Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In balancing oxidation–reduction reactions, why is it permissible to add water to either side of the equation?

Short Answer

Expert verified
It is permissible to add water to either side of the equation when balancing oxidation-reduction reactions because water is a common species in redox reactions and does not interfere with the original reaction. Adding water helps balance the hydrogen and oxygen atoms and maintains charge neutrality when balancing the half-reactions separately. This ensures that the final balanced redox reaction contains an equal amount of oxidizing and reducing agents, and the number of electrons lost during oxidation equals the number of electrons gained during reduction.

Step by step solution

01

Write the given redox reaction and identify oxidation and reduction

Write the given redox reaction, and identify which substances are undergoing oxidation (loss of electrons) and reduction (gain of electrons).
02

Split the reaction into half-reactions

Divide the redox reaction into two separate half-reactions for oxidation and reduction. Write each half-reaction with all the reactants and products involved in oxidation and reduction processes.
03

Balance the half-reactions separately

Balance each half-reaction, first by balancing the atoms other than hydrogen and oxygen, and then balancing hydrogen and oxygen atoms using water molecules. In other words, we add water molecules to either side of the half-reactions, adjusting the number of H and O atoms accordingly. Adding water in this step is permissible because water is a common species in redox reactions and forms naturally during the transfer of electrons. The presence of water does not interfere with the original reaction, as it often functions as a solvent or a combination of H⁺ and OH⁻ ions that ensure charge neutrality.
04

Balance the charges in each half-reaction

Balance the charges in each half-reaction by adding appropriate numbers of electrons to the side that needs them. This ensures that the net charges on each side of the half-reaction are equal.
05

Combine the balanced half-reactions

Combine the balanced half-reactions by multiplying each half-reaction by an appropriate integer to ensure that the number of electrons lost during oxidation equals the number of electrons gained during reduction. Add the half-reactions together and simplify by canceling out any ions or molecules that appear on both sides of the equation. The balanced redox reaction will now contain an equal amount of oxidizing and reducing agents, along with any added water molecules required for balancing the atoms and charges in the equation. Since the water molecules only serve to balance the equation and maintain charge neutrality, it is permissible to add them during the balancing process.

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.

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.

Balancing Chemical Equations
Balancing chemical equations in redox reactions is crucial to maintaining the principle of the conservation of mass. In a chemical reaction, the number of atoms of each element must remain the same on both sides of the equation. This ensures that no atom is lost or gained during the reaction, as chemicals transform from reactants to products.

To achieve a balanced chemical equation, follow these steps:
  • Identify all reactants and products present in the reaction.
  • Count the number of atoms for each element on both sides of the equation.
  • Adjust the coefficients of the reactants or products to equalize the number of atoms for each element across the arrow.
  • Finally, double-check your balancing to ensure each atom, as well as the overall charge, is balanced.
Balancing chemical equations ensures that we accurately represent the matter involved, allowing us to understand the detailed chemistry behind the reaction.
Half-Reactions
In balancing redox reactions, splitting the equation into separate half-reactions is a powerful technique. Each half-reaction represents either an oxidation or a reduction process, allowing us to focus individually on these two critical components.

The process of deriving half-reactions involves the following:
  • Identify which substances in the reaction are oxidized (losing electrons) and which are reduced (gaining electrons).
  • Write separate equations for the oxidation and reduction processes, detailing all reactants and products involved.
  • In each half-reaction, balance the atoms other than hydrogen and oxygen first.
Once these half-reactions are balanced individually, they can be combined to represent the overall balanced redox reaction. This method simplifies complex reactions, allowing you to tackle each part systematically.
Oxidation and Reduction
At the heart of redox reactions are the concepts of oxidation and reduction, each signifying a specific change in electron count for atoms or molecules involved. Understanding these processes can greatly help in visualizing how redox reactions proceed.

Oxidation is defined as the loss of electrons from a substance, typically resulting in an increase in oxidation state or becoming more positive. In a chemical equation, this takes place with certain elements or compounds losing electrons to another substance. Meanwhile, reduction is the gain of electrons, leading to a decrease in oxidation state or a more negative charge. During a redox reaction, both processes occur simultaneously— one substance is oxidized while another is reduced.
A simple rule to remember is OIL RIG: "Oxidation Is Loss" and "Reduction Is Gain." Understanding which substances give up or take in electrons is vital in accurately writing half-reactions and balancing redox equations.
Water's Role in Reactions
Water is often introduced into redox reactions to help balance equations, particularly in aqueous solutions. When balancing redox equations, water molecules are frequently used to adjust the number of hydrogen and oxygen atoms. This is because water is a familiar substance in most reactions and does not alter the fundamental nature of the chemical process.

Here's why the inclusion of water is acceptable:
  • Water acts as a medium or solvent for many reactions, allowing ions to move freely and react.
  • It provides a source of hydrogen ( H^{+} ) and hydroxide ( OH^{-} ) ions which helps to maintain equilibrium in terms of both charge and atom count.
  • Water's neutrality in reactions means it can balance species without introducing new reactants or altering product formation.
Understanding water's dual role as both a constituent in reactions and as a tool for balancing helps simplify the technical process of achieving balanced redox equations.

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

Three students were asked to find the identity of the metal in a particular sulfate salt. They dissolved a 0.1472-g sample of the salt in water and treated it with excess barium chloride, resulting in the precipitation of barium sulfate. After the precipitate had been filtered and dried, it weighed 0.2327 g. Each student analyzed the data independently and came to different conclusions. Pat decided that the metal was titanium. Chris thought it was sodium. Randy reported that it was gallium. What formula did each student assign to the sulfate salt? Look for information on the sulfates of gallium, sodium, and titanium in this text and reference books such as the CRC Handbook of Chemistry and Physics. What further tests would you suggest to determine which student is most likely correct?

What mass of solid AgBr is produced when 100.0 \(\mathrm{mL}\) of 0.150 \(\mathrm{MAgNO}_{3}\) is added to 20.0 \(\mathrm{mL}\) of 1.00 \(\mathrm{M} \mathrm{NaBr} ?\)

What mass of iron(III) hydroxide precipitate can be produced by reacting 75.0 mL of 0.105 M iron(III) nitrate with 125 mL of 0.150 M sodium hydroxide?

Consider a 1.50-g mixture of magnesium nitrate and magnesium chloride. After dissolving this mixture in water, 0.500 M silver nitrate is added dropwise until precipitate formation is complete. The mass of the white precipitate formed is 0.641 g. a. Calculate the mass percent of magnesium chloride in the mixture. b. Determine the minimum volume of silver nitrate that must have been added to ensure complete formation of the precipitate.

You are given a 1.50-g mixture of sodium nitrate and sodium chloride. You dissolve this mixture into 100 mL of water and then add an excess of 0.500 M silver nitrate solution. You produce a white solid, which you then collect, dry, and measure. The white solid has a mass of 0.641 g. a. If you had an extremely magnified view of the solution (to the atomic- molecular level), list the species you would see (include charges, if any). b. Write the balanced net ionic equation for the reaction that produces the solid. Include phases and charges. c. Calculate the mass percent of sodium chloride in the original unknown mixture.

See all solutions

Recommended explanations on Chemistry Textbooks

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