   Chapter 19, Problem 55PS

Chapter
Section
Textbook Problem

In the presence of oxygen and water two half-reactions responsible for the corrosion of iron areFe(s) → Fe2+(aq) + 2 e−O2(g) + 2 H2O(ℓ) + 4 e− → 4 OH−(aq)Calculate the the standard potential E°, and decide whether the reaction is product-favored at equilibrium. Will decreasing the pH make the reaction less thermodynamically product-favored at equilibrium?

Interpretation Introduction

Interpretation:

The Eo value for the given reactions has to be determined and have to decide whether each one 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 (Ecello).

Standard potential (Ecello) can be calculated by the following formula.

Ecello=Ecathodeo-Eanodeo

The Ecello value is positive, the reaction is predicted to be product favoured at equilibrium.

The Ecello value is negative, the reaction is predicted to be reactant favoured at equilibrium

Explanation

The given reaction is as follows.

Fe(s)   Fe2+(aq) + 2e-O2(g) + 2H2O(l) + 4e-   4OH-(aq)

Let’s write the half reactions:

At anode:Oxidation:  2Fe(s) 2Fe2+(s) +24e-At cathode:Reduction: O2(g) +2H2O + 4e-  4OH-(aq)

Let’s calculate the Ecello value

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