Question

(i) Define molar conductance of the solution of an electrolyte and discuss its change with the concentration.
(ii) Calculate the electromotive force (EMF) of the following cell:
\( \text{Ni(s)} \mid \text{Ni}^{2+}(0.16\,M) \parallel \text{Ag}^{+}(0.002\,M) \mid \text{Ag(s)} \)
(Given: \( E^{\circ}_{cell} = 1.05\,V \), \( \log_{10}2 = 0.3010 \))

Hint:

Molar conductance \( \Lambda_m = 1000\kappa/c \), rises on dilution. For EMF use Nernst \( E = E^{\circ} - (0.0591/n)\log([\text{Ni}^{2+}]/[\text{Ag}^{+}]^2) \) with \( n=2 \).

Answer 1

Part (i): Molar conductance
Step 1 (Definition): Molar conductance (\( \Lambda_m \)) is the conducting power of all the ions produced by dissolving one mole of an electrolyte in solution. It is related to conductivity (\( \kappa \)) by
\[ \Lambda_m = \frac{\kappa \times 1000}{c} \]
where \( c \) is the molar concentration (mol L\(^{-1}\)) and \( \kappa \) is the conductivity (S cm\(^{-1}\)). Its unit is S cm\(^2\) mol\(^{-1}\).
Step 2 (Variation with concentration): Molar conductance increases as concentration decreases (i.e. on dilution), because the total volume containing one mole of electrolyte increases and more ions conduct freely.
• For a strong electrolyte the increase is small and \( \Lambda_m \) follows Debye-Huckel-Onsager equation \( \Lambda_m = \Lambda^{\circ}_m - A\sqrt{c} \); \( \Lambda^{\circ}_m \) is obtained by extrapolating the straight line to zero concentration.
• For a weak electrolyte the increase is very sharp near infinite dilution because the degree of dissociation rises steeply, so \( \Lambda^{\circ}_m \) cannot be found by extrapolation (it is found using Kohlrausch's law).

Part (ii): EMF of the cell
Step 1 (Electrode reactions): Anode (oxidation): \( \text{Ni} \rightarrow \text{Ni}^{2+} + 2e^{-} \). Cathode (reduction): \( \text{Ag}^{+} + e^{-} \rightarrow \text{Ag} \) (multiply by 2). Net reaction: \( \text{Ni} + 2\text{Ag}^{+} \rightarrow \text{Ni}^{2+} + 2\text{Ag} \), so number of electrons \( n = 2 \).
Step 2 (Reaction quotient):
\[ Q = \frac{[\text{Ni}^{2+}]}{[\text{Ag}^{+}]^{2}} = \frac{0.16}{(0.002)^{2}} = \frac{0.16}{4\times10^{-6}} = 4\times10^{4} \]
Step 3 (Nernst equation):
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n}\log Q \]
Step 4 (Evaluate log): \( \log(4\times10^{4}) = 2\log 2 + 4 = 2(0.3010) + 4 = 4.6020 \).
Step 5 (Substitute):
\[ E_{cell} = 1.05 - \frac{0.0591}{2}(4.6020) = 1.05 - (0.02955)(4.6020) \]
\[ E_{cell} = 1.05 - 0.136 = 0.914\,V \]
\[ \boxed{E_{cell} \approx 0.914\,V} \]