JEE Advanced (One or More Correct Option): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. Which of the following compound (s) do not conduct electricity in their molten state?
(a) MgCl₂ 
(b) BeCl₂
(c) BeF₂
(d) MgF₂

Correct Answer is option (b and c)
The values of E₀ of some reactions are given
I₂ +   2e^- → 2I^-   E₀ = 0.54 volts
Sn⁺⁴  +   2e^- → Sn²⁺   E₀ = 0.152 volts
Cl₂  +   2e^- → 2Cl^-   E₀ = 1.36 volts
Fe⁺³ +   e^- → Fe²⁺   E₀ = 0.76 volts
Ce⁺⁴ +   e^- → Ce³⁺ E₀ = 1.6 volts
Hence
(A) Fe³⁺ oxidizes Ce⁺³ 
(B) Ce⁴⁺ can oxidize Fe²⁺ 
(C) Sn²⁺ will reduce Fe³⁺ to Fe²⁺ 
(D) Cl₂ will be liberated from KCl by passing I₂.

Q.2. Select the correct statement(s) for the concentration cell.
(a) For the cell Cu/CuSO₄ (xM) : CuSO₄(yM)/Cu material transfers from a more concentrated to diluted solution
(b) ΔG for the overall reaction of concentration cell is the measure of diffusion of electrolyte
(c) Solubility product of a sparingly soluble salt can be measured by setting up a concentration cell
(d) The E.M.F of cell Ag/AgNO₃ (0.01M) : AgNO₃(1N)/Ag is approx. 0.059V.

Correct Answer is option (a, b and c)
In conc. cell electrical energy is obtained by concentration difference alone without any chemical reaction.
For the cell A/AgNO₃(0.01M) : : : Ag(NO₃(1M)/Ag
Net cell reaction is
Ag⁺ (1M) → Ag⁺ (0.01M)

Q.3. As dilution takes place
(a) Molar conductivity increases
(b) Conductance increases
(c) Ionic mobility increases
(d) Specific conductance increases

Correct Answer is option (a, b and c)
(A), (B) and (C) explain: On dilution the molarity of the solution decreases and hence molar conductivity increases. So choice (A) is correct.
(B) As dilution increases, the conductance of the solution increases because interionic forces of attraction are broken. Hence, choice (b) is correct.
(C) On dilution, ionic mobility increases due to decrease in interionic forces of attraction, hence (C) is correct.
(D) But the conducting power of ions per cubic centimetre decreases, hence specific conductance decreases. As volume of solution increases.
Hence, choice (d) is incorrect.

Q.4. In the following electrochemical cell
Zn | Zn²⁺ || H⁺ | Pt(H₂)
E^o_cell = E^o_cell. This will be when
(a) [Zn²⁺] = [H⁺] = 1 M and ^PH₂ = 1atm
(b) [Zn²⁺] = 0.01 M, [H⁺] = 0.1M and ^PH₂ = 1atm
(c) [Zn²⁺] = 1 M, [H⁺] = 0.1 M and ^PH₂ = 0.01atm
(d) [Zn²⁺] = [H+] = 0.1 M and ^PH₂ = 0.1atm

Correct Answer is option (a, b, c and d)

Q.5. In Zn-Cu cell containing K₂SO₄ in the salt bridge
(a) Electrons flow from Cu to Zn
(b) Conventional current flows from Cu to Zn
(c) SO^2–₄ ions flow from Cu to Zn
(d) The weight of zinc rod decreases and copper rod increases

Correct Answer is option (b, c and d)
(B), (C) and (D) Explanation: (B), (C) and (D) are correct while (A) is not correct.
(B) is correct because electrons flow from Zn to Cu and conventional current flows from Cu to Zn.
(C) is correct because SO^2–₄ ions are in excess, therefore, moves from Cu to Zn.
(D) is correct because zinc metal will dissolve and its weight will keep on decreasing and weight of copper rod will keep on increasing.

Q.6. For the reduction of MnO^-₄ to Mn²⁺. E^° is + 1.51V. The reduction potentials for Zn²⁺, Ag⁺, Au⁺ and  Na⁺ are given as:
 and

Which of these metals will be oxidized by MnO^-₄ ion?
(a) Zn
(b) Ag
(c) Au
(d) Na

Correct Answer is option (a, b and d)
Zn: 1.51V + 0.761 V = 2.27 V > 0
Ag: 1.51V – 0.7956V = 0.710V > 0
Au: 1.51V – 1.692V = –0.18V < 0
Na : 1.51 V + 3.0 V = 4.51 > 0

Q.7. 200 ml of 0.1 M NaOH is mixed with 300 ml of 0.2 M (BaOH )₂ then the resulting solution has
(a) molarity = 0.16 M
(b) normality 0.28 N
(c) normality = 1.6 N
(d) molarity = 3.2 N

Correct Answer is option (a and b)

Q.8. A sample of water has a hardness expressed as 77.5 ppm Ca²⁺. This sample is passed through an ion exchange column and the Ca²⁺ is replaced by H⁺. Select correct statement(s)–
(a) pH of the water after it has been so treated is 2.4
(b) Every Ca²⁺ ion is replaced by one H+ ion
(c) Every Ca²⁺ ion is replaced by two H+ ions
(d) pH of the solution remains unchanged

Correct Answer is option (a and c)
Ca²⁺ + RH₂ → CaR + 2H⁺
77.5 ppm   = 77.5 g Ca²⁺ per 10⁶ mL
= 77.5 × 10^–3 g L^–1 Ca²⁺

∴ [H⁺] = 3.875 × 10^-3
∴ pH = -log (H⁺) = 2.4

Q.9. Standard electrode potential of two metals A and B are E⁰₁ and E⁰₂ respectively. When half cells A²⁺/A and B²⁺/B are connected electrons flow from A to B but when A is connected with standard hydrogen electrode electrons flow from A to standard hydrogen electrode. If magnitude of E⁰₁ and E⁰₂ are 0.25 and 0.34 volt respectively the among the following the correct statement(s) is/are -
(a) E⁰₁ for A²⁺ + 2e → A is -0.25 V
(b) E⁰₂ for B²⁺ + 2e → B is -0.34 V
(c) E⁰_cell of the cell connecting these electrodes is 0.59 V
(d) E⁰₂ for B²⁺ + 2e → B is 0.34 V

Correct Answer is option (a, c and d)
The A is more oxidising than B and A is more oxidising than H₂
∴ A → A²⁺ + 2e E⁰₁ = 0.25 V
A²⁺ + 2e → A E⁰₁ = -0.25 V
A → A²⁺ + 2e
B²⁺ + 2e → B
A + B²⁺ → A²⁺ + B
E⁰_cell = 0.34 + 0.25 = 0.59 V

Q.10. Which of the following expression(s) represent the voltage of cell at 298 K:
Ag (s) | AgI (saturated solution) || Ag₂C₂O₄ (saturated solution) | Ag(s)
(a)
(b)
(c)
(d)

Correct Answer is option (b and d)
The given cell is concentration cell:
Ag  → (Ag⁺) LHE + e
(Ag⁺) RHE + e → Ag
Net reaction: (Ag⁺)RHE → (Ag⁺)LHE


In AgI solution,        [Ag⁺] = Ksp (AgI)^1/2
In Ag₂C₂O₄ solution [Ag⁺]

= [2 K_sp (Ag₂C₂O₄)]^1/3