Problem 123
Question
The best reducing agent of the following substances is (a) \(\mathrm{H}_{2} \mathrm{S} ;\) (b) \(\mathrm{O}_{3} ;\) (c) \(\mathrm{H}_{2} \mathrm{SO}_{4} ;\) (d) \(\mathrm{NaF}\); (e) \(\mathrm{H}_{2} \mathrm{O}\).
Step-by-Step Solution
Verified Answer
The best reducing agent among the given substances is \(\mathrm{H}_{2}\mathrm{S}\).
1Step 1: Identify the oxidations states of the substances
The substances given are \(\mathrm{H}_{2}\mathrm{S}\), \(\mathrm{O}_{3}\), \(\mathrm{H}_{2} \mathrm{SO}_{4}\), \(\mathrm{NaF}\) and \(\mathrm{H}_{2} \mathrm{O}\). Their oxidation states are -2, 0, +6, -1, and -2 respectively. Remember that the oxidation state indicates the degree of oxidation of an atom in a chemical compound. The most electronegative element in a compound is assigned the negative oxidation number.
2Step 2: Analyze the standard reduction potentials of the substances
Knowing the oxidation states, the standard reduction potentials of the substances can be checked from a standard reduction potentials table. The substance with the most positive standard reduction potential is the strongest oxidizing agent, while the one with the most negative standard reduction potential is the strongest reducing agent.
3Step 3: Identify the strongest reducing agent
By comparing the standard reduction potentials, the substance with the highest ability to lose electrons (i.e., the one with the most negative or least positive standard reduction potential) will be the best reducing agent. Among the given substances, \( \mathrm{H}_{2}\mathrm{S} \) is the best reducing agent as it has the most negative standard reduction potential compared to the others.
Key Concepts
Understanding Oxidation StatesDeciphering Standard Reduction PotentialRole of Electronegativity
Understanding Oxidation States
Oxidation states, also known as oxidation numbers, help us understand how electrons are distributed in a compound. They represent the hypothetical charge an atom would have if all bonds were ionic. For example, in
- a compound like \( \mathrm{H}_{2}\mathrm{S}\), the oxidation state of hydrogen is typically +1, while sulfur is -2.
- In \( \mathrm{O}_{3} \) or ozone, the oxygen atoms have an oxidation state of 0 as it is a molecule made up of the same element.
- In complex molecules like \( \mathrm{H}_{2} \mathrm{SO}_{4} \) (sulfuric acid), hydrogen is +1, sulfur is usually considered to be +6, and oxygen is -2.
Deciphering Standard Reduction Potential
Standard reduction potential is a measure of the tendency of a chemical species to acquire electrons and be reduced. These potentials are typically measured in volts.
- A substance with a high positive standard reduction potential readily gains electrons, making it a strong oxidizing agent.
- Conversely, a substance with a low or negative standard reduction potential tends to lose electrons easily, acting as a good reducing agent.
Role of Electronegativity
Electronegativity is an atom's ability to attract and hold onto electrons in a chemical bond. The more electronegative an element, the stronger its pull on electrons. This concept plays a crucial role in determining oxidation states, as the more electronegative element often takes on a negative oxidation state, while the less electronegative one takes on a positive state.
- In \( \mathrm{NaF} \), fluorine is more electronegative than sodium, so it exhibits a -1 oxidation state.
- Electronegativity also helps explain why, in compounds like \( \mathrm{H}_{2}\mathrm{O} \), oxygen carries a -2 oxidation state because it's more electronegative than hydrogen.
Other exercises in this chapter
Problem 121
All of the following compounds yield \(\mathrm{O}_{2}(\mathrm{g})\) when heated to about \(1000 \mathrm{K}\) except (a) \(\mathrm{KClO}_{3} ;\) (b) \(\mathrm{KC
View solution Problem 122
All of the following substances are bases except for (a) \(\mathrm{H}_{2} \mathrm{NNH}_{2} ;\) (b) \(\mathrm{NH}_{3} ;\) (c) \(\mathrm{HN}_{3} ;\) (d) \(\mathrm
View solution Problem 125
All of the following have a tetrahedral shape except (a) \(\mathrm{SO}_{4}^{2-} ;\) (b) \(\mathrm{XeF}_{4} ;\) (c) \(\mathrm{CCl}_{4} ;\) (d) \(\mathrm{XeO}_{4}
View solution Problem 126
Two of the following, through a reaction occurring in a weakly acidic solution, produce the same gaseous product. They are (a) \(\mathrm{CaH}_{2}(\mathrm{s}) ;\
View solution