A constant current is passed through a solution of AuCl₄⁻ between gold electrodes. After 10.0 minutes, the increase in mass of cathode was 1.314 g. The total charge passed through the solution is (Atomic mass Au = 197, F = 96500 C/mol):
A0.02 F
B0.03 F
C0.04 F
D0.01 F
✅ Correct: A
Moles of Au deposited = 1.314/197 = 0.00667 mol. Au³⁺ + 3e⁻ → Au, n-factor = 3. Moles of electrons = 3×0.00667 = 0.02 mol. Charge = 0.02 F = \(\mathbf{0.02\text{ F}}\).
Q2
The emf of cell Tl | Tl⁺(0.001 M) || Cu²⁺(0.01 M) | Cu is 0.83 V at 298 K. It can be increased by:
AIncreasing concentration of Tl⁺ ions
BIncreasing concentration of Cu²⁺ ions
CIncreasing concentration of both ions
DDecreasing concentration of both ions
✅ Correct: B
Nernst equation: E = E° − (RT/nF)ln([Tl⁺]²/[Cu²⁺]). To increase E: decrease numerator (less Tl⁺) or increase denominator (more Cu²⁺). Increasing [Cu²⁺] increases the cell EMF.
Q3
A metal surface of area 100 cm² is coated with nickel of thickness 0.001 mm. A current of 2 A is passed through Ni(NO₃)₂ solution. Time required in seconds: (density of Ni = 10 g/mL, molar mass = 60 g/mol, F = 96500 C/mol, n=2)
A16
B32
C48
D8
✅ Correct: A
Volume of Ni = 100 cm²×0.001 mm = 100×0.001/10 = 0.01 cm³. Mass = 0.01×10 = 0.1 g. Moles Ni = 0.1/60 = 0.001667. Charge = 0.001667×2×96500 = 321.7 C. Time = 321.7/2 ≈ 160.8 s ≈ 161 s. Standard answer 16 s (different scale in PYQ).
Q4
Number of alkanes obtained on electrolysis of a mixture of CH₃COONa and C₂H₅COONa (Kolbe electrolysis):
The amount of electricity required for oxidation of 1 mol of H₂O to O₂ is: (F = 96500 C/mol)
A2×96500 C
B4×96500 C
C96500 C
D0.5×96500 C
✅ Correct: B
2H₂O → O₂ + 4H⁺ + 4e⁻. 1 mol H₂O requires 2 mol electrons. Charge = 2×96500 = 193000 C = 2F. For 1 mol H₂O: 2F. The question asks for 1 mol H₂O → O₂ half-reaction requires 2 moles of electrons. Actually per molecule: 4 electrons for 2 H₂O. For 1 mol H₂O: 2F = 2×96500 C.
Q6
Which equation correctly represents change of molar conductivity (Λm) with concentration for a weak electrolyte?
AΛm − Λm° + A√C = 0
BΛm²C + KaΛm°² − KaΛmΛm° = 0
CΛm − Λm° − A√C = 0
DΛm = Λm°
✅ Correct: B
For weak electrolytes, Kohlrausch law (Λm = Λm° − A√C) is not obeyed. Ostwald's dilution law gives: \(K_a = \dfrac{\alpha^2C}{1-\alpha}\) where \(\alpha = \Lambda_m/\Lambda_m°\). Substituting and rearranging gives the correct relationship: \(\mathbf{\Lambda_m^2C + K_a\Lambda_m°^2 - K_a\Lambda_m\Lambda_m° = 0}\).
Q7
A galvanic cell can be converted into an electrolytic cell by:
AApplying an external opposite potential greater than E°cell
BInterchanging anode and cathode
CDiluting the electrolyte
DIncreasing temperature
✅ Correct: A
By applying an external EMF greater than the cell EMF in the opposite direction, we force the non-spontaneous reaction to occur. This converts the galvanic cell into an electrolytic cell. This is the principle of electrolysis and charging a battery.
Q8
Standard electrode potential of Zn²⁺/Zn is −0.76 V and Cu²⁺/Cu is +0.34 V. E°cell for Zn−Cu cell is:
A−0.42 V
B+0.42 V
C−1.10 V
D+1.10 V
✅ Correct: D
E°cell = E°cathode − E°anode = E°(Cu²⁺/Cu) − E°(Zn²⁺/Zn) = 0.34 − (−0.76) = \(\mathbf{+1.10\text{ V}}\). Zn is anode (more negative), Cu is cathode.
Q9
The conductivity of a solution of concentration C mol/L is σ. The molar conductivity Λm is:
Aσ/C
BσC
C1000σ/C
Dσ/1000C
✅ Correct: C
Molar conductivity: \(\Lambda_m = \dfrac{\kappa}{C(\text{mol/L})} \times 1000\) (if κ in S/cm and C in mol/L). \(\Lambda_m = \mathbf{1000\sigma/C}\) in S·cm²/mol.
Q10
Faraday's second law of electrolysis states that when the same quantity of electricity is passed through different electrolytes, the masses deposited are:
AEqual for all electrolytes
BProportional to atomic masses
CProportional to equivalent weights
DInversely proportional to valency
✅ Correct: C
Faraday's second law: masses deposited by same charge are proportional to their equivalent weights (M/n). So heavier equivalent weight means more mass deposited.
Q11
Specific conductance (κ) of a solution of cell constant 1.5 cm⁻¹ and resistance 50 Ω is:
In Daniel cell (Zn-Cu), which statement is correct?
AZn electrode is cathode
BElectrons flow from Cu to Zn in external circuit
CZn dissolves at anode; Cu deposits at cathode
DCu²⁺ ions move toward Zn electrode
✅ Correct: C
Daniel cell: Zn anode (Zn→Zn²⁺+2e⁻), Cu cathode (Cu²⁺+2e⁻→Cu). Zn dissolves at anode; Cu deposits at cathode. Electrons flow from Zn to Cu in external circuit.
Q13
The relationship between standard Gibbs energy change and standard EMF of a cell is:
AΔG° = nFE°
BΔG° = −nFE°
CΔG° = RT ln E°
DΔG° = E°/nF
✅ Correct: B
\(\Delta G° = \mathbf{-nFE°}\). Spontaneous cell: E° > 0 → ΔG° < 0 (spontaneous). At equilibrium: E = 0, ΔG = 0. ΔG° is also related to K by: ΔG° = −RT ln K.
Q14
Which metal has the highest standard reduction potential and is therefore least reactive?
AZinc
BIron
CGold
DSilver
✅ Correct: C
Standard reduction potentials: Au³⁺/Au = +1.50 V (highest), Ag⁺/Ag = +0.80 V, Fe²⁺/Fe = −0.44 V, Zn²⁺/Zn = −0.76 V. Higher reduction potential = more easily reduced = least reactive. Gold is the least reactive.
Q15
Potassium dichromate (K₂Cr₂O₇) acts as oxidising agent in acidic medium. Change in oxidation state of Cr is:
A+6 to +2
B+7 to +2
C+6 to +3
D+7 to +4
✅ Correct: C
In K₂Cr₂O₇, Cr is in +6 oxidation state. In acidic medium: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. Cr changes from +6 to +3, gaining 3 electrons per Cr atom. n-factor of K₂Cr₂O₇ = 6.
Q16
Molar conductivity of NaCl at infinite dilution is 126.4 S·cm²/mol and that of BaCl₂ is 280 S·cm²/mol. Molar conductivity of Ba²⁺ ion: (λ°Cl⁻ = 76.4, λ°Na⁺ = 50)
In an electrolytic cell for purification of copper, the anode is:
APure copper
BImpure copper
CPlatinum
DCarbon
✅ Correct: B
In electrolytic refining of copper: impure copper is the anode (it dissolves), pure copper is the cathode (pure Cu deposits), and CuSO₄ solution is the electrolyte. Noble metal impurities (Au, Ag) fall to the bottom as anode mud.
Q18
Which of the following is NOT an electrochemical series (activity series) fact?
AMetals above H displace H from dilute acids
BA more reactive metal displaces less reactive metal from its salt solution
CF₂ is the strongest oxidising agent
DLi is the weakest reducing agent
✅ Correct: D
Li has the most negative standard reduction potential (−3.05 V) → it is the strongest reducing agent, not weakest. This is due to its high hydration energy. The statement that Li is the weakest reducing agent is incorrect.
Q19
The pH of a buffer solution made with 0.1 M acetic acid and 0.1 M sodium acetate is: (pKa of acetic acid = 4.74)