Chapter 19, Problem 56PS

Interpretation Introduction

Interpretation:

The ${\text{E}}^{\text{o}}$ value for the given reactions has to be determined and has to decide whether each is product favoured at equilibrium and also has to check whether decreasing $pH$ makes the reaction less thermodynamically product-favoured at equilibrium.

Concept introduction:

Electrochemical cells:

Therese are chemical energy is converted into electrical energy.

In all electrochemical cells, oxidation occurs at anode and reduction occurs at cathode.

An anode is indicated by negative sign and cathode is indicated by the positive sign.

Electrons flow in the external circuit from the anode to the cathode.

In the electrochemical cells two half cells are connected with salt bridge. It allows the cations and anions to move between the two half cells.

Under certain conditions a cell potential is measured it is called as standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$.

Standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$ can be calculated by the following formula.

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

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is positive, the reaction is predicted to be product favoured at equilibrium.

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is negative, the reaction is predicted to be reactant favoured at equilibrium.