Chapter 18, Problem 18.103SP

Interpretation Introduction

Interpretation:

The $pH$ of the solution in cathode compartment should be determined.

Concept introduction:

• $pH$: - It is defined as the negative logarithm of the hydrogen ion concentration of the aqueous solution.
Thus

  $\begin{array}{l}pH=-\log \left[H^{+}\right]\\ \left[H^{+}\right]={10}^{-pH}\end{array}$
• Electrode potential: -The tendency of an electrode to lose or gain electrons when it is in interaction with solution of its own ion. If in the half cell, the rod of metal is suspended in a solution of $\text{1}\text{M}$ concentration and the temperature is kept at $298K$, the electrode potential is called the standard electrode potential.
• The standard electrode potential of many electrodes has been determined using the standard electrode potential for hydrogen, for which the electrode potential has been fixed to zero.
• EMF of the cell or cell potential: - The potential difference between two half-cell which constitute an electrochemical cell when no current is flowing through it.

  $E_{cell}=E^0{}_{cathode}-E^0{}_{anode}$
• Nernst equation: - It gives the quantitative relationship between the concentration of ions, temperature and electrode potential or cell potential
The electrode potential of an electrode is said to be standard electrode potential if it is measured under standard condition i.e. electrolytic concentration of $1M$ and temperature $298K$.However if the electrolytic concentration is different from $1M$ and temperature is different from $298K$ , the electrode potential has a different value. The value is obtained using Nernst equation.
Let us write the reaction
  $aA+bB\rightleftharpoons xX+yY$

Then Nernst equation is applied as follows:
  $E_{cell}=E^0{}_{cell}-\frac{RT}{nF}\ln \frac{{\left[X\right]}^x{\left[Y\right]}^y}{{\left[A\right]}^a{\left[B\right]}^b}$

where, $E^0$ is the standard electrode potential
$R$ is the universal gas constant= $8.314J/K/mol$

  $\mathrm{F}=96500coulomb$ =Faraday’s constant
  $T=298K$

  $n$ is the number of electrons in the electrode reactions
Now,
  $\begin{array}{l}{E}_{cell}=E^0{}_{cell}-\frac{RT}{nF}\ln \frac{{\left[X\right]}^x{\left[Y\right]}^y}{{\left[A\right]}^a{\left[B\right]}^b}\\ E_{cell}=E^0{}_{cell}-\frac{2.303RT}{nF}\log \frac{{\left[X\right]}^x{\left[Y\right]}^y}{{\left[A\right]}^a{\left[B\right]}^b}\\ E_{cell}=E^0{}_{cell}-\frac{0.0591}{n}\log \frac{{\left[X\right]}^x{\left[Y\right]}^y}{{\left[A\right]}^a{\left[B\right]}^b}\end{array}$

To determine:

The $pH$ of the solution in the cathode compartment