Chapter 19, Problem 19.133SP

Interpretation Introduction

(a)

Interpretation:

The sketch of the cell should be drawn that shows the anode and cathode, the sign of the electrodes and the direction of electrons and ion flow for the electrolysis of aqueous sulfuric acid.

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 known 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:

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

Interpretation Introduction

(b)

Interpretation:

The balanced equation for anode, cathode and overall cell reaction for the electrolysis of aqueous sulfuric acid should be determined.

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 known 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:

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