Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 54

Write the balanced half-reaction for the change of \(\mathrm{SO}_{2}(a q)\) to \(\mathrm{HSO}_{4}^{-}(a q)\) in acidic solution.

Short Answer

Expert verified
The balanced half-reaction for the change of \(\mathrm{SO}_{2}(a q)\) to \(\mathrm{HSO}_{4}^{-}(a q)\) in acidic solution is \(\mathrm{SO}_{2}(a q) + 2H_2O \rightarrow \mathrm{HSO}_{4}^{-}(a q) + 4H^+ + 2e^-\)

Step by step solution

01

Balance Sulfur Atoms

Start by balancing the atoms that change oxidation state, in this case sulfur. This gives us \(\mathrm{SO}_{2}(a q)\) to \(\mathrm{HSO}_{4}^{-}(a q)\). No balancing is needed for Sulfur because there's one in each side.
02

Balance Oxygen Atoms

We balance the oxygen atoms by adding the appropriate number of water molecules on the side deficient in oxygen. In this case, add 2 H\(_2\)O to the left side: \(\mathrm{SO}_{2}(a q) + 2H_2O \rightarrow \mathrm{HSO}_{4}^{-}(a q)\)
03

Balance Hydrogen Atoms

Hydrogen atoms are balanced by adding the appropriate number of H\(^+\) ions on the side that needs hydrogen. Add 4 H\(^+\) to the right side: \(\mathrm{SO}_{2}(a q) + 2H_2O \rightarrow \mathrm{HSO}_{4}^{-}(a q) + 4H^+\)
04

Balance Charges

To balance the charge, we add electrons to the more positive side. In this case, add 2 electrons to the right side: \(\mathrm{SO}_{2}(a q) + 2H_2O \rightarrow \mathrm{HSO}_{4}^{-}(a q) + 4H^+ + 2e^-\). Now, the charge is balanced.

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!

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

Think of ammonium nitrite as \(\left(\mathrm{NH}_{4}^{+}\right)\left(\mathrm{NO}_{2}^{-}\right)\) and assign oxidation numbers within each ion. Now think of ammonium nitrate as \(\mathrm{N}_{2} \mathrm{H}_{4} \mathrm{O}_{2},\) and assign oxidation numbers. Which assignment makes more sense?

Write the balanced half-reaction for the change of \(\mathrm{O}_{2}(a q)\) to \(\mathrm{H}_{2} \mathrm{O}(l)\) in acidic solution.

Write the half-reaction for the production of silver metal from silver ions. Write the half-reaction for the production of gold metal from gold(III) ions. Using these half- reactions, provide an explanation for the difference in the rates of production of the metals.

Identify which of the following reactions (as written) is an anodic reaction and which is a cathodic reaction. Write the balanced over- all ionic equation for the redox reaction of the cell. $$\begin{array}{l}{\mathrm{Cd}(s) \rightarrow \mathrm{Cd}^{2+}(a q)+2 e^{-}} \\\ {\mathrm{Ag}^{+}(a q)+e^{-} \rightarrow \mathrm{Ag}(s)}\end{array}$$

Using half-reactions, balance the equation for the redox reaction when \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)\) and \(\mathrm{Fe}^{2+}(a q)\) react to form \(\mathrm{Cr}^{3+}(a q)\) and \(\mathrm{Fe}^{3+}(a q)\) in acidic solution.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

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