What Are the Rules for Assigning Oxidation Numbers?

Redox Reactions and Electrochemistry

andriano_cz / Getty Images

  • Chemical Laws
  • Periodic Table
  • Projects & Experiments
  • Scientific Method
  • Biochemistry
  • Physical Chemistry
  • Medical Chemistry
  • Chemistry In Everyday Life
  • Famous Chemists
  • Activities for Kids
  • Abbreviations & Acronyms
  • Weather & Climate
  • Ph.D., Biomedical Sciences, University of Tennessee at Knoxville
  • B.A., Physics and Mathematics, Hastings College

Electrochemical reactions involve the transfer of electrons . Mass and charge are conserved when balancing these reactions, but you need to know which atoms are oxidized and which atoms are reduced during the reaction. Oxidation numbers are used to keep track of how many electrons are lost or gained by each atom. These oxidation numbers are assigned using the following rules.

Rules for Assigning Oxidation Numbers

  • The convention is that the cation is written first in a formula, followed by the anion . For example, in NaH, the H is H-; in HCl, the H is H+.
  • The oxidation number of a free element is always 0. The atoms in He and N 2 , for example, have oxidation numbers of 0.
  • The oxidation number of a monatomic ion equals the charge of the ion. For example, the oxidation number of Na + is +1; the oxidation number of N 3- is -3.
  • The usual oxidation number of hydrogen is +1. The oxidation number of hydrogen is -1 in compounds containing elements that are less ​ electronegative than hydrogen, as in CaH 2 .
  • The oxidation number of oxygen in compounds is usually -2. Exceptions include OF 2 because F is more electronegative than O, and BaO 2 , due to the structure of the peroxide ion, which is [O-O] 2- .
  • The oxidation number of a Group IA element in a compound is +1.
  • The oxidation number of a Group IIA element in a compound is +2.
  • The oxidation number of a Group VIIA element in a compound is -1, except when that element is combined with one having a higher electronegativity. The oxidation number of Cl is -1 in HCl, but the oxidation number of Cl is +1 in HOCl.
  • The sum of the oxidation numbers of all of the atoms in a neutral compound is 0.
  • The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion. For example, the sum of the oxidation numbers for SO 4 2- is -2.
  • Assigning Oxidation States Example Problem
  • How to Balance Net Ionic Equations
  • Balance Redox Reaction Example Problem
  • The Difference Between Oxidation State and Oxidation Number
  • 5 Steps for Balancing Chemical Equations
  • Chemistry Vocabulary Terms You Should Know
  • Reduction Definition in Chemistry
  • Learn About Redox Problems (Oxidation and Reduction)
  • Oxidation Reduction Reactions—Redox Reactions
  • How to Neutralize a Base With an Acid
  • Net Ionic Equation Definition
  • Oxidation Definition and Example in Chemistry
  • Types of Chemical Reactions
  • Reactions in Water or Aqueous Solution
  • How Many Protons, Neutrons, and Electrons in an Atom?
  • Valence Definition in Chemistry

Library homepage

  • school Campus Bookshelves
  • menu_book Bookshelves
  • perm_media Learning Objects
  • login Login
  • how_to_reg Request Instructor Account
  • hub Instructor Commons
  • Download Page (PDF)
  • Download Full Book (PDF)
  • Periodic Table
  • Physics Constants
  • Scientific Calculator
  • Reference & Cite
  • Tools expand_more
  • Readability

selected template will load here

This action is not available.

Chemistry LibreTexts

19.2: Balancing Oxidation-Reduction Equations

  • Last updated
  • Save as PDF
  • Page ID 47058

Learning Objectives

  • To identify oxidation–reduction reactions in solution.

We described the defining characteristics of oxidation–reduction, or redox, reactions. Most of the reactions we considered there were relatively simple, and balancing them was straightforward. When oxidation–reduction reactions occur in aqueous solution, however, the equations are more complex and can be more difficult to balance by inspection. Because a balanced chemical equation is the most important prerequisite for solving any stoichiometry problem, we need a method for balancing oxidation–reduction reactions in aqueous solution that is generally applicable. One such method uses oxidation states , and a second is referred to as the half-reaction method.

Balancing Redox Equations Using Oxidation States

To balance a redox equation using the oxidation state method, we conceptually separate the overall reaction into two parts: an oxidation—in which the atoms of one element lose electrons—and a reduction—in which the atoms of one element gain electrons. Consider, for example, the reaction of Cr 2 + (aq) with manganese dioxide (MnO 2 ) in the presence of dilute acid. Equation \(\ref{20.2.1}\) is the net ionic equation for this reaction before balancing; the oxidation state of each element in each species has been assigned using the procedure described previously (in red above each element):

\[\overset{\color{red}{+2}}{\ce{Cr^{2+}} ( aq }) + \overset{\color{red}{+4}}{\ce{Mn}} \overset{\color{red}{-2}}{\ce{O_2} ( aq )} + \overset{\color{red}{+1}} {\ce{H^{+}} ( aq )} \rightarrow \overset{\color{red}{+3}}{\ce{Cr^{3+}} ( aq )} + \overset{\color{red}{+2}}{\ce{Mn^{2+}}( aq )} + \overset{\color{red}{+1}} {\ce{H_2}} \overset{\color{red}{-2}} {\ce{O} (l)} \label{20.2.1} \]

Notice that chromium is oxidized from the +2 to the +3 oxidation state, while manganese is reduced from the +4 to the +2 oxidation state. We can write an equation for this reaction that shows only the atoms that are oxidized and reduced (ignoring the oxygen and hydrogen atoms):

\[\ce{Cr^{2+} + Mn^{4+} -> Cr^{3+} + Mn^{2+}} \label{20.2.2} \]

The oxidation can be written as

\[\underbrace{\ce{Cr^{2+} -> Cr^{3+} + e^{-}}}_{\text{oxidation with 1 electron lost}} \label{20.2.3} \]

and the reduction as

\[\underbrace{\ce{Mn^{4+} + 2e^{-} \rightarrow Mn^{2+}}}_{\text{reduction with 2 electrons gained}} \label{20.2.4} \]

For the overall chemical equation to be balanced, the number of electrons lost by the reductant must equal the number gained by the oxidant. We must therefore multiply the oxidation and the reduction equations by appropriate coefficients to give us the same number of electrons in both. In this example, we must multiply the oxidation (Equation \ref{20.2.3}) by 2 to give

\[\ce{2Cr^{2+} -> 2Cr^{3+} + 2e^{-}} \label{20.2.5} \]

The number of electrons lost in the oxidation now equals the number of electrons gained in the reduction (Equation \ref{20.2.4}):

\[\begin{align*} \ce{2Cr^{2+}} &\rightarrow \ce{2Cr^{3+}} + \ce{2e^{-}} \label{20.2.6} \\[8pt] \ce{Mn^{4+}} + \ce{2e^{-}} &\rightarrow \ce{Mn^{2+}} \end{align*} \]

We then add the equations for the oxidation and the reduction and cancel the electrons on both sides of the equation, using the actual chemical forms of the reactants and products:

\[\begin{align*} \ce{2Cr^{2+}} &\rightarrow \ce{2Cr^{3+}} + \cancel{\ce{2e^{-}}} \\[8pt] \ce{Mn^{4+}} + \cancel{\ce{2e^{-}}} &\rightarrow \ce{Mn^{2+}} \end{align*} \nonumber \]

to result in the balanced redox reaction (metals only)

\[\ce{ Mn^{4+} +2Cr^{2+} \rightarrow 2Cr^{3+} + Mn^{2+}} \label{20.2.7} \]

now we can add the non-redox active atoms back into the equation (ignoring water and hydronium for now)

\[\ce{MnO2(aq) + 2Cr^{2+}(aq) -> 2Cr^{3+}(aq) + Mn^{2+}(aq)} \label{20.2.7b} \]

In a balanced redox reaction, the number of electrons lost by the reductant equals the number of electrons gained by the oxidant.

Although the electrons cancel and the metal atoms are balanced, the total charge on the left side of Equation \ref{20.2.7b} (+4) does not equal the charge on the right side (+8). Because the reaction is carried out in the presence of aqueous acid, we can add \(\ce{H^{+}}\) as necessary to either side of the equation to balance the charge. By the same token, if the reaction were carried out in the presence of aqueous base, we could balance the charge by adding \(\ce{OH^{−}}\) as necessary to either side of the equation to balance the charges.

In this case, adding four \(\ce{H^{+}}\) ions to the left side of Equation \ref{20.2.7b} to give

\[\ce{ MnO2(s) + 2Cr^{2+}(aq) + 4H^{+}(aq) -> 2Cr^{3+}(aq) + Mn^{2+}(aq)} \label{20.2.8} \]

Although the charges are now balanced in Equation \ref{20.2.8}, we have two oxygen atoms on the left side of the equation and none on the right. We can balance the oxygen atoms without affecting the overall charge balance by adding \(\ce{H2O}\) as necessary to either side of the equation. Here, we need to add two \(\ce{H2O}\) molecules to the right side of Equation \ref{20.2.8}:

\[\ce{MnO2(s) + 2Cr^{2+}(aq) + 4H^{+}(aq) -> 2Cr^{3+}(aq) + Mn^{2+}(aq) + 2H_2O(l)} \label{20.2.9} \]

Although we did not explicitly balance the hydrogen atoms, we can see by inspection that the overall chemical equation is now balanced with respect to all atoms and charge. All that remains is to check to make sure that we have not made a mistake. This procedure for balancing reactions is summarized below and illustrated in Example \(\PageIndex{1}\) below.

Procedure for Balancing Oxidation–Reduction Reactions by the Oxidation State Method

  • Write the unbalanced chemical equation for the reaction, showing the reactants and the products.
  • Assign oxidation states to all atoms in the reactants and the products and determine which atoms change oxidation state.
  • Write separate equations for oxidation and reduction, showing (a) the atom(s) that is (are) oxidized and reduced and (b) the number of electrons accepted or donated by each.
  • Multiply the oxidation and reduction equations by appropriate coefficients so that both contain the same number of electrons.
  • Write the oxidation and reduction equations showing the actual chemical forms of the reactants and the products, adjusting the coefficients as necessary to give the numbers of atoms in step 4.
  • Add the two equations and cancel the electrons.
  • Balance the charge by adding \(\ce{H^{+}}\) or \(\ce{OH^{−}}\) ions as necessary for reactions in acidic or basic solution, respectively.
  • Balance the oxygen atoms by adding \(\ce{H2O}\) molecules to one side of the equation.
  • Check to make sure that the equation is balanced in both atoms and total charges.

Example \(\PageIndex{1}\): Balancing in Acid Solutions

Arsenic acid (\(\ce{H3AsO4}\)) is a highly poisonous substance that was once used as a pesticide. The reaction of elemental zinc with arsenic acid in acidic solution yields arsine (\(\ce{AsH3}\), a highly toxic and unstable gas) and \(\ce{Zn^{2+}(aq)}\). Balance the equation for this reaction using oxidation states:

\[\ce{H3AsO4(aq) + Zn(s) -> AsH3(g) + Zn^{2+}(aq)} \nonumber \]

Given: reactants and products in acidic solution

Asked for: balanced chemical equation using oxidation states

Follow the procedure given above for balancing a redox equation using oxidation states. When you are done, be certain to check that the equation is balanced.

  • Write a chemical equation showing the reactants and the products. Because we are given this information, we can skip this step.
  • Assign oxidation states and determine which atoms change oxidation state. The oxidation state of arsenic in arsenic acid is +5, and the oxidation state of arsenic in arsine is −3. Conversely, the oxidation state of zinc in elemental zinc is 0, and the oxidation state of zinc in \(Zn^{2+}(aq)\) is +2: \[H_3\overset{\color{red}{+5}}{As}O_4(aq) + \overset{\color{red}{0}}{Zn}(s) \rightarrow \overset{\color{red}{-3}}{As}H_3(g) + \overset{\color{red}{+2}}{Zn^{2+}}(aq) \nonumber \]

\[ \underbrace{ \overset{\color{red}{+5}}{As} + 8e^- \rightarrow \overset{\color{red}{-3}}{As}}_{\text{Reduction with gain of 8 electrons}} \nonumber \]

Each zinc atom in elemental zinc is oxidized from 0 to +2, which requires the loss of two electrons per zinc atom:

\[ \underbrace{ \overset{\color{red}{0}} {Zn} \rightarrow \overset{\color{red}{+2}} {Zn^{2+}} + 2e^- }_{\text{Oxidation with loss of 2 electrons}}\nonumber \]

  • Multiply the oxidation and reduction equations by appropriate coefficients so that both contain the same number of electrons. The reduction equation has eight electrons, and the oxidation equation has two electrons, so we need to multiply the oxidation equation by 4 to obtain \[\begin{align*} \overset{\color{red}{+5}}{\ce{As}} + \ce{8e^{-}} & \rightarrow \overset{\color{red}{-3}}{\ce{As}} \nonumber \\ \overset{\color{red}{0}} {\ce{4Zn}} & \rightarrow \overset{\color{red}{+2}} {\ce{4Zn^{2+}}} + \ce{8e^{-}} \end{align*} \nonumber \]
  • Reduction: \[\ce{H3AsO4(aq) + 8e^{-} \rightarrow AsH3(g)} \nonumber \]
  • Oxidation: \[\ce{4Zn(s) -> 4Zn^{2+}(aq) + 8e^{-}} \nonumber \]
  • Add the two equations and cancel the electrons. The sum of the two equations in step 5 is \[\ce{H3AsO4(aq) + 4Zn(s)} + \cancel{\ce{8e^{-}}} \rightarrow \ce{AsH3(g)} + \ce{4Zn^{2+}(aq)} + \cancel{\ce{8e^{-}}} \nonumber \] which then yields after canceling electrons \[\ce{H3AsO4(aq) + 4Zn(s) \rightarrow AsH3(g) + 4Zn^{2+}(aq)} \nonumber \]
  • Balance the charge by adding \(\ce{H^{+}}\) or \(\ce{OH^{−}}\) ions as necessary for reactions in acidic or basic solution, respectively. Because the reaction is carried out in acidic solution, we can add H + ions to whichever side of the equation requires them to balance the charge. The overall charge on the left side is zero, and the total charge on the right side is 4 × (+2) = +8. Adding eight H + ions to the left side gives a charge of +8 on both sides of the equation: \[\ce{H3AsO4(aq) + 4Zn(s) + 8H^{+}(aq) \rightarrow AsH3(g) + 4Zn^{2+}(aq)} \nonumber \]
  • Balance the oxygen atoms by adding \(\ce{H2O}\) molecules to one side of the equation. There are 4 \(\ce{O}\) atoms on the left side of the equation. Adding 4 \(\ce{H2O}\) molecules to the right side balances the \(\ce{O}\) atoms: \[\ce{H3AsO4(aq) + 4Zn(s) + 8H^{+}(aq) \rightarrow AsH3(g) + 4Zn^{2+}(aq) + 4H2O(l)} \nonumber \] Although we have not explicitly balanced \(\ce{H}\) atoms, each side of the equation has 11 \(\ce{H}\) atoms.
  • Atoms: \[\ce{1As + 4Zn + 4O + 11H} \overset{\checkmark}{=} \ce{1As + 4Zn + 4O + 11H} \nonumber \]
  • Charge: \[ 8(+1) \overset{\checkmark}{=} 4(+2) \nonumber \]

The balanced chemical equation (both for charge and for atoms) for this reaction is therefore:

\[\ce{H3AsO4(aq) + 4Zn(s) + 8H^{+}(aq) -> AsH3(g) + 4Zn^{2+}(aq) + 4H2O(l)} \nonumber \]

Exercise \(\PageIndex{1}\): Oxidizing Copper

Copper commonly occurs as the sulfide mineral \(\ce{CuS}\). The first step in extracting copper from \(\ce{CuS}\) is to dissolve the mineral in nitric acid, which oxidizes the sulfide to sulfate and reduces nitric acid to \(\ce{NO}\). Balance the equation for this reaction using oxidation states:

\[\ce{CuS(s) + H^{+}(aq) + NO^{-}3(aq) -> Cu^{2+}(aq) + NO(g) + SO^{2-}4(aq)} \nonumber \]

\[\ce{3CuS(s) + 8H^{+}(aq) + 8NO^{-}3(aq) \rightarrow 3Cu^{2+}(aq) + 8NO(g) + 3SO^{2-}4(aq) + 4H2O(l)} \nonumber \]

Reactions in basic solutions are balanced in exactly the same manner. To make sure you understand the procedure, consider Example \(\PageIndex{2}\).

Example \(\PageIndex{2}\): Balancing in Basic Solution

The commercial solid drain cleaner, Drano, contains a mixture of sodium hydroxide and powdered aluminum. The sodium hydroxide dissolves in standing water to form a strongly basic solution, capable of slowly dissolving organic substances, such as hair, that may be clogging the drain. The aluminum dissolves in the strongly basic solution to produce bubbles of hydrogen gas that agitate the solution to help break up the clogs. The reaction is as follows:

\[\ce{Al(s) + H2O(l) \rightarrow [Al(OH)4]^{-}(aq) + H2(g)} \nonumber \]

Balance this equation using oxidation states.

Given: reactants and products in a basic solution

Asked for: balanced chemical equation

Follow the procedure given above for balancing a redox reaction using oxidation states. When you are done, be certain to check that the equation is balanced.

We will apply the same procedure used in Example \(\PageIndex{1}\), but in a more abbreviated form.

  • The equation for the reaction is given, so we can skip this step.
  • The oxidation state of aluminum changes from 0 in metallic \(Al\) to +3 in \(\ce{[Al(OH)4]^{−}}\). The oxidation state of hydrogen changes from +1 in \(\ce{H_2O}\) to 0 in \(\ce{H2}\). Aluminum is oxidized, while hydrogen is reduced: \[ \overset{\color{red}{0}}{Al}_{(s)} + \overset{\color{red}{+1}}{H}_2 O_{(aq)} \rightarrow [ \overset{\color{red}{+3}}{Al} (OH)_4 ]^- _{(aq)} + \overset{\color{red}{0}}{H_2}_{(g)} \nonumber \]
  • Reduction: \[\overset{\color{red}{+1}}{H} + e^- \rightarrow \overset{\color{red}{0}}{H} \: (in\: H_2 ) \nonumber \]
  • Oxidation: \[\overset{\color{red}{0}}{Al} \rightarrow \overset{\color{red}{+3}}{Al} + 3e^- \nonumber \]
  • Reduction: \[\ce{3H^{+} + 3e^{-} -> 3H^0}\: (in\: \ce{H2}) \nonumber \]
  • Oxidation: \[\ce{Al^0 -> Al^{3+} + 3e^{-}} \nonumber \]
  • Reduction: \[\ce{3/2 H2O + 3e^{-} -> 3/2 H2} \nonumber \]
  • Oxidation: \[\ce{Al -> [Al(OH)4]^{-} + 3e^{-}} \nonumber \]
  • Adding the equations and canceling the electrons gives \[ \ce{Al} + \ce{3/2 H2O} + \cancel{\ce{3e^{-}}} \ce{->} \ce{[Al(OH)4]^{-}} + \ce{3/2 H2} + \cancel{\ce{3e^{-}}} \nonumber \] \[ \ce{Al} + \ce{3/2 H2O} \ce{->} \ce{[Al(OH)4]^{-}} + \ce{3/2 H2} \nonumber \] To remove the fractional coefficients, multiply both sides of the equation by 2: \[\ce{2Al + 3H2O \rightarrow 2[Al(OH)4]^{-} + 3H2} \nonumber \]
  • The right side of the equation has a total charge of −2, whereas the left side has a total charge of 0. Because the reaction is carried out in basic solution, we can balance the charge by adding two \(\ce{OH^{−}}\) ions to the left side: \[\ce{2Al + 2OH^{-} + 3H2O -> 2[Al(OH)4]^{-} + 3H2} \nonumber \]
  • The left side of the equation contains five O atoms, and the right side contains eight O atoms. We can balance the O atoms by adding three H 2 O molecules to the left side: \[\ce{2Al + 2OH^{-} + 6H2O -> 2[Al(OH)4]^{-} + 3H2} \nonumber \]
  • Atoms: \[\ce{2Al + 8O + 14H} \overset{\checkmark}{=} \ce{2Al + 8O + 14H} \nonumber \]
  • Charge: \[ (2)(0) + (2)(-1) + (6)(0) \overset{\checkmark}{=} (2)(-1) + (3)(0) \nonumber \]

The balanced chemical equation is therefore

\[\ce{ 2Al(s) + 2OH^{-}(aq) + 6H2O(l) \rightarrow 2[Al(OH)4]^{-}(aq) + 3H2(g)} \nonumber \]

Thus 3 mol of \(\ce{H2}\) gas are produced for every 2 mol of \(\ce{Al}\) consumed.

Exercise \(\PageIndex{2}\): Reducing Manganese in permanganate

The permanganate ion reacts with nitrite ion in basic solution to produce manganese (IV) oxide and nitrate ion. Write a balanced chemical equation for the reaction.

\[\ce{2MnO4^{-}(aq) + 3NO2^{-}(aq) + H2O(l) -> 2MnO2(s) + 3NO3^{-}(aq) + 2OH^{-}(aq)} \nonumber \]

As suggested in Examples \(\PageIndex{1}\) and \(\PageIndex{2}\), a wide variety of redox reactions are possible in aqueous solutions. The identity of the products obtained from a given set of reactants often depends on both the ratio of oxidant to reductant and whether the reaction is carried out in acidic or basic solution, which is one reason it can be difficult to predict the outcome of a reaction. Because oxidation–reduction reactions in solution are so common and so important, however, chemists have developed two general guidelines for predicting whether a redox reaction will occur and the identity of the products:

  • Compounds of elements in high oxidation states (such as \(\ce{ClO4^{−}}\), \(\ce{NO3^{−}}\), \(\ce{MnO4^{−}}\), \(\ce{Cr2O7^{2−}}\), and \(\ce{UF6}\)) tend to act as oxidants and become reduced in chemical reactions.
  • Compounds of elements in low oxidation states (such as \(\ce{CH4}\), \(\ce{NH3}\), \(\ce{H2S}\), and \(\ce{HI}\)) tend to act as reductants and become oxidized in chemical reactions.

When an aqueous solution of a compound that contains an element in a high oxidation state is mixed with an aqueous solution of a compound that contains an element in a low oxidation state, an oxidation–reduction reaction is likely to occur.

Species in high oxidation states act as oxidants, whereas species in low oxidation states act as reductants.

Balancing a Redox Reaction in Acidic Conditions: Balancing a Redox Reaction in Acidic Conditions (opens in new window) [youtu.be]

Oxidation–reduction reactions are balanced by separating the overall chemical equation into an oxidation equation and a reduction equation. In oxidation–reduction reactions, electrons are transferred from one substance or atom to another. We can balance oxidation–reduction reactions in solution using the oxidation state method , in which the overall reaction is separated into an oxidation equation and a reduction equation.

Contributors and Attributions

Modified by Joshua Halpern ( Howard University )

assign oxidation numbers to each element on each side of the equation

Snapsolve any problem by taking a picture. Try it in the Numerade app?

CHEMISTRY COMMUNITY

Created by Dr. Laurence Lavelle

Skip to content

  • Register Alias and Password (Only available to students enrolled in Dr. Lavelle’s classes.)
  • Board index Chem 14B Electrochemistry Balancing Redox Reactions

assign oxidation numbers to each element on each side of the equation

Moderators: Chem_Mod , Chem_Admin

Post by Rhea Thielbar 3A » Thu Feb 24, 2022 7:22 pm

Re: Achieve 1

Post by Gabrielle Malte 2G » Thu Feb 24, 2022 7:26 pm

Post by RossLechner3E » Thu Feb 24, 2022 8:46 pm

Post by 905695298 » Thu Feb 24, 2022 8:48 pm

Post by Ruben Adamov 1E » Thu Feb 24, 2022 8:50 pm

Post by Rio Gagnon 1G » Sun Feb 27, 2022 1:24 am

Post by Emily Hou 1H » Sun Feb 27, 2022 1:31 am

Return to “Balancing Redox Reactions”

  •     NEWS & RESOURCES
  • About The Forum
  • Forum Rules and Helpful Hints
  • How to make a New Post (submit a question) and use Equation Editor (click for details)
  • Email Notification (click for details)
  • How to Subscribe to a Forum, Subscribe to a Topic, and Bookmark a Topic (click for details)
  • Endorsed Post (click for details)
  • Multimedia Attachments (click for details)
  • Strikethrough (click for details)
  •     Review of Chemical & Physical Principles
  •        SI Units, Unit Conversions
  •        Significant Figures
  •        Accuracy, Precision, Mole, Other Definitions
  •        Molarity, Solutions, Dilutions
  •        Empirical & Molecular Formulas
  •        Balancing Chemical Reactions
  •        Limiting Reactant Calculations
  •     The Quantum World
  •        Properties of Light
  •        Properties of Electrons
  •        Einstein Equation
  •        *Black Body Radiation
  •        Photoelectric Effect
  •        Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
  •        DeBroglie Equation
  •        Heisenberg Indeterminacy (Uncertainty) Equation
  •        *Shrodinger Equation
  •        *Particle in a Box
  •        Wave Functions and s-, p-, d-, f- Orbitals
  •        Quantum Numbers and The H-Atom
  •        Electron Configurations for Multi-Electron Atoms
  •        Trends in The Periodic Table
  •     Chemical Bonds
  •        Ionic & Covalent Bonds
  •        Sigma & Pi Bonds
  •        Lewis Structures
  •        Resonance Structures
  •        Formal Charge and Oxidation Numbers
  •        Octet Exceptions
  •        Coordinate Covalent Bonds
  •        Polarisability of Anions, The Polarizing Power of Cations
  •        Electronegativity
  •        Dipole Moments
  •        Bond Lengths & Energies
  •     Forces and Liquid Structure
  •        Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
  •        *Liquid Structure (Viscosity, Surface Tension, Liquid Crystals, Ionic Liquids)
  •     Molecular Shape and Structure
  •        Determining Molecular Shape (VSEPR)
  •        Hybridization
  •        *Molecular Orbital Theory (Bond Order, Diamagnetism, Paramagnetism)
  •     Coordination Compounds and their Biological Importance
  •        Naming
  •        Shape, Structure, Coordination Number, Ligands
  •        Biological Examples
  •        Industrial Examples
  •        *Stereochemistry
  •        *Crystal Field Theory
  •        *Molecular Orbital Theory Applied To Transition Metals
  •     Acids and Bases
  •        Properties & Structures of Inorganic & Organic Acids
  •        Properties & Structures of Inorganic & Organic Bases
  •        Amphoteric Compounds
  •        Lewis Acids & Bases
  •        Bronsted Acids & Bases
  •        Conjugate Acids & Bases
  •        Acidity & Basicity Constants and The Conjugate Seesaw
  •        Calculating pH or pOH for Strong & Weak Acids & Bases
  •        Polyprotic Acids & Bases
  •        Identifying Acidic & Basic Salts
  •        Calculating the pH of Salt Solutions
  •        Air Pollution & Acid Rain
  •     Chem 14A Uploaded Files (Worksheets, etc.)
  •     Student Social/Study Group
  •     Administrative Questions and Class Announcements
  •     General Science Questions
  •     *Aqueous Equilibria
  •        *Making Buffers & Calculating Buffer pH (Henderson-Hasselbalch Equation)
  •        *Biological Importance of Buffer Solutions
  •        *Titrations & Titration Calculations
  •        *Indicators
  •     Chemical Equilibrium
  •        Ideal Gases
  •        Equilibrium Constants & Calculating Concentrations
  •        Non-Equilibrium Conditions & The Reaction Quotient
  •        Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
  •     Thermochemistry
  •        Phase Changes & Related Calculations
  •        Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
  •        Heat Capacities, Calorimeters & Calorimetry Calculations
  •     Thermodynamics
  •        Thermodynamic Systems (Open, Closed, Isolated)
  •        Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
  •        Calculating Work of Expansion
  •        Concepts & Calculations Using First Law of Thermodynamics
  •        Concepts & Calculations Using Second Law of Thermodynamics
  •        Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
  •        Entropy Changes Due to Changes in Volume and Temperature
  •        Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
  •        Gibbs Free Energy Concepts and Calculations
  •        Van't Hoff Equation
  •        Environment, Fossil Fuels, Alternative Fuels
  •        Biological Examples (*DNA Structural Transitions, etc.)
  •     Electrochemistry
  •        Balancing Redox Reactions
  •        Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
  •        Work, Gibbs Free Energy, Cell (Redox) Potentials
  •        Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
  •        Interesting Applications: Rechargeable Batteries (Cell Phones, Notebooks, Cars), Fuel Cells (Space Shuttle), Photovoltaic Cells (Solar Panels), Electrolysis, Rust
  •     Chemical Kinetics
  •        Kinetics vs. Thermodynamics Controlling a Reaction
  •        General Rate Laws
  •        Method of Initial Rates (To Determine n and k)
  •        Zero Order Reactions
  •        First Order Reactions
  •        Second Order Reactions
  •        Reaction Mechanisms, Reaction Profiles
  •        Arrhenius Equation, Activation Energies, Catalysts
  •        *Enzyme Kinetics
  •        Experimental Details
  •        Environment, Ozone, CFCs
  •     Chem 14B Uploaded Files (Worksheets, etc.)
  •     *Thermodynamics and Kinetics of Organic Reactions
  •        *Electrophiles
  •        *Nucleophiles
  •        *Organic Reaction Mechanisms in General
  •        *Electrophilic Addition
  •        *Nucleophilic Substitution
  •        *Free Energy of Activation vs Activation Energy
  •        *Complex Reaction Coordinate Diagrams
  •     *Names and Structures of Organic Molecules
  •        *Alkanes
  •        *Cycloalkanes
  •        *Alkenes
  •        *Cycloalkenes
  •        *Alkynes
  •        *Constitutional and Geometric Isomers (cis, Z and trans, E)
  •        *Haloalkanes
  •        *Haloalkenes
  •        *Alcohols
  •        *Ethers
  •        *Aldehydes
  •        *Ketones
  •        *Carboxylic Acids
  •        *Amines
  •        *Identifying Primary, Secondary, Tertiary, Quaternary Carbons, Hydrogens, Nitrogens
  •     *Conformations of Organic Molecules
  •        *Alkanes and Substituted Alkanes (Staggered, Eclipsed, Gauche, Anti, Newman Projections)
  •        *Cyclopropanes and Cyclobutanes
  •        *Cyclopentanes
  •        *Cyclohexanes (Chair, Boat, Geometric Isomers)
  •        *Calculations Using ΔG° = -RT ln K
  •        *ChemDraw
  •        *Chem3D
  • Chem 14C/D Topics
  •     Resonance in Organic Compounds
  •     Stereochemistry in Organic Compounds (Chirality, Stereoisomers, R/S, d/l, Fischer Projections)

Who is online

Users browsing this forum: No registered users and 5 guests

  • Board index
  • All times are UTC

IMAGES

  1. How to Assign Oxidation Numbers

    assign oxidation numbers to each element on each side of the equation

  2. How To Calculate Oxidation Numbers

    assign oxidation numbers to each element on each side of the equation

  3. Solved For the reaction KCIOKCI02 assign oxidation numbers

    assign oxidation numbers to each element on each side of the equation

  4. (Get Answer)

    assign oxidation numbers to each element on each side of the equation

  5. Assign oxidation states to each element on each side of the following

    assign oxidation numbers to each element on each side of the equation

  6. Solved For the reaction KCIO, KCl + O2 assign oxidation

    assign oxidation numbers to each element on each side of the equation

VIDEO

  1. Determining Oxidation Number 3D Animation

  2. Oxidation Numbers : how to find oxidation number

  3. Rules for Assigning Oxidation Number #electrochemistry #chemistry of class 9th # @FatimaCHEMAX

  4. Module 5-4: Redox Reactions/Oxidation Numbers

  5. Assign oxidation number to the underlined elements in each of the following species: a.NaH_(2)PO

  6. Oxidation number OR Oxidation state

COMMENTS

  1. Solved For the reaction KClO KCl+1/2O2 assign oxidation

    assign oxidation numbers to each element on each side of the equation. K in KClO: K in KCl: Cl in KClO:: Cl in KCl: O in KClO: O in O2: Which element is oxidized? K O Cl Which element is reduced? O Cl K There are 2 steps to solve this one. Expert-verified 100% (8 ratings) Step 1 KClO KCl + 1 2 O A 2 Oxidation Number K in KClO =+1

  2. 22.6: Assigning Oxidation Numbers

    In the chlorate ion (ClO−3) ( ClO 3 −), the oxidation number of Cl Cl is +5 + 5, and the oxidation number of O O is −2 − 2. In a neutral atom or molecule, the sum of the oxidation numbers must be 0. In a polyatomic ion, the sum of the oxidation numbers of all the atoms in the ion must be equal to the charge on the ion. Example 22.6.1 22 ...

  3. Using oxidation numbers to identify oxidation and reduction (worked

    By assigning oxidation numbers to the atoms of each element in a redox equation, we can determine which element is oxidized and which element is reduced during the reaction. In this video, we'll use this method to identify the oxidized and reduced elements in the reaction that occurs between I⁻ and MnO₄⁻ in basic solution. Created by Sal Khan.

  4. Rules for Assigning Oxidation Numbers

    The oxidation number of a Group IIA element in a compound is +2. The oxidation number of a Group VIIA element in a compound is -1, except when that element is combined with one having a higher electronegativity. The oxidation number of Cl is -1 in HCl, but the oxidation number of Cl is +1 in HOCl.

  5. 4.3: Oxidation Numbers and Redox Reactions

    Rules for Assigning Oxidation Numbers. The oxidation state of an uncombined element is zero. This applies regardless of the structure of the element: Xe, Cl 2, S 8, and large structures of carbon or silicon each have an oxidation state of zero.(Since atoms of the same element always form pure covalent bonds, they share electrons equally, neither losing nor gaining, e.g., Cl 2.)

  6. Oxidation States (Oxidation Numbers)

    The positive oxidation state is the total number of electrons removed from the elemental state. It is possible to remove a fifth electron to form another the VO+2 VO 2 + ion with the vanadium in a +5 oxidation state. Each time the vanadium is oxidized (and loses another electron), its oxidation state increases by 1.

  7. Oxidation numbers calculator

    To calculate oxidation numbers of elements in the chemical compound, enter it's formula and click 'Calculate' (for example: Ca2+, HF2^-, Fe4 [Fe (CN)6]3, NH4NO3, so42-, ch3cooh, cuso4*5h2o ). Formula: The oxidation state of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds.

  8. Oxidation-reduction (redox) reactions (article)

    Oxidation numbers can be assigned to the atoms in a reaction using the following guidelines: An atom of a free element has an oxidation number of 0 . For example, each Cl atom in Cl A 2 has an oxidation number of 0 . The same is true for each H atom in H A 2

  9. KClO3 --->KCl+3/2O2 assign oxidation states to each element on each

    Final answer: In this given chemical reaction, KClO3 decomposes into KCl and O2. The Chlorine (Cl) is the element being oxidized, changing from an oxidation state of +5 in KClO3 to -1 in KCl. Explanation: In the given chemical equation, KClO3 decomposes into KCl and O2. The oxidation states of the elements involved are as follows:

  10. 19.2: Balancing Oxidation-Reduction Equations

    2Cr2 + 2Cr3 + + 2e −. The number of electrons lost in the oxidation now equals the number of electrons gained in the reduction (Equation 19.2.4 ): 2Cr2 + → 2Cr3 + + 2e − Mn4 + + 2e − → Mn2 +. We then add the equations for the oxidation and the reduction and cancel the electrons on both sides of the equation, using the actual chemical ...

  11. For the following reaction , KClO $---\gt$ KCl + 1/2 O2 , as

    Chemistry Question For the following reaction , KClO ---\gt − −− > KCl + 1/2 O2 , assign oxidation states to each element on each side of the equation. K: Reactants___?, Products____? Cl: Reactants___?, Products____? O: Reactants___?, Products____? Which element is oxidized? Which element is reduced? Solution Verified

  12. For the reaction KClO3 ⟶ KCl + 3/2O2, assign oxidation numbers to

    VIDEO ANSWER: Everyone in this question, assign oxidation numbers to the elements on each side of the equation. Let's understand the concept in the question. It is oxidation and reduction. ... For the reaction KClO3 ⟶ KCl + 3/2O2, assign oxidation numbers to each element on each side of the equation. K in KClO3 = +1 Cl in KClO3 = +5 O in ...

  13. Sapling HW 7/8 #1

    Now we need to determine it for Cl but since is neturally charged (all charges add up to 0) we do (-2*3)+1= -5, this means that the oxidation number of Cl has to be +5 to make the compound netural. KCl would follow the same process, Cl = -1 from periodic table, it's neturally charged so K has to equal +1.

  14. Achieve 1

    assign oxidation numbers to each element on each side of the equation. Does someone have an easy way to remember which would be oxidized and which would be reduced? I know oxidation is loss of electrons but how can you tell from this equation alone which one loses electrons and which one gains? Top Gabrielle Malte 2G Posts: 102

  15. For the following reaction KClO4 --> KCl + 2O2 Assign oxidation states

    For the following reaction, KClO_2 ---> KCl + O_2, assign oxidation states to each element on each side of the equation. BOTH Reactants AND Products Reactants K Cl O WHICH ELEMENT IS OXIDIZED Produc; A). For the following reaction K C I O ? K C I + 1 / 2 O 2 assign oxidation states to each element on each side of the equation.

  16. Solved For the reaction KClO3 KCl+32O2 assign oxidation

    Science Chemistry Chemistry questions and answers For the reaction KClO3 KCl+32O2 assign oxidation numbers to each element on each side of the equation. K in KClO3: K in KCl: Cl in KClO3: Cl in KCl: O in KClO3: O in O2: Which element is oxidized? Cl O K Which element is reduced? K O Convert 1444.0 mmHg into pa This problem has been solved!