Transcript Redox
Redox Assign oxidation numbers to reactant and product species. •Define oxidation and reduction. Reduction of Iron Ore •Explain oxidationreduction reactions (redox reactions). Oxidation of Copper Redox LEO the lion says GER! Loss of Electrons is Oxidation. Gain of Electrons is Reduction. OIL RIG Oxidation Is the Loss of electrons. Reduction Is the Gain of electrons. Connection to Electrochemistry • RED CAT • AN OX REDuction occurs at the CAThode. OXidation occurs at the ANode. Half Reactions • Show what happens to electrons in the two parts of a redox reaction: the reduction and the oxidation • Steel hulls in ships have zinc blocks attached because they oxidize and release electrons. These electrons get consumed by the steel and prevent corrosion. Rules for Writing Half-Reactions Equations 1. The number of electrons gained must equal the number of electrons lost. 2. The equation must balance with respect to both atoms and charge. (Charge and mass are conserved in redox equations.) Assigning Oxidation Numbers •An oxidation number is not based on any real charge on the atom. •It is based on the atom’s electronegativity relative to the other atoms to which it is bonded in a given molecule. •Oxidation numbers are always reported for one individual atom or ion and not for groups of atoms or ions. Write the half-reaction equations for the redox reaction Mg (s) + Br2 (l) → MgBr2 (s) • Step 1: Use oxidation numbers to determine what is oxidized and what is reduced 0 0 2+ 1Mg (s) + Br2 (l) → MgBr2 (s) What happened to Mg? Mg gained a positive charge so it lost electrons. It is oxidized. 0 0 2+ 1Mg (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: one showing oxidation and another showing reduction. Use coefficients to balance the number of atoms. Mg → Mg2+ Br2 → 2Br- Step 3: Make the charge balance in each equation by adding electrons. Mg → Mg2+ + 2eBr2 + 2e- → 2BrNote that electrons lost in oxidation appear on the product side of the oxidation half-reaction. Check: 1. The number of electrons gained must equal the number of electrons lost. 2. The equation must balance with respect to both atoms and charge. Practice Assigning Oxidation Numbers • http://www.occc.edu/kmbailey/chem1115tu torials/oxidation_numbers.htm Practice Identifying the Elements Oxidized and Reduced • http://www.occc.edu/kmbailey/chem111 5tutorials/Element_Oxidized.htm Ba(NO3)2 1. Is the substance elemental? No, three elements are present. 2. Is the substance ionic? Yes, metal + non-metal. 3. Are there any monatomic ions? Yes, barium ion is monatomic. Barium ion = Ba2+ Oxidation # for Ba = +2 4. Which elements have specific rules? Oxygen has a rule....-2 in most compounds Oxidation # for O = -2 5. Which element does not have a specific rule? N does not have a specific rule. Use rule 8 to find the oxidation # of N Let N = Oxidation # for nitrogen (# Ba) (Oxid. # of Ba) + (# N) (Oxid. # N) + (# of O) (Oxid. # of O) =0 1(+2) + 2(N) + 6(-2) = 0 N = +5 Redox Reactions and Covalent Bonds • Substances with covalent bonds also undergo redox reactions. When hydrogen burns in chlorine, a covalent bond forms from the sharing of two electrons. 0 0 +1 –1 H2 + Cl2 2HCl • The pair of electrons is more strongly attracted to the chlorine atom because of its higher electronegativity. • Neither atom has totally lost or totally gained any electrons. • Hydrogen has donated a share of its bonding electron to the chlorine but has not completely transferred that electron. Objectives • Balance redox equations by using the half-reaction method. • Identify oxidizing and reducing agents. • Explain the concept of disproportionation. Balancing Redox Equations • Simple redox equations can be balanced by inspection (the method you previously learned), but most redox equations cannot be balanced by inspection. • A more systematic approach called the halfreaction method, or ionelectron method, is required. Seven Steps of the Half-Reaction Method 1. Write the formula equation if it is not given in the problem. Then write the ionic equation. 2. Assign oxidation numbers. Delete substances containing only elements that do not change oxidation 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 the coefficients in front of the electrons so that the number lost in oxidation equals the number gained in reduction. • 6. Combine the half-reactions, and cancel out anything common to both sides of the equation. • 7. Combine ions to form the compounds shown in the original formula equation. Check to ensure that all other ions balance. Balancing Redox Equations Problem • A deep purple solution of potassium permanganate is titrated with a colorless solution of iron(II) sulfate and sulfuric acid. The products are iron(III) sulfate, manganese(II) sulfate, potassium sulfate, and water—all of which are colorless. • Do Step1 Write the formula equation if it is not given in the problem. Then write the ionic equation. KMnO4 + FeSO4 + H2SO4 Fe2 (SO4 )3 + MnSO4 + K 2SO4 + H2O K + + MnO4– +Fe2+ + SO24– + 2H+ + SO24– 2Fe3+ + 3SO24– + Mn2+ + SO24– + 2K + + SO24– + H2O • Do Step 2 Assign oxidation numbers to each element and ion. Delete substances containing an element that does not change oxidation state. +7 – 2 +1 + – 4 +2 K + MnO +Fe +3 2Fe 3+ +6 – 2 + 3SO 2+ 2– 4 +6 – 2 + SO +2 + Mn +1 2– 4 2+ +6 – 2 2– 4 + + 2H + SO +6 – 2 2– 4 + SO +1 + +6 – 2 2– 4 + 2K + SO +1 – 2 + H2O Only ions or molecules whose oxidation numbers change are retained. +7 – 2 – 4 +2 MnO + Fe 2+ +3 3+ +2 Fe + Mn2+ • Do Step 3 Write the half-reaction for oxidation. The iron shows the increase in oxidation number. Therefore, it is oxidized. +2 +3 3+ Fe2+ Fe Balance the mass. The mass is already balanced. Balance the charge. +2 +3 3+ Fe Fe + e– 2+ • Do Step 4 Write the half-reaction for reduction. Manganese is reduced. +7 +2 MnO4– Mn2+ Balance the mass. Water and hydrogen ions must be added to balance the oxygen atoms in the permanganate ion. +7 +2 MnO4– + 8H+ Mn2+ + 4H2O Balance the charge. +7 +2 MnO + 8H + 5e Mn2+ + 4H2O – 4 + – • Do Step 5 Adjust the coefficients to conserve charge. e– lost in oxidation 1 – e gained in reduction 5 5(Fe2+ Fe3+ + e – ) 1(MnO4– + 8H+ + 5e– Mn2+ + 4H2O) • Do Step 6 Combine the half-reactions and cancel. 5(Fe2+ Fe3+ + e – ) MnO4– + 8H+ + 5e– Mn2+ + 4H2O MnO4– + 5Fe2+ + 8H+ + 5e– Mn2+ + 5Fe3+ + 4H2O + 5e– • Do Step 7 Combine ions to form compounds from the original equation. 2(5Fe2+ + MnO4– + 8H+ 5Fe3+ + Mn2+ + 4H2O) 10Fe2+ + 2MnO4 + 16H+ 10Fe3+ + 2Mn2+ + 8H2O 10FeSO4 + 2KMnO4 + 8H2 SO4 5Fe2 (SO4 )3 + 2MnSO4 + K 2SO4 + 8H2O Oxidizing and Reducing Agents • Label a small object (such as an empty box) “electrons.” • Ask a classmate to take the electrons from you. • The other student was the agent of your losing the electrons and you were the agent of the other student’s gaining the electrons. • By causing you to lose your electrons, the other student is the oxidizing agent. • You are the reducing agent because you caused the student to gain electrons. • The student is reduced by you, and you are oxidized by the other student. A reducing agent is a substance that has the potential to cause another substance to be reduced. An oxidizing agent is a substance that has the potential to cause another substance to be oxidized. Disproportionation • A process in which a substance acts as both an oxidizing agent and a reducing agent is called disproportionation. • A substance that undergoes disproportionation is both self-oxidizing and self-reducing. Example: Hydrogen peroxide is both oxidized and reduced