Question

(a) Assuming standard conditions, arrange the following in order of increasing strength as oxidizing agents in acidic solution: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Cu}^{2+}, \mathrm{C}_{2}, \mathrm{O}_{2}\) (b) Arrange the following in order of increasing strength as reducing agents in acidic solution: \(\mathrm{Zn}, \mathrm{I}^{-}, \mathrm{Sn}^{2+}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{AL}\). 20.48 Based on the data in Appendix E, (a) which of the following is the strongest oxidizing agent and which is the weakest in acidic solution: \(\mathrm{Br}_{2}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Zn}, \mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-} ?\) (b) Which of the following is the strongest reducing agent, and which is the weaket in acidic solution: \(\mathrm{F}^{-}, \mathrm{Zn}, \mathrm{N}_{1}{ }^{+}\), \(\mathrm{I}_{\mathrm{n}} \mathrm{NO}\) ?

Short answer

(a) In increasing strength as oxidizing agents: \(\mathrm{Cu}^{2+}<\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}<\mathrm{H}_{2} \mathrm{O}_{2}\). (b) In increasing strength as reducing agents: \(\mathrm{H}_{2} \mathrm{O}_{2}<\mathrm{I}^{-}<\mathrm{Sn}^{2+}<\mathrm{Zn}<\mathrm{Al}\). 20.48 (a) The strongest oxidizing agent is \(\mathrm{H}_{2} \mathrm{O}_{2}\) and the weakest is \(\mathrm{Zn}\). (b) The strongest reducing agent is \(\mathrm{Zn}\) and the weakest is \(\mathrm{F}^{-}\).

Step-by-step solution

(a) Arrange in order of increasing strength as oxidizing agents

To solve this, first, we need to find the standard reduction potentials of the given species in acidic solution from the appendix E or standard electrochemistry reference. The reduction potential values are for half-reactions where the species are reduced. We can find the following reduction potentials: 1. \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\): \(E^\circ = +1.33 \, \mathrm{V}\) 2. \(\mathrm{H}_{2} \mathrm{O}_{2}\): \(E^\circ = +1.776\,\mathrm{V}\) 3. \(\mathrm{Cu}^{2+}\): \(E^\circ = +0.342\,\mathrm{V}\) 4. \(\mathrm{C}_{2} \mathrm{O}_{2}\): Not Pervasive/Uncommon Since higher reduction potential corresponds with better oxidizing agent, we can arrange them in increasing strength as oxidizing agents: \(\mathrm{Cu}^{2+}<\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}<\mathrm{H}_{2} \mathrm{O}_{2}\). As \(\mathrm{C}_{2} \mathrm{O}_{2}\) is not a common redox active species, we disregard it.

(b) Arrange in order of increasing strength as reducing agents

For reducing agents, we should also find the standard reduction potentials for each species in their corresponding half-reactions. However, the lower reduction potential corresponds with a better reducing agent in this case. We can find the following reduction potentials from appendix E: 1. \(\mathrm{Zn}\): \(E^\circ = -0.762\,\mathrm{V}\) 2. \(\mathrm{I}^{-}\): \(E^\circ = +0.535\,\mathrm{V}\) 3. \(\mathrm{Sn}^{2+}\): \(E^\circ = -0.137\,\mathrm{V}\) 4. \(\mathrm{H}_{2} \mathrm{O}_{2}\): \(E^\circ = +1.776\,\mathrm{V}\) (as it is given in part (a)) 5. \(\mathrm{Al}\): \(E^\circ = -1.66\,\mathrm{V}\) Now, we can arrange them in increasing strength as reducing agents: \(\mathrm{H}_{2} \mathrm{O}_{2}<\mathrm{I}^{-}<\mathrm{Sn}^{2+}<\mathrm{Zn}<\mathrm{Al}\).

20.48 (a) Find the strongest and weakest oxidizing agent

To solve this, we need to find the standard reduction potentials of the given species in acidic solution. The higher the standard reduction potential, the better an oxidizing agent the species is. We can find the following reduction potentials: 1. \(\mathrm{Br}_{2}\): \(E^\circ = +1.087\,\mathrm{V}\) 2. \(\mathrm{H}_{2} \mathrm{O}_{2}\): \(E^\circ = +1.776\,\mathrm{V}\) (as given in part (a)) 3. \(\mathrm{Zn}\): \(E^\circ = -0.762\,\mathrm{V}\) 4. \(\mathrm{Cr}_{2}\mathrm{O}_{7}{ }^{2-}\): \(E^\circ = +1.33\,\mathrm{V}\) (as given in part (a)) Comparing these values, we find the strongest oxidizing agent is \(\mathrm{H}_{2} \mathrm{O}_{2}\) and the weakest oxidizing agent is \(\mathrm{Zn}\).

20.48 (b) Find the strongest and weakest reducing agent

To solve this, we need to find the standard reduction potentials of the given species in acidic solution. The lower the standard reduction potential, the better reducing agent the species is. We can find the following reduction potentials: 1. \(\mathrm{F}^{-}\): \(E^\circ = +2.866\,\mathrm{V}\) 2. \(\mathrm{Zn}\): \(E^\circ = -0.762\,\mathrm{V}\) (as given in part 20.48 (a)) 3. \(\mathrm{N}_{1}{ }^{+}\): This species should be \(\mathrm{Ni}^{2+}\) (typo) - \(E^\circ = -0.257\,\mathrm{V}\) 4. \(\mathrm{I}_{\mathrm{n}} \mathrm{NO}\): This species should be \(\mathrm{In}^{3+}\) (typo) - \(E^\circ = -0.342\,\mathrm{V}\) Comparing these values, we find that the strongest reducing agent is \(\mathrm{Zn}\) and weakest reducing agent is \(\mathrm{F}^{-}\).

Key concepts

Standard Reduction Potentials

Understanding standard reduction potentials is key to mastering redox reactions. In essence, a standard reduction potential, often denoted as E^{\( ^\circ \)}, is a measure of the tendency of a species to gain electrons and thus to be reduced. Each half-reaction has a different E^{\( ^\circ \)} value, which is measured in volts (V) under standard conditions (25°C, 1M concentration, and 1 atm pressure).

The importance of these values comes into play when determining the strength of oxidizing and reducing agents. In electrochemistry, a more positive E^{\( ^\circ \)} value indicates a better oxidizing agent because it's more likely to gain electrons and become reduced. On the other hand, a more negative E^{\( ^\circ \)} value suggests a stronger reducing agent, indicating a greater tendency to lose electrons and oxidize.

Electrochemical Series

The electrochemical series, also known as the activity series, is a list of elements and compounds arranged in order of their standard reduction potentials. At one end of the series are the strongest oxidizers, which have the highest E^{\( ^\circ \)} values, and at the other end are the strongest reducers, which have the lowest E^{\( ^\circ \)} values.

By referencing the electrochemical series, you can predict the outcome of redox reactions and determine which species will be oxidized and which will be reduced. This makes it an invaluable tool for chemists and students alike, especially when dealing with complex redox systems.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are processes where electrons are transferred between chemical species. The species that loses electrons is said to be oxidized, while the one that gains electrons is reduced. These two simultaneous processes—one species undergoing oxidation and another undergoing reduction—are what characterize redox reactions.

An oxidizing agent, or oxidant, is a substance that has the ability to oxidize other substances by accepting electrons from them. Conversely, a reducing agent, or reductant, donates electrons to another substance, causing it to reduce. An easy tip to remember which is which: the oxidizing agent itself gets reduced, and the reducing agent gets oxidized during the course of a redox reaction.

Half-Reaction Method

The half-reaction method is a systematic approach for balancing redox equations. This method breaks down the overall redox reaction into two separate half-reactions—one for oxidation and one for reduction. Each half-reaction is balanced separately for mass and charge, and then they're combined to form the balanced equation for the entire redox process.

To ensure that both half-reactions are balanced, you may need to add electrons to one side of the half-reaction to account for the change in oxidation state. Then, the number of electrons lost in the oxidation half-reaction is matched with the number gained in the reduction half-reaction, ensuring that the electrons cancel out when the two half-reactions are added back together. This technique not only systematically balances the equation but also provides insight into the movement of electrons during the reaction.