Chapter 19, Problem 19.33QP

Interpretation Introduction

Interpretation:

The ratio of $\left[{\text{Cu}}^{\text{2+}}\right]\text{ and }{\left[{\text{Zn}}^{\text{2+}}\right]}^{}$ ion needed for spontaneity of the reaction was to be identified by using Figure 18.1.

Concept introduction:

For an electrochemical reaction to take place spontaneously the change in free energy must be negative. Change in free energy ΔG represents the maximum amount of useful work that can be obtained from the reaction and can be given by the following formula

$\Delta G=w_{\text{max}}$

$\Delta G=-nF{E}_{cell}$

Both n and F are positive quantities and ΔG is negative for a spontaneous process, so E_cell must be positive for a spontaneous process.

E_cell of the given cell can be calculated from the simplified Nernst equation.

${\text{E}}^{\text{o}}{}_{\text{cell}}{\text{=E}}^{\text{o}}{}_{\text{reduction}}{\text{-E}}^{\text{o}}{}_{\text{oxidation}}$

${\text{E}}_{\text{cell}}{\text{= E}}^{\text{o}}{}_{\text{cell}}\text{-(}\frac{\text{0}\text{.0257}}{n}\text{)ln}\frac{\text{[ox]}}{\text{[red]}}$

E^o_cell = standard potential of a cell

n = Number of electrons transferred in the cell reaction

[ox] = Concentration of the oxidation half

[red] = Concentration of the reduction half

In order to calculate the ratio of [Cu²⁺]/[Zn²⁺] for a spontaneous cell reaction E_cell was fixed as zero, because this is the point at which the reaction become spontaneous and Nernst equation can be modified as

${\text{0= E}}^{\text{o}}{}_{\text{cell}}\text{-(}\frac{\text{0}\text{.0257}}{n}\text{)ln}\frac{{\text{[Cu}}^{\text{2+}}\text{]}}{{\text{[Zn}}^{\text{2+}}\text{]}}$

To find: The ratio of $\left[{\text{Cu}}^{\text{2+}}\right]\text{/}\left[{\text{Zn}}^{\text{2+}}\right]$ at which the given reaction can takes spontaneously