CBSE Class 12 Chemistry Electrochemistry MCQs Set 08

Question 1. In an electrochemical process, a salt bridge is used
(a) as a reducing agent
(b) as an oxidising agent
(c) to complete the circuit so that current can flow
(d) None of the options
Answer: (c) to complete the circuit so that current can flow
In simple words: The salt bridge serves as an electrical connection between the two halves of a battery, allowing the current to complete its loop and flow freely.

Exam Tip: Remember that a salt bridge completes the electrical circuit and keeps the solutions electrically neutral by providing counter-ions.

Question 2. Four half reactions I to IV are shown below. I. \( 2\text{Cl}^{-} \longrightarrow \text{Cl}_2 + 2e^{-} \) II. \( 4\text{OH}^{-} \longrightarrow \text{O}_2 + 2\text{H}_2\text{O} + 2e^{-} \) III. \( \text{Na}^{+} + e^{-} \longrightarrow \text{Na} \) IV. \( 2\text{H}^{+} + 2e^{-} \longrightarrow \text{H}_2 \). Which two of these reactions are most likely to occur when concentrated brine is electrolysed?
(a) I and III
(b) I and IV
(c) II and III
(d) II and IV
Answer: (b) I and IV
In simple words: When you electrolyze saltwater, chlorine gas is produced at the anode, and hydrogen gas is produced at the cathode because hydrogen is much easier to reduce than sodium.

Exam Tip: Understand the concept of overpotential of oxygen to explain why chlorine gas is liberated at the anode instead of oxygen during brine electrolysis.

Question 3. Standard electrode potential for \( \text{Sn}^{4+} / \text{Sn}^{2+} \) couple is +0.15 V and that for the \( \text{Cr}^{3+} / \text{Cr} \) couple is -0.74 V. The two couples in their standard states are connected to make a cell. The cell potential will be
(a) +1.19 V
(b) +0.89 V
(c) +0.18 V
(d) +1.83 V
Answer: (b) +0.89 V
In simple words: The electrode with the larger positive number is the cathode, and the other is the anode. Subtract the anode's potential from the cathode's to get the total cell potential.

Exam Tip: Always use the formula \( E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \), and remember that the electrode with the higher reduction potential acts as the cathode.

Question 4. At the time of equilibrium inside the Daniell cell, Nernst equation can be expressed as
(a) \( 0 = E^{\circ}_{\text{cell}} - \frac{2.303 RT}{nF} \log \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \)
(b) \( E_{\text{cell}} = 0 - \frac{2.303 RT}{nF} \log \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \)
(c) \( 0 = E^{\circ}_{\text{cell}} - \frac{2.303 RT}{nF} \log \frac{[\text{Cu}^{2+}]}{[\text{Zn}^{2+}]} \)
(d) \( E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{2.303 RT}{nF} \log \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \)
Answer: (a) \( 0 = E^{\circ}_{\text{cell}} - \frac{2.303 RT}{nF} \log \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \)
In simple words: At equilibrium, a battery stops producing voltage, meaning \( E_{\text{cell}} = 0 \). Plugging this into the Nernst equation gives the correct formula.

Exam Tip: Remember that at equilibrium, the cell potential \( E_{\text{cell}} \) is always 0, while the standard cell potential \( E^{\circ}_{\text{cell}} \) remains non-zero.

Question 5. Which of the following cell was used in Apollo space programme?
(a) Mercury cell
(b) \( \text{H}_2-\text{O}_2 \) fuel cell
(c) Dry cell
(d) Ni-Cd cell
Answer: (b) \( \text{H}_2-\text{O}_2 \) fuel cell
In simple words: The hydrogen-oxygen fuel cell was chosen for the Apollo space missions because it creates electricity and produces water as a byproduct.

Exam Tip: Fuel cells are highly efficient and eco-friendly because their only byproduct is pure water

Question 6. Calculate the equilibrium constant for the reaction, \( \text{Zn}(s) + \text{Cu}^{2+}(aq) \longrightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s) \) [Given, \( E^{\circ}_{\text{cell}} = 1.1 \text{ V} \)]
(a) \( 2 \times 10^{32} \)
(b) \( 2 \times 10^{34} \)
(c) \( 2 \times 10^{37} \)
(d) \( 2 \times 10^{39} \)
Answer: (c) \( 2 \times 10^{37} \)
In simple words: Using the equilibrium constant formula, we find \( \log K_c \) is about 37.3. Taking the antilog gives \( 2 \times 10^{37} \).

Exam Tip: Be careful with the number of electrons \( n = 2 \) transferred in this reaction when calculating equilibrium constants.

Question 7. In the given reaction, \( 2\text{Cu}^{+}(aq) \rightleftharpoons \text{Cu}^{2+}(aq) + \text{Cu}(s) \) \( E^{\circ}_{\text{Cu}^{+}/\text{Cu}} = 0.62 \text{ V} \) and \( E^{\circ}_{\text{Cu}^{2+}/\text{Cu}^{+}} = 0.26 \text{ V} \). Find out the equilibrium constant.
(a) \( 2.76 \times 10^2 \)
(b) \( 1.2 \times 10^4 \)
(c) \( 1.2 \times 10^6 \)
(d) \( 2.76 \times 10^8 \)
Answer: (c) \( 1.2 \times 10^6 \)
In simple words: Determine the cell potential first by subtracting 0.26 V from 0.62 V. Since one electron moves in this reaction, use the formula to find the equilibrium constant is \( 1.2 \times 10^6 \).

Exam Tip: For this reaction, the number of moles of electrons transferred \( n \) is 1, not 2, which is a common point of confusion.

Question 8. A voltaic cell is made by connecting two half cells represented by half equations below \( \text{Sn}^{2+}(aq) + 2e^{-} \longrightarrow \text{Sn}(s); E^{\circ} = -0.14 \text{ V} \) \( \text{Fe}^{3+}(aq) + e^{-} \longrightarrow \text{Fe}^{2+}(aq); E^{\circ} = +0.77 \text{ V} \). Which statement is correct about this voltaic cell?
(a) \( \text{Fe}^{2+} \) is oxidised and the voltage of the cell is -0.91 V.
(b) Sn is oxidised and the voltage of the cell is 0.91 V.
(c) \( \text{Fe}^{2+} \) is oxidised and the voltage of the cell is 0.91 V.
(d) Sn is oxidised and the voltage of the cell is 0.63 V.
Answer: (b) Sn is oxidised and the voltage of the cell is 0.91 V.
In simple words: Since iron has a higher positive potential, it gets reduced while tin gets oxidized. Subtracting the lower potential from the higher potential gives a positive cell voltage of 0.91 V.

Exam Tip: Always identify the cathode as the half-cell with the higher reduction potential to ensure a positive, spontaneous cell voltage.

Question 9. Which metal cannot replace \( \text{H}_2 \) from hydrochloric acid?
(a) Zn
(b) Cu
(c) Mg
(d) Al
Answer: (b) Cu
In simple words: Metals with standard reduction potentials greater than zero, like copper, cannot react with dilute acids to release hydrogen gas.

Exam Tip: Remember that any metal with a positive standard reduction potential (below hydrogen in the activity series) cannot displace hydrogen gas from acids.

Question 10. Arrange the following species in the increasing order of their reducing power on the basis of standard electrode potential values given below. (I) \( \text{K}^{+} / \text{K} = -2.93 \text{ V} \) (II) \( \text{Ni}^{2+} / \text{Ni} = -0.25 \text{ V} \) (III) \( \text{Ag}^{+} / \text{Ag} = 0.80 \text{ V} \)
(a) I < II < III
(b) II < III < I
(c) III < II < I
(d) III < I < II
Answer: (c) III < II < I
In simple words: The more negative the standard electrode potential, the easier it is for the metal to lose electrons and act as a strong reducing agent.

Exam Tip: A lower (more negative) reduction potential always corresponds to a stronger reducing power.

Question 11. Which of the following option will be the limiting molar conductivity of \( \text{CH}_3\text{COOH} \) if the limiting molar conductivity of \( \text{CH}_3\text{COONa} \) is \( 91 \text{ S cm}^2\text{ mol}^{-1} \)? Limiting molar conductivity for individual ions are given in the following table.

(a) \( 350 \text{ S cm}^2\text{ mol}^{-1} \)
(b) \( 375.3 \text{ S cm}^2\text{ mol}^{-1} \)
(c) \( 390.5 \text{ S cm}^2\text{ mol}^{-1} \)
(d) \( 340.4 \text{ S cm}^2\text{ mol}^{-1} \)
Answer: (c) \( 390.5 \text{ S cm}^2\text{ mol}^{-1} \)
In simple words: Using Kohlrausch's law, we add the conductivity of hydrogen ions and subtract sodium ions from sodium acetate to find the value for acetic acid.

Exam Tip: Kohlrausch's law allows us to determine the limiting molar conductivity of weak electrolytes from those of strong electrolytes by adding and subtracting appropriate ionic conductivities.

Question 12. Dilution affects both conductivity as well as molar conductivity. Effect of dilution on both is as follows.
(a) Both increase with dilution.
(b) Both decreases with dilution.
(c) Conductivity increases whereas molar conductivity decreases on dilution.
(d) Conductivity decreases whereas molar conductivity increases on dilution.
Answer: (d) Conductivity decreases whereas molar conductivity increases on dilution.
In simple words: Diluting a solution reduces the number of ions in each milliliter, lowering regular conductivity, but it lets ions move more freely, which increases molar conductivity.

Exam Tip: Always remember that conductivity is defined per unit volume, so it decreases on dilution, while molar conductivity depends on the total number of ions produced from one mole of electrolyte, which increases.

Question 13. There are two beakers ‘A’ and ‘B’ containing KCl and \( \text{CH}_3\text{COOH} \) solutions, respectively. On adding water to beakers A and B, which of the following change in \( \Lambda_{\text{m}} \) of the solutions will be correct?
(a) It increases sharply in beaker A and slowly in beaker B.
(b) It increases slowly in beaker A and sharply in beaker B.
(c) It decreases in beaker A but no change in beaker B.
(d) There is no change in beaker A but it decreases slowly in beaker B.
Answer: (b) It increases slowly in beaker A and sharply in beaker B.
In simple words: Weak electrolytes dissociate much more when diluted, causing a sharp rise in conductivity, while strong electrolytes are already fully broken down and show only a small increase.

Exam Tip: Identify KCl as a strong electrolyte (linear Debye-Huckel-Onsager plot) and acetic acid as a weak electrolyte (steep curve near zero concentration) to understand their dilution behaviors.

Question 14. Kohlrausch give the following relation for strong electrolytes. \( \Lambda_{\text{m}} = \Lambda^{\circ}_{\text{m}} - A\sqrt{C} \). Which of the following equality holds?
(a) \( \Lambda_{\text{m}} = \Lambda^{\circ}_{\text{m}} \) as \( C \longrightarrow \sqrt{A} \)
(b) \( \Lambda_{\text{m}} = \Lambda^{\circ}_{\text{m}} \) as \( C \longrightarrow \infty \)
(c) \( \Lambda_{\text{m}} = \Lambda^{\circ}_{\text{m}} \) as \( C \longrightarrow 0 \)
(d) \( \Lambda_{\text{m}} = \Lambda^{\circ}_{\text{m}} \) as \( C \longrightarrow 1 \)
Answer: (c) \( \Lambda_{\text{m}} = \Lambda^{\circ}_{\text{m}} \) as \( C \longrightarrow 0 \)
In simple words: As concentration drops to zero, molar conductivity reaches its maximum value, called the limiting molar conductivity.

Exam Tip: The extrapolation of molar conductivity to zero concentration to find limiting molar conductivity is valid only for strong electrolytes.

Question 15. Which of the following solutions of KCl will have the highest value of molar conductivity?
(a) 0.01 M
(b) 1 M
(c) 0.5 M
(d) 0.1 M
Answer: (a) 0.01 M
In simple words: Molar conductivity increases as a solution becomes more dilute. Thus, the solution with the smallest concentration, 0.01 M, has the highest molar conductivity.

Exam Tip: Molar conductivity always increases with a decrease in concentration for both strong and weak electrolytes.

Question 16. The molar ionic conductivities of \( \text{Al}^{3+} \) and \( \text{SO}_4^{2-} \) are \( 189 \text{ S cm}^2\text{ mol}^{-1} \) and \( 160 \text{ S cm}^2\text{ mol}^{-1} \) respectively. The value of limiting molar conductivity of \( \text{Al}_2(\text{SO}_4)_3 \) will be
(a) \( 198 \text{ S cm}^2\text{ mol}^{-1} \)
(b) \( 858 \text{ S cm}^2\text{ mol}^{-1} \)
(c) \( 588 \text{ S cm}^2\text{ mol}^{-1} \)
(d) \( 891 \text{ S cm}^2\text{ mol}^{-1} \)
Answer: (b) \( 858 \text{ S cm}^2\text{ mol}^{-1} \)
In simple words: Multiply the aluminium ion conductivity by 2 and the sulfate ion conductivity by 3, then add them together to find the total limiting molar conductivity.

Exam Tip: Always write down the balanced dissociation equation of the salt to determine the correct stoichiometric coefficients for each ion.

Question 17. Match the following Column I with Column II and choose the correct option using the codes given below.

Choose the correct option among the given options.
Codes:

Answer: (a) A-(ii), B-(i), C-(iv), D-(iii)
In simple words: First law connects weight and charge (\( W \propto Q \)), second law connects weight ratios to equivalents, Faraday's constant is \( 96500 \text{ C/mol} \), and charge \( Q \) equals current times time.

Exam Tip: Understanding the basic definitions of electrochemical terms makes matching and direct numerical calculations very straightforward.

Question 18. The quantity of charge required to obtain one mole of aluminium from \( \text{Al}_2\text{O}_3 \) is
(a) 1 F
(b) 3 F
(c) 6 F
(d) 2 F
Or
The charge required for the reduction of 1 mol of \( \text{Al}^{3+} \) to Al will be
(a) 1 F
(b) 3 F
(c) 27 F
(d) 13 F
Answer: (b) 3 F
In simple words: Reducing one aluminium ion requires three electrons. Therefore, producing one mole of aluminium requires three moles of electrons, which equals 3 Faradays of charge.

Exam Tip: The charge required for reducing any ion to its elemental state is always equal to \( n \times \text{F} \), where \( n \) is the valency of the metal ion.

Question 19. An electrolytic cell contains alumina. If we have to obtain 50 g Al by using 105 A of current, the time required is
(a) 1.54 h
(b) 2.15 h
(c) 1.42 h
(d) 1.32 h
Answer: (c) 1.42 h
In simple words: Using Faraday's law with the equivalent weight of aluminium (9), we calculate the total seconds needed to deposit 50 g of Al, which comes out to 1.42 hours.

Exam Tip: Always double-check your units when calculating electrolysis time, ensuring you convert seconds to hours by dividing by 3600.

Question 20. 1.5 A current is flowing through a metallic wire. If it flows for 3 h, how many electrons would flow through the wire?
(a) \( 2.05 \times 10^{22} \) electrons
(b) \( 1.0 \times 10^{23} \) electrons
(c) \( 10^{24} \) electrons
(d) \( 4.5 \times 10^{23} \) electrons
Answer: (b) \( 1.0 \times 10^{23} \) electrons
In simple words: First, multiply the current by the total seconds in 3 hours to find the total charge in Coulombs. Then, divide by the charge of one electron to find the total number of electrons.

Exam Tip: Ensure that time is converted into seconds before using the charge equation \( Q = It \).

Question 21. 6 A current with 75% efficiency is passed through a cell for 6 h? (Z = \( 4 \times 10^{-4} \)). The amount of metal deposited will be
(a) 22.4 g
(b) 36.0 g
(c) 32.4 g
(d) 38.8 g
Answer: (d) 38.8 g
In simple words: Multiply the electrochemical equivalent by the actual current used (75% of 6 A) and the total seconds in 6 hours to find the weight of the metal deposited.

Exam Tip: When current efficiency is less than 100%, always multiply the total current by the efficiency percentage before calculating the product.

Question 22. Anode in the Leclanche cell is
(a) zinc container
(b) graphite electrode
(c) carbon
(d) \( \text{MnO}_2 + \text{C} \)
Answer: (a) zinc container
In simple words: In a standard dry cell battery, the outer zinc metal case serves as the anode.

Exam Tip: In dry cells, remember that zinc undergoes oxidation at the anode, while the graphite rod surrounded by manganese dioxide serves as the cathode.

Question 23. In a lead storage battery
(a) \( \text{PbO}_2 \) is reduced to \( \text{PbSO}_4 \) at the cathode
(b) Pb is oxidised to \( \text{PbSO}_4 \) at the anode
(c) Both electrodes are immersed in the same aqueous solution of \( \text{H}_2\text{SO}_4 \)
(d) All of the options
Answer: (d) All of the options
In simple words: A lead storage battery uses sulfuric acid as its electrolyte, where lead is oxidized at the anode and lead dioxide is reduced at the cathode, both forming lead sulfate.

Exam Tip: During discharging of a lead storage battery, sulfuric acid is consumed, which is why its density decreases.

Question 24. How many Faradays are required to reduce 1 mole of \( \text{MnO}_4^{-} \) to \( \text{Mn}^{2+} \)?
(a) 4
(b) 3
(c) 6
(d) 5
Answer: (d) 5
In simple words: The oxidation state of manganese changes from +7 in permanganate to +2. This five-electron gain requires exactly 5 Faradays of charge.

Exam Tip: To find the number of Faradays required, write down the oxidation states of the key element on both sides and calculate the difference.

Assertion-Reason

Directions (Q. Nos. 27-39) In the following questions an Assertion (A) is followed by a corresponding Reason (R). Use the following keys to choose the appropriate answer.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.

Question. Assertion (A) 96500 C charge is required for the reduction of one mole of silver ions.
Reason (R) The amount of electricity (or charge) required for oxidation or reduction depends on the stoichiometry of the electrode reaction.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
In simple words: One silver ion needs one electron, so one mole of silver ions needs one mole of electrons, which is 96500 Coulombs. The stoichiometry directly explains this charge requirement.

Exam Tip: Always write down the balanced ionic equation to determine the stoichiometry and calculate the Faraday charge needed.

Question. Assertion (A) \( E_{\text{Ag}^{+}/\text{Ag}} \) increases with increase in concentration of \( \text{Ag}^{+} \) ions.
Reason (R) \( E_{\text{Ag}^{+}/\text{Ag}} \) has a positive value.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
In simple words: The potential increases because of the logarithmic relation with the concentration in the Nernst equation, not simply because the standard potential is positive.

Exam Tip: Remember that the change in potential with concentration is governed by the Nernst equation, regardless of whether the standard potential is positive or negative.

Question. Assertion (A) \( E_{\text{cell}} \) should have positive value for the cell to function.
Reason (R) \( E_{\text{anode}} > E_{\text{cathode}} \).
Answer: (c) (A) is true, but (R) is false.
In simple words: A cell can only work spontaneously if its overall potential is positive, which requires the cathode's potential to be higher than the anode's potential.

Exam Tip: For any working galvanic cell, remember that \( \Delta G < 0 \), which mathematically dictates that \( E_{\text{cell}} \) must be positive.

Question. Assertion (A) Copper is less reactive than zinc and hydrogen.
Reason (R) The value of \( E^{\circ}_{\text{Cu}^{2+}/\text{Cu}} \) is negative.
Answer: (c) (A) is true, but (R) is false.
In simple words: Copper is less reactive because it does not lose electrons as easily as zinc or hydrogen. Its standard reduction potential is positive, so the reason is incorrect.

Exam Tip: Remember that a positive standard reduction potential indicates that the species is easily reduced and is less reactive as a metal compared to hydrogen.

Question. Assertion (A) For a Daniell cell, \( \text{Zn} | \text{Zn}^{2+}(1 \text{ M}) || \text{Cu}^{2+}(1 \text{ M}) | \text{Cu} \) with \( E^{\circ}_{\text{cell}} = 1.1 \text{ V} \), if the external opposing potential is more than 1.1 V, the electrons flow from Cu to Zn.
Reason (R) Cell acts like a galvanic cell.
Answer: (c) (A) is true, but (R) is false.
In simple words: An opposing voltage higher than 1.1 V forces the chemical reaction to run backward, turning the galvanic cell into an electrolytic cell.

Exam Tip: When \( E_{\text{ext}} > E_{\text{cell}} \), the cell no longer generates electricity but consumes it, behaving as an electrolytic cell.

Question. Assertion (A) For measuring resistance of an ionic solution an AC source is used.
Reason (R) Concentration of ionic solution will change if DC source is used.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
In simple words: Using direct current would cause a chemical reaction (electrolysis) that changes the solution's concentration. Alternating current prevents this, allowing us to measure resistance accurately.

Exam Tip: An AC source in a Wheatstone bridge setup prevents electrolysis and polarization at the electrodes when measuring ionic solution resistance.

Question. Assertion (A) For strong electrolytes, there is a slow increase in molar conductivity with dilution and can be represented by the equation \( \Lambda^{\circ}_{\text{m}} = \Lambda_{\text{m}} - AC^{1/2} \).
Reason (R) The value of the constant ‘A’ for NaCl, \( \text{CaCl}_2 \) and \( \text{MgSO}_4 \) in a given solvent at a given temperature is different.
Answer: (d) (A) is false, but (R) is true.
In simple words: The formula given in the assertion is written backward; the correct equation is \( \Lambda_{\text{m}} = \Lambda^{\circ}_{\text{m}} - AC^{1/2} \). The constant \( A \) depends on the ion charges, which are different for these salts, so the reason is correct.

Exam Tip: Be very careful with the positions of \( \Lambda_{\text{m}} \) and \( \Lambda^{\circ}_{\text{m}} \) in the Debye-Huckel-Onsager equation.

Question. Assertion (A) Conductivity decreases with decrease in concentration of electrolyte.
Reason (R) Number of ions per unit volume that carry the current in a solution decreases on dilution.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
In simple words: Lowering the concentration means there are fewer ions in each milliliter of solution to carry the current, which directly causes the conductivity to decrease.

Exam Tip: Conductivity always decreases with dilution because the number of ions per unit volume decreases, regardless of whether the electrolyte is strong or weak.

Question. Assertion (A) \( \Lambda_{\text{m}} \) for weak electrolytes shows a sharp increase when the electrolytic solution is diluted.
Reason (R) For weak electrolytes, degree of dissociation increases with dilution of solution.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
In simple words: Diluting a weak electrolyte makes it break down into many more ions, which causes a sudden, sharp rise in its molar conductivity.

Exam Tip: Remember that weak electrolytes do not dissociate completely at higher concentrations, but approach 100% dissociation as dilution approaches infinity.

Question. Assertion (A) Electrolysis of aqueous solution of NaCl gives chlorine gas at anode instead of oxygen gas.
Reason (R) Formation of oxygen gas at anode requires overpotential.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
In simple words: Making oxygen gas at the anode is slow and needs extra voltage (overpotential), so chlorine gas is produced instead.

Exam Tip: Overpotential (or overvoltage) is the extra potential required to overcome kinetic barriers for slow gas evolution reactions like that of oxygen.

Question. Assertion (A) Mercury cells give a constant voltage throughout its life.
Reason (R) Electrolyte KOH is not involved in the reaction.
Answer: (c) (A) is true, but (R) is false.
In simple words: The overall reaction does not change the concentration of any ions in the cell, which keeps the voltage steady. Since KOH is the electrolyte used in the reaction, the reason is false.

Exam Tip: Understand that the constant voltage of a mercury cell is due to the absence of any ionic species in the overall cell reaction that can change concentration during discharge.

Question. Assertion (A) \( \text{H}_2-\text{O}_2 \) fuel cell gives a constant voltage throughout its life.
Reason (R) In this fuel cell, \( \text{H}_2 \) reacts with \( \text{OH}^{-} \) ions yet the overall concentration of \( \text{OH}^{-} \) ions does not change.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
In simple words: Hydroxide ions are consumed at one electrode and produced at the other in equal amounts, keeping the concentration constant and the voltage steady.

Exam Tip: The continuous supply of reactants from outside in a fuel cell, combined with a constant electrolyte concentration, ensures uninterrupted and stable power generation.

Question. Assertion (A) Galvanised iron does not rust.
Reason (R) Zinc has a more negative electrode potential than iron.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
In simple words: Zinc is more reactive than iron and has a more negative potential, so it rusts first (sacrifices itself) to keep the iron underneath safe.

Exam Tip: Cathodic protection or sacrificial protection works because the coating metal has a lower (more negative) reduction potential than the protected metal.

Case-Study

40. Read the passage given below and answer the following questions.

Oxidation-reduction reactions are commonly known as redox reactions. They involve transfer of electrons from one species to another. In a spontaneous reaction, energy is released which can be used to do useful work. The reaction is split into two half reactions. Two different containers are used and a wire is used to drive the electrons from one side to the other and a Voltaic/Galvanic cell is created. In this cell, chemical energy of a redox reaction is converted into electrical energy. The simplest galvanic cell is in which Zn rod is dipped in a solution of \( \text{ZnSO}_4 \) and Cu rod is placed in a solution of \( \text{CuSO}_4 \). The two rods are connected by a metallic wire through a voltmeter. The two solutions are joined by a salt bridge. The difference between the two electrode potentials of the two electrodes is known as electromotive force. If \( E^{\circ}_{\text{cell}} \) is positive, the reaction is spontaneous and if it is negative, the reaction is non-spontaneous and is referred to as electrolytic cell. The process of electrolysis refers to the decomposition of a substance by an electric current. One mole of electric charge when passed through a cell will discharge half a mole of a divalent metal ion such as \( \text{Cu}^{2+} \). This was first formulated by Faraday in the form of laws of electrolysis. The conductance is the property of materials due to which a material allows the flow of ions through itself and thus conducts electricity. Conductivity is represented by \( \kappa \) and it depends upon nature and concentration of electrolyte, temperature, etc.

Question (i) (a) What will happen to the concentration of \( \text{Zn}^{2+} \) and \( \text{Cu}^{2+} \), when \( E_{\text{cell}} = 0 \)?
(b) Why does conductivity of a solution decreases with dilution?
Answer:
(a) The concentration of \( \text{Zn}^{2+} \) will increase while the concentration of \( \text{Cu}^{2+} \) will decrease until the system reaches chemical equilibrium.
(b) The conductivity of an electrolyte decreases upon dilution because the total number of current-carrying ions per unit volume in the solution is reduced.
In simple words: (a) Zn2+ concentration goes up and Cu2+ concentration goes down until equilibrium is established. (b) Dilution reduces the number of ions in each unit of volume, which lowers conductivity.

Exam Tip: Remember that concentration of ions per unit volume directly determines conductivity, which is why dilution always decreases conductivity.

Question (ii) Which plate is going to act as cathode of the cell?
(a) Zn
(b) Cu
(c) Both of these
(d) None of these
Answer: (b) Cu
In simple words: The copper electrode has a higher standard reduction potential, meaning it undergoes reduction and serves as the cathode of the cell.

Exam Tip: In a Zn-Cu galvanic cell, zinc always acts as the anode (undergoes oxidation) and copper acts as the cathode (undergoes reduction).

Question (iii) When does electrochemical cell behaves like an electrolytic cell?
(a) \( E_{\text{ext}} > E_{\text{cell}} \)
(b) \( E_{\text{ext}} < E_{\text{cell}} \)
(c) \( E_{\text{ext}} = E_{\text{cell}} \)
(d) \( E_{\text{ext}} = 0 \)
Answer: (a) \( E_{\text{ext}} > E_{\text{cell}} \)
In simple words: When the external voltage applied is greater than the cell's own voltage, the current reverses, and the cell operates as an electrolytic cell.

Exam Tip: If the external voltage matches the cell voltage, the current stops completely; if it exceeds it, the reaction is forced backward.

Question (iii) Or How much charge in terms of Faraday required for the conversion of 1 mol of \( \text{H}_2\text{O} \) to \( \text{O}_2 \)?
(a) 1 F
(b) 2 F
(c) 3 F
(d) 4 F
Answer: (b) 2 F
In simple words: Oxidation of water to form oxygen releases 2 moles of electrons for every mole of water, which requires 2 Faradays of charge.

Exam Tip: Write the balanced oxidation half-reaction of water to easily find the number of electrons transferred per mole of reactant.

41. Read the passage given below and answer the following questions.

Rahul set-up an experiment to find resistance of aqueous KCl solution for different concentrations at 298 K using a conductivity cell connected to a Wheatstone bridge. He fed the Wheatstone bridge with a.c. power in the audio frequency range 550 to 5000 cycles per second. Once the resistance was calculated from null point he also calculated the conductivity (\( \kappa \)) and molar conductivity (\( \Lambda_{\text{m}} \)) and recorded his readings in tabular form.

Question (i) Why does conductivity decrease with dilution?
Answer: Conductivity decreases upon dilution because the number of current-carrying ions present per unit volume of the solution decreases.
In simple words: Diluting the solution reduces the concentration of ions in any given volume, which directly lowers the solution's conductivity.

Exam Tip: Molar conductivity increases with dilution, but regular conductivity always decreases because it is defined per unit volume.

Question (ii) If \( \Lambda^{\circ}_{\text{m}} \) of KCl is \( 150.0 \text{ S cm}^2\text{ mol}^{-1} \). Calculate the degree of dissociation of 0.01 M KCl.
Answer: Given data:
Molarity of KCl \( = 0.01 \text{ M} \)
From the table, the conductivity \( \kappa \) of 0.01 M KCl is \( 1.41 \times 10^{-3} \text{ S cm}^{-1} \).
First, we calculate the molar conductivity \( \Lambda_{\text{m}} \):
\( \Lambda_{\text{m}} = \frac{1000 \times \kappa}{\text{Molarity}} = \frac{1000 \times 1.41 \times 10^{-3} \text{ S cm}^{-1}}{0.01 \text{ M}} = 141.0 \text{ S cm}^2\text{ mol}^{-1} \)
Now, the degree of dissociation \( \alpha \) is:
\( \alpha = \frac{\Lambda_{\text{m}}}{\Lambda^{\circ}_{\text{m}}} = \frac{141.0}{150.0} = 0.94 \)
In simple words: First find the molar conductivity using the concentration and the table value. Then, divide this by the limiting molar conductivity to get a degree of dissociation of 0.94.

Exam Tip: Ensure that the conductivity value from the table is matched correctly with the given concentration before running your calculations.

Question (iii) If Rahul had used HCl instead of KCl, then would you expect the \( \Lambda_{\text{m}} \) values to be more or less than those per KCl for a given concentration. Justify.
Answer: We would expect the molar conductivity \( \Lambda_{\text{m}} \) values for HCl to be higher than those for KCl. This is because the hydrogen ion \( (\text{H}^{+}) \) is much smaller in size and has a significantly higher ionic mobility and velocity compared to the potassium ion \( (\text{K}^{+}) \).
In simple words: The values would be higher because hydrogen ions are smaller and move much faster through the solution than potassium ions.

Exam Tip: Small ions with higher charge density (like \( \text{H}^{+} \)) have exceptionally high ionic mobilities in water due to Grotthuss proton-hopping mechanisms.

Question (iii) Or Amit, a classmate of Rahul repeated the same experiment with \( \text{CH}_3\text{COOH} \) solution instead of KCl solution. Give one point that would be similar and one that would be different in his observations as compared to Rahul.
Answer:
- Similarity: For both acetic acid and potassium chloride, the conductivity decreases with dilution, while the molar conductivity increases with dilution.
- Difference: Acetic acid is a weak electrolyte and does not dissociate completely. Thus, its molar conductivity will show a sharp, steep increase at very low concentrations, and its values will be significantly lower compared to the strong electrolyte KCl at higher concentrations.
In simple words: Both will show a decrease in conductivity as they are diluted. However, since acetic acid is weak, its molar conductivity will rise much more steeply at very high dilution compared to the strong electrolyte KCl.

Exam Tip: Be ready to compare strong and weak electrolytes using the Debye-Huckel-Onsager equation and the Kohlrausch law graphical trends.