Chapter 8, Problem 17Q

In addition to hydrogen, methane also has been studied for use in PEM fuel cells. Balance the given oxidation and reduction half-reactions, and write the overall equation for a methane-based fuel cell.

Oxidation half-reaction:

$\begin{array}{l}\_\_{\text{CH}}_4(g)+\_\_{\text{OH}}^{-}(aq)\to \\ \_\_{\text{CO}}_2(g)+\_\_{\text{H}}_2\text{O}(l)+\_\_{\text{e}}^{-}\end{array}$

Reduction half-reaction:

$\_\_{\text{O}}_2(g)+\_\_{\text{H}}_2\text{O}(l)+\_\_{\text{e}}^{-}\to {\text{OH}}^{-}(aq)$

Interpretation Introduction

Interpretation:

The given oxidation and reduction reactions have to be balanced and the overall equation for a methane-based fuel cell has to be written.

Concept Introduction:

Fuel cell is an electrochemical cell which converts chemical energy contained in readily available fuel oxidant system into electrical energy.

The basic principle of the fuel cell is same as that of electrochemical cell.

Example: $Fuel+Oxidant\underset{}{\to }Oxidationproduct+electricity$

Balanced equation: A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.

The equation for a reaction, which has same number of atoms and charge of the ions in both reactants and product sides, is known as balanced equation.

Oxidation half-reaction: The oxidation half-reaction is the part of a redox reaction that shows only for the oxidized species with the transferred electrons and its oxidation state increases with the loss of electrons.

Reduction half-reaction: The reduction half-reaction is also the part of a redox reaction (the counter part of oxidation half-reaction) that shows only for the reduced species with the transferred electrons and its oxidation state decreases with the gain of electrons.

Explanation of Solution

Given oxidation and reduction equations are,

$\begin{array}{l}{\text{CH}}_4\text{(g)}\text{+}{\text{OH}}^{-}\text{(}\text{aq) }\stackrel{}{\to }{\text{CO}}_{\text{2}}\text{(g)}\text{+}{\text{H}}_{\text{2}}\text{O(l)}+{\text{e}}^{\text{-}}\\ {\text{O}}_2\text{(g)}\text{+}{\text{H}}_{\text{2}}\text{O(l)}+{\text{e}}^{\text{-}}\underset{}{\to }{\text{OH}}^{-}\text{(}\text{aq)}\end{array}$

Balancing the equation:

A balanced equation will have same elements and same number of atoms of each side of the reaction.

Oxidation half-reaction

The complete and balanced equation for given combustion.

${\text{CH}}_4\text{(g)}\text{+}8{\text{OH}}^{-}\text{(}\text{aq) }\stackrel{}{\to }{\text{CO}}_{\text{2}}\text{(g)}\text{+}6{\text{H}}_{\text{2}}\text{O(l)}+8{\text{e}}^{\text{-}}----[1]$

This reaction is oxidation half reaction, the reactant ($C{H}_4$) is lost eight electrons this is an oxidation half-reaction. The balanced equation is given above.

Reduction half-reaction

${\text{2O}}_2\text{(g)}\text{+}4{\text{H}}_{\text{2}}\text{O(l)}+8{\text{e}}^{\text{-}}\underset{}{\to }8{\text{OH}}^{-}\text{(}\text{aq)---------[2]}$

This reaction is a reduction half reaction, because the eight electrons are on the reactants side, then the reactant is gaining these eight electrons. So, this is a reduction reaction.

Added for reaction (1) and (2) we get final reaction it is a overall reaction.

$\begin{array}{l}{\text{CH}}_4\text{(g)}\text{+}\cancel{8{\text{OH}}^{-}\text{(}\text{aq) }}\stackrel{}{\to }{\text{CO}}_{\text{2}}\text{(g)}\text{+}\cancel{6{\text{H}}_{\text{2}}\text{O(l)}}+\cancel{8{\text{e}}^{\text{-}}}---Oxidationhalfreaction\\ {\text{2O}}_2\text{(g)}\text{+}\cancel{4{\text{H}}_{\text{2}}\text{O(l)}}+\cancel{8{\text{e}}^{\text{-}}}\underset{}{\to }\cancel{8{\text{OH}}^{-}\text{(}\text{aq)}}----\mathrm{Re}ductionhalf-reaction\\ ----------------------\\ {\text{CH}}_4\text{(g)}+{\text{2O}}_2\text{(g)}\stackrel{}{\to }{\text{CO}}_{\text{2}}\text{(g)}\text{+}2{\text{H}}_{\text{2}}\text{O(l)}---Overallreaction\\ ----------------------\end{array}$