/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 When balancing equations in Chap... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 20

When balancing equations in Chapter \(5,\) we did not mention that reactions must be charge balanced as well as mass balanced. What do charge balanced and mass balanced mean? How are redox equations charge balanced?

Short Answer

Expert verified
In balancing chemical equations, both mass and charge must be conserved. Mass balanced means that the number of atoms of each element is conserved on both sides of the equation, following the law of conservation of mass. Charge balanced means that the total charge on both sides of the equation is equal, conserving the net charge in the system. Redox reactions involve electron transfer between species, with one undergoing oxidation (losing electrons) and another undergoing reduction (gaining electrons). To balance redox equations while conserving mass and charge, follow these steps: 1) Identify oxidation and reduction species and determine oxidation numbers, 2) Split the reaction into half-reactions for oxidation and reduction, 3) Balance mass in each half-reaction, 4) Balance charge by adding electrons, 5) Equalize electron transfer between half-reactions, and 6) Combine half-reactions to obtain the balanced redox equation.

Step by step solution

01

1. Understanding charge and mass balance

When balancing a chemical equation, it is crucial to ensure that both the mass and charge are balanced. Mass balanced means that the number of atoms of each element on both sides of the equation is conserved. This ensures that the law of conservation of mass is followed, meaning no atoms are created or destroyed during the reaction. Charge balanced means that the total charge on both sides of the equation is equal. In other words, the sum of the charges of reactants equals the sum of the charges of the products. This ensures that the law of conservation of charge is followed, meaning that the net charge in a closed system remains constant.
02

2. Understanding redox reactions

Redox reactions (short for reduction-oxidation reactions) are a class of chemical reactions where electrons are transferred between species. In the reaction, one species loses electrons (undergoes oxidation), while another species gains electrons (undergoes reduction).
03

3. Balancing redox equations

To balance redox equations with charge and mass conservation, we follow these steps: Step 1: Identify the species that undergo oxidation and reduction. Determine the oxidation numbers of the various species in the reaction. Step 2: Split the overall reaction into half-reactions for the oxidation and reduction processes. Step 3: Balance the half-reactions for mass by making sure each half-reaction has the same number of each element on both sides. Step 4: Balance the half-reactions for charge by adding a suitable number of electrons to the more charged side of the half-reaction. Step 5: Equalize the electron transfer by multiplying each half-reaction with a suitable integer so that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction. Step 6: Combine the half-reactions to obtain the balanced redox equation. By following these steps, you will achieve a balanced redox equation, ensuring both mass and charge conservation.

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 91影视!

Key Concepts

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

Charge Balance
Charge balance in chemical reactions is a critical aspect that ensures the law of conservation of charge is followed. It's the principle that the total charge before a reaction must equal the total charge after the reaction. But why is this important? Well, imagine if you had more charge after a reaction than before - mysterious charge appearing out of the blue? That just doesn't happen in a closed system! The universe likes its books balanced.

So, when we say a chemical equation is 'charge balanced,' it means that the algebraic sum of charges on the reactant side equals the sum on the product side. It's like having the same amount of money before and after you go shopping, except with charges on ions and molecules. This concept is especially central to redox reactions, where electrons are hopping from one atom to another like social butterflies.
Mass Balance
Equally vital as charge balance is mass balance - an application of the law of conservation of mass that ensures no atoms are unfairly duplicated or go missing during chemical reactions. In more formal terms, mass balance means that for each element in the reactants, there must be an equal number of atoms in the products.

Envision your atoms as guests at a party. When the party ends (the reaction), all your guests (atoms) should leave the venue (the reactants) and arrive at their homes (the products) without anyone getting lost in the hustle. It鈥檚 essential to count all our atoms, otherwise it's like we are creating matter out of nothing, and that's a big no-no in chemistry!
Redox Reactions
Redox reactions, also known as reduction-oxidation reactions, are the bread and butter of transferring electrons in chemistry. They are a family of reactions where electrons take center stage by moving between atoms, thereby changing the oxidation states of these atoms.

One key player in a redox reaction loses electrons, which we call 'oxidation.' The other, the receiver of these electrons, goes through 'reduction.' It may sound a tad backward, but just remember: 'OIL RIG'鈥攐xidation is loss, reduction is gain. Redox reactions are everywhere - from the rusting of an old iron fence to the cellular respiration in our bodies that helps us break down food to release energy.
Oxidation Numbers
Oxidation numbers are like the tracking devices in the spy movies of chemistry. They help us keep tabs on the electrons during a chemical reaction, indicating the degree of oxidation (electron loss) or reduction (electron gain) of an atom within a compound. These are assigned following certain rules that regard the element's position on the periodic table, electronegativity, and molecular structure.

The oxidation number helps us understand the role of different atoms in redox reactions. For instance, when the oxidation number of an atom increases during a reaction, we know it has lost electrons and has been oxidized. Conversely, if it decreases, our atom has gained electrons and has been reduced. Keeping track of these numbers is crucial when orchestrating a redox equation to balance!
Half-Reactions
Half-reactions are the solos in the symphony of a redox reaction. They are the individual parts of the reactants and products that show either the oxidation or reduction process alone. In simpler terms, half-reactions let us peek at just one part of the electron shuffle: one half-reaction exclusively deals with oxidation, while the other half grapples with reduction.

By dividing the full reaction into these two halves, balancing becomes a more orderly process. We ensure each half is mass and charge balanced by adding electrons, hydrogen ions, or water molecules as needed. It's a divide-and-conquer strategy that helps us sum up to a fully balanced redox equation when the half-reactions are recombined appropriately with their electron count in perfect harmony.

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

You have a concentration cell with Cu electrodes and \(\left[\mathrm{Cu}^{2+}\right]\) \(=1.00 M\) (right side) and \(1.0 \times 10^{-4} M\) (left side). a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\) b. The \(\mathrm{Cu}^{2+}\) ion reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) by the following equation: $$\begin{aligned} &\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) & K=1.0 \times 10^{13} \end{aligned}$$ Calculate the new cell potential after enough \(\mathrm{NH}_{3}\) is added to the left cell compartment such that at equilibrium \(\left[\mathrm{NH}_{3}\right]=2.0 \mathrm{M}\)

How can one construct a galvanic cell from two substances, each having a negative standard reduction potential?

The solubility product for \(\operatorname{CuI}(s)\) is \(1.1 \times 10^{-12} .\) Calculate the value of \(\mathscr{E}^{\circ}\) for the half-reaction $$ \operatorname{CuI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)+\mathrm{I}^{-}(a q) $$

The equation \(\Delta G^{\circ}=-n F \mathscr{E}^{\circ}\) also can be applied to halfreactions. Use standard reduction potentials to estimate \(\Delta G_{\mathrm{r}}^{\circ}\) for \(\mathrm{Fe}^{2+}(a q)\) and \(\mathrm{Fe}^{3+}(a q) .\left(\Delta G_{\mathrm{f}}^{\circ} \text { for } \mathrm{e}^{-}=0 .\right)\)

You have a concentration cell in which the cathode has a silver electrode with 0.10 \(M\) Ag \(^{+}\). The anode also has a silver electrode with \(\mathrm{Ag}^{+}(a q), 0.050 \space \mathrm{M}\space \mathrm{S}_{2} \mathrm{O}_{3}^{2-},\) and \(1.0 \times 10^{-3} \mathrm{M}\) \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-} .\) You read the voltage to be 0.76 \(\mathrm{V}\) a. Calculate the concentration of \(\mathrm{Ag}^{+}\) at the anode. b. Determine the value of the equilibrium constant for the formation of \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}\) $$\mathrm{Ag}^{+}(a q)+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}(a q) \quad K=?$$
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

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.