Chapter 20.2, Problem 20SSC

Interpretation Introduction

Interpretation:

The cell potential for hydrogen-oxygen fuel is to be calculated.

Concept introduction Electrode potential is capacity of electrode to gain or lose electron when it is dipped in solution of its own ions. The absolute magnitude of cell potential of an electrode cannot be determined as oxidation half reaction or reduction half reaction cannot occur alone.

Fuel cells are those in which electrical energy is produced by combustion of fuels. Example of fuel cell is hydrogen fuel cells. It consists 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 overall Hydrogen fuel cell reaction is as follows:

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

Answer to Problem 20SSC

The cell potential for hydrogen-oxygen fuel is $1.2287\text{V}$.

Explanation of Solution

The reaction at anode and cathode in a hydrogen-oxygen fuel 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:

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

As per the latest convention of sign, the electrode at which reduction occurs with respect to standard hydrogen electrode is assigned positive sign or has higher reduction potential and the electrode at which oxidation occurs with respect to standard hydrogen electrode is assigned negative sign or has lower reduction potential.

As per table 20.1, standard potential for the half cell reactions are as follows:

$\begin{array}{l}{\text{2H}}_2\left(\text{g}\right)+{\text{OH}}^{-}\left(\text{aq}\right)\to 4{\text{H}}_2\text{O}\left(\text{l}\right)+4{\text{e}}^{-};-0.8277\text{V}\\ {\text{O}}_2\left(\text{g}\right)+4{\text{H}}_2\text{O}\left(\text{l}\right)+4{\text{e}}^{-}\to 4{\text{OH}}^{-}\left(\text{aq}\right);0.401\text{V}\end{array}$

Since copper has positive electrode potential then aluminum so reduction occurs at copper electrode and oxidation occurs at aluminum electrode.

The formula for cell potential is as follows:

$E_{\text{cell}}^{\text{0}}=E_{\text{Reduction}}^{\text{0}}-E_{\text{oxidation}}^{\text{0}}$

Where,

$E_{\text{cell}}^{\text{0}}$ is standard cell potential.

$E_{\text{Reduction}}^{\text{0}}$ is half cell potential for reduction.

$E_{\text{oxidation}}^{\text{0}}$ is half cell potential for oxidation.

Substitute $-0.8277\text{V}$ for $E_{\text{oxidation}}^{\text{0}}$ and $0.401\text{V}$ for $E_{\text{Reduction}}^{\text{0}}$ in above formula.

$\begin{array}{c}{E}_{\text{cell}}^{\text{0}}=E_{\text{Reduction}}^{\text{0}}-E_{\text{oxidation}}^{\text{0}}\\ =0.401\text{V}-\left(-0.8277\text{V}\right)\\ =1.2287\text{V}\end{array}$

Conclusion

The cell potential for hydrogen-oxygen fuel is $1.2287\text{V}$.