Every Book Back multiple-choice question from Electro Chemistry (12th Standard Chemistry, Samacheer Kalvi) — each with the correct option highlighted and a clear, worked explanation. Free to read in English and Tamil.
Q1
The number of electrons that have a total charge of 9650 coulombs is
- A. \(6.22 \times 10^{23}\)
- B. \(6.022 \times 10^{24}\)
- C. \(6.022 \times 10^{22}\)Correct
- D. \(6.022 \times 10^{-34}\)
Explanation. One Faraday (96500 C) corresponds to the charge of one mole of electrons (\(6.022 \times 10^{23}\) electrons). Dividing 9650 C by the charge of a single electron (\(1.602 \times 10^{-19}\) C) yields \(6.022 \times 10^{22}\) electrons.
Q2
Consider the following half cell reactions:
\(Mn^{2+} + 2e^- \rightarrow Mn\) \(E^\circ = -1.18V\)
\(Mn^{3+} + e^- \rightarrow Mn^{2+}\) \(E^\circ = 1.51V\)
The \(E^\circ\) for the reaction \(3Mn^{2+} \rightarrow Mn + 2Mn^{3+}\) and the possibility of the forward reaction are respectively.
- A. 2.69V and spontaneous
- B. -2.69V and non spontaneousCorrect
- C. 0.33V and Spontaneous
- D. 4.18V and non spontaneous
Explanation. The target cell potential is calculated by adding the reduction potential of \(Mn^{2+}/Mn\) (-1.18V) and the oxidation potential of \(Mn^{2+}/Mn^{3+}\) (-1.51V), resulting in -2.69V. A negative cell potential indicates a non-spontaneous reaction.
Q3
The button cell used in watches function as follows
\(Zn (s) + Ag_2O (s) + H_2O (l) \rightleftharpoons 2 Ag (s) + Zn^{2+} (aq) + 2OH^- (aq)\)
the half cell potentials are \(Ag_2O (s) + H_2O (l) + 2e^- \rightarrow 2Ag (s) + 2 OH^- (aq)\) \(E^\circ = 0.34V\) and \(Zn^{2+} (aq) + 2e^- \rightarrow Zn (s)\) \(E^\circ = -0.76V\). The cell potential will be
- A. 0.84V
- B. 1.34V
- C. 1.10VCorrect
- D. 0.42V
Explanation. The standard cell potential is the sum of the standard reduction potential of the cathode (\(Ag_2O\) reduction, 0.34V) and the standard oxidation potential of the anode (Zinc oxidation, +0.76V). Thus, \(E^\circ_{cell} = 0.34 + 0.76 = 1.10V\).
Q4
The molar conductivity of a \(0.5 \text{ mol dm}^{-3}\) solution of \(AgNO_3\) with electrolytic conductivity of \(5.76 \times 10^{-3} \text{ S cm}^{-1}\) at 298 K is
- A. \(2.88 \text{ S cm}^2 \text{ mol}^{-1}\)
- B. \(11.52 \text{ S cm}^2 \text{ mol}^{-1}\)Correct
- C. \(0.086 \text{ S cm}^2 \text{ mol}^{-1}\)
- D. \(28.8 \text{ S cm}^2 \text{ mol}^{-1}\)
Explanation. Molar conductivity (\(\Lambda_m\)) is calculated using the formula \(\Lambda_m = (\kappa \times 1000) / M\). Substituting the given electrolytic conductivity (\(\kappa = 5.76 \times 10^{-3}\)) and molarity (M = 0.5) gives \((5.76 \times 10^{-3} \times 1000) / 0.5 = 11.52 \text{ S cm}^2 \text{ mol}^{-1}\).
Q5
Using the appropriate molar conductances at infinite dilution in water at 25 \(^\circ\)C: \(\Lambda_{KCl}^\circ = 149.9\), \(\Lambda_{KNO_3}^\circ = 145.0\), \(\Lambda_{HCl}^\circ = 426.2\), \(\Lambda_{NaOAc}^\circ = 91.0\), and \(\Lambda_{NaCl}^\circ = 126.5 \text{ S cm}^2 \text{ mol}^{-1}\). Calculate \(\Lambda_{HOAc}^\circ\).
- A. 517.2
- B. 552.7
- C. 390.7Correct
- D. 217.5
Explanation. Applying Kohlrausch's law, the molar conductivity of acetic acid (HOAc) is determined by the combination \(\Lambda_{HCl}^\circ + \Lambda_{NaOAc}^\circ - \Lambda_{NaCl}^\circ\). Using the values provided: \(426.2 + 91.0 - 126.5 = 390.7 \text{ S cm}^2 \text{ mol}^{-1}\).
Q6
Faraday constant is defined as
- A. charge carried by 1 electron
- B. charge carried by one mole of electronsCorrect
- C. charge required to deposit one mole of substance
- D. charge carried by \(6.22 \times 10^{10}\) electrons
Explanation. The Faraday constant (F) represents the total electric charge carried by one mole of electrons. It is approximately equal to 96485 coulombs per mole.
Q7
How many faradays of electricity are required for the following reaction to occur: \(MnO_4^- \rightarrow Mn^{2+}\)
- A. 5FCorrect
- B. 3F
- C. 1F
- D. 7F
Explanation. In the reduction of \(MnO_4^-\) to \(Mn^{2+}\), the oxidation state of Manganese changes from +7 to +2. This process requires a transfer of 5 electrons per manganese ion, corresponding to 5 Faradays per mole.
Q8
A current strength of 3.86 A was passed through molten Calcium oxide for 41 minutes and 40 seconds. The mass of Calcium in grams deposited at the cathode is (atomic mass of Ca is \(40 \text{ g/mol}\) and 1F = 96500 C).
- A. 4
- B. 2Correct
- C. 8
- D. 6
Explanation. Total charge Q = It = 3.86 A \(\times\) 2500 s = 9650 C. Since depositing 1 mole (40 g) of \(Ca^{2+}\) requires 2 Faradays (193000 C), the mass deposited by 9650 C is \((9650 / 193000) \times 40 = 2 \text{ g}\).
Q9
During electrolysis of molten sodium chloride, the time required to produce 0.1 mole of chlorine gas using a current of 3A is
- A. 55 minutes
- B. 107.2 minutesCorrect
- C. 220 minutes
- D. 330 minutes
Explanation. Producing 1 mole of \(Cl_2\) gas requires 2 Faradays (193000 C). Thus, 0.1 mole requires 19300 C. Using t = Q / I, time = 19300 C / 3 A = 6433.3 seconds, which is approximately 107.2 minutes.
Q10
The number of electrons delivered at the cathode during electrolysis by a current of 1A in 60 seconds is (charge of electron = \(1.6 \times 10^{-19} \text{ C}\))
- A. \(236.22 \times 10^{...}\)
- B. \(6.022 \times 10^{20}\)
- C. \(3.75 \times 10^{20}\)Correct
- D. \(7.48 \times 10^{23}\)
Explanation. Total charge Q = It = 1 A \(\times\) 60 s = 60 C. The number of electrons is Q divided by the elementary charge: 60 / (\(1.6 \times 10^{-19}\)) = \(3.75 \times 10^{20}\) electrons.
Q11
Which of the following electrolytic solution has the least specific conductance
- A. 2N
- B. 0.002NCorrect
- C. 0.02N
- D. 0.2N
Explanation. Specific conductance measures the ability of a volume of solution to conduct electricity. Since it depends on the number of ions per unit volume, increasing dilution reduces the ion concentration, resulting in lower specific conductance for the most dilute solution.
Q12
While charging lead storage battery
- A. \(PbSO_4\) on cathode is reduced to Pb
- B. \(PbSO_4\) on anode is oxidised to \(PbO_2\)
- C. \(PbSO_4\) on anode is reduced to PbCorrect
- D. \(PbSO_4\) on cathode is oxidised to Pb
Explanation. Charging a lead storage battery involves reversing the spontaneous discharge reactions using an external power source. At the electrode that functioned as the anode during discharge, lead sulphate is reduced back to metallic lead to restore the original state.
Q13
Among the following cells I) Leclanche cell II) Nickel – Cadmium cell III) Lead storage battery IV) Mercury cell Primary cells are
- A. I and IVCorrect
- B. I and III
- C. III and IV
- D. II and III
Explanation. Primary cells are batteries where the electrochemical reaction is irreversible, meaning they cannot be recharged. The Leclanche cell and the mercury button cell are typical primary cells, while lead storage and nickel-cadmium batteries are secondary, rechargeable types.
Q14
Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because
- A. Zinc is lighter than iron
- B. Zinc has lower melting point than iron
- C. Zinc has lower negative electrode potential than iron
- D. Zinc has higher negative electrode potential than ironCorrect
Explanation. Galvanization relies on zinc having a more negative standard reduction potential than iron. This makes zinc a stronger reducing agent that oxidizes preferentially, protecting the underlying iron. Iron cannot coat zinc because zinc is more chemically reactive.
Q15
Assertion : pure iron when heated in dry air is converted with a layer of rust. Reason : Rust has the composition \(Fe_3O_4\)
- A. if both assertion and reason are true and reason is the correct explanation of assertion.
- B. if both assertion and reason are true but reason is not the correct explanation of assertion.
- C. assertion is true but reason is false
- D. both assertion and reason are false.Correct
Explanation. The assertion is false because rusting requires moisture and carbon dioxide alongside oxygen; dry air alone won't produce rust. The reason is also false because the chemical composition of rust is hydrated ferric oxide, not magnetite.
Q16
In \(H_2-O_2\) fuel cell the reaction occurs at cathode is
- A. \(O_2(g) + 2H_2O(l) + 4e^- \rightarrow 4OH^-(aq)\)Correct
- B. \(H^+(aq) + OH^-(aq) \rightarrow H_2O(l)\)
- C. \(2H_2(g) + O_2(g) \rightarrow 2H_2O(g)\)
- D. \(H^+ + e^-- \rightarrow \frac{1}{2}H_2\)
Explanation. In an alkaline hydrogen-oxygen fuel cell, reduction takes place at the cathode. Gaseous oxygen molecules from the oxidant supply react with water and incoming electrons to produce hydroxide ions, which then migrate through the electrolyte to complete the circuit.
Q17
The equivalent conductance of M/36 solution of a weak monobasic acid is 6 mho \(cm^2\) equivalent –1 and at infinite dilution is 400 mho \(cm^2\) equivalent –1. The dissociation constant of this acid is
- A. \(1.25 \times 10^{-6}\)
- B. \(6.25 \times 10^{-6}\)Correct
- C. \(1.25 \times 10^{-4}\)
- D. \(56.25 \times 10^{-6}\)
Explanation. The dissociation constant of a weak acid is found using Ostwald's dilution law. First determine the degree of dissociation by taking the ratio of equivalent conductance to its value at infinite dilution, then calculate the constant using the provided molar concentration.
Q18
A conductivity cell has been calibrated with a 0.01M, 1:1 electrolytic solution (specific conductance (\(\kappa = 1.25 \times 10^{-3} S cm^{-1}\)) in the cell and the measured resistance was 800 \(\Omega\) at \(25^\circ C\). The cell constant is,
- A. \(10^{-1} cm^{-1}\)
- B. \(10^1 cm^{-1}\)
- C. \(1 cm^{-1}\)Correct
- D. \(125.7 \times 10^{-1}\)
Explanation. The cell constant is defined as the product of the solution's specific conductance and its measured resistance. Multiplying the given conductance value by the resistance measured at a specific temperature yields a constant value of unity for this particular cell setup.
Q19
Conductivity of a saturated solution of a sparingly soluble salt AB (1:1 electrolyte) at 298K is \(1.85 \times 10^{-5} S m^{-1}\). Solubility product of the salt AB at 298K \(\Lambda^\circ_m(AB) = 14 \times 10^{-3} S m^2 mol^{-1}\)
- A. \(125.7 \times 10^{-6}\)
- B. \(1.32 \times 10^{-12}\)
- C. \(7.5 \times 10^{-12}\)
- D. \(1.74 \times 10^{-12}\)Correct
Explanation. Solubility product calculation requires finding the molar solubility first by dividing the specific conductance by the molar conductivity at infinite dilution. For a simple 1:1 binary electrolyte, the solubility product constant equals the square of this molar concentration.
Q20
In the electrochemical cell: \(Zn | ZnSO_4 (0.01M) || CuSO_4 (1.0M) | Cu\), the emf of this Daniel cell is \(E_1\). When the concentration of \(ZnSO_4\) is changed to 1.0M and that \(CuSO_4\) changed to 0.01M, the emf changes to \(E_2\). From the above, which one is the relationship between \(E_1\) and \(E_2\)?
- A. \(E_1 < E_2\)
- B. \(E_1 > E_2\)Correct
- C. \(E_1 \geq E_2\)
- D. \(E_1 = E_2\)
Explanation. Applying the Nernst equation shows that cell potential depends on the concentration ratio of anode to cathode ions. Higher cathode concentration and lower anode concentration maximize voltage, making the first condition produce a significantly higher potential than the second.
