Question

In a hydrogen fuel cell, the steps of the chemical reaction are
$$\begin{gathered} \text{at}-\text{electrode:}{\mathrm{H}}_2+2{\mathrm{OH}}^{-}⟶2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-} \\ \text{at}+\text{electrode:}\frac{1}{2}{\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}⟶2{\mathrm{OH}}^{-} \end{gathered}$$

Calculate the voltage of the cell. What is the minimum voltage required for electrolysis of water? Explain briefly.

Short answer

The minimum voltage required is W_e=1.23 eV

Step-by-step solution

Given Information

Given reaction is

$$\begin{gathered} \text{at}-\text{electrode:}{\mathrm{H}}_2+2{\mathrm{OH}}^{-}⟶2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-} \\ \text{at}+\text{electrode:}\frac{1}{2}{\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}⟶2{\mathrm{OH}}^{-} \end{gathered}$$

And Gibbs energy is +237 kJ.

Explanation

Gibbs free energy can be written as work done as

ΔG = W ....................................(1)

The expression for the work done by each electron is written as

W_e = W /(2N_A) ...............................(2)

Where W_e is the work done per electron and N_A is Avogadro's number.

As ΔG= 237kJ so W= 237kJ.

Substitute W=237 kJ and N_A=6.023 x 10²³

^{$$\begin{gathered} W_e=\frac{237\mathrm{kJ}}{2\left(6.623\times {10}^{23}\right)} \\ =\frac{237\mathrm{kJ}\left(\frac{1000\mathrm{J}}{1\mathrm{kJ}}\right)}{2\left(6.623\times {10}^{23}\right)} \\ =1.968\times {10}^{-19}\mathrm{J}\left(\frac{{10}^{19}\mathrm{eV}}{1.6(1\mathrm{J})}\right) \\ =1.23\mathrm{eV} \end{gathered}$$}