   Chapter 19.9, Problem 1.1ACP

Chapter
Section
Textbook Problem

A lithium ion battery produces a voltage of 3.6 V.a. What is the energy released for each mole of lithium oxidized during the discharge of the battery?b. What is the energy released per kilogram of lithium?

(a)

Interpretation Introduction

Interpretation:

The energy released for each mole of lithium oxidised during the discharge of the battery has to be determined if the battery produces a voltage of 3.6V.

Concept introduction:

In voltaic cell, the maximum cell potential is directly related to the free energy difference between the reactants and products in the cell.

ΔG0= -nFE0n = Number of moles transferred per mole of reactant and productsF = Faradayconstant=96485C/mol  E0= Volts = Work(J)/Charge(C)

The relation between standard cell potential and equilibrium constant is as follows.

lnK = nE00.0257 at 298K

Explanation

Given data:

E°=3.6V

Let’s calculate the ΔG0 for the reaction.

ΔG0= -nFE0n = 1F = 96485C/mol  E0= 3.6 V

Substitute the values and calculate the ΔG0.

ΔG0= -nFE0n = -(1)(96485  C/mol)(3.6 V)=-3

(b)

Interpretation Introduction

Interpretation:

The energy released per kilogram of lithium has to be determined.

Concept introduction:

In voltaic cell, the maximum cell potential is directly related to the free energy difference between the reactants and products in the cell.

ΔG0= -nFE0n = Number of moles transferred per mole of reactant and productsF = Faradayconstant=96485C/mol  E0= Volts = Work(J)/Charge(C)

The relation between standard cell potential and equilibrium constant is as follows.

lnK = nE00.0257 at 298K

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