Question

a. Calculate e.m.f of cell for the reaction:
$M{g}_{\left(s\right)}+C{u}^{2+}\left(0.0001M\right)\to M{g}^{2+}\left(0.001M\right)+C{u}_{\left(s\right)}$
Given that: $E_{M{g}^{2+}/Mg}^o=-2.37$V
$E_{C{u}^{2+}/Cu}^o=+0.34$V
b. i) State Kohlrausch law.
ii) What is meant by limiting molar conductance.

Answer 1

a. $E_{cell}=[E_{C{u}^{+2}|Cu}^o-E_{M{g}^{+2}|Mg}^o]-[\frac{0.0592}{n}\times log\frac{\left[M{g}^{+2}\right]}{\left[C{u}^{+2}\right]}]$
$E_{cell}=[0.34-(-2.37\left)\right]-[\frac{0.0592}{2}\times lo{g}_{10}\frac{{10}^{-3}}{{10}^{-4}}]$
$E_{cell}=2.71-[0.0296\times lo{g}_{10}10]$
$E_{cell}=2.71-[0.0296\times 1]$
$E_{cell}=2.71-0.0296$
$E_{cell}=2.68$V
b. i) Limiting molar conductivity of an electrolyte can be represented as sum of the individual contribution of the anion and cation of the electrolyte.
${\Lambda }^0={\lambda }_{+}^0+{\lambda }_{-}^0$
In other words, at infinite dilution, each ion migrates independently of its co ion and makes its own contribution to the total molar conductivity of an electrolyte irrespective of the nature of the other ion with which it is associated.
ii) When concentration approaches zero, the molar conductance is known as limiting molar conductance.
As $C\to 0$, $\Lambda \to {\Lambda }^0$