Chapter 19.7, Problem 1PPB

Interpretation Introduction

Interpretation:

The emf of theconcentration cell at ${\text{25}}^{\circ }\text{C}$ isto be determined.

Concept introduction:

The solubility product, $K_{\text{sp}}$, is shown as the product of the dissolved ions concentrations raised to the power of their stoichiometric coefficients. For example:

$\begin{array}{l}\text{MyXz }\underset{}{\rightleftharpoons }{\text{ yM}}^{\text{z+}}{\text{(aq) + zX}}^{\text{y-}}(aq)\\ K_{\text{sp}}={[}^{{\text{M}}^{\text{z+}}}{\text{ [X}}^{\text{y-}}{\text{]}}^{\text{z}}\end{array}$

Here, $[{\text{M}}^{\text{z+}}\text{]}$ and ${\text{[X}}^{\text{y-}}\text{]}$ are the molar solubilities of the ions.

The standard reduction potential of a galvanic cell may be calculated in terms of the standard reduction potential of cathode and anode as the relation mentioned below.

$E_{cell}^o=E_{cathode}^o-E_{anode}^o$

Here, $E_{cell}^{\circ }$ is the standard cell potential, $E_{anode}^{\circ }$ is the standard potential of anode, and $E_{cathode}^{\circ }$ is the standard potential of cathode.

Reaction quotient is the ratio of the molar concentration of the ions in the products to that of the reactants, raised to the power of their stoichiometric coefficients.

$Q=\frac{\text{concentration}\text{of}\text{product}}{\text{concentration}\text{of}\text{reactant}}$

According to the Nernst equation, the relation between emf, standard cell potential, and reaction quotient, at $298\text{K}$, is as follows:

$E\text{ = }{E}_{\text{cell}}^{\text{o}}-\frac{\text{0}\text{.0592V}}{\text{n}}\log Q$

Here, $E$ is the cell potential, $E_{cell}^{\circ }$ is the standard cell potential, Q is the reaction quotient, and $n$ is the transfer of electrons in the half-cell reactions.