Question

Two half-reactions proposed for the corrosion of iron in the absence of oxygen are

$\begin{array}{l}\mathbf{Fe}\left(s\right)\mathbf{\to }{\mathbf{Fe}}^{\mathbf{2}\mathbf{+}}\left(\mathrm{aq}\right)\mathbf{+}\mathbf{2}{\mathbf{e}}^{\mathbf{-}}\\ \mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(l\right)\mathbf{+}\mathbf{2}{\mathbf{e}}^{\mathbf{-}}\mathbf{\to }\mathbf{2}{\mathbf{OH}}^{\mathbf{-}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\left(g\right)\end{array}$

Calculate the standard cell potential generated by a galvanic cell running this pair of half-reactions. Is the overall reaction spontaneous under standard conditions? As the pH falls from 14, will the reaction become spontaneous?

Short answer

The reaction potential of the reaction is -0.409 V. The standard cell potential is negative, so the reaction is nonspontaneous. The equilibrium shift towards the side hydroxyl ion (-OH) concentration increases, so the reaction will move to the product side as the pH value drops to 14, and the reaction becomes spontaneous.

Step-by-step solution

Calculation of the standard cell potential

Anode half-cell reaction is,

$\mathrm{Fe}\left(\mathrm{s}\right)\to {\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}{\mathrm{E}}_{\mathrm{anode}}^{\mathrm{o}}=-0.409\mathrm{V}$

Cathode half-cell reaction is,

$$\begin{gathered} 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+2{\mathrm{e}}^{-}\to 2{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right) \\ {\mathrm{E}}^{\mathrm{o}}{=\mathrm{E}}_{\mathrm{cathode}}^{\mathrm{o}}{-\mathrm{E}}_{\mathrm{anode}}^{\mathrm{o}} \end{gathered}$$

Now, calculate the standard cell potential.

$\begin{array}{c}{\mathrm{E}}^{\mathrm{o}}=-0.828\mathrm{V}-\left(-0.409\mathrm{V}\right)\\ =-0.419\mathrm{V}\end{array}$

Hence the standard cell potential is -0.409 V.

The reaction is nonspontaneous

The standard cell potential is -0.409 V

Since standard cell potential comes out to be negative, the given cell reaction is a nonspontaneous one under standard conditions.

pH of the reaction

$\frac{\begin{array}{l}\mathrm{Fe}\left(\mathrm{s}\right)\to {\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\left(\mathrm{anode}\right)\\ 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+2{\mathrm{e}}^{-}\to 2{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)\left(\mathrm{cathode}\right)\end{array}}{\mathrm{Fe}\left(\mathrm{s}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)}$

The pH value drops from 14, the concentration of${\mathrm{H}}_3{\mathrm{O}}^{+}$increases, and the${\mathrm{OH}}^{-}$concentrationdecreases.

As the concentration of${\mathrm{OH}}^{-}$decreases, the equilibrium shifts towards the side where the concentration of${\mathrm{OH}}^{-}$increases. So the reaction will move towards the product's side as the pH value drops from 14.

Therefore, the reaction will be spontaneous as the pH value drops from 14.