Chapter 20.2, Problem 17SSC

Interpretation Introduction

Interpretation:

The half-cell as well as the overall reaction that occur in hydrogen fuel cell is to be explained.

Concept introduction:

Electrochemical cell is a device used to convert chemical energy produced in redox reaction into electrical energy. A battery is a series of electrochemical cells connected in series which produce electric current.

There are majorly three types of commercial cells which are as follows:

1. Primary cells
2. Secondary cells
3. Fuel cells

Primary cells are those in which redox reaction occurs once and then they cannot be used again, become dead. Example: Dry cell.

Secondary cells are those in which can be recharged again and again and can be re-used. Example is lead storage battery.

Fuel cells are those in which electrical energy is produced by combustion of fuels.

Answer to Problem 17SSC

The reaction at anode and cathode occur as follows:

$\begin{array}{l}{\text{2H}}_2\left(\text{g}\right)+{\text{OH}}^{-}\left(\text{aq}\right)\to 4{\text{H}}_2\text{O}\left(\text{l}\right)+4{\text{e}}^{-}\left(\text{at anode}\right)\\ {\text{O}}_2\left(\text{g}\right)+4{\text{H}}_2\text{O}\left(\text{l}\right)+4{\text{e}}^{-}\to 4{\text{OH}}^{-}\left(\text{aq}\right)\left(\text{at cathode}\right)\end{array}$

So the overall reaction turns out to be as follows:

${\text{2H}}_2\left(\text{g}\right)+{\text{O}}_2\left(\text{g}\right)\to 2{\text{H}}_2\text{O}\left(\text{l}\right)$

Explanation of Solution

Hydrogen fuel cell consist of porous carbon electrode which contains catalyst like $\text{Pt/Pb}$.

Here generally $\text{KOH/NaOH}\left(\text{concentrated}\right)$ are used as electrolyte. ${\text{H}}_2$ and ${\text{O}}_{\text{2}}$ gases are bubbled out of porous electrode into electrolyte solution.

The reaction at anode and cathode occur as follows:

$\begin{array}{l}{\text{2H}}_2\left(\text{g}\right)+{\text{OH}}^{-}\left(\text{aq}\right)\to 4{\text{H}}_2\text{O}\left(\text{l}\right)+4{\text{e}}^{-}\left(\text{at anode}\right)\\ {\text{O}}_2\left(\text{g}\right)+4{\text{H}}_2\text{O}\left(\text{l}\right)+4{\text{e}}^{-}\to 4{\text{OH}}^{-}\left(\text{aq}\right)\left(\text{at cathode}\right)\end{array}$

The hydrogen at anode utilizes $\text{KOH/NaOH}\left(\text{concentrated}\right)$ to get oxidized and electron released in process via external circuit reaches to cathode and reduction reaction occurs. Oxygen gets reduced to produce four hydroxide ions. Hence electrolytic ions consumed at anode get replenished.

So the overall reaction turns out to be as follows:

${\text{2H}}_2\left(\text{g}\right)+{\text{O}}_2\left(\text{g}\right)\to 2{\text{H}}_2\text{O}\left(\text{l}\right)$

Conclusion

The reaction at anode and cathode occur as follows:

$\begin{array}{l}{\text{2H}}_2\left(\text{g}\right)+{\text{OH}}^{-}\left(\text{aq}\right)\to 4{\text{H}}_2\text{O}\left(\text{l}\right)+4{\text{e}}^{-}\left(\text{at anode}\right)\\ {\text{O}}_2\left(\text{g}\right)+4{\text{H}}_2\text{O}\left(\text{l}\right)+4{\text{e}}^{-}\to 4{\text{OH}}^{-}\left(\text{aq}\right)\left(\text{at cathode}\right)\end{array}$

So the overall reaction turns out to be as follows:

${\text{2H}}_2\left(\text{g}\right)+{\text{O}}_2\left(\text{g}\right)\to 2{\text{H}}_2\text{O}\left(\text{l}\right)$