   Chapter 18, Problem 123AE

Chapter
Section
Textbook Problem

# A fuel cell designed to react grain alcohol with oxygen has the following net reaction: C 2 H 5 OH ( l )   +   3 O 2 ( g )   →   2 CO 2 ( g )   +   3 H 2 O ( l ) The maximum work that 1 mole of alcohol can do is 1.32 × 103 kJ. What is the theoretical maximum voltage this cell can achieve at 25°C?

Interpretation Introduction

Interpretation:

The reaction taking place in grain alcohol and Oxygen in a fuel cell is given. The theoretical maximum voltage that given cell can achieve at 25°C for the given amount of maximum work is to be calculated.

Concept introduction:

Gibbs free energy is basically the maximum amount of non-expansion work done. Therefore, it is represented as,

Wmax=ΔG°

The maximum amount of work that is done by an electrochemical cell is given as,

Wmax=nFE°cell

The relationship between Gibbs free energy change and equilibrium constant is given by the formula,

ΔG°=nFE°cell

To determine: The value of theoretical maximum voltage that given cell can achieve at 25°C .

Explanation

Given,

The reaction between grain alcohol and oxygen is,

C2H5OH(l)+3O2(g)2CO2(g)+3H2O(l)

The maximum amount of work done is 1.32×103kJ .

Conversion of kJ to J is done as,

1kJ=1000J

Conversion of 1.32×103kJ to J is,

1.32×103kJ=1.32×103×1000J=132×104J

The number of moles of alcohol is 1

The reaction taking place at the cathode is,

3O2+12H++12e6H2O

The reaction taking place at the anode is,

C2H5OH+3H2O12H++12e+2CO2

The reaction involves the transfer of 12 moles of electrons.

The relationship between Gibbs free energy change and cell potential is given by the formula,

ΔG°=nFE°cell

Where,

• ΔG° is the Gibbs free energy change at the standard conditions.
• n is the number of moles of electrons that are involved in the reaction

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