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):
Step 1: Molar conductivity \(\Lambda_m\) measures how well one mole of an electrolyte conducts electricity when placed between two electrodes 1 cm apart, with enough solution to hold that whole mole. Formula: \(\Lambda_m = \dfrac{1000\,\kappa}{c}\), unit S cm\(^2\) mol\(^{-1}\).
Step 2: On dilution the number of free ions per mole effectively rises, so \(\Lambda_m\) goes up. Strong electrolytes (e.g. KCl) already fully ionise, so their rise is gentle and linear in \(\sqrt{c}\); extrapolating gives the limiting value \(\Lambda^{\circ}_m\). Weak electrolytes (e.g. CH\(_3\)COOH) ionise only partly, so at low \(c\) their \(\Lambda_m\) shoots up steeply and \(\Lambda^{\circ}_m\) must be got indirectly through Kohlrausch's law of independent migration of ions.

Part (ii):
Step 1: Silver is the stronger oxidiser, so Ag\(^{+}\) is reduced at the cathode and Ni is oxidised at the anode. Balanced cell reaction: Ni \(+\) 2Ag\(^{+}\) \(\rightarrow\) Ni\(^{2+}\) \(+\) 2Ag, with \(n = 2\) electrons transferred.
Step 2: Use the base-10 Nernst form at 298 K, \(E = E^{\circ} - \dfrac{0.0591}{n}\log\dfrac{[\text{Ni}^{2+}]}{[\text{Ag}^{+}]^{2}}\).
Step 3: Ratio \(= \dfrac{0.16}{(2\times10^{-3})^{2}} = \dfrac{0.16}{4\times10^{-6}} = 40000\). Its log \(= \log 4 + \log 10^{4} = 0.6020 + 4 = 4.6020\).
Step 4: \(E = 1.05 - 0.02955\times4.6020 = 1.05 - 0.1360\).
\[ \boxed{E_{cell} = 0.914\,V} \]
The positive EMF confirms the reaction is spontaneous as written.