   Chapter 18, Problem 127AE

Chapter
Section
Textbook Problem

# An experimental fuel cell has been designed that uses carbon monoxide as fuel. The overall reaction is 2 CO ( g )   +   O 2 ( g )   →   2 CO 2 ( g ) The two half-cell reactions are CO + O 2 −   →   CO 2 + 2 e − O 2 + 4 e − → 2 O 2 − The two half-reactions are carried out in separate compartments connected with a solid mixture of CeO2 and Gd2O3. Oxide ions can move through this solid at high temperatures (about 800°C). ∆G for the overall reaction at 800°C under certain concentration conditions is −380 kJ. Calculate the cell potential for this fuel cell at the same temperature and concentration conditions.

Interpretation Introduction

Interpretation:

The reactions taking place in a fuel cell in a given mixture of CeO2 and Gd2O3 and having Carbon monoxide as a fuel is given. The cell potential for this fuel at the given temperature conditions 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 cell potential for the given fuel cell at the given temperature conditions

Explanation

Given,

The value of ΔG is 380kJ .

The given temperature is 800°C .

Conversion of kJ to J is done as,

1kJ=1000J

Conversion of 380kJ to J is,

380kJ=380×1000J=-380×104J

The two half-cell reactions are,

CO+O2CO2+2e (1)

O2+4e2O2 (2)

Multiply equation (1) with a coefficient of 2 and then add both the equations,

2CO+2O22CO2+4eO2+4e2O2

The final equation is,

2CO+O22CO2

The reaction involves the transfer of 4 moles of electrons.

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

ΔG=nFEcell

Where,

• ΔG is the Gibbs free energy change at the given temperature conditions

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