Question

If the half cell reactions are given as (i) \(\mathrm{Fe}^{2+}(\mathrm{aq})+2 \mathrm{e} \longrightarrow \mathrm{Fe}(\mathrm{s}) ; \mathrm{E}^{0}=-0.44 \mathrm{~V}\) (ii) \(2 \mathrm{H}^{*}(\mathrm{aq})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{e} \longrightarrow \mathrm{H}_{2} \mathrm{O}(\ell)\); \(\mathrm{E}^{\circ}=+1.23 \mathrm{~V}\) the \(E^{\circ}\) for the reaction, \(\mathrm{Fe}(\mathrm{s})+2 \mathrm{H}^{+}+\) \(\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)\) is: \((1)+1.67 \mathrm{~V}\) (2) \(1.67 \mathrm{~V}\) (3) \(+0.79 \mathrm{~V}\) (4) \(-0.79 \mathrm{~V}\) \((5)+1.58 \mathrm{~V}\)

Short answer

+0.79 V

Step-by-step solution

Write down the given half-cell reactions and their standard electrode potentials

Given half-cell reactions are: (i) \[\text{Fe}^{2+}(\text{aq}) + 2\text{e}^{-} \rightarrow \text{Fe}(\text{s}) \; \text{E}^{0} = -0.44 \text{ V} \] (ii) \[\text{2H}^{+}(\text{aq}) + \frac{1}{2} \text{O}_2 (\text{g}) + 2\text{e}^{-} \rightarrow \text{H}_2 \text{O}(\text{l}) \; \text{E}^{0} = +1.23 \text{ V} \]

Identify which reaction is reduction and which is oxidation

The first reaction \[\text{Fe}^{2+}(\text{aq}) + 2\text{e}^{-} \rightarrow \text{Fe}(\text{s}) \] is a reduction reaction with \(\text{E}^{0} = -0.44 \text{ V}\). The second reaction \[\text{2H}^{+}(\text{aq}) + \frac{1}{2} \text{O}_2 (\text{g}) + 2\text{e}^{-} \rightarrow \text{H}_2 \text{O}(\text{l}) \] can be reversed to \[\text{H}_2 \text{O}(\text{l}) \rightarrow \text{2H}^{+}(\text{aq}) + \frac{1}{2} \text{O}_2 (\text{g}) + 2\text{e}^{-} \text{E}^{0} = -1.23 \text{ V} \] for oxidation.

Combine the cell potentials

To find the standard electrode potential for the overall reaction, add the standard electrode potentials of the reduction and oxidation reactions: \[\text{E}^{0}_{\text{cell}} = \text{E}^{0}_{\text{reduction}} + \text{E}^{0}_{\text{oxidation}} = -0.44\text{ V} + 1.23\text{ V} = 0.79\text{ V}\]

Determine the final cell potential and select the answer

The calculated \(\text{E}^{0}_{\text{cell}}\) for the reaction \[\text{Fe}(\text{s}) + 2\text{H}^{+}(\text{aq}) + \frac{1}{2} \text{O}_2(\text{g}) \rightarrow \text{Fe}^{2+}(\text{aq}) + \text{H}_2 \text{O}(\text{l})\] is \(+0.79 \text{ V}\). Thus the correct answer is option (3).

Key concepts

half-cell reactions

Half-cell reactions are the individual reactions that occur at each electrode in an electrochemical cell. Each half-cell has its own set of chemical species and an electrode where either reduction or oxidation occurs.
In our given problem, we have two half-cell reactions:

• The first half-cell reaction involves iron: \(\text{Fe}^{2+}(\text{aq}) + 2\text{e}^{-} \rightarrow \text{Fe}(\text{s})\) This represents the reduction of iron ions (Fe^{2+}) to solid iron (Fe).


• The second half-cell reaction involves hydrogen and oxygen: \(\text{2H}^{+}(\text{aq}) + \frac{1}{2} \text{O}_2(\text{g}) + 2\text{e}^{-} \rightarrow \text{H}_2 \text{O}(\text{l})\) This represents the reduction of hydrogen ions and oxygen to form water (H_2O).

Half-cell reactions are fundamental in understanding how electrochemical cells work. By splitting the overall reaction into these manageable pieces, you can analyze and calculate the cell potential.

standard electrode potential

Standard electrode potential (\text{E}^\text{0}) is a measure of the tendency of a chemical species to be reduced, and it is measured in volts (V). This value is determined under standard conditions: 1 M concentration, 1 atm pressure, and 25°C temperature.
In our problem, the standard electrode potentials are given for two half-cell reactions:

• For the reduction of iron: \(\text{E}^\text{0} = -0.44 \text{ V}\)


• For the reduction of hydrogen ions and oxygen: \(\text{E}^\text{0} = +1.23 \text{ V}\)

The standard electrode potential helps in predicting the direction of the overall cell reaction. A positive standard electrode potential implies a strong tendency to gain electrons (reduction), while a negative value indicates a tendency to lose electrons (oxidation).
In the given reaction, we use these standard electrode potentials to determine the overall cell potential, helping us understand the feasibility of the electrochemical reaction.

oxidation-reduction reactions

Oxidation-reduction (redox) reactions involve the transfer of electrons between chemical species. In an electrochemical cell, one species undergoes oxidation (loses electrons), while the other undergoes reduction (gains electrons).
In our example, the half-cell reactions are:

• Reduction: \(\text{Fe}^{2+}(\text{aq}) + 2\text{e}^{-} \rightarrow \text{Fe}(\text{s})\)


• Oxidation: Reversing the given reaction for calculation, it becomes: \(\text{H}_2 \text{O}(\text{l}) \rightarrow \text{2H}^{+}(\text{aq}) + \frac{1}{2} \text{O}_2 (\text{g}) + 2\text{e}^{-}\)

These half-reactions combine to give the overall redox reaction:\br> \(\text{Fe}(\text{s}) + 2\text{H}^{+}(\text{aq}) + \frac{1}{2} \text{O}_2 (\text{g}) \rightarrow \text{Fe}^{2+}(\text{aq}) + \text{H}_2 \text{O}(\text{l})\)
By identifying which species is oxidized and which is reduced, you can calculate the standard cell potential (\text{E}^\text{0}_{\text{cell}}) of the overall reaction. In this case, the standard cell potential is the sum of the potentials for the reduction and oxidation reactions, giving:\br> \(\text{E}^\text{0}_{\text{cell}} = -0.44 \text{ V} + 1.23 \text{ V} = 0.79 \text{ V}\)
Hence, understanding oxidation-reduction reactions allows us to calculate the energy produced by electrochemical cells, which is crucial for applications like batteries and fuel cells.