Appeared in: 2026 (3 sets)
(A) (a) The conductivity of $0.1\text{ mol L}^{-1}$ solution of $\text{NaCl}$ is $1.06 \times 10^{-2}\text{ S cm}^{-1}$. Calculate its molar conductivity and degree of dissociation.
$\lambda^\circ_{\text{Na}^+} = 50.1\text{ S cm}^2\text{ mol}^{-1}$
$\lambda^\circ_{\text{Cl}^-} = 76.5\text{ S cm}^2\text{ mol}^{-1}$
(b) (i) Following cell reaction occurs in a galvanic cell :
$2\text{Ag}^+_{(\text{aq})} + \text{Zn}_{(\text{s})} \longrightarrow 2\text{Ag}_{(\text{s})} + \text{Zn}^{2+}_{(\text{aq})}\text{ E}^\circ_{(\text{cell})} = +1.56\text{ V}$
Predict the direction of flow of current.
(ii) Differentiate between a primary battery and a secondary battery.
OR
(B) (a) Resistance of a conductivity cell filled with $0.1\text{ M KCl}$ solution is $100\text{ }\Omega$. If the resistance of the same cell when filled with $0.01\text{ mol L}^{-1}\text{ KCl}$ solution is $300\text{ }\Omega$, calculate the conductivity and molar conductivity of $0.01\text{ mol L}^{-1}\text{ KCl}$ solution. The conductivity of $0.1\text{ M KCl}$ solution is $1.29 \times 10^{-2}\text{ S cm}^{-1}$.
(b) (i) Write any two advantages of $\text{H}_2 - \text{O}_2$ fuel cell.
(ii) Why does the cell potential of mercury cell remain constant throughout the life ?
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Solution (A)
(a) Molar Conductivity and Degree of Dissociation
Step 1 (Formula & Given Data): [1 Mark]
- Given: $\kappa = 1.06 \times 10^{-2} \text{ S cm}^{-1}$, $C = 0.1 \text{ mol L}^{-1} = 0.1 \times 10^{-3} \text{ mol cm}^{-3}$
- Formula for Molar Conductivity: $\Lambda_m = \frac{\kappa \times 1000}{M}$
- Formula for Limiting Molar Conductivity: $\Lambda_m^\circ = \lambda^\circ_{\text{Na}^+} + \lambda^\circ_{\text{Cl}^-}$
Step 2 (Mathematical Working): [1 Mark]
Calculation of $\Lambda_m$:
$$\Lambda_m = \frac{1.06 \times 10^{-2} \text{ S cm}^{-1} \times 1000 \text{ cm}^3 \text{ L}^{-1}}{0.1 \text{ mol L}^{-1}} = 106 \text{ S cm}^2 \text{ mol}^{-1}$$Calculation of $\Lambda_m^\circ$:
$$\Lambda_m^\circ = 50.1 + 76.5 = 126.6 \text{ S cm}^2 \text{ mol}^{-1}$$Calculation of Degree of Dissociation ($\alpha$):
$$\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{106}{126.6} \approx 0.837$$Step 3 (Final Answer): [0.5 Mark]
$\Lambda_m = 106 \text{ S cm}^2 \text{ mol}^{-1}$ and $\alpha = 0.837$ (or $83.7\%$).
(b) Galvanic Cell Concepts
(i) Direction of current: [0.5 Mark]
Current flows from the cathode (Ag electrode) to the anode (Zn electrode) in the external circuit.
(ii) Primary vs Secondary Battery: [2 Marks]
- Primary Battery: The cell reaction occurs only once; the battery becomes dead after use and cannot be recharged (e.g., Dry cell).
- Secondary Battery: The cell reaction can be reversed by passing an external current, allowing the battery to be recharged and used multiple times (e.g., Lead storage battery).
Solution (B)
(a) Conductivity and Molar Conductivity
Step 1 (Formula & Given Data): [1 Mark]
- Cell constant ($G^*$) = $\kappa \times R = 1.29 \times 10^{-2} \text{ S cm}^{-1} \times 100 \text{ }\Omega = 1.29 \text{ cm}^{-1}$
- Conductivity ($\kappa$) = $\frac{G^*}{R_2} = \frac{1.29}{300}$
Step 2 (Mathematical Working): [1.5 Marks]
Calculation of $\kappa$:
$$\kappa = 4.3 \times 10^{-3} \text{ S cm}^{-1}$$Calculation of $\Lambda_m$:
$$\Lambda_m = \frac{\kappa \