen in the series and do not liberate H₂ from dilute acids.

Answer: (b) 0.89 V

Explanation:
1. Sn⁴⁺/Sn²⁺ (higher E°) is the cathode and Cr³⁺/Cr is the anode.
2. E°_cell = 0.15 − (−0.74) = 0.89 V

Teacher's Note:
1. Never multiply E° values by stoichiometric coefficients; E°_cell is simply E°_cathode − E°_anode.

Answer: (b) Mn²⁺ < Cl⁻ < Cr³⁺ < Cr

Explanation:
1. Reducing power of the reduced forms increases as E° decreases: 1.51 V (Mn²⁺) > 1.36 V (Cl⁻) > 1.33 V (Cr³⁺) > −0.74 V (Cr).
2. Hence, Mn²⁺ < Cl⁻ < Cr³⁺ < Cr.

Teacher's Note:
1. Cr₂O₇²⁻ and MnO₄⁻ are oxidising agents, so any order listing them as reducing agents can be rejected at once.

Answer: (a) Cr³⁺

Explanation:
1. The oxidised species with the lowest reduction potential has the least tendency to gain electrons, so it is the most stable.
2. Cr³⁺ (in Cr³⁺/Cr, E° = −0.74 V) is therefore the most stable oxidised species.

Teacher's Note:
1. Stability of an oxidised form means low tendency to be reduced – look for the smallest E° among the oxidised forms.

Answer: (a) Br⁻ < Fe²⁺ < Al

Explanation:
1. Lower the reduction potential, greater the reducing power of the reduced form.
2. E°: Br₂/Br⁻ (+1.80) > Fe³⁺/Fe²⁺ (+0.77) > Al³⁺/Al (−1.66), so reducing power: Br⁻ < Fe²⁺ < Al.

Teacher's Note:
1. Reducing power order is exactly the reverse of the E° order.

Answer: (d) E°_cell > 0, ΔG° < 0, Q < K_c

Explanation:
1. For a spontaneous cell reaction, E_cell > 0 and ΔG < 0 (ΔG = −nFE).
2. The reaction proceeds forward only while Q < K_c.

Teacher's Note:
1. ΔG and E always have opposite signs; Q < K means the reaction still has to move forward to reach equilibrium.

Answer: (c) 10³⁰

Explanation:
1. log K_c = \( \frac{nE^\circ_{cell}}{0.059} \) = \( \frac{3 \times 0.59}{0.059} \) = 30
2. K_c = 10³⁰

Teacher's Note:
1. Use log K = nE°/0.059 at 298 K; forgetting n gives 10¹⁰, a common wrong choice.

Answer: (a) S m⁻¹

Explanation:
1. κ = \( \frac{1}{R} \) × \( \frac{l}{A} \); in SI units l is in m and A in m², so the unit is S m⁻¹.

Teacher's Note:
1. S cm⁻¹ is also a unit of conductivity, but it is not the SI unit.

Answer: (a) ionic conductance increases and electronic conductance decreases.

Explanation:
1. On heating, ions move faster (viscosity falls, dissociation increases), so ionic conductance increases.
2. In metals, vibrating kernels obstruct the flow of electrons, so electronic conductance decreases.

Teacher's Note:
1. Metals and electrolytes respond to temperature in opposite ways – a frequent 1-mark question.

Answer: (b) Λ⁰_m(NH4Cl) + Λ⁰_m(NaOH) − Λ⁰_m(NaCl)

Explanation:
1. By Kohlrausch’s law: (λ°_NH4⁺ + λ°_Cl⁻) + (λ°_Na⁺ + λ°_OH⁻) − (λ°_Na⁺ + λ°_Cl⁻) = λ°_NH4⁺ + λ°_OH⁻

Teacher's Note:
1. Choose strong electrolytes so that the unwanted ions (Na⁺, Cl⁻) cancel and only NH₄⁺ and OH⁻ remain.

Answer: (b) 9.65 × 10⁴ coulomb

Explanation:
1. Charge on 1 mol of electrons = 6.022 × 10²³ × 1.602 × 10⁻¹⁹ C ≈ 96500 C = 9.65 × 10⁴ C = 1 F.

Teacher's Note:
1. 1.6 × 10⁻¹⁹ C is the charge on one electron; multiply by Avogadro’s number for a mole.

Answer: (c) 3F

Explanation:
1. Al³⁺ + 3e⁻ → Al; one mole of Al needs 3 mol of electrons, i.e., 3F.

Teacher's Note:
1. Count electrons per atom of Al, not per formula unit of Al₂O₃ (which would give 6F for 2 mol Al).

Answer: (d) Both (a) and (b)

Explanation:
1. At the cathode H⁺ ions are reduced to H₂.
2. At the anode, oxidation of water (1.23 V) is easier than oxidation of SO₄²⁻ (1.96 V) in dilute acid, so water is oxidised to O₂.

Teacher's Note:
1. In concentrated H₂SO₄ the result changes: SO₄²⁻ is oxidised to peroxodisulphate (S₂O₈²⁻).

Answer: (c) paste of KOH and ZnO

Explanation:
1. In the mercury cell, the anode is zinc–mercury amalgam, the cathode is a paste of HgO and carbon, and the electrolyte is a paste of KOH and ZnO.

Teacher's Note:
1. Do not confuse the cathode paste (HgO + carbon) with the electrolyte paste (KOH + ZnO).

Answer: (c) fuel cell

Explanation:
1. Galvanic cells designed to convert the energy of combustion of fuels such as H₂, CH₄ and CH₃OH directly into electrical energy are called fuel cells.

Teacher's Note:
1. Fuel cells run as long as reactants are supplied, unlike batteries which store a fixed amount of reactants.

Answer: (b) Pt

Explanation:
1. Platinum is an inert electrode; it does not take part in the cell reaction and only provides a surface for the electrode reaction and for conduction of electrons.

Teacher's Note:
1. Pt is used in the standard hydrogen electrode for exactly this reason.

Answer: (b) Graph (b)

Explanation:
1. \( E_{Mg^{2+}/Mg} = E^{0}_{Mg^{2+}/Mg} + \frac{0.059}{2} \log [Mg^{2+}] \)
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 and intercept E^°_Mg²⁺/Mg.

Teacher's Note:
1. Rewrite −log(1/x) as +log x before deciding the sign of the slope; the slope here is +0.059/2.

Answer: (a) PbSO₄ anode is reduced to Pb.

Explanation:
1. While charging the lead storage battery the reaction occurring on cell is reversed and lead sulphate (PbSO₄) on anode and cathode is converted into (lead) Pb and lead dioxide (PbO₂), respectively. The electrode reactions are as follows :
2. At cathode : PbSO₄(s) + 2e⁻ → Pb(s) + SO₄²⁻(aq) (Reduction)
3. At anode : PbSO₄(s) + 2H₂O → PbO₂(s) + SO₄²⁻ + 4H⁺ + 2e⁻ (Oxidation)
4. Overall reaction : 2PbSO₄(s) + 2H₂O → Pb(s) + PbO₂(s) + 4H⁺(aq.) + 2SO₄²⁻(aq.)

Teacher's Note:
1. “Anode” here means the Pb electrode (the anode during discharge); on charging its PbSO₄ is reduced back to Pb.

Answer: (c) H₂SO₄ is consumed

Explanation:
1. On discharge: Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l).
2. H₂SO₄ is used up and PbSO₄ is formed at both electrodes.

Teacher's Note:
1. During discharge the density of the acid falls; this is how the state of charge of a car battery is checked.

Answer: (c) 1.93 × 10⁵ C

Explanation:
1. H₂O₂ → O₂ + 2H⁺ + 2e⁻ (O changes from −1 to 0).
2. Charge = 2F = 2 × 96500 C = 1.93 × 10⁵ C

Teacher's Note:
1. Each O atom in H₂O₂ loses one electron, so 1 mol H₂O₂ needs 2 mol of electrons.

Answer: (d) Migration factor of K⁺ and Cl⁻ ions are almost equal

Explanation:
1. K⁺ and Cl⁻ ions have almost equal ionic mobilities (transport numbers), so they carry equal shares of current and no liquid junction potential develops.

Teacher's Note:
1. KNO₃ and NH₄NO₃ are used for the same reason when Cl⁻ would react with the electrolyte (e.g. Ag⁺).

Answer: (a) a straight line

Explanation:
1. E°_cell = \( \frac{0.059}{n} \) log K_c, which is of the form y = mx.
2. So the plot is a straight line through the origin with slope 0.059/n.

Teacher's Note:
1. Whenever one quantity is directly proportional to another (y = mx), the graph is a straight line through the origin.

Answer: (c) −ve, > 1, +ve

Explanation:
1. ΔG° = −nFE°_cell = −2.303RT log K.
2. For a spontaneous reaction ΔG is negative, so E°_cell is positive and log K > 0, i.e. K > 1.

Teacher's Note:
1. Remember the trio: ΔG < 0, E° > 0 and K > 1 always go together.

Answer: (a) 0.1773

Explanation:
1. n = 2; E°_cell = \( \frac{0.0591}{n} \) log K_c = \( \frac{0.0591}{2} \) × log 10⁶
2. = 0.02955 × 6 = 0.1773 V

Teacher's Note:
1. Find n from the change in oxidation state (Sn²⁺ → Sn⁴⁺ loses 2 electrons) before using the formula.

Answer: (d) I⁻

Explanation:
1. Among the halogens, I₂ has the lowest reduction potential, so I⁻ is most easily oxidised.
2. Hence I⁻ is the strongest reducing agent.

Teacher's Note:
1. The weaker the oxidising agent (I₂), the stronger its conjugate reducing agent (I⁻).

Answer: (a) drops to zero

Explanation:
1. Without the salt bridge the internal circuit is broken and the solutions cannot remain electrically neutral, so no current flows and the voltage drops to zero.

Teacher's Note:
1. The salt bridge completes the inner circuit; the external wire alone cannot keep a galvanic cell working.

Answer: (c) equivalent weight of the electrolyte

Explanation:
1. By Faraday’s laws, the mass deposited W = Z × Q, and Z = E/96500, where E is the equivalent weight.

Teacher's Note:
1. The second law: the same charge deposits masses in the ratio of their equivalent weights.

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

Explanation:
1. E_cell = E_cathode − E_anode. To have positive value of E_cell, E_cathode should be greater than E_anode.

Teacher's Note:
1. A positive E_cell makes ΔG = −nFE_cell negative, which is the condition for a working (spontaneous) cell.

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

Explanation:
1. Cu is less reactive than hydrogen because E°_Cu²⁺/Cu is positive (+0.34 V).

Teacher's Note:
1. Metals with positive E° lie below hydrogen in the series and do not displace H₂ from dilute acids.

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

Explanation:
1. Zinc will get dissolved in CuSO₄ solution, since zinc is more reactive than copper.

Teacher's Note:
1. Rule of thumb: a solution can be stored in a vessel only if the vessel metal is less reactive (higher E°) than the metal ion in the solution.

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

Explanation:
1. Ag⁺ + e⁻ → Ag,   E_Ag⁺/Ag = E°_Ag⁺/Ag − \( \frac{RT}{nF} \) ln \( \frac{1}{[Ag^{+}]} \)
2. On increasing [Ag⁺], E_Ag⁺/Ag will increase and it has a positive value.

Teacher's Note:
1. The increase with [Ag⁺] comes from the Nernst equation, not from the sign of E° – it would happen even for a couple with negative E°.

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

Explanation:
1. Formation of oxygen has lower value of E° than formation of chlorine even then it is not formed because of kinetically slow oxidation of O₂; it requires over voltage.

Teacher's Note:
1. Overpotential is a kinetic effect – it can override the prediction made from E° values alone.

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

Explanation:
1. The overall reaction of the mercury cell, Zn(Hg) + HgO(s) → ZnO(s) + Hg(l), involves no ions in solution, so its potential stays constant (about 1.35 V) throughout its life.
2. The Assertion is incorrect and the Reason is correct.

Teacher's Note:
1. Cell potential changes only when concentrations in the Nernst expression change; no ions in the overall reaction means a steady voltage.

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

Explanation:
1. When E_cell = 0, ΔG = 0 and the cell reaction has reached equilibrium, so there is no net reaction and no current.

Teacher's Note:
1. A “dead” cell is simply a cell whose reaction has reached equilibrium.

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

Explanation:
1. A DC source would cause electrolysis and change the concentration of ions near the electrodes, giving a wrong resistance.
2. An AC source prevents electrolysis, so the Reason correctly explains the Assertion.

Teacher's Note:
1. That is also why conductivity cells use platinised electrodes and AC in a Wheatstone bridge.

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

Explanation:
1. Conductivity is the conductance of unit volume of solution; on dilution the number of ions per unit volume decreases, so conductivity decreases for all electrolytes.

Teacher's Note:
1. Conductivity (κ) falls on dilution, but molar conductivity (Λ_m) rises – keep the two separate.