Redox Reactions and Electrochemistry MCQs with Answers – Part 2 (NEET Chemistry)

Explanation: Metallic conduction and electrolytic conduction involve different mobile charge carriers. In a copper wire, conduction occurs primarily through mobile electrons. The positively charged metal-ion cores remain approximately fixed in the metallic lattice. An applied electric field produces a net drift of conduction electrons through the wire. Therefore, Arrangement P exhibits metallic or electronic conduction. Molten sodium chloride contains mobile sodium and chloride ions. The molten state allows these ions to move under the applied electric field. Sodium ions move towards the negative electrode, while chloride ions move towards the positive electrode. Their oppositely directed motions both contribute to electric current. Therefore, Arrangement Q exhibits electrolytic or ionic conduction. In solid sodium chloride, the sodium and chloride ions are held at approximately fixed positions within the crystal lattice. They cannot migrate freely through the solid as they can in the molten state. Under the stated ordinary conditions, the solid crystal therefore exhibits no appreciable electrical conduction. The important distinction is not the presence or absence of charged particles. Both molten and solid sodium chloride contain ions. The difference is whether those ions possess sufficient mobility to transport charge through the material. Similarly, copper contains positively charged metal-ion cores but conducts mainly through mobile electrons. The correct classification must therefore consider both the identity and mobility of the available charge carriers.

Explanation: The effect of temperature depends on the mechanism of charge transport. In an ordinary metal, mobile electrons are responsible for conduction. As temperature increases, vibrations of the metal lattice generally become more pronounced. These vibrations increase scattering of conduction electrons. Greater scattering tends to oppose the orderly drift of electrons under an electric field. Consequently, the electrical resistance of an ordinary metallic conductor generally increases with temperature. For a metallic specimen of fixed geometry: \[ R=\rho\frac{l}{A} \] An increase in resistivity \(\rho\) causes an increase in resistance. Conductivity is the reciprocal of resistivity: \[ \kappa=\frac{1}{\rho} \] Thus, ordinary metallic conductivity generally decreases as temperature increases. In an electrolytic solution, mobile ions are responsible for conduction. Heating generally decreases solvent viscosity and increases ionic mobility. The ions can consequently move more readily under an applied electric field. The conductivity of an ordinary aqueous electrolyte therefore generally increases with temperature over a suitable interval. The question explicitly excludes chemical decomposition and precipitation, which could introduce additional effects. The opposing trends arise from the different mechanisms limiting charge transport. In metals, increased lattice scattering generally dominates. In electrolytic solutions, increased ionic mobility generally dominates. It would be incorrect to assume that higher temperature must produce the same conductivity trend in every type of conductor. The distinction is an important link between microscopic carrier behaviour and measurable electrical properties.

Explanation: The left electrode is connected to the negative terminal of the DC power supply. It is therefore the cathode of this electrolytic cell. The right electrode is connected to the positive terminal and acts as the anode. Molten sodium chloride contains mobile \(\mathrm{Na^+}\) and \(\mathrm{Cl^-}\) ions. Positive ions experience an electric force towards the negatively charged electrode. Thus, sodium ions migrate towards the left cathode. Negative ions experience an electric force towards the positively charged electrode. Therefore, chloride ions migrate towards the right anode. The two ion types move in opposite physical directions. However, their motions contribute to conventional electric current in the same effective direction because their charges have opposite signs. Within the external metallic conductors, mobile electrons carry electric current. In the molten electrolyte, charge transport occurs primarily through movement of ions rather than free electrons travelling through the bulk liquid. At the cathode, sodium ions can gain electrons and undergo reduction. At the anode, chloride ions can lose electrons and undergo oxidation. These electrode electron-transfer processes are distinct from ionic charge transport through the molten salt. The cell therefore requires different mechanisms in different regions: electronic conduction in the metallic wires and ionic conduction in the electrolyte. The migration direction of each ion is determined by the electrode polarity, not by the direction of electron flow in the external circuit.

Explanation: \(\textbf{Identify the invariant material property:}\) Both wires are made from the same material and maintained at the same temperature. Their resistivities are therefore equal. Resistivity is a property of the material at specified conditions. It does not change merely because the conductor is made longer or thinner. \(\textbf{Resistance of wire P:}\) For a uniform conductor: \[ R=\rho\frac{l}{A} \] Hence: \[ R_{\mathrm{P}}=\rho\frac{l}{A} \] \(\textbf{Resistance of wire Q:}\) The length is doubled, while the cross-sectional area is halved. Therefore: \[ R_{\mathrm{Q}}=\rho\frac{2l}{A/2} \] Simplifying: \[ R_{\mathrm{Q}}=4\rho\frac{l}{A} \] Thus: \[ R_{\mathrm{Q}}=4R_{\mathrm{P}} \] \(\textbf{Connect resistance with current:}\) Both wires are subjected separately to the same potential difference. For an ohmic conductor: \[ I=\frac{V}{R} \] Therefore: \[ I_{\mathrm{P}}=\frac{V}{R_{\mathrm{P}}} \] \[ I_{\mathrm{Q}}=\frac{V}{4R_{\mathrm{P}}} \] \(\textbf{Calculate the current ratio:}\) \[ \frac{I_{\mathrm{P}}}{I_{\mathrm{Q}}}=\frac{V/R_{\mathrm{P}}}{V/(4R_{\mathrm{P}})} \] \[ \frac{I_{\mathrm{P}}}{I_{\mathrm{Q}}}=4 \] \(\textbf{Interpret the result:}\) The longer wire offers greater resistance because the charge carriers traverse a greater conducting length. The smaller cross-sectional area further increases the resistance by providing a narrower conducting path. Both geometrical changes increase resistance, producing a combined factor of four. Since the same potential difference is applied, current varies inversely with resistance. The two wires still possess identical material conductivity because their material and temperature are unchanged. A change in conductor geometry affects measured resistance and current, not the intrinsic conductivity of the material. \(\textbf{Final answer:}\) \[ \boxed{I_{\mathrm{P}}:I_{\mathrm{Q}}=4:1} \]

Explanation: \(\textbf{Identify the conducting ions in P:}\) Sodium chloride dissociates according to: \[ \mathrm{NaCl\rightarrow Na^++Cl^-} \] One formula unit supplies one sodium ion and one chloride ion. At the stated concentration: \[ [\mathrm{Na^+}]=0.10\ \mathrm{mol\,L^{-1}} \] \[ [\mathrm{Cl^-}]=0.10\ \mathrm{mol\,L^{-1}} \] \(\textbf{Calculate the relative conductivity contribution:}\) Each ion type contributes one arbitrary unit per unit molar concentration. Therefore: \[ \kappa_{\mathrm{P}}\propto(0.10)(1)+(0.10)(1) \] \[ \kappa_{\mathrm{P}}\propto0.20 \] \(\textbf{Identify the conducting ions in Q:}\) Calcium chloride dissociates according to: \[ \mathrm{CaCl_2\rightarrow Ca^{2+}+2Cl^-} \] One formula unit supplies one calcium ion and two chloride ions. Thus: \[ [\mathrm{Ca^{2+}}]=0.10\ \mathrm{mol\,L^{-1}} \] \[ [\mathrm{Cl^-}]=2(0.10)=0.20\ \mathrm{mol\,L^{-1}} \] \(\textbf{Calculate the relative contribution in Q:}\) Under the given simplified contribution model, each calcium ion contributes two units per unit molar concentration. Each chloride ion contributes one. Therefore: \[ \kappa_{\mathrm{Q}}\propto(0.10)(2)+(0.20)(1) \] \[ \kappa_{\mathrm{Q}}\propto0.20+0.20 \] \[ \kappa_{\mathrm{Q}}\propto0.40 \] \(\textbf{Compare the conductivities:}\) \[ \frac{\kappa_{\mathrm{Q}}}{\kappa_{\mathrm{P}}}=\frac{0.40}{0.20} \] \[ =2 \] \(\textbf{Interpret the assumptions:}\) The comparison requires both the number of ions supplied by each formula unit and the specified relative contributions of those ions. It would be incorrect to compare conductivity using only the analytical electrolyte concentrations, since those concentrations are equal. The calculation uses the explicitly stated idealised contribution model. Actual ionic mobilities and their concentration dependence must be considered when comparing real electrolyte solutions. \(\textbf{Final answer:}\) \[ \boxed{\kappa_{\mathrm{Q}}:\kappa_{\mathrm{P}}=2:1} \]

Explanation: Statement I is correct because ionic compounds contain charged particles even in their crystalline states. However, the constituent ions occupy positions in a crystal lattice and generally cannot migrate freely through the solid. For this reason, ordinary solid ionic compounds often have very low electrical conductance. When the ionic compound melts, its ions become mobile. An applied electric field can then produce directed migration of cations and anions. The molten ionic compound can consequently conduct electricity by ionic charge transport. The increased conductance arises mainly from increased carrier mobility rather than the creation of ions that were completely absent in the solid. Statement II is incorrect because metallic conduction is associated with mobile electrons. Melting a metal changes its physical structure, but it does not necessarily convert its conducting mechanism into the transport of positive and negative ions characteristic of an ordinary molten ionic salt. Electrons remain important charge carriers in liquid metals. The physical state of a conductor alone is therefore insufficient to identify its conduction mechanism. A molten ionic compound and a molten metal may both conduct electricity while using different dominant mobile charge carriers. The decisive distinction is the electronic and ionic nature of the material, together with the mobility of its charge carriers. Thus, Statement I is correct and Statement II is incorrect.

Explanation: The system contains two different conducting media joined through electrode interfaces. In the metallic wire, the dominant mobile charge carriers are electrons. An electric field causes an ordered drift of these electrons in a direction opposite to conventional current. In the electrolytic solution, dissolved ions carry electric charge. Cations and anions migrate in opposite directions under the applied electric field. Because their charges are also opposite, both ion-migration processes contribute to conventional electric current. Increasing the applied potential generally increases the current in the stated ohmic range. However, increasing the potential does not change the fundamental identity of the charge carriers in either region. The metallic conductor remains an electronic conductor. The electrolyte remains an ionic conductor. At an electrode–solution interface, appropriate electrochemical reactions can transfer charge between the electronic and ionic conducting regions. Electrons do not need to migrate freely through the bulk electrolyte for ionic current to flow. Likewise, the metal's positive ion cores do not need to migrate along the external metallic wire. The current can be continuous around the conducting arrangement even though different charge carriers are responsible in different regions. The important concept is continuity of charge transfer across connected regions, not identity of the microscopic charge carriers. Thus, metallic and electrolytic conduction are complementary mechanisms in the complete electrical system.

Explanation: \(\textbf{Determine the cell constant from calibration:}\) The resistance of a conductivity cell is related to the conductivity of the solution through: \[ R=\frac{1}{\kappa}\frac{l}{A} \] The cell constant is: \[ G^*=\frac{l}{A}=\kappa R \] For the standard solution: \[ G^*=(0.0120\ \mathrm{S\,cm^{-1}})(150\ \Omega) \] Since \(\mathrm{S}=\Omega^{-1}\): \[ G^*=1.80\ \mathrm{cm^{-1}} \] \(\textbf{Calculate the unknown conductivity:}\) The same conductivity cell is used without changing its geometry. Therefore, its cell constant remains \(1.80\ \mathrm{cm^{-1}}\). \[ \kappa_{\mathrm{unknown}}=\frac{G^*}{R_{\mathrm{unknown}}} \] Substituting: \[ \kappa_{\mathrm{unknown}}=\frac{1.80\ \mathrm{cm^{-1}}}{300\ \Omega} \] \[ \kappa_{\mathrm{unknown}}=0.00600\ \mathrm{S\,cm^{-1}} \] \(\textbf{Convert conductivity into molar conductivity:}\) When conductivity is expressed in \(\mathrm{S\,cm^{-1}}\) and concentration in \(\mathrm{mol\,L^{-1}}\): \[ \Lambda_m=\frac{1000\kappa}{c} \] The factor \(1000\) converts litres into cubic centimetres. For the unknown solution: \[ \Lambda_m=\frac{1000(0.00600)}{0.040} \] \[ \Lambda_m=150\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Check the physical interpretation:}\) The unknown solution has twice the resistance of the standard solution in the same cell. Its conductivity must consequently be half that of the standard. Indeed: \[ \frac{0.00600}{0.0120}=\frac12 \] This confirms the conductivity calculation. The molar conductivity additionally depends on the concentration of the unknown electrolyte, so it cannot be obtained from the resistance ratio alone. \(\textbf{Final answer:}\) \[ \boxed{\Lambda_m=150\ \mathrm{S\,cm^2\,mol^{-1}}} \]

Explanation: Conductance and conductivity are related but physically distinct quantities. Conductance refers to the electrical response of the particular solution-filled cell. Conductivity describes the conducting property of the solution at the specified temperature and composition. \(\textbf{Compare measured conductances:}\) Conductance is the reciprocal of resistance: \[ G=\frac1R \] For solution P: \[ G_{\mathrm P}=\frac{1}{150} \] \[ G_{\mathrm P}=0.00667\ \mathrm S \] For solution Q: \[ G_{\mathrm Q}=\frac{1}{80} \] \[ G_{\mathrm Q}=0.0125\ \mathrm S \] Therefore: \[ G_{\mathrm P}\lt G_{\mathrm Q} \] \(\textbf{Account for the different cell constants:}\) Conductivity is calculated from: \[ \kappa=\frac{G^*}{R} \] For P: \[ \kappa_{\mathrm P}=\frac{1.80}{150} \] \[ \kappa_{\mathrm P}=0.0120\ \mathrm{S\,cm^{-1}} \] For Q: \[ \kappa_{\mathrm Q}=\frac{0.80}{80} \] \[ \kappa_{\mathrm Q}=0.0100\ \mathrm{S\,cm^{-1}} \] Consequently: \[ \kappa_{\mathrm P}\gt \kappa_{\mathrm Q} \] \(\textbf{Interpret the reversal:}\) Solution P has greater intrinsic conductivity despite exhibiting smaller measured conductance. Its conductivity cell has a larger cell constant, which contributes to a greater measured resistance. Comparing resistance or conductance readings from different cells without considering their geometry can therefore produce the wrong conductivity ranking. The cell constant corrects for the effective electrode separation and conducting cross-sectional area. The correct conclusion requires both the resistance readings and the corresponding cell constants.

Explanation: \(\textbf{Extract the resistance relation from the graph:}\) For a uniform column of electrolyte: \[ R=\rho\frac{l}{A} \] At constant cross-sectional area, resistance varies linearly with length. Therefore, the slope of an \(R\) versus \(l\) graph is: \[ m=\frac{R}{l}=\frac{\rho}{A} \] Rearranging: \[ \rho=mA \] \(\textbf{Determine the resistivity of P:}\) \[ \rho_{\mathrm P}=(100\ \Omega\,\mathrm{cm^{-1}})(2.0\ \mathrm{cm^2}) \] \[ \rho_{\mathrm P}=200\ \Omega\,\mathrm{cm} \] Conductivity is the reciprocal of resistivity. \[ \kappa_{\mathrm P}=\frac{1}{200} \] \[ \kappa_{\mathrm P}=0.00500\ \mathrm{S\,cm^{-1}} \] \(\textbf{Determine the resistivity of Q:}\) \[ \rho_{\mathrm Q}=(150\ \Omega\,\mathrm{cm^{-1}})(2.0\ \mathrm{cm^2}) \] \[ \rho_{\mathrm Q}=300\ \Omega\,\mathrm{cm} \] Therefore: \[ \kappa_{\mathrm Q}=\frac{1}{300} \] \[ \kappa_{\mathrm Q}=0.00333\ \mathrm{S\,cm^{-1}} \] \(\textbf{Convert to molar conductivities:}\) For solution P: \[ \Lambda_{m,\mathrm P}=\frac{1000\kappa_{\mathrm P}}{c_{\mathrm P}} \] \[ \Lambda_{m,\mathrm P}=\frac{1000(0.00500)}{0.10} \] \[ \Lambda_{m,\mathrm P}=50.0\ \mathrm{S\,cm^2\,mol^{-1}} \] For solution Q: \[ \Lambda_{m,\mathrm Q}=\frac{1000\kappa_{\mathrm Q}}{c_{\mathrm Q}} \] \[ \Lambda_{m,\mathrm Q}=\frac{1000(1/300)}{0.050} \] \[ \Lambda_{m,\mathrm Q}=\frac{200}{3}\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Calculate the requested ratio:}\) \[ \frac{\Lambda_{m,\mathrm Q}}{\Lambda_{m,\mathrm P}}=\frac{200/3}{50} \] \[ =\frac43 \] \(\textbf{Interpret the graph correctly:}\) The steeper resistance graph corresponds to greater resistivity and lower conductivity when electrode area is unchanged. However, molar conductivity also incorporates electrolyte concentration. Although Q has lower conductivity, its concentration is sufficiently smaller that its molar conductivity exceeds that of P. A ranking based only on graph slopes would therefore be inadequate. \(\textbf{Final answer:}\) \[ \boxed{\Lambda_{m,\mathrm Q}:\Lambda_{m,\mathrm P}=4:3} \]

Explanation: The Assertion is true because electrolytic solutions conduct electricity through the movement of ions. When a steady direct current passes through a suitable electrolyte, oxidation and reduction reactions may occur at the electrodes. Such reactions can consume or produce ions near the electrode surfaces. The ionic composition close to an electrode may consequently become different from that of the bulk solution. Additional electrode polarisation effects can also influence the electrical measurement. The observed resistance may then contain contributions associated with electrode processes rather than representing only the bulk solution resistance. The Reason correctly identifies this difficulty. An alternating-current source reverses the direction of the applied electrical field repeatedly. With suitable frequency and measurement conditions, it helps minimise net electrolysis and electrode polarisation during the resistance measurement. This makes the measurement more representative of the conducting solution. The method does not change the fundamental charge carriers in the electrolyte. Cations and anions still participate in ionic conduction. Nor does alternating current make the cell constant unnecessary; the measured resistance must still be combined with the calibrated cell constant to obtain conductivity. Thus, both statements are true, and the Reason provides the correct explanation of the Assertion.

Explanation: \(\textbf{Determine the conductivity change:}\) For an unchanged conductivity cell: \[ \kappa=\frac{G^*}{R} \] The cell constant \(G^*\) remains fixed. Therefore, conductivity is inversely proportional to measured resistance. \[ \frac{\kappa_2}{\kappa_1}=\frac{R_1}{R_2} \] Substituting: \[ \frac{\kappa_2}{\kappa_1}=\frac{200}{400} \] \[ \frac{\kappa_2}{\kappa_1}=\frac12 \] The conductivity decreases to half its initial value. \(\textbf{Account for the concentration change:}\) Molar conductivity is: \[ \Lambda_m=\frac{1000\kappa}{c} \] Taking the ratio: \[ \frac{\Lambda_{m,2}}{\Lambda_{m,1}}=\frac{\kappa_2}{\kappa_1}\frac{c_1}{c_2} \] The concentration ratio is: \[ \frac{c_1}{c_2}=\frac{0.100}{0.040} \] \[ =2.5 \] \(\textbf{Combine the two effects:}\) \[ \frac{\Lambda_{m,2}}{\Lambda_{m,1}}=\frac12\times2.5 \] \[ =1.25 \] Thus: \[ \frac{\Lambda_{m,2}}{\Lambda_{m,1}}=\frac54 \] \(\textbf{Interpret the dilution behaviour:}\) The conductivity decreases because dilution reduces the number of conducting ions per unit volume. However, the concentration decreases by a factor of \(2.5\), whereas conductivity decreases by only a factor of two. After normalising conductivity to one mole of electrolyte, the molar conductivity is therefore greater. For strong electrolytes, reduced interionic interactions generally allow greater ionic mobility on dilution. The measured data are consistent with that behaviour. It would be incorrect to conclude that molar conductivity must decrease merely because the measured resistance increases. \(\textbf{Final answer:}\) \[ \boxed{\frac{\Lambda_{m,2}}{\Lambda_{m,1}}=\frac54} \]

Explanation: \(\textbf{Express molar conductivity in terms of measured resistance:}\) For a conductivity cell: \[ \kappa=\frac{G^*}{R} \] Molar conductivity is: \[ \Lambda_m=\frac{1000\kappa}{c} \] Combining: \[ \Lambda_m=\frac{1000G^*}{Rc} \] \(\textbf{Identify the quantities common to all measurements:}\) All four solutions are examined using the same conductivity cell. The cell constant \(G^*\) is therefore identical for every measurement. The factor \(1000G^*\) is common to all four molar-conductivity expressions. Consequently: \[ \Lambda_m\propto\frac{1}{Rc} \] The solution with the smallest product \(Rc\) has the greatest molar conductivity. \(\textbf{Evaluate solution P:}\) \[ (Rc)_{\mathrm P}=100(0.100) \] \[ (Rc)_{\mathrm P}=10.0 \] \(\textbf{Evaluate solution Q:}\) \[ (Rc)_{\mathrm Q}=250(0.025) \] \[ (Rc)_{\mathrm Q}=6.25 \] \(\textbf{Evaluate solution R:}\) \[ (Rc)_{\mathrm R}=180(0.040) \] \[ (Rc)_{\mathrm R}=7.20 \] \(\textbf{Evaluate solution S:}\) \[ (Rc)_{\mathrm S}=120(0.080) \] \[ (Rc)_{\mathrm S}=9.60 \] \(\textbf{Compare the products:}\) \[ 6.25\lt 7.20\lt 9.60\lt 10.0 \] Thus, the decreasing molar-conductivity order is Q, R, S and P. Solution Q has the greatest measured resistance, so it has the lowest conductivity among these solutions in the same cell. Nevertheless, its concentration is sufficiently small that its molar conductivity is the greatest. This distinction illustrates why measured resistance or conductivity alone cannot establish a molar-conductivity ranking across solutions of different concentrations. The concentration normalisation is essential.

Explanation: \(\textbf{Conductance:}\) Conductance represents the reciprocal of the resistance of a particular conductor or conductivity cell. Its defining relation is: \[ G=\frac1R \] Its SI unit is the siemens, equivalent to reciprocal ohm. Therefore, P matches 2. \(\textbf{Resistivity:}\) The resistance of a uniform conducting column is: \[ R=\rho\frac{l}{A} \] Rearranging: \[ \rho=\frac{RA}{l} \] Resistivity describes the material's opposition to electrical conduction at specified conditions. Its SI unit is \(\Omega\,\mathrm m\). Therefore, Q matches 3. \(\textbf{Cell constant:}\) The cell constant is determined by the effective electrode separation divided by the conducting cross-sectional area. \[ G^*=\frac{l}{A} \] It has the dimension of reciprocal length. For a fixed cell geometry, it does not depend on which solution occupies the cell. Therefore, R matches 4. \(\textbf{Molar conductivity:}\) Molar conductivity normalises the conductivity of a solution to its electrolyte concentration. When coherent SI units are used: \[ \Lambda_m=\frac{\kappa}{c} \] Here, conductivity is expressed in \(\mathrm{S\,m^{-1}}\) and concentration in \(\mathrm{mol\,m^{-3}}\). The resulting molar conductivity has units of \(\mathrm{S\,m^2\,mol^{-1}}\). Therefore, S matches 1. Conductance and conductivity must not be confused: the former depends on the conducting geometry, whereas the latter is a property of the electrolyte solution at specified conditions. Similarly, resistivity is a material property, while resistance depends additionally on geometry. The correct matching is P–2, Q–3, R–4, S–1.

Explanation: \(\textbf{Initial conductance:}\) For a uniform conducting column of electrolyte: \[ R=\frac{l}{\kappa A} \] Conductance is the reciprocal of resistance: \[ G=\frac{\kappa A}{l} \] Therefore, the initial conductance is: \[ G_1=\frac{\kappa A}{l} \] \(\textbf{Account for the conductivity change:}\) The diluted solution has: \[ \kappa_2=0.60\kappa \] If the original cell geometry were retained, this change alone would reduce conductance to \(0.60\) times its original value. However, the second cell has different electrode dimensions. \(\textbf{Account for the new cell geometry:}\) The new area is: \[ A_2=2A \] The new electrode separation is: \[ l_2=\frac l2 \] The final conductance is: \[ G_2=\frac{\kappa_2A_2}{l_2} \] \(\textbf{Substitute both changes:}\) \[ G_2=\frac{(0.60\kappa)(2A)}{l/2} \] Simplifying: \[ G_2=\frac{2.40\kappa A}{l} \] \(\textbf{Compare with the initial conductance:}\) \[ \frac{G_2}{G_1}=\frac{2.40\kappa A/l}{\kappa A/l} \] \[ \frac{G_2}{G_1}=2.40 \] Therefore: \[ G_2:G_1=12:5 \] \(\textbf{Interpret the competing effects:}\) Dilution reduces the intrinsic conductivity to \(60\%\) of its original value. However, doubling electrode area increases the conducting cross-section. Halving the separation further reduces the resistance. The combined geometrical change increases conductance by a factor of four for a solution of unchanged conductivity. After accounting for dilution: \[ 4\times0.60=2.40 \] Thus, the measured conductance increases even though the electrolyte's conductivity decreases. This demonstrates why a change in measured conductance does not necessarily imply a corresponding change in the intrinsic conducting ability of the solution. \(\textbf{Final answer:}\) \[ \boxed{G_2:G_1=12:5} \]

Explanation: \(\textbf{Relate conductivity to molar conductivity:}\) For an electrolyte solution: \[ \Lambda_m=\frac{\kappa}{c} \] when coherent concentration and conductivity units are used. Equivalently: \[ \kappa=\Lambda_m c \] Thus, the conductivity depends on both molar conductivity and electrolyte concentration. \(\textbf{Establish the common dilution factor:}\) Each solution is diluted until: \[ c_2=\frac{c_1}{4} \] Therefore: \[ \frac{c_2}{c_1}=\frac14 \] \(\textbf{Determine the conductivity change for S:}\) The final molar conductivity is: \[ \Lambda_{m,2}=1.10\Lambda_{m,1} \] The conductivity ratio is: \[ \frac{\kappa_2}{\kappa_1}=\frac{\Lambda_{m,2}}{\Lambda_{m,1}}\frac{c_2}{c_1} \] Substituting: \[ \frac{\kappa_{2,\mathrm S}}{\kappa_{1,\mathrm S}}=1.10\times\frac14 \] \[ =0.275 \] Hence, the strong-electrolyte solution retains \(27.5\%\) of its initial conductivity. \(\textbf{Determine the conductivity change for W:}\) For the weak electrolyte: \[ \Lambda_{m,2}=1.90\Lambda_{m,1} \] Therefore: \[ \frac{\kappa_{2,\mathrm W}}{\kappa_{1,\mathrm W}}=1.90\times\frac14 \] \[ =0.475 \] The weak-electrolyte solution retains \(47.5\%\) of its initial conductivity. \(\textbf{Interpret the different responses:}\) Both conductivities decrease because the electrolyte concentration becomes much smaller. However, the weak electrolyte shows a larger relative increase in molar conductivity. For a weak electrolyte, dilution increases the degree of ionisation appreciably. For a strong electrolyte, which is already extensively ionised, the increase in molar conductivity is generally more gradual and is mainly associated with changes in interionic interactions and ionic mobility. The greater molar-conductivity increase for W partly compensates for the decrease in concentration. Nevertheless, neither electrolyte shows an increase in conductivity under the given numerical conditions. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{\kappa_{2,\mathrm S}}{\kappa_{1,\mathrm S}},\frac{\kappa_{2,\mathrm W}}{\kappa_{1,\mathrm W}}\right)=(0.275,\ 0.475)} \]

Explanation: \(\textbf{Identify the governing concentration relation:}\) For the specified dilute strong-electrolyte range: \[ \Lambda_m=\Lambda_m^0-A\sqrt c \] Here, \(\Lambda_m^0\) represents limiting molar conductivity. The constant \(A\) determines the magnitude of the negative slope of the \(\Lambda_m\) versus \(\sqrt c\) graph. \(\textbf{Use the first data point:}\) At: \[ c_1=0.0100\ \mathrm{mol\,L^{-1}} \] \[ \sqrt{c_1}=0.100 \] The equation becomes: \[ 150=\Lambda_m^0-0.100A \] \(\textbf{Use the second data point:}\) At: \[ c_2=0.0400\ \mathrm{mol\,L^{-1}} \] \[ \sqrt{c_2}=0.200 \] Therefore: \[ 130=\Lambda_m^0-0.200A \] \(\textbf{Determine the slope constant:}\) Subtract the second equation from the first: \[ 150-130=0.100A \] \[ 20=0.100A \] \[ A=200 \] Here, the numerical value of \(A\) corresponds to the concentration units specified in the question. \(\textbf{Determine limiting molar conductivity:}\) Substitute into the first equation: \[ 150=\Lambda_m^0-200(0.100) \] \[ 150=\Lambda_m^0-20 \] \[ \Lambda_m^0=170\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Predict molar conductivity at the required concentration:}\) \[ c_3=0.00250\ \mathrm{mol\,L^{-1}} \] \[ \sqrt{c_3}=0.0500 \] Therefore: \[ \Lambda_{m,3}=170-200(0.0500) \] \[ \Lambda_{m,3}=160\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Convert to conductivity:}\) When conductivity is in \(\mathrm{S\,cm^{-1}}\) and concentration in \(\mathrm{mol\,L^{-1}}\): \[ \kappa=\frac{\Lambda_m c}{1000} \] Substituting: \[ \kappa_3=\frac{160(0.00250)}{1000} \] \[ \kappa_3=4.00\times10^{-4}\ \mathrm{S\,cm^{-1}} \] \(\textbf{Check the dilution trend:}\) The predicted molar conductivity of \(160\ \mathrm{S\,cm^2\,mol^{-1}}\) exceeds the value at \(0.0100\ \mathrm{mol\,L^{-1}}\). It remains below the limiting value of \(170\ \mathrm{S\,cm^2\,mol^{-1}}\). Both observations are consistent with the stated strong-electrolyte relation. The conductivity is smaller at the lower concentration because the number of conducting ions per unit volume is reduced. \(\textbf{Final answer:}\) \[ \boxed{\kappa=4.00\times10^{-4}\ \mathrm{S\,cm^{-1}}} \]

Explanation: \(\textbf{Calculate the molar conductivity of each solution:}\) For conductivity expressed in \(\mathrm{S\,cm^{-1}}\) and concentration in \(\mathrm{mol\,L^{-1}}\): \[ \Lambda_m=\frac{1000\kappa}{c} \] For the strong-electrolyte solution P: \[ \Lambda_{m,\mathrm{P}}=\frac{1000(1.10\times10^{-3})}{0.0100} \] \[ \Lambda_{m,\mathrm{P}}=110\ \mathrm{S\,cm^2\,mol^{-1}} \] For the weak-acid solution Q: \[ \Lambda_{m,\mathrm{Q}}=\frac{1000(1.20\times10^{-3})}{0.0100} \] \[ \Lambda_{m,\mathrm{Q}}=120\ \mathrm{S\,cm^2\,mol^{-1}} \] Thus, solution Q has a higher measured molar conductivity than solution P. \(\textbf{Determine the degree of ionisation of HA:}\) For the specified weak electrolyte: \[ \alpha=\frac{\Lambda_m}{\Lambda_m^0} \] Substituting the values: \[ \alpha=\frac{120}{400} \] \[ \alpha=0.30 \] Therefore, the degree of ionisation of the weak acid is: \[ \alpha\times100=30\% \] Only \(30\%\) of the weak-acid molecules are ionised under the stated conditions. \(\textbf{Interpret the conductivity comparison:}\) The conductivity of an electrolyte solution depends on both the concentrations of mobile ions and their ionic mobilities. A highly mobile ion can make a large contribution to conductivity even when the parent electrolyte is only partially ionised. In particular, hydrogen ions have unusually high ionic mobility in aqueous solution. Consequently, a weak acid can possess a higher measured conductivity than a particular strong-electrolyte solution at the same analytical concentration. The observation: \[ \kappa_{\mathrm{Q}}>\kappa_{\mathrm{P}} \] does not establish: \[ \alpha_{\mathrm{Q}}>\alpha_{\mathrm{P}} \] For a strong electrolyte, the measured-to-limiting molar-conductivity ratio must not automatically be interpreted as its degree of ionisation because interionic interactions also affect molar conductivity. \(\textbf{Evaluate the alternatives:}\) Option A is correct because the weak acid is \(30\%\) ionised, while its higher conductivity does not establish a greater degree of ionisation. Option B incorrectly equates higher measured conductivity with more extensive ionisation. Option C gives an incorrect degree of ionisation and incorrectly infers the number of ions from conductivity alone. Option D gives an incorrect degree of ionisation and ignores the important contribution of ionic mobility. The correct conclusion is that conductivity and degree of ionisation are related but are not interchangeable measures of electrolyte behaviour.

Explanation: \(\textbf{Use the SI molar-conductivity relation:}\) When conductivity is expressed in \(\mathrm{S\,m^{-1}}\) and concentration in \(\mathrm{mol\,m^{-3}}\): \[ \Lambda_m=\frac{\kappa}{c} \] The resulting unit is: \[ \frac{\mathrm{S\,m^{-1}}}{\mathrm{mol\,m^{-3}}}=\mathrm{S\,m^2\,mol^{-1}} \] \(\textbf{Calculate the initial molar conductivity:}\) The initial conditions are: \[ \kappa_1=0.200\ \mathrm{S\,m^{-1}} \] \[ c_1=25.0\ \mathrm{mol\,m^{-3}} \] Therefore: \[ \Lambda_{m,1}=\frac{0.200}{25.0} \] \[ \Lambda_{m,1}=0.00800\ \mathrm{S\,m^2\,mol^{-1}} \] \(\textbf{Convert to centimetre-based units:}\) The relevant conversion is: \[ 1\ \mathrm{S\,m^2\,mol^{-1}}=10^4\ \mathrm{S\,cm^2\,mol^{-1}} \] Thus: \[ \Lambda_{m,1}=0.00800\times10^4 \] \[ \Lambda_{m,1}=80.0\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Calculate the diluted-solution value:}\) The final conditions are: \[ \kappa_2=0.120\ \mathrm{S\,m^{-1}} \] \[ c_2=12.5\ \mathrm{mol\,m^{-3}} \] Therefore: \[ \Lambda_{m,2}=\frac{0.120}{12.5} \] \[ \Lambda_{m,2}=0.00960\ \mathrm{S\,m^2\,mol^{-1}} \] Convert to the requested units: \[ \Lambda_{m,2}=0.00960\times10^4 \] \[ \Lambda_{m,2}=96.0\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Check the dilution trend:}\) The conductivity decreases from \(0.200\) to \(0.120\ \mathrm{S\,m^{-1}}\). However, the concentration decreases to half its original value. Therefore: \[ \frac{\Lambda_{m,2}}{\Lambda_{m,1}}=\frac{96}{80} \] \[ =1.20 \] Molar conductivity increases by \(20\%\) despite the reduction in conductivity. The units are consistent throughout because concentration was supplied directly in \(\mathrm{mol\,m^{-3}}\). Inserting an additional factor of \(1000\) into the SI relation would be incorrect. \(\textbf{Final answer:}\) \[ \boxed{(\Lambda_{m,1},\Lambda_{m,2})=(80,\ 96)\ \mathrm{S\,cm^2\,mol^{-1}}} \]

Explanation: \(\textbf{Determine the concentration-axis values:}\) The strong-electrolyte relation is linear in \(\sqrt c\), not directly in \(c\). For the lower concentration: \[ \sqrt{0.0100}=0.100 \] For the higher concentration: \[ \sqrt{0.0400}=0.200 \] \(\textbf{Determine the slope constant for P:}\) \[ \Lambda_{m,\mathrm P}=\Lambda_{m,\mathrm P}^0-A_{\mathrm P}\sqrt c \] Using both data points: \[ 135=\Lambda_{m,\mathrm P}^0-0.100A_{\mathrm P} \] \[ 115=\Lambda_{m,\mathrm P}^0-0.200A_{\mathrm P} \] Subtract: \[ 135-115=0.100A_{\mathrm P} \] \[ A_{\mathrm P}=200 \] \(\textbf{Determine the limiting value for P:}\) \[ 135=\Lambda_{m,\mathrm P}^0-200(0.100) \] \[ \Lambda_{m,\mathrm P}^0=155\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Determine the slope constant for Q:}\) \[ 110=\Lambda_{m,\mathrm Q}^0-0.100A_{\mathrm Q} \] \[ 105=\Lambda_{m,\mathrm Q}^0-0.200A_{\mathrm Q} \] Subtract: \[ 110-105=0.100A_{\mathrm Q} \] \[ A_{\mathrm Q}=50 \] \(\textbf{Determine the limiting value for Q:}\) \[ 110=\Lambda_{m,\mathrm Q}^0-50(0.100) \] \[ \Lambda_{m,\mathrm Q}^0=115\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Compare the intercepts:}\) \[ \Delta\Lambda_m^0=155-115 \] \[ \Delta\Lambda_m^0=40\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Interpret the different slopes:}\) The two electrolytes have different values of \(A\), so their molar conductivities decrease at different rates as \(\sqrt c\) increases. Consequently, the difference between their molar conductivities at a finite concentration need not equal the difference between their limiting values. The limiting conductivities must be determined from their separate intercepts. The calculation is valid under the explicitly stated linear strong-electrolyte assumption. \(\textbf{Final answer:}\) \[ \boxed{\Lambda_{m,\mathrm P}^0-\Lambda_{m,\mathrm Q}^0=40\ \mathrm{S\,cm^2\,mol^{-1}}} \]