Conductors of electricity. Materials fall into two classes. Electronic (metallic) conductors carry current by the flow of electrons; their conductance falls as temperature rises. Electrolytic conductors — solutions of salts, acids and bases, or molten salts — carry current by the movement of ions; their conductance rises with temperature and depends on ion concentration. Electrochemistry deals mainly with the second class.
Resistance and conductance. For any conductor, $R = \rho\frac{l}{A}$, where $\rho$ is the resistivity. The reciprocal of resistance is conductance $G = 1/R$, measured in siemens (S). The reciprocal of resistivity is the conductivity $\kappa$ (kappa): $\kappa = \frac{1}{\rho} = \frac{l}{A}\cdot\frac{1}{R} = G\cdot\frac{l}{A}$. The factor $\frac{l}{A}$ is the cell constant, $G^* $, fixed for a given conductivity cell. The unit of $\kappa$ is $\text{S m}^{-1}$ (often $\text{S cm}^{-1}$).
Conductivity, $\kappa$. Conductivity is the conductance of a solution held between two electrodes of unit area, one metre (or one cm) apart. It measures conductance per unit volume, so it falls on dilution — there are simply fewer ions in each unit volume.
Molar conductivity, $\Lambda_m$. To compare electrolytes fairly we use the molar conductivity, the conducting power of all the ions from one mole of electrolyte: $\Lambda_m = \frac{\kappa \times 1000}{C}$, where $C$ is the concentration in $\text{mol L}^{-1}$ and $\kappa$ is in $\text{S cm}^{-1}$, giving $\Lambda_m$ in $\text{S cm}^2\,\text{mol}^{-1}$. Unlike $\kappa$, molar conductivity rises on dilution, because the same one mole of ions spreads through more solution and the ions move more freely.
Variation with concentration. The way $\Lambda_m$ changes with dilution distinguishes the two kinds of electrolyte:
- Strong electrolytes (e.g. $\text{KCl}$, $\text{HCl}$, $\text{NaCl}$) are almost fully ionised. As concentration falls, $\Lambda_m$ increases only slightly and linearly with $\sqrt{C}$. Debye, Hückel and Onsager showed $\Lambda_m = \Lambda_m^0 - A\sqrt{C}$. Extrapolating the straight line to $C \rightarrow 0$ gives the limiting molar conductivity $\Lambda_m^0$.
- Weak electrolytes (e.g. $\text{CH}_3\text{COOH}$, $\text{NH}_4\text{OH}$) are only partly ionised. As the solution is diluted, the degree of dissociation rises sharply, so $\Lambda_m$ increases steeply and the curve shoots up near zero concentration — it cannot be extrapolated to find $\Lambda_m^0$ directly.
Kohlrausch's law of independent migration. At infinite dilution, where ions move independently of one another, the limiting molar conductivity is the sum of the individual ionic contributions: $\Lambda_m^0 = \nu_+\lambda_+^0 + \nu_-\lambda_-^0$, where $\lambda^0$ are the limiting molar conductivities of the ions and $\nu$ their numbers per formula unit.
Applications of Kohlrausch's law. (i) It gives $\Lambda_m^0$ for weak electrolytes indirectly — e.g. $\Lambda_m^0(\text{CH}_3\text{COOH}) = \Lambda_m^0(\text{CH}_3\text{COONa}) + \Lambda_m^0(\text{HCl}) - \Lambda_m^0(\text{NaCl})$. (ii) It gives the degree of dissociation $\alpha = \frac{\Lambda_m}{\Lambda_m^0}$ and hence the dissociation constant $K_a$ of weak acids.
The resistance of a conductivity cell filled with $0.1\,\text{M}\;\text{KCl}$ is $100\,\Omega$. If the conductivity of $0.1\,\text{M}\;\text{KCl}$ is $1.29\,\text{S m}^{-1}$, find the cell constant.
Solution- Step 1: Conductivity $\kappa = \dfrac{1}{R}\times G^* = \dfrac{G^*}{R}$, so cell constant $G^* = \kappa \times R$.
- Step 2: Substitute $\kappa = 1.29\,\text{S m}^{-1}$ and $R = 100\,\Omega$.
- Step 3: $G^* = 1.29 \times 100 = 129\,\text{m}^{-1}$.
- Step 4: Converting, $G^* = 1.29\,\text{cm}^{-1}$.
Answer: The cell constant is $129\,\text{m}^{-1}$ (i.e. $1.29\,\text{cm}^{-1}$).
A $0.02\,\text{M}$ solution of an electrolyte has a conductivity of $0.0248\,\text{S cm}^{-1}$. Calculate its molar conductivity.
Solution- Step 1: $\Lambda_m = \dfrac{\kappa \times 1000}{C}$ with $\kappa$ in $\text{S cm}^{-1}$ and $C$ in $\text{mol L}^{-1}$.
- Step 2: Substitute $\kappa = 0.0248$, $C = 0.02$.
- Step 3: $\Lambda_m = \dfrac{0.0248 \times 1000}{0.02} = \dfrac{24.8}{0.02}$.
- Step 4: $\Lambda_m = 1240\,\text{S cm}^2\,\text{mol}^{-1}$.
Answer: The molar conductivity is $1240\,\text{S cm}^2\,\text{mol}^{-1}$.
Given $\lambda^0(\text{Na}^+) = 50.1$ and $\lambda^0(\text{Cl}^-) = 76.3\,\text{S cm}^2\,\text{mol}^{-1}$, find $\Lambda_m^0$ for $\text{NaCl}$.
Solution- Step 1: By Kohlrausch's law, $\Lambda_m^0(\text{NaCl}) = \lambda^0(\text{Na}^+) + \lambda^0(\text{Cl}^-)$.
- Step 2: Substitute the values: $50.1 + 76.3$.
- Step 3: $\Lambda_m^0 = 126.4\,\text{S cm}^2\,\text{mol}^{-1}$.
Answer: $\Lambda_m^0(\text{NaCl}) = 126.4\,\text{S cm}^2\,\text{mol}^{-1}$.
The molar conductivity of $0.1\,\text{M}$ acetic acid is $5.2\,\text{S cm}^2\,\text{mol}^{-1}$ and $\Lambda_m^0 = 390.5\,\text{S cm}^2\,\text{mol}^{-1}$. Find its degree of dissociation.
Solution- Step 1: Degree of dissociation $\alpha = \dfrac{\Lambda_m}{\Lambda_m^0}$.
- Step 2: Substitute $\Lambda_m = 5.2$ and $\Lambda_m^0 = 390.5$.
- Step 3: $\alpha = \dfrac{5.2}{390.5} = 0.0133$.
- Step 4: As a percentage, $\alpha \approx 1.33\%$.
Answer: The degree of dissociation is about $0.0133$ (1.33%).
Using $\Lambda_m^0(\text{CH}_3\text{COONa}) = 91.0$, $\Lambda_m^0(\text{HCl}) = 426.2$ and $\Lambda_m^0(\text{NaCl}) = 126.5\,\text{S cm}^2\,\text{mol}^{-1}$, find $\Lambda_m^0$ of acetic acid.
Solution- Step 1: Kohlrausch: $\Lambda_m^0(\text{CH}_3\text{COOH}) = \Lambda_m^0(\text{CH}_3\text{COONa}) + \Lambda_m^0(\text{HCl}) - \Lambda_m^0(\text{NaCl})$.
- Step 2: Substitute: $91.0 + 426.2 - 126.5$.
- Step 3: $= 517.2 - 126.5$.
- Step 4: $\Lambda_m^0 = 390.7\,\text{S cm}^2\,\text{mol}^{-1}$.
Answer: $\Lambda_m^0(\text{CH}_3\text{COOH}) = 390.7\,\text{S cm}^2\,\text{mol}^{-1}$.
Explain why molar conductivity of a weak electrolyte increases sharply on dilution while that of a strong electrolyte increases only slightly.
Solution- Step 1: Strong electrolytes are nearly fully ionised at all concentrations, so dilution barely changes the number of ions.
- Step 2: Dilution only reduces interionic attractions, raising ion mobility a little, so $\Lambda_m$ rises slowly along $\Lambda_m = \Lambda_m^0 - A\sqrt{C}$.
- Step 3: Weak electrolytes are only partly ionised; dilution drives the dissociation equilibrium forward (Le Chatelier).
- Step 4: The number of free ions rises steeply, so $\Lambda_m$ shoots up rapidly near infinite dilution.
Answer: Strong electrolytes gain ions slowly (mobility only); weak electrolytes gain many more ions through increased dissociation on dilution.