Problem 96 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

(a) The oxidation of Sn to \(\mathrm{Sn}^{2+}\) by \(\mathrm{I}_{2}\) (to form \(\mathrm{I}^{-}\)) is spontaneous with \(E_{tot} = 0.68 \,\text{V}\). (b) The reduction of \(\mathrm{Ni}^{2+}\) to Ni by \(\mathrm{I}^{-}\) (to form \(\mathrm{I}_{2}\)) is not spontaneous with \(E_{tot} = -0.79 \,\text{V}\). (c) The reduction of \(\mathrm{Ce}^{4+}\) to \(\mathrm{Ce}^{3+}\) by \(\mathrm{H}_{2}\mathrm{O}_{2}\) is spontaneous with \(E_{tot} = 0.85 \,\text{V}\). (d) The reduction of \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\) by \(\mathrm{Sn}^{2+}\) (to form \(\mathrm{Sn}^{4+}\)) is spontaneous with \(E_{tot} = 0.19 \,\text{V}\).

Step 1: 1. Identify the half-reactions

First, let's write down the half-reactions for each given reaction: (a) \(\mathrm{I}_{2} + 2e^{-} -> 2\mathrm{I}^{-}\) \(\mathrm{Sn} -> \mathrm{Sn}^{2+} + 2e^{-}\) (b) \(\mathrm{Ni}^{2+} + 2e^{-} -> \mathrm{Ni}\) \(2\mathrm{I}^{-} -> \mathrm{I}_{2} + 2e^{-}\) (c) \(\mathrm{Ce}^{4+} + e^{-} -> \mathrm{Ce}^{3+}\) \(\mathrm{H}_{2}\mathrm{O}_{2} + 2e^{-} -> 2\mathrm{OH}^{-} + 2\mathrm{H}^{+}\) (d) \(\mathrm{Cu}^{2+} + 2e^{-} -> \mathrm{Cu}\) \(\mathrm{Sn}^{2+} -> \mathrm{Sn}^{4+} + 2e^{-}\)

Step 2: 2. Determine the standard reduction potentials

Using a table of standard reduction potentials, we can identify the standard reduction potentials for each half-reaction: (a) \(\mathrm{E_{I_{2}/I^{-}}} = +0.54 \,\text{V}\) \(\mathrm{E_{Sn/Sn^{2+}}} = -0.14 \,\text{V}\) (b) \(\mathrm{E_{Ni^{2+}/Ni}} = -0.25 \,\text{V}\) \(\mathrm{E_{I_{2}/I^{-}}} = +0.54 \,\text{V}\) (c) \(\mathrm{E_{Ce^{4+}/Ce^{3+}}} = +1.72\, \text{V}\) \(\mathrm{E_{H_{2}O_{2}/OH^{-}}} = +0.87\, \text{V}\) (d) \(\mathrm{E_{Cu^{2+}/Cu}} = +0.34\, \text{V}\) \(\mathrm{E_{Sn^{2+}/Sn^{4+}}} = +0.15\, \text{V}\)

Step 3: 3. Calculate the standard cell potentials

To calculate the standard cell potential (E°) for each reaction, add or subtract the standard reduction potentials for each half-reaction: (a) \(E_{tot} = E_{I_{2}/I^{-}} - E_{Sn/Sn^{2+}} = 0.54 - (-0.14) = 0.68 \,\text{V}\) (b) \(E_{tot} = E_{Ni^{2+}/Ni} - E_{I_{2}/I^{-}} = -0.25 - 0.54 = -0.79 \,\text{V}\) (c) \(E_{tot} = E_{Ce^{4+}/Ce^{3+}} - E_{H_{2}{O}_{2}/OH^{-}} = 1.72 - 0.87 = 0.85 \,\text{V}\) (d) \(E_{tot} = E_{Cu^{2+}/Cu} - E_{Sn^{2+}/Sn^{4+}} = 0.34 - 0.15 = 0.19 \,\text{V}\)

Step 4: 4. Determine spontaneity

A positive value for E° indicates that the reaction is spontaneous, whereas a negative value means it is non-spontaneous. Therefore: (a) The oxidation of Sn to \(\mathrm{Sn}^{2+}\) by \(\mathrm{I}_{2}\) (to form \(\mathrm{I}^{-}\)) is spontaneous. (b) The reduction of \(\mathrm{Ni}^{2+}\) to Ni by \(\mathrm{I}^{-}\) (to form \(\mathrm{I}_{2}\)) is not spontaneous. (c) The reduction of \(\mathrm{Ce}^{4+}\) to \(\mathrm{Ce}^{3+}\) by \(\mathrm{H}_{2}\mathrm{O}_{2}\) is spontaneous. (d) The reduction of \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\) by \(\mathrm{Sn}^{2+}\) (to form \(\mathrm{Sn}^{4+}\)) is spontaneous.

Key Concepts

Standard Reduction Potentials

Understanding standard reduction potentials is crucial for predicting the spontaneity of redox reactions. These are measures of the tendency of a chemical species to gain electrons and be reduced. They are tabulated under standard conditions, which include a solute concentration of 1 M, a gas pressure of 1 atm, and a temperature of 25°C (298 K).

A higher reduction potential means that the species is more likely to gain electrons and be reduced. We use standard reduction potentials to calculate the overall standard cell potential by taking the difference between the potentials for the reduction half-reaction and the oxidation half-reaction. This is the key in determining whether a reaction will occur spontaneously under standard conditions.

For instance, the step-by-step solution for the exercise showed us specific standard reduction potentials and how we use them to ascertain spontaneity for each reaction. In short, the positive total standard cell potential (\(E_{tot}\)) in reactions (a), (c), and (d) indicates that these reactions are spontaneous.

Galvanic Cell Spontaneity

In the realm of electrochemistry, the spontaneity of a redox reaction is often exemplified in a galvanic cell. A galvanic cell generates electric current as a result of a spontaneous redox reaction. The calculation of the standard cell potential, \(E°\text{cell}\), is vital for determining spontaneity.

A positive \(E°\text{cell}\) implies that the galvanic cell will produce electrical energy spontaneously. If \(E°\text{cell}\) is negative, external energy is required to drive the reaction, characteristic of an electrolytic cell. Thus, \(E°\text{cell}\) provides a quantitative measure of the driving force behind the reaction.

Using the data from the exercise, the reactions with a positive E° are spontaneous, making them suitable for constructing a galvanic cell. These reactions offer insights into how batteries operate since they are built upon the principles of galvanic cells.

Electrochemical Series

The electrochemical series is a ranking of elements or compounds based on their standard reduction potentials. It serves as a guide for predicting the outcome of redox reactions. At the top of the series are the strongest oxidizing agents, which have the highest positive reduction potentials and are readily reduced. Conversely, at the bottom, you'll find strong reducing agents with more negative reduction potentials prone to lose electrons.

To predict whether a given redox reaction will proceed spontaneously, one can compare the reacting species on the electrochemical series. The species with the higher reduction potential will generally take electrons from the species with the lower potential. In the exercise provided, identifying where the half-reactions stand in the electrochemical series helps to intuitively understand why some reactions are spontaneous while others are not.

In essence, familiarity with the electrochemical series not only aids in quick predictions about reaction spontaneity but also enriches comprehension of the chemical behavior of different elements and their ions in redox processes.