Question

Consider the following \(E^{\circ}\) values \(E\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)=+0.77 \mathrm{~V}\) \(E\left(\mathrm{Sn}^{2+} / \mathrm{Sn}\right)=-0.14 \mathrm{~V}\) Under standard conditions, the potential for the reaction \(\mathrm{Sn}(\mathrm{s})+2 \mathrm{Fe}^{3+}(\mathrm{aq}) \longrightarrow 2 \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Sn}^{2+}(\mathrm{aq})\) is (a) \(1.68 \mathrm{~V}\) (b) \(0.91 \mathrm{~V}\) (c) \(0.63 \mathrm{~V}\) (d) \(1.46 \mathrm{~V}\)

Short answer

The potential for the reaction is 0.91 V (b).

Step-by-step solution

Identify the half-cell reactions

For the given overall reaction, we first need to identify the half-cell reactions involved. The reduction half-reaction for iron is \( \mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+} \) with \( E^{\circ} = 0.77 \ \mathrm{V} \). The oxidation half-reaction for tin is \( \mathrm{Sn} \rightarrow \mathrm{Sn}^{2+} + 2e^- \) with \( E^{\circ} = -0.14 \ \mathrm{V} \).

Define the overall reaction

The overall redox reaction \( \mathrm{Sn(s)} + 2\mathrm{Fe}^{3+}(aq) \rightarrow 2\mathrm{Fe}^{2+}(aq) + \mathrm{Sn}^{2+}(aq) \) involves the oxidation of tin and the reduction of iron.

Calculate the standard potential of the overall reaction

The standard cell potential, \( E^{\circ}_{\text{cell}} \), is given by the difference in standard reduction potentials: \( E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \). Here, \( E^{\circ}_{\text{cathode}} \) is \( 0.77 \ \mathrm{V} \) and \( E^{\circ}_{\text{anode}} \) is \( -(-0.14) \ \mathrm{V} \). So, the calculation becomes \( E^{\circ}_{\text{cell}} = 0.77 \ - (-0.14) = 0.77 + 0.14 = 0.91 \ \mathrm{V} \).

Identify the correct answer

Compare the calculated potential to the given choices. The value closest to \( 0.91 \ \mathrm{V} \) is option (b) \( 0.91 \ \mathrm{V} \). Therefore, the correct answer for the standard potential of the reaction is \( 0.91 \ \mathrm{V} \).

Key concepts

Standard Electrode Potential

The standard electrode potential, denoted as \(E^{\circ}\), represents the tendency of a chemical species to gain electrons and hence be reduced. It is measured under standard conditions, which include a concentration of 1 M for each ion, a pressure of 1 atm, and a temperature of 25°C (298 K). The standard electrode potential is a fundamental concept in electrochemistry, as it helps predict the direction of redox reactions.When comparing two half-reactions, the one with the higher \(E^{\circ}\) value is more likely to undergo reduction. Conversely, the half-reaction with the lower \(E^{\circ}\) value tends to be oxidized. For example, in the exercise above, \(E^{\circ}(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}) = +0.77 \, \mathrm{V}\) indicates a strong tendency for iron (III) ions to gain electrons and be reduced to iron (II). On the other hand, \(E^{\circ}(\mathrm{Sn}^{2+} / \mathrm{Sn}) = -0.14 \, \mathrm{V}\) shows that Sn tends not to gain electrons as strongly, making it more likely to lose electrons (get oxidized). These differences in potential are crucial in determining which direction a redox reaction will proceed under standard conditions.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between two species. In these reactions, one substance is reduced (gains electrons), and the other is oxidized (loses electrons).The given exercise is an example of a redox reaction involving tin and iron. Here, the overall redox process can be divided into two half-reactions:

• The reduction half-reaction: \( \mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+} \)
• The oxidation half-reaction: \( \mathrm{Sn} \rightarrow \mathrm{Sn}^{2+} + 2e^- \)

Understanding each half-reaction simplifies the study of the overall process. Balancing the charges and the number of electrons transferred is a vital part of analyzing redox reactions. The electrons lost in oxidation must be equal to those gained in reduction. This balancing ensures that electrical neutrality is maintained throughout the reaction.In the practice problem, tin is oxidized to form \( \mathrm{Sn}^{2+} \) while iron (III) ions are reduced to iron (II) ions. This results in the balanced overall equation provided.

Cell Potential Calculation

The cell potential is the measure of the voltage, or emf (electromotive force), produced by a galvanic cell under standard conditions. This potential is calculated using the standard reduction potentials of the two half-reactions involved, based on the formula:\[E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\]For the cell in the exercise, the cathode half-reaction, where reduction occurs, is \( \mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+} \) with \(E^{\circ} = +0.77 \, \mathrm{V}\). The anode half-reaction, where oxidation occurs, is \( \mathrm{Sn} \rightarrow \mathrm{Sn}^{2+} + 2e^- \) with \(E^{\circ} = -0.14 \, \mathrm{V}\).When placing these into the formula, the anode potential is reversed to become positive for the oxidation process, giving:\[E^{\circ}_{\text{cell}} = 0.77 \, \mathrm{V} - (-0.14 \, \mathrm{V}) = 0.77 \, \mathrm{V} + 0.14 \, \mathrm{V} = 0.91 \, \mathrm{V}\]This positive \(E^{\circ}_{\text{cell}}\) indicates that the redox reaction is spontaneous under standard conditions. The correct answer in the original exercise was correctly identified as 0.91 V, reflecting the natural tendency for these specific redox processes to occur.