Reduction-Oxidation Reactions Redox Reactions Chapter 4 Reduction-Oxidation Reactions Redox Reactions

Sodium chloride

Sodium chloride Na  Na+ + e Cl2 + 2e  2Cl

Oxidation–Reduction Reactions Involves 2 processes: Oxidation = Loss of Electrons Na  Na+ + e Oxidation Half-Reaction Reduction = Gain of electrons Cl2 + 2e  2Cl Reduction Half-Reaction Net reaction: 2Na + Cl2  2Na+ + 2Cl Oxidation & reduction always occur together Can't have one without the other

Oxidation Reduction Reaction Oxidizing Agent - Substance that accepts e's Accepts e's from another substance Substance that is reduced Cl2 + 2e  2Cl– Reducing Agent - Substance that donates e's Releases e's to another substance Substance that is oxidized Na  Na+ + e–

Your Turn! Which species functions as the oxidizing agent in the following oxidation-reduction reaction? Zn(s) + Pt2+(aq)  Pt(s) + Zn2+(aq) Pt(s) Zn2+(aq) Pt2+(aq) Zn(s) None of these, as this is not a redox reaction.

Redox Reactions Very common Batteries—car, flashlight, cell phone, computer Metabolism of food Combustion Chlorine Bleach Dilute NaOCl solution Cleans through redox reaction Oxidizing agent Destroys stains by oxidizing them

Redox Reaction Ex. Fireworks displays Net: 2Mg + O2  2MgO Oxidation: Mg  Mg2+ + 2e Loses electrons = Oxidized Reducing agent Reduction: O2 + 4e  2O2 Gains electrons = Reduced Oxidizing agent

Redox Reaction? S + O2  SO2 Combustion: Oxidation in old sense, reaction with oxygen But n no ions in SO2 How can we decide which loses and which gains!! Oxidation number (state)! If compound were ionic, what would the charges have been.

Rules for Oxidation States (Numbers) The sum of Oxidation numbers equals to the charge on molecule, formula unit or ion. The oxidation state of elements is zero. Oxidation state for monoatomic ions are the same as their charge. In its compounds fluorine is always –1. Hydrogen is assigned the oxidation state +1. Oxygen is assigned an oxidation state of -2 in its covalent compounds (except as a peroxide). If two rules conflict, apply higher rule.

Oxidation States Assign the oxidation states to each element in the following. CO2 NO3- H2SO4 Fe2O3 Fe3O4 Cr2O72- O2F2 H2O2 LiH BaO2

Oxidation-Reduction Transfer electrons, so the oxidation states change. Ox Red Oxidation Increase in Oxid. number Reduction decrease in Oxid. number Ox Red

Ox Red Ox Red Red C (CH4) oxidized → CH4 reducing agent O2 reduced → O2 oxidizing agent

PbS has been oxidized, PbS is the reducing agent. O2 has been reduced, O2 is the oxidizing agent.

PbO has been reduced, PbO is the oxidizing agent. CO has been oxidized, CO is the reducing agent.

Identify the 1) Oxidizing agent 2) Reducing agent 3) Substance oxidized 4) Substance reduced in the following reactions Fe (s) + O2(g) ® Fe2O3(s) Fe2O3(s)+ 3 CO(g) ® 2 Fe(l) + 3 CO2(g) SO3- + H+ + MnO4- ® SO4- + H2O + Mn+2

Balancing Redox Reactions Ion-Electron Method – Acidic Solution 1. Divide equation into 2 half-reactions 2. Balance atoms other than H & O 3. Balance O by adding H2O to side that needs O 4. Balance H by adding H+ to side that needs H 5. Balance net charge by adding e– 6. Make e– gain equal e– loss; then add half- reactions 7. Cancel anything that is the same on both sides

Balance in Acidic Solution Cr2O72– + Fe2+  Cr3+ + Fe3+ 1. Break into half-reactions Cr2O72  Cr3+ Fe2+  Fe3+ 2. Balance atoms other than H & O Cr2O72  2Cr3+ Put in 2 coefficient to balance Cr Fe already balanced Start Mon 1110

3. Balance O by adding H2O to the side that needs O. Cr2O72  2Cr3+ Right side has 7 O atoms Left side has none Add 7 H2O to left side Fe2+  Fe3+ No O to balance + 7 H2O

Cr2O72  2Cr3+ + 7H2O 14H+ + Fe2+  Fe3+ 4. Balance H by adding H+ to side that needs H Cr2O72  2Cr3+ + 7H2O Left side has 14 H atoms Right side has none Add 14 H+ to right side Fe2+  Fe3+ No H to balance 14H+ +

5. Balance net charge by adding electrons. 14H+ + Cr2O72  2Cr3+ + 7H2O 6 electrons must be added to reactant side Fe2+  Fe3+ 1 electron must be added to product side Now both half-reactions balanced for mass & charge 6e + Net Charge = 14(+1) (–2) = 12 Net Charge = 2(+3)+7(0) = 6 + e

7. Cancel anything that's the same on both sides 6. Make e– gain equal e– loss; then add half-reactions 6e + 14H+ + Cr2O72–  2Cr3+ + 7H2O Fe2+  Fe3+ + e 7. Cancel anything that's the same on both sides 6[ ] 6e + 6Fe2+ + 14H+ + Cr2O72  6Fe3+ + 2Cr3+ + 7H2O + 6e 6Fe2+ + 14H+ + Cr2O72  6Fe3+ + 2Cr3+ + 7H2O

Practice The following reactions occur in acidic aqueous solution. Balance them: MnO4- + Fe2+ ® Mn2+ + Fe3+ Cu + NO3- ® Cu2+ + NO(g) Pb + PbO2 + SO42- ® PbSO4 Mn2+ + NaBiO3 ® Bi3+ + MnO4- Cr2O72- + C2H5OH ® Cr3+ + CO2

Ion-Electron method in Basic Solution The simplest way to balance an equation in basic solution Use steps 1-7 above, then 8. Add the same number of OH– to both sides of the equation as there are H+. 9. Combine H+ & OH– to form H2O 10. Cancel any H2O that you can from both sides

Basic Solution Ag + CN- +O2 ® Ag(CN)2- Cr(OH)3 + OCl- + OH- ® CrO42- + Cl- + H2O CrI3 + Cl2 ® CrO4- + IO4- + Cl-