Chapter 19, Problem 19.96SP

Interpretation Introduction

Interpretation:

The cell potential at 25°C for the following galvanic cell and the concentration of ${\text{Cu}}^{\text{2+}}$ when the cell potential drops to zero to maintain the concentration of ${\text{Pb}}^{\text{2+}}$ at 1.0 M should be determined:

${\text{Pb}}_{\text{(s)}}{\text{|Pb}}^{\text{2+}}\text{(1}{\text{.0 M) || Cu}}^{\text{2+}}\text{(1}\text{.0}\times {\text{10}}^{\text{-4}}{\text{M)|Cu}}_{\text{(s)}}$

Concept introduction:

In the electrochemical cell, the reactions at cathode and anode occur due to the difference in their reduction electrode potential value. The EMF of the cell can be calculated with the help of electrode reduction potential values. The reaction at each electrode is called as half-reaction and the combination of both half-reactions gives the cell reaction of given electrochemical cell. The standard cell potential for an electrochemical cell can be calculated as:

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{°}}{\text{ = E}}_{\text{cathode}}^{\text{°}}\text{}{\text{ - E}}_{\text{anode}}^{\text{°}}\\ {\text{E}}_{\text{cell}}^{\text{°}}{\text{ = E}}_{\text{reduction}}^{\text{°}}\text{}{\text{ - E}}_{\text{oxidation}}^{\text{°}}\end{array}$

The potential of the cell can be calculated with the help of the Nernst equation that can be shown as:

$\begin{array}{l}{\text{E° = E°}}_{\text{cell}}\text{ - }\frac{\text{0}\text{.0592 V}}{\text{n}}\text{ log Q }\\ \text{n = number of electrons}\\ \text{Q = reaction quotient}\end{array}$