   Chapter 19, Problem 62GQ

Chapter
Section
Textbook Problem

Which of the following reactions is (arc) product favored at equilibrium? (a) Zn(s) + I2(s) →Zn2+(aq) + 2 I−(aq) (b) 2 Cl−(aq) + I2(s) → Cl2(g) + 2I−(aq) (c) 2 Na+(aq) + 2 Cl−(aq) → 2 Na(s) + Cl2(g) (d) 2 K(s) + 2 H2O(ℓ) → 2 K+(aq) + H2(g) + 2 OH−(aq)

Interpretation Introduction

Interpretation:

The reactions which are product favoured has to be determined.

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

Reason for correct answers:

(a)

The given reaction is as follows.

Zn(s) + I2(s) Zn2+(aq) + 2I-(aq)

Let’s calculate the Ecello value.

Ecello= Ecathodeo-Eanodeo= 0.535 V- (-0.763 V)= +1.298 V

The value is positive, therefore it is a product favoured.

(d)

The given reaction is as follows.

2K(s) + 2H2O(l) 2K+(aq) + H2(g) + 2OH-(aq)

Let’s calculate the Ecello value.

Ecello= Ecathodeo-Eanodeo= -0.8277  V - (-2.925 V)= 2.0973 V

The value is positive, therefore it is a product favoured

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