Chapter 19, Problem 19.34QP

Interpretation Introduction

Interpretation:

Need to calculate the emf of ${\text{Mg}}_{\text{(s)}}│{\text{Mg}}^{\text{2+}}{}_{\text{(0}\text{.24M)}}\parallel {\text{Mg}}^{\text{2+}}{}_{\text{(0}\text{.53M)}}│{\text{Mg}}_{\text{(s)}}\text{cell}$.

Concept Introduction:

Since the electrode potential depends on the concentration of the ion, galvanic cell with two half-cell made up of same materials with different concentration can be constructed, these cells are called concentration cells. The given galvanic cell belongs to such category in which both anode and cathode are made of magnesium, but differ in concentration.  

Cell reaction for the above cell can written as

$\begin{array}{l}\text{Anode (Oxidation)}{\text{Mg}}_{\text{(s)}}\stackrel{}{\to }{\text{Mg}}^{\text{2+}}{}_{\text{(0}\text{.24}\text{M)}}{\text{+2e}}^{\text{-}}\\ \text{Cathode (Reduction)}{\text{Mg}}^{\text{2+}}{}_{\text{(0}\text{.53}\text{M)}}{\text{+2e}}^{\text{-}}\stackrel{}{\to }{\text{Mg}}_{\text{(s)}}\\ \text{Overall}\text{Reaction}{\text{ Mg}}^{\text{2+}}{}_{\text{(0}\text{.53}\text{M)}}\stackrel{}{\to }{\text{Mg}}^{\text{2+}}{}_{\text{(0}\text{.24M)}}\end{array}$

In order to calculate the emf of the cell, standard potential of a cell (E^o_cell) of the cell need to be calculated using the formula, In the present case, since cathode and anode are made up of same material, so the standard potential of the given cell can be calculated by the below equation.

${\text{E}}^{\text{o}}{}_{\text{Anode}}{\text{=E}}^{\text{o}}{}_{\text{cathode}}$

Electro motive force (emf) of the cell can be calculated by Nernst equation:

${\text{E}}_{\text{cell}}{\text{= E}}^{\text{o}}{}_{\text{cell}}\text{-}\frac{\text{RT}}{\text{nF}}\text{lnQ}$

E^o_cell = Standard potential of a cell

R = Universal gas constant (8.31 J mole^-1 K^-1)

T = Absolute temperature in Kelvin

F = Faraday constant, number of coulombs per mole of electron (96845 C/mole)

n = Number of electrons transferred in the cell reaction

Q = Reaction quotient

Since the cell reaction takes place at room temperature (298K), so the Nernst equation was simplified as

${\text{E}}_{\text{cell}}{\text{= E}}^{\text{o}}{}_{\text{cell}}\text{-(}\frac{\text{0}\text{.0257}}{n}\text{)lnQ}$

Further the above equation can be written in terms of log₁₀

${\text{E}}_{\text{cell}}{\text{= E}}^{\text{o}}{}_{\text{cell}}\text{-(}\frac{\text{0}\text{.0592}}{n}\text{)logQ}$

Reaction quotient can be rewritten in terms of oxidation and reduction concentration 

${\text{E}}_{\text{cell}}{\text{= E}}^{\text{o}}{}_{\text{cell}}\text{-(}\frac{\text{0}\text{.0592}}{n}\text{)log}\frac{\text{[ox]}}{\text{[red]}}$

So Nernst equation for the concentration cell can be given as follows

${\text{E}}_{\text{cell}}{\text{= E}}^{\text{o}}{}_{\text{cell}}\text{-(}\frac{\text{0}\text{.0592}}{n}\text{)log}\frac{\text{[dilute solution]}}{\text{[Concentrated solution]}}$

To find: Calculation of emf of the given concentration cell.