Problem 99 - Chemistry: The Central Science

Question

Predict whether the following reactions will be spontaneous in acidic solution under standard conditions: (a) oxidation of \(\mathrm{Sn}\) to \(\mathrm{Sn}^{2+} b_{y} \mathrm{I}_{2}\) (to form IT), (b) reduction of \(\mathrm{Ni}^{2+}\) to \(\mathrm{Ni}\) by \(\mathrm{I}\) (to form \(\mathrm{I}_{2}\) ), (c) reduction of \(\mathrm{Ce}^{4+}\) to \(\mathrm{Ce}^{3+}\) by \(\mathrm{H}_{2} \mathrm{O}_{2}\), (d) reduction of \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cuby} \mathrm{Sn}^{2+}\) (to form \(\mathrm{Sn}^{47}\) ).

Step-by-Step Solution

Answer

The reactions under standard conditions for the given cases are as follows: (a) spontaneous, (b) not spontaneous, (c) spontaneous, and (d) likely to be spontaneous.

Step 1: (Reaction a: Sn to Sn²⁺ by I₂)

For this reaction, we have the following half-reactions: Oxidation: Sn -> Sn²⁺ + 2e⁻ Reduction: I₂ + 2e⁻ -> 2I⁻ Now we look up the standard reduction potentials (E°) of these half-reactions: E° (Sn -> Sn²⁺) = -0.14 V E° (I₂ -> 2I⁻) = 0.54 V Add the two half-reactions and sum up the standard reduction potentials: Sn + I₂ -> Sn²⁺ + 2I⁻ E° (cell) = -0.14 V + 0.54 V = 0.40 V Since E° (cell) > 0, the reaction is spontaneous under standard conditions.

Step 2: (Reaction b: Ni²⁺ reduction to Ni by I₂)

For this reaction, we have the following half-reactions: Oxidation: 2I⁻ -> I₂ + 2e⁻ Reduction: Ni²⁺ + 2e⁻ -> Ni Standard reduction potentials: E° (2I⁻ -> I₂) = -0.54 V E° (Ni²⁺ -> Ni) = -0.25 V Combine the two half-reactions and sum up the standard reduction potentials: 2I⁻ + Ni²⁺ -> I₂ + Ni E° (cell) = -0.54 V + (-0.25 V) = -0.79 V Since E° (cell) < 0, the reaction is not spontaneous under standard conditions.

Step 3: (Reaction c: Ce⁴⁺ reduction to Ce³⁺ by H₂O₂)

Given half-reactions: Oxidation: H₂O₂ -> O₂ + 2e⁻ Reduction: Ce⁴⁺ + e⁻ -> Ce³⁺ Standard reduction potentials: E° (H₂O₂ -> O₂) = 1.77 V E° (Ce⁴⁺ -> Ce³⁺) = -1.72 V Combine the two half-reactions and sum up the standard reduction potentials: H₂O₂ + Ce⁴⁺ -> O₂ + Ce³⁺ E° (cell) = 1.77 V + (-1.72 V) = 0.05 V Since E° (cell) > 0, the reaction is spontaneous under standard conditions.

Step 4: (Reaction d: Cu²⁺ reduction to Cu by Sn²⁺)

Given half-reactions: Oxidation: Sn²⁺ -> Sn⁴⁷ + 2e⁻ Reduction: Cu²⁺ + 2e⁻ -> Cu Standard reduction potentials: E° (Sn²⁺ -> Sn⁴⁷) = (?) E° (Cu²⁺ -> Cu) = 0.34 V We don't have the standard reduction potential for Sn²⁺ -> Sn⁴⁷. However, we can use the fact that Sn oxidation to Sn²⁺ is -0.14 V. This means that the oxidation of Sn²⁺ to Sn⁴⁷ will have a lower potential than that, making E° (Sn²⁺ -> Sn⁴⁷) < -0.14 V. Combined and approximate cell potential: Sn²⁺ + Cu²⁺ -> Sn⁴⁷ + Cu E° (cell) ≈ (-0.14 V) + 0.34 V = 0.20 V (approximate) Since E° (cell) > 0 (approximate), the reaction is likely to be spontaneous under standard conditions.

Key Concepts

Standard Reduction Potential

In electrochemistry, the standard reduction potential, denoted as \(E^°\), is a measure of the tendency of a chemical species to gain electrons and be reduced. Defined under standard conditions, \(E^°\) is usually measured in volts and it provides valuable insights into the likelihood of a reaction occurring.

These potentials are determined by comparing the reduction process of a species with the standard hydrogen electrode (SHE), which has an assigned value of 0.00 V.

A positive \(E^°\) indicates a strong tendency to be reduced, while a negative \(E^°\) suggests a weaker tendency.

• For example, a half-reaction with an \(E^°\) of 0.54 V (e.g., \(I_2\) reducing to \(2I^-\)) is more likely to occur spontaneously compared to one with an \(E^°\) of -0.14 V (e.g., Sn reducing to \(\text{Sn}^{2+}\)).
• By summing the standard reduction potentials of the oxidation and reduction half-reactions in a redox reaction, we can determine the cell potential \(E^°_{\text{cell}}\) and predict if the entire reaction is spontaneous.

Half-Reactions

Half-reactions are a crucial part of understanding the full flow of electrons in redox reactions. In such reactions, one species is oxidized and loses electrons, while another is reduced and gains electrons.

These half-reactions help to break down the complex redox processes into manageable parts, making them easier to analyze and understand.

• The oxidation half-reaction describes the loss of electrons, such as \(\text{Sn} \rightarrow \text{Sn}^{2+} + 2e^-\).
• The reduction half-reaction refers to the gain of electrons, for example, \(\text{I}_2 + 2e^- \rightarrow 2\text{I}^-\).

To determine if the overall redox reaction is spontaneous, we combine the standard reduction potentials of these half-reactions.

Summing them helps to compute the cell potential, thus giving hints on the spontaneity of the reaction.

Spontaneous Reaction

In the context of redox reactions, a spontaneous reaction is one that tends to occur on its own without any external influence. To determine the spontaneity, we rely on the calculated cell potential \(E^°_{\text{cell}}\).

When examining reactions under standard conditions:

• If \(E^°_{\text{cell}} > 0\), the reaction is spontaneous; it naturally progresses from reactants to products.
• In contrast, if \(E^°_{\text{cell}} < 0\), the reaction is non-spontaneous and requires an external source of energy to proceed.

Consider the case of \(\text{Sn} + \text{I}_2 \rightarrow \text{Sn}^{2+} + 2\text{I}^-\), which has an \(E^°_{\text{cell}}\) of 0.40 V, confirming that it can proceed spontaneously under standard conditions.

Understanding the conditions and calculations that lead to spontaneity is pivotal in electrochemistry, allowing predictions about reaction feasibility.