Concentration Cells - Practice Questions & MCQ

The device in which both the half cells contain the same electrode but differ in the concentration (or activity ) of the species involved. Oxidation and reduction occur at respective electrodes until the concentration becomes equal in both the half cells. In other words, the concentration cell is one in which emf arises as a result of different concentrations of the same electrolyte in the component half-cells.

• The two solutions are connected by a salt bridge and the electrodes are joined by a piece of metallic wire.
• A concentration cell dilutes the concentrated solution and concentrates the more dilute solution and generate potential till the cell reaches an equilibrium.
• Potential is generated due to a decrease in Gibb's energy of cell till the attainment of equilibrium.
• E^o of a concentration cell is equal to zero where the E_cell depends upon the concentration (on acting of species involves).

Let us consider the given concentration cell,

$\begin{array}{ccccc}Cu(s)& C{u}^{2+}& \Vert & C{u}^{2+}(\text{ aq. })& \mid \\ & C_{1}M& & C_{2}M& \end{array}$

For the above cell to be working, E_cell  > 0

Therefore, From Nernst Equation:

$\begin{array}{rl}& {\mathrm{E}}_{\text{cell }}={\mathrm{E}}_{\text{cell }}^{\mathrm{o}}-\frac{0.059}{2}\log \frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}\\ & {\mathrm{E}}_{\text{cell }}={\mathrm{E}}_{\text{cell }}^{\mathrm{o}}-\frac{0.059}{2}\log \frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}\\ & {\mathrm{E}}_{\text{cell }}=0+\frac{0.059}{2}\log \frac{{\mathrm{c}}_2}{{\mathrm{c}}_1}>0\\ & \Rightarrow {\mathrm{c}}_2>{\mathrm{c}}_1\end{array}$

The cell representation of the concentration cell with respect to hydrogen is given as follows:

$\underline{\begin{array}{cccccl}Pt\mid & H_2(g)& \mid H^{+}(aq)& \Vert & H^{+}(aq)& \mid H_2(g)\\ P_1& C_1& C_2& P_2& \mid Pt\end{array}}$

At anode:

$\frac{1}{2}{\mathrm{H}}_2\to {\mathrm{H}}^{+}+1{\mathrm{e}}^{-}$

At cathode:

${\mathrm{H}}^{+}+1{\mathrm{e}}^{-}⟶\frac{1}{2}{\mathrm{H}}_2$

The complete cell reaction is the addition of both of these reactions and is given as follows:

$\frac{1}{2}{\mathrm{H}}_2(\mathrm{g})+{\mathrm{H}}^{+}(\mathrm{aq})\to {\mathrm{H}}^{+}(\mathrm{aq})+\frac{1}{2}{\mathrm{H}}_2(\mathrm{g})$

Thus, the reaction quotient is given as follows:

$\mathrm{Q}=\frac{\left[{\mathrm{H}}^{+}\right]{\left[{\mathrm{P}}_{\mathrm{H}2}\right]}^{1/2}}{\left[{\mathrm{H}}^{+}\right]{\left[{\mathrm{P}}_{\mathrm{H}}2\right]}^{1/2}}=\frac{{\mathrm{c}}_1{\left({\mathrm{P}}_2\right)}^{1/2}}{{\mathrm{c}}_2{\left({\mathrm{P}}_1\right)}^{1/2}}$
where P₁ and P₂ are the pressures of hydrogen gas at anode and cathode, respectively.

Now, we have:

$\begin{array}{rl}& {\mathrm{E}}_{\text{cell }}^{\mathrm{o}}=0\\ & \mathrm{n}=1\end{array}$

Thus, at T = 298K, the cell equation can be given as follows:

$\begin{array}{rl}& {\mathrm{E}}_{\text{cell }}=-\frac{0.059}{\mathrm{n}}{\log }_{10}\mathrm{Q}\\ & {\mathrm{E}}_{\text{cell }}=\frac{0.059}{1}{\log }_{10}\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}\mathrm{x}{\left(\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}\right)}^{1/2}\\ & {\mathrm{E}}_{\text{cell }}=-\frac{0.059}{1}[{\log }_{10}{\left[{\mathrm{H}}^{+}\right]}_{\mathrm{A}}-{\log }_{10}{\left[{\mathrm{H}}^{+}\right]}_{\mathrm{c}}+{\log }_{10}{\left(\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}\right)}^{1/2}]\\ & {\mathrm{E}}_{\text{cell }}=0.059[-{\log }_{10}{\left[{\mathrm{H}}^{+}\right]}_{\mathrm{A}}-(-{\log }_{10}{\left[{\mathrm{H}}^{+}\right]}_{\mathrm{c}})-{\log }_{10}{\left(\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}\right)}^{1/2}]\end{array}$

$\begin{array}{rl}& {\mathrm{E}}_{\text{cell }}=0.059[{\mathrm{pH}}_{\text{(anode) })}-{\mathrm{pH}}_{\text{(cathode })}-\frac{1}{2}{\log }_{10}\left(\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}\right)]\\ & {\mathrm{E}}_{\text{cell }}=0.059[{\mathrm{pH}}_{\text{(anode) })}-{\mathrm{pH}}_{\text{(cathode) }}+\frac{1}{2}{\log }_{10}\left(\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}\right)]\end{array}$

This is the final equation for the value of E_cell for concentration cell with respect to standard hydrogen electrode.

It is necessary to define a few terms before we consider the subject of conductance of electricity through electrolytic solutions. The electrical resistance is represented by the symbol ‘R’ and it is measured in ohm (Ω). It can be measured with the help of a Wheatstone bridge with which you are familiar with your study of physics. The electrical resistance of any object is directly proportional to its length, l, and inversely proportional to its area of cross-section, A. That is,
$\mathrm{R}\propto \frac{\mathrm{l}}{\mathrm{A}}$ or $\mathrm{R}=\rho \frac{\mathrm{l}}{\mathrm{A}}$
The constant of proportionality, ρ (Greek, rho), is called resistivity (specific resistance). Its SI units are ohm meter (Ω m) and quite often its submultiple, ohm centimeter (Ω cm) is also used.

The inverse of resistance, R, is called conductance, G, and we have the relation:
$\mathrm{G}=\frac{1}{\mathrm{R}}=\frac{\mathrm{A}}{\rho \mathrm{l}}=\kappa \frac{\mathrm{A}}{\mathrm{l}}$
The SI unit of conductance is siemens, represented by the symbol ‘S’ and is equal to ohm^–1 (also known as mho) or Ω^–1. The inverse of resistivity, called conductivity (specific conductance) is represented by the symbol, κ (Greek, kappa). The SI units of conductivity are S m^–1 but quite often, κ is expressed in S cm^–1. The conductivity of a material in S m^–1 is its conductance when it is 1 m long and its area of cross-section is 1 m². It may be noted that 1 S cm^–1 = 100 S m^–1.

The quantity l/A is called cell constant and is denoted by the symbol G*. It depends upon the distance between electrodes and the area of cross-section. On the basis of their definition, the conductance(G), Conductivity() and cell constant (G*) are related as:

$\kappa =\mathrm{G}\times {\mathrm{G}}^{\ast }=\mathrm{G}\times \frac{\mathrm{l}}{\mathrm{A}}$

Now, when l=1 cm² then v=1 cm³ = 1 ml.

So,  can also be represented as the conductance of 1 ml of the electrolytic solution in the consistent set of units.

It has been observed that the magnitude of conductivity varies a great deal and depends on the nature of the material. It also depends on the temperature and pressure at which the measurements are made. Materials are classified into conductors, insulators, and semiconductors depending on the magnitude of their conductivity.

Electrical conductance through metals is called metallic or electronic conductance and is due to the movement of electrons. The electronic conductance depends on:
(i) the nature and structure of the metal
(ii) the number of valence electrons per atom
(iii) temperature (it decreases with the increase in temperature).

The conductance of electricity by ions present in the solutions is called electrolytic or ionic conductance. The conductivity of electrolytic (ionic) solutions depends on:
(i) the nature of the electrolyte added
(ii) size of the ions produced and their solvation
(iii) the nature of the solvent and its viscosity
(iv) the concentration of the electrolyte
(v) temperature (it increases with the increase in temperature).
Passage of direct current through an ionic solution over a prolonged period can lead to change in its composition due to electrochemical reactions and hence alternating current is used for the measurement of conductance.