Chapter 19, Problem 36E

Interpretation Introduction

(a)

Interpretation:

The value of $E_{\text{cell}}$ should be determined if the unknown solution in the half-cell on the left side contains the solution with pH =5.25.

Concept introduction:

For a solution, pH can be calculated as follows:

$\text{pH}=-{\text{log}}_{\text{10}}\left[{\text{H}}^{+}\right]$

Here, $\left[{\text{H}}^{+}\right]$ is hydrogen ion concentration in a solution.

In an electrochemical cell, there are two electrodes that can be observed.

If oxidation takes place on an electrode, that electrode is called anode. The species in that electrode remove electrons anditself gets oxidized.

If reduction takes place on an electrode, that electrode is called cathode. The species in that electrode absorbs electrons and itself gets reduced.

The electrode potential of cell can be calculated as follows:

$E^0{}_{\text{cell}}=E^0{}_{\text{cathode}} -E^0{}_{\text{anode}}$

Nernst equation for a cell can be represented as follows:

${\text{E}}_{cell}={\text{E}}^0{}_{cell}-\frac{0.0592\text{ V}}{\text{z}}\text{ log Q}$

Here,

z = number of moles of electrons transferred in the cell and Q is ratio of concentration of products to reactant in a cell.

Interpretation Introduction

(b)

Interpretation:

The value of $E_{\text{cell}}$ should be determined if the unknown solution in the half-cell on the left side contains the 0.0103 M HCl.

Concept introduction:

In an electrochemical cell, there are two electrodes that can be observed.

If oxidation takes place on an electrode, that electrode is called anode. The species in that electrode remove electrons and itself gets oxidized.

If reduction takes place on an electrode, that electrode is called cathode. The species in that electrode absorbs electrons and itself gets reduced.

The electrode potential of cell can be calculated as follows:

$E^0{}_{\text{cell}}=E^0{}_{\text{cathode}} -E^0{}_{\text{anode}}$

Nernst equation for a cell can be represented as follows:

${\text{E}}_{cell}={\text{E}}^0{}_{cell}-\frac{0.0592\text{ V}}{\text{z}}\text{ log Q}$

Here,

z = number of moles of electrons transferred in the cell and Q is ratio of concentration of products to reactant in a cell.

Interpretation Introduction

(c)

Interpretation:

The value of $E_{\text{cell}}$ should be determined if the unknown solution in the half-cell on the left side contains the 0.158 M CH₃COOH.

Concept introduction:

The acetic acid, CH₃COOH is weak acid. Therefore, partially dissociates into ions in water.

${\text{CH}}_{\text{3}}\text{COOH(aq)}\rightleftharpoons {\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{\text{(aq) + H}}^{\text{+}}\text{(aq)}$

In an electrochemical cell, there are two electrodes that can be observed.

If oxidation takes place on an electrode, that electrode is called anode. The species in that electrode remove electrons and itself gets oxidized.

If reduction takes place on an electrode, that electrode is called cathode. The species in that electrode absorbs electrons and itself gets reduced.

The electrode potential of cell can be calculated as follows:

$E^0{}_{\text{cell}}=E^0{}_{\text{cathode}} -E^0{}_{\text{anode}}$

Nernst equation for a cell can be represented as follows:

${\text{E}}_{cell}={\text{E}}^0{}_{cell}-\frac{0.0592\text{ V}}{\text{z}}\text{ log Q}$

Here,

z = number of moles of electrons transferred in the cell and Q is ratio of concentration of products to reactant in a cell.