Question

The conductivity of $0.1$ mol $L^{-1}$ solution of NaCl is $1.06 \times 10^{-2}$ S $cm^{-1}$. Calculate its molar conductivity and degree of dissociation. ($\lambda^\circ_{Na^+} = 50.1, \lambda^\circ_{Cl^-} = 76.5$ S $cm^2$ $mol^{-1}$). (b) (i) Predict current flow direction for $2Ag^+ + Zn \rightarrow 2Ag + Zn^{2+}$. (ii) Differentiate between primary and secondary battery.

Hint:

Always check the units of conductivity ($\kappa$). If it's in S $cm^{-1}$, use the factor of $1000$ in the numerator to get $\Lambda_m$ in S $cm^2$ $mol^{-1}$.

Answer 1

Concept:

• Molar conductivity: $\Lambda_m = \frac{\kappa \times 1000}{C}$

• Degree of dissociation: $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$

• Current flows in the direction opposite to electron flow.
Step 1: Calculating Conductivity and Dissociation.
Molar Conductivity ($\Lambda_m$): \[ \Lambda_m = \frac{1.06 \times 10^{-2} \times 1000}{0.1} = 106 \text{ S cm}^2 \text{ mol}^{-1} \] Limiting Molar Conductivity ($\Lambda_m^\circ$): \[ \Lambda_m^\circ = \lambda^\circ_{Na^+} + \lambda^\circ_{Cl^-} = 50.1 + 76.5 = 126.6 \text{ S cm}^2 \text{ mol}^{-1} \] Degree of dissociation ($\alpha$): \[ \alpha = \frac{106}{126.6} \approx 0.837 \]

Step 2: Cell Logic and Batteries.
(b)(i) In the reaction, $Zn$ is oxidized (anode) and $Ag^+$ is reduced (cathode). Electrons flow from $Zn \rightarrow Ag$. Thus, current flows from Silver (Ag) to Zinc (Zn). (b)(ii) Primary batteries cannot be recharged (reaction occurs once), e.g., Dry cell. Secondary batteries can be recharged by passing current, e.g., Lead storage battery.