Chapter 19, Problem 19.75QP

Interpretation Introduction

Interpretation:

The electrode potential of the given concentration cell has to be calculated.

Concept Introduction:

Concentration cell is an electrochemical cell in which the two half cells are the same.  The potential produced in the concentration cell is due to the different concentrations of the reacting species.  The cell potential of a concentration cell can be found by Nernst equation.

The standard electrode potential of a cell ${\text{(E}}^{°}{}_{\text{cell}})$ is the difference in electrode potential of the cathode and anode.

  ${\text{E}}^{\text{°}}{}_{\text{cell}}={\text{E}}^{\text{°}}{}_{\text{cathode}}-{\text{E}}^{\text{°}}{}_{\text{anode}}$

Nernst equation is one of the important equations in electrochemistry.  In Nernst equation the electrode potential of a cell reaction is related to the standard electrode potential, concentration or activities of the species that is involved in the chemical reaction and temperature.

  ${\text{E}}_{\text{cell}}{\text{=E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{\text{RT}}{\text{2}\text{.303nF}}\text{log}\frac{\left[\text{Red}\right]}{\left[\text{Oxd}\right]}$

Where,

  ${\text{E}}_{\text{cell}}$ is the potential of the cell at a given temperature

  ${\text{E}}^{\text{°}}{}_{\text{cell}}$ is the standard electrode potential

  $\text{R}$ is the universal gas constant $\text{(R=8}{\text{.314JK}}^{\text{-1}}{\text{mol}}^{\text{-1}})$

  $\text{T}$ is the temperature

  $\text{n}$ is the number of electrons involved in a reaction

  $\text{F}$ is the Faraday constant $\text{(F=9}\text{.64853399}\times {\text{10}}^4{\text{Cmol}}^{\text{-1}})$

  $\left[\text{Red}\right]$ is the concentration of the reduced species

  $\left[\text{Oxd}\right]$ is the concentration of the oxidised species

At room temperature $({25}^{°}C)$, after substituting the values of all the constants the equation can be written as

  ${\text{E}}_{\text{cell}}{\text{=E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{0.0591}{\text{n}}\text{log}\frac{\left[\text{Red}\right]}{\left[\text{Oxd}\right]}$