Question

In the presence of oxygen, the cathode half-reaction written in the preceding problem is replaced by

$\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\left(\mathbf{g}\right)\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\left(\mathbf{aq}\right)\mathbf{+}\mathbf{2}{\mathbf{e}}^{\mathbf{-}}\to \mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(\mathbf{l}\right)$

but the anode half-reaction is unchanged. Calculate the standard cell potential for this pair of reactions operating as a galvanic cell. Is the overall reaction spontaneous under standard conditions? As the water becomes more acidic, does the driving force for the rusting of iron increase or decrease?

Short answer

The standard cell potential of the reaction is 1.26 V. The delta G value is negative, so the overall reaction is spontaneous. The more acidic water increases the driving force for rusting of iron.

Step-by-step solution

Rusting of iron

The iron reacts with oxygen and water to form iron oxide. The iron oxide formation is called rusting of iron or corrosion of iron. In rusting reaction, oxidation occurs on iron, and reduction occurs on oxygen.

 Step 2: Rusting reaction of iron

The rusting reaction of iron is shown below.

$2\mathrm{Fe}\left(\mathrm{s}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)+\left(6+\mathrm{x}\right){\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{Fe}}_2{\mathrm{O}}_3.{\mathrm{xH}}_2\mathrm{O}+4{\mathrm{H}}_3{\mathrm{O}}^{+}\left(\mathrm{aq}\right)$

${\mathrm{E}}^{\mathrm{o}}$ is calculated by using standard cell potential as follows,

${\mathrm{E}}^{\mathrm{o}}={{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cathode}}-{{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{anode}}$ --- (a)

Here

${{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cathode}}$= reduction potential of cathode

${{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{anode}}$= reduction potential for anode

The anode half-reaction for the oxidation of Fe to Fe3+

$\mathrm{Fe}\left(\mathrm{s}\right)\to {\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right)+3{\mathrm{e}}^{-}{{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{anode}}=0.0367\mathrm{V}$

Reverse the above half-cell reaction to get reduction potential at the anode

${\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right)+3{\mathrm{e}}^{-}\to \mathrm{Fe}\left(\mathrm{s}\right){{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{anode}}=-0.0367\mathrm{V}$

The cathode half-cell reaction for the reduction of${\mathrm{O}}_2$ is shown below,

$\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_3{\mathrm{O}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to 3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right){{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{Cathode}}=1.229\mathrm{V}$

Now, calculate the standard cell potential using equation (a)

$$\begin{gathered} {\mathrm{E}}^{\mathrm{o}}={{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cathode}}-{{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{anode}} \\ \begin{array}{c}=1.229\mathrm{V}-\left(-0.0367\mathrm{V}\right)\\ =1.26\mathrm{V}\end{array} \end{gathered}$$

Thus, the cell potential is for the reaction is 1.26 V.

Rusting of iron increases along with moisture in the atmosphere

The standard cell potential is positive. So, the value of ${\mathbf{\Delta G}}^{\mathbf{o}}$ is negative.

Therefore, the overall reaction is spontaneous.

Higher acidity also increases corrosion because ${\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}$produced in rusting reaction and used as a reactant in the reduction process at the cathode, which continues the corrosion cycle.

Hence, the driving force for the rusting of iron increases as water becomes more acidic.