Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 78

Describe how to determine the oxidation number of sulfur in (a) \(\mathrm{H}_{2} \mathrm{~S}\) and \((\mathrm{b}) \mathrm{SO}_{3}^{2-}\)

Short Answer

Expert verified
In \(\text{H}_2 \text{S}\), sulfur's oxidation number is -2. In \(\text{SO}_3^{2-}\), it is +4.

Step by step solution

01

Understanding Oxidation Numbers

Oxidation numbers are assigned to atoms to help track electron transfer in redox reactions. The sum of oxidation numbers for all atoms in a molecule must equal the overall charge of the molecule.
02

Assigning Known Oxidation Numbers

Identify and assign known oxidation numbers to other atoms in the molecules. For hydrogen, the oxidation number is typically +1. For oxygen, it is typically -2.
03

Determine Oxidation Number in \(\text{H}_2 \text{S}\)

For \(\text{H}_2 \text{S}\), hydrogen has an oxidation number of +1. Since there are two hydrogen atoms, their total contribution is \(2 \times +1 = +2\). Let the oxidation number of sulfur be \(x\). The sum of oxidation numbers should equal zero: \(2(+1) + x = 0\). Solving for \(x\) gives \(x = -2\).
04

Determine Oxidation Number in \(\text{SO}_3^{2-}\)

For \(\text{SO}_3^{2-}\), oxygen has an oxidation number of -2. Since there are three oxygen atoms, their total contribution is \(3(-2) = -6\). Let the oxidation number of sulfur be \(x\). The sum of oxidation numbers should equal the charge of the molecule (-2): \(x + 3(-2) = -2\). Solving for \(x\) gives \(x = +4\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation States
Oxidation states, also known as oxidation numbers, help us understand how electrons are distributed among atoms in a molecule. This is crucial for tracking electron transfer in chemical reactions. Each atom in a molecule is assigned an oxidation state, reflecting its theoretical charge.

Here are the basic rules to assign oxidation states:
  • Elements in their natural state (like O2, H2) have an oxidation state of 0.
  • The sum of oxidation states in a neutral molecule must be 0, while in an ion, it must equal the ion charge.
  • Hydrogen usually has an oxidation state of +1, and oxygen usually has an oxidation state of -2.
  • More electronegative elements are assigned negative oxidation states.

By following these rules, we can determine the distribution of electrons in complex molecules and ions.
Redox Reactions
Redox reactions, or oxidation-reduction reactions, involve the transfer of electrons between atoms. These reactions are essential in various chemical processes. In a redox reaction, one substance gets oxidized (loses electrons), and another gets reduced (gains electrons).

To identify redox reactions, follow these steps:
  • Assign oxidation states to all atoms in the reactants and products.
  • Identify changes in oxidation states: an increase indicates oxidation, while a decrease indicates reduction.
  • Ensure the total number of electrons lost equals the total number of electrons gained, maintaining charge balance.
Redox reactions are fundamental in energy production, metabolism, and various industrial processes. Understanding them enables us to control and harness these chemical transformations effectively.
Electron Transfer
Electron transfer is a critical concept in oxidation and reduction processes. It involves the movement of electrons from a donor (reducing agent) to an acceptor (oxidizing agent).

Let's look at the step-by-step process:
  • Identify the substances involved and their respective oxidation states.
  • Determine which atoms are losing electrons (oxidized) and which are gaining electrons (reduced).
  • Balance the overall electron transfer to ensure the reaction is balanced in terms of both mass and charge.
Electron transfer mechanisms are central to many scientific fields, including electrochemistry, biology (cellular respiration), and industrial chemistry (batteries, corrosion). Mastery of electron transfer concepts lays the groundwork for understanding and leveraging redox reactions, furthering both scientific knowledge and practical applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

How many milliliters of \(0.383 M \mathrm{HCl}\) are needed to react with \(16.2 \mathrm{~g}\) of \(\mathrm{CaCO}_{3} ?\) $$ 2 \mathrm{HCl}(a q)+\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaCl}_{2}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$

A sample of concentrated nitric acid has a density of \(1.41 \mathrm{~g} / \mathrm{mL}\) and contains \(70.0 \%\) HNO \(_{3}\) by mass. (a) What mass \((\mathrm{g})\) of \(\mathrm{HNO}_{3}\) is present per liter of solution? (b) What is the molarity of the solution?

Why are some ionic compounds soluble in water and others are not?

Is the following a redox reaction? Explain. $$ \mathrm{NH}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q) $$

In 1997 and \(2009,\) at United Nations conferences on climate change, many nations agreed to expand their research efforts to develop renewable sources of carbon-based fuels. For more than a quarter century, Brazil has been engaged in a program to replace gasoline with ethanol derived from the root crop manioc (cassava). (a) Write separate balanced equations for the complete combustion of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and of gasoline (represented by the formula \(\mathrm{C}_{8} \mathrm{H}_{18}\) ). (b) What mass (g) of oxygen is required to burn completely \(1.00 \mathrm{~L}\) of a mixture that is \(90.0 \%\) gasoline \((d=0.742 \mathrm{~g} / \mathrm{mL})\) and \(10.0 \%\) ethanol \((d=0.789 \mathrm{~g} / \mathrm{mL})\) by volume? (c) If \(1.00 \mathrm{~mol}\) of \(\mathrm{O}_{2}\) occupies \(22.4 \mathrm{~L}\), what volume of \(\mathrm{O}_{2}\) is needed to burn \(1.00 \mathrm{~L}\) of the mixture? (d) Air is \(20.9 \% \mathrm{O}_{2}\) by volume. What volume of air is needed to burn \(1.00 \mathrm{~L}\) of the mixture?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free