RedoxAssign oxidationnumbers to reactantand product species.•Define oxidation andreduction.•Explain oxidation-reduction reactions(redox reactions).Oxidation of CopperReduction of Iron Ore
Oxidation and reduction reactions:what the gosh darn heck are they?Oxidation: the LOSS of electronsReduction: the GAIN of electrons
RedoxLEO the lion says GER! OIL RIGLoss of Electrons is Oxidation.Gain of Electrons is Reduction.Oxidation Is the Loss of electrons.Reduction Is the Gain of electrons.
Connection to Electrochemistry• RED CAT • AN OXREDuction occurs at the CAThode.OXidation occurs at the ANode.
Half Reactions• Show what happens to electrons in the two parts ofa redox reaction: the reduction and the oxidation• Steel hulls in ships have zinc blocks attachedbecause they oxidize and release electrons. Theseelectrons get consumed by the steel and preventcorrosion.
Rules for Writing Half-ReactionsEquations1. The number of electrons gained must equalthe number of electrons lost.2. The equation must balance with respect toboth atoms and charge.(Charge and mass are conserved in redox equations.)
•An oxidation number isnot based on any realcharge on the atom.•It is based on theatom’s electronegativityrelative to the otheratoms to which it isbonded in a givenmolecule.•Oxidation numbers arealways reported for oneindividual atom or ion andnot for groups of atoms orions.
Write the half-reaction equations for the redoxreactionMg (s) + Br2 (l) → MgBr2 (s)• Step 1: Use oxidation numbers to determinewhat is oxidized and what is reduced0 0 2+ 1-Mg (s) + Br2 (l) → MgBr2 (s)What happened to Mg?Mg gained a positive charge so it lost electrons.It is oxidized.
0 0 2+ 1-Mg (s) + Br2 (l) → MgBr2 (s)What happened to Br2?Br2 got a negative charge so it gained electrons.It is reduced.• Step 2 : Write two separate equations: oneshowing oxidation and another showingreduction. Use coefficients to balance thenumber of atoms.Mg → Mg2+Br2 → 2Br-
Step 3: Make the charge balance in eachequation by adding electrons.Mg → Mg2++ 2e-Br2 + 2e-→ 2Br-Check:1. The number of electrons gained mustequal the number of electrons lost.2. The equation must balance with respectto both atoms and charge.
Practice Assigning OxidationNumbers• http://www.occc.edu/kmbailey/chem1115tutoriPractice Identifying the ElementsOxidized and Reduced• http://www.occc.edu/kmbailey/chem1115tu
Redox Reactions and CovalentBonds• Substances with covalent bonds also undergo redoxreactions. When hydrogen burns in chlorine, a covalentbond forms from the sharing of two electrons.• The pair of electrons is more strongly attracted to thechlorine atom because of its higher electronegativity.2 20 0 +1 1H + Cl 2HCl→–• Neither atom has totally lost or totally gained any electrons.• Hydrogen has donated a share of its bonding electron to thechlorine but has not completely transferred that electron.
Objectives• Balance redox equations by using thehalf-reaction method.• Identify oxidizing and reducing agents.• Explain the concept of disproportionation.
Balancing Redox Equations• Simple redox equationscan be balanced byinspection (the methodyou previously learned),but most redox equationscannot be balanced byinspection.• A more systematicapproach called the half-reaction method, or ion-electron method, isrequired.
Seven Steps of the Half-ReactionMethod1. Write the formula equation if it is not given in theproblem. Then write the ionic equation.2. Assign oxidation numbers. Delete substancescontaining only elements that do not changeoxidation state.3.Write the half-reaction for oxidation.Balance the atoms.Balance the charge.4.Write the half-reaction for reduction.Balance the atoms.Balance the charge.
• 5. Conserve charge by adjusting thecoefficients in front of the electrons sothat the number lost in oxidation equalsthe number gained in reduction.• 6. Combine the half-reactions, andcancel out anything common to bothsides of the equation.• 7. Combine ions to form the compoundsshown in the original formula equation.Check to ensure that all other ionsbalance.
Problem• A deep purple solution ofpotassium permanganate istitrated with a colorlesssolution of iron(II) sulfateand sulfuric acid. Theproducts are iron(III)sulfate, manganese(II)sulfate, potassium sulfate,and water—all of which arecolorless.
• Do Step1Write the formula equation if it is not givenin the problem. Then write the ionicequation.4 4 2 42 4 3 4 2 4 2KMnO + FeSO + H SOFe (SO ) + MnSO + K SO + H O→+ 2+ 2 + 24 4 43+ 2 2+ 2 + 24 4 4 2K + MnO +Fe + SO + 2H + SO2Fe + 3SO + Mn + SO + 2K + SO + H O→– – –– – –
• Do Step 2Assign oxidation numbers to each elementand ion. Delete substances containing anelement that does not change oxidationstate.+1 +7 2 +2 +6 2 +1 +6 2+3 +6+ 2+ 2 + 2–4 4 43+ 2– 2+ 2– + 24 42 +2 +6 2 +1 +6 2 +4 21 2K + MnO +Fe + SO + 2H + SO2Fe + 3SO + Mn + SO + 2K + SO + H O→– – –– ––– –– –Only ions or molecules whose oxidation numberschange are retained.+7 2 +2 +3 +22+ 3+ 2+4MnO + Fe Fe + Mn→––
• Do Step 3Write the half-reaction for oxidation. Theiron shows the increase in oxidationnumber. Therefore, it is oxidized.+2 +2 3+3+Fe Fe→Balance the mass.The mass is already balanced.Balance the charge.2+2+ 3++3Fe Fe + e→ –
• Do Step 4Write the half-reaction for reduction.Manganese is reduced.+74+22+MnO Mn→–Balance the mass.Water and hydrogen ions must be added tobalance the oxygen atoms in thepermanganate ion.++7 +22+4 2MnO + 8H Mn + 4H O→–Balance the charge.+7+ 24 22++MnO + 8H + 5 Mn + 4H Oe →– –
• Do Step 5Adjust the coefficients to conserve charge.lost in oxidation 1gained in reduction 5ee=––2+ 3+5(Fe Fe + )e→ –+ 2+4 21(MnO + 8H + 5 Mn + 4H O)e →– –
• Do Step 6Combine the half-reactions and cancel.2+ 3+Fe Fe + e→ –+ 2+4 2MnO + 8H + 5 Mn + 4H Oe →– –2+ + 2+ 3+4 2MnO + 5Fe + 8H + 5 Mn + 5Fe + 4H O + 5e e→– – –
• Do Step 7Combine ions to form compounds from theoriginal equation.2+ + 3+ 2+4 22(5Fe + MnO + 8H 5Fe + Mn + 4H O)→–→2+ + 3+ 2+4 210Fe + 2MnO + 16H 10Fe + 2Mn + 8H On→4 4 2 42 4 3 4 2 4 210FeSO + 2KMnO + 8H SO5Fe (SO ) + 2MnSO + K SO + 8H O
Oxidizing and Reducing Agents• Label a small object (suchas an empty box)“electrons.”• Ask a classmate to take theelectrons from you.• The other student was theagent of your losing theelectrons and you were theagent of the other student’sgaining the electrons.
• By causing you to lose your electrons, theother student is the oxidizing agent.• You are the reducing agent because youcaused the student to gain electrons.• The student is reduced by you, and youare oxidized by the other student.
A reducing agent is asubstance that has thepotential to causeanother substance tobe reduced.An oxidizing agent is asubstance that has thepotential to causeanother substance tobe oxidized.
Disproportionation• A process in which a substance acts as both anoxidizing agent and a reducing agent is calleddisproportionation.• A substance that undergoes disproportionation isboth self-oxidizing and self-reducing.Example: Hydrogen peroxide is both oxidizedand reduced -1 -1 02 2 2 22H O 2H O O→ +