Q.Resistance of a conductivity cell filled with $0.1\ mol\ L^{-1}$ KCl solution is $100\ \Omega$. If the resistance of the same cell when filled with $0.02\ mol\ L^{-1}$ KCl solution is $520\ \Omega$, calculate the conductivity and molar conductivity of $0.02\ mol\ L^{-1}$ KCl solution. The conductivity of $0.1\ mol\ L^{-1}$ KCl solution is $1.29\ S\ m^{-1}$.

Concept and Intuition

A conductivity cell has a fixed geometric cell constant $G^*=\ell/A$ (independent of the solution inside it). Once $G^*$ is found using one known solution (KCl, with known $\kappa$), the SAME cell can be used to find the conductivity of any other solution by simply measuring its conductance and multiplying by $G^*$.

Step-by-Step Solution

1. Conductivity $\kappa$ = conductance $G$ × cell constant $G^*$, i.e. $G^* = \kappa/G$.
2. For 0.01 M KCl: $\kappa_{KCl}=p$, conductance $=q$, so $G^* = p/q$.
3. This $G^*$ is a property of the cell itself, so it is unchanged when the cell is refilled with 0.01 M $H_2SO_4$.
4. For $H_2SO_4$: conductance $=r$, so $\kappa_{H_2SO_4} = G^*\times r = \dfrac{p}{q}\times r = \dfrac{pr}{q}$.

Common Mistakes

• Assuming the cell constant changes with the solution — it does not; it is purely geometric.
• Multiplying/dividing $p$, $q$, $r$ in the wrong order (e.g. giving $pq$ or $qr$, which don't even have correct units for conductivity).
• Forgetting to check units: conductivity (ohm⁻¹cm⁻¹) = conductance (ohm⁻¹) × cell constant (cm⁻¹), confirming $pr/q$ is dimensionally consistent.

ANSWER: D

Concept and Intuition

Conductivity ($\kappa$) is the conducting ability per unit volume of solution — it depends on the number of ions present in that volume. On dilution, although each ion becomes individually more mobile (which is why molar conductivity increases), the total number of ions per unit volume drops sharply, so the overall conductivity (specific conductance) decreases. Separately, a fuel cell is a galvanic (voltaic) device: it spontaneously combines fuel and oxidant to release electrical energy, so the $H_2$–$O_2$ fuel cell's net reaction must be the exergonic combination of hydrogen and oxygen to form water, not its reverse (which would require electrical energy input, as in electrolysis).

Step-by-Step Solution

1. Statement I: as a solution is diluted, ions per unit volume decrease, so specific conductance $\kappa$ decreases with dilution — this is correct (a standard NCERT electrochemistry fact; it's molar conductivity $\Lambda_m$ that increases with dilution, not $\kappa$).
2. Statement II: the actual $H_2$–$O_2$ fuel cell reaction is $2H_2(g) + O_2(g) \to 2H_2O(l)$ — hydrogen is oxidized at the anode and oxygen is reduced at the cathode, releasing electrical energy. The statement writes the reverse ($2H_2O(l) \to 2H_2(g)+O_2(g)$), which is the electrolysis of water, not the fuel cell's spontaneous discharge reaction — so Statement II is incorrect.
3. Hence Statement I is correct, Statement II is not correct.

Common Mistakes

• Confusing specific conductivity ($\kappa$, decreases on dilution) with molar conductivity ($\Lambda_m$, increases on dilution) — a very common source of error in this topic.
• Not noticing the reaction in Statement II is written backwards (water splitting instead of water formation), since fuel cells by definition run spontaneously in the water-forming direction.

ANSWER: C

1. Understanding the Assertion (A): "Conductivity of an electrolyte decreases on dilution."

Conductivity, also called specific conductance, is the conductance of a solution held between two electrodes 1 cm apart and of cross-sectional area 1 cm². It measures how well the solution conducts electricity per unit volume.

• When you dilute an electrolyte solution, you add more water. The total number of ions in the solution remains the same (if the electrolyte is fully dissociated) or even increases slightly (for weak electrolytes, due to greater dissociation), but the volume increases.
• Because conductivity depends on the number of ions per unit volume, dilution spreads the same (or slightly more) ions over a larger volume. Thus, the concentration of charge carriers drops.
• Result: Conductivity (specific conductance) decreases on dilution. So Assertion (A) is correct.

2. Examining the Reason (R): "On dilution number of ions per unit volume increases."

• This statement is the opposite of what actually happens. On dilution, the number of ions per unit volume (i.e., concentration of ions) decreases.
• For example: If you have 1 liter of 1 M NaCl solution, there are about $6.02 \times 10^{23}$ ions per liter. Dilute it to 2 liters — now the same number of ions is spread over twice the volume, so the number per unit volume is halved.
• Even for weak electrolytes like acetic acid, where dilution increases the degree of dissociation, the increase in total ions is not enough to offset the volume increase — the concentration of ions still falls.
• Result: Reason (R) is false.

3. Relating (A) and (R)

• (A) is correct: conductivity decreases on dilution.
• (R) is incorrect: number of ions per unit volume decreases, not increases.
• Since (R) is false, it cannot be the correct explanation of (A). So the only possible answer is that (A) is correct but (R) is not correct.

Concept and Intuition

It's important to distinguish specific conductivity ($\kappa$) from molar conductivity ($\Lambda_m$). Molar conductivity decreases with increasing concentration (due to increased ion-ion interactions reducing ionic mobility per mole), but specific conductivity — the actual conductance per unit volume — increases with concentration for a strong electrolyte like KCl, because the sheer number of charge carriers (ions) per unit volume grows faster than any mobility decrease.

Step-by-Step Solution

1. Recognize X, Y, Z as specific conductivities at 0.1 M, 0.01 M, and 1.0 M respectively.
2. For a strong electrolyte, specific conductivity increases with molar concentration.
3. Order of concentration: 1.0 M (Z) $>$ 0.1 M (X) $>$ 0.01 M (Y).
4. Hence order of conductivity: $Z > X > Y$.

Common Mistakes

• Confusing specific conductivity's concentration trend with molar conductivity's (which decreases with concentration) — a very common trap in this exact question style.

ANSWER: B

Concept and Intuition

A bare alkali-metal cation's size increases down the group ($Li^+<Na^+<K^+<Cs^+$). But in water, ions are surrounded by shells of water molecules held by ion-dipole attraction; the smaller and more highly charged-density the bare ion, the more strongly and extensively it attracts water, forming a larger hydrated shell. This means $Li^+$, despite being the smallest bare ion, has the largest hydrated radius, making it move sluggishly through solution under an applied field. Conversely, $Cs^+$, the largest bare ion, holds its (few) water molecules loosely, giving it the smallest hydrated radius and hence the highest ionic mobility (and molar conductivity).

Step-by-Step Solution

1. Order of bare ionic radii: $Li^+ < Na^+ < K^+ < Cs^+$.
2. Degree of hydration (number/tightness of water molecules held) follows the reverse trend: $Li^+$ is hydrated most, $Cs^+$ least.
3. Hydrated radius order (largest to smallest): $Li^+ > Na^+ > K^+ > Cs^+$.
4. Ionic mobility is inversely related to hydrated radius (a bulkier hydrated ion moves more slowly through the solvent).
5. So mobility order (highest to lowest): $Cs^+ > K^+ > Na^+ > Li^+$.

Common Mistakes

• Assuming mobility follows the bare ionic radius trend directly (would incorrectly give $Li^+$ as least mobile due to being smallest, without realising hydration reverses this reasoning correctly — actually here it does end up matching since $Li^+$ is indeed least mobile, but for the hydration reason, not because it's the smallest bare ion).
• Forgetting that hydration, not bare size, is what actually governs aqueous mobility.

ANSWER: C

Concept and Intuition

A conductivity cell has a fixed geometric cell constant $G^* = \dfrac{l}{A}$, which relates the measured resistance to the conductivity of whatever solution fills it: $\kappa = G^*/R$, i.e. $G^* = \kappa R$. Since the cell constant doesn't change between measurements (same cell, same electrodes), you can find it from one filling and reuse it for the other.

Step-by-Step Solution

1. From the 0.1 M NaCl filling: $G^* = \kappa_1 \times R_1 = 1.29\ \text{S m}^{-1} \times 100\ \Omega = 129\ \text{m}^{-1}$.
2. This $G^*$ is a property of the cell, unchanged when refilled with 0.02 M NaCl.
3. For the 0.02 M solution: $\kappa_2 = \dfrac{G^*}{R_2} = \dfrac{129}{258} = 0.5\ \text{S m}^{-1}$.

Common Mistakes

• Trying to scale conductivity directly with concentration (conductivity isn't simply proportional to concentration once dilution effects on molar conductivity are considered) — always go through the cell constant instead.
• Mixing up $R$ and $\kappa$ in the ratio (conductivity is inversely proportional to resistance for fixed cell constant).

ANSWER: D

Concept and Intuition

The cell constant ($G^* = l/A$) relates the specific conductance (a property of

the solution alone) to the conductance measured for a particular cell:

$\kappa = G^* \times \dfrac{1}{R}$ (since conductance $=1/R$), so

$G^* = \kappa \times R$.

Step-by-Step Solution

1. Relationship: $\kappa = \dfrac{G^*}{R} \Rightarrow G^* = \kappa \times R$.
2. Substitute the given values: $G^* = 0.024\ ohm^{-1}.cm^{-1} \times 85\ ohm$.
3. $G^* = 2.04\ cm^{-1}$.

Common Mistakes

• Dividing $\kappa$ by $R$ instead of multiplying (mixing up conductance and resistance in the relation).

ANSWER: A

Concept and Intuition

When two different electrolyte solutions are in contact (as at the ends of a salt bridge), a liquid-junction potential can develop if the cation and anion of the bridge electrolyte diffuse into the surrounding solutions at different rates — the faster ion races ahead, separating charge and creating an extra potential difference that corrupts the cell's measured EMF. To make a salt bridge that essentially eliminates this junction potential, the bridge electrolyte must be chosen so that its cation and anion have nearly equal ionic mobilities (transport numbers close to 0.5 each), so that neither ion diffuses preferentially and no significant charge separation builds up.

Step-by-Step Solution

1. Recall the purpose of a salt bridge: to maintain electrical neutrality between two half-cells while minimising the liquid-junction potential at the boundaries.
2. The junction potential arises specifically from unequal diffusion rates of the bridge's own cations and anions into the adjoining solutions.
3. $K^+$ and $NO_3^-$ (like $K^+$ and $Cl^-$) have transport numbers very close to each other (~0.49 and ~0.51), meaning they diffuse at nearly the same velocity.
4. Because the two ions move essentially together, there is no significant net charge separation at either junction, so the liquid-junction potential is minimised (nearly eliminated) — this is precisely why KNO₃/KCl with agar-agar (which also immobilises the solution to prevent bulk mixing/convection) is the standard salt-bridge choice.
5. Options based on ion size rather than ionic mobility (options A, D) are not the operative reason — mobility (velocity) is what determines the junction potential, not physical size alone (though the two are related, it's the mobility match that matters).

Common Mistakes

• Attributing the choice to ionic size differences rather than to matched ionic mobilities (diffusion velocities), which is the actual electrochemical reason.
• Thinking one ion must move faster to "balance" charge — in fact, it is precisely equal (not unequal) velocities that minimise the junction potential.

ANSWER: C