Chapter 20: Oxidation – Reduction Reactions

Oxidation and Reduction Oxidation involves the loss of electrons reduction involves the gain of electrons O xidation L osing I s E lectrons L osing O xidation R eduction G aining G aining R eduction

Oxidation and Reduction Oxidation states/number: the distribution of electrons or the charge of an element Any chemical process in which elements undergo changes in oxidation number is an oxidation-reduction, or redox reaction

Oxidation and Reduction The part of the reaction involving oxidation or reduction alone can be written as a half-reaction Cu → Cu+2 + 2e – oxidation half-reaction 2NO3 – + 2e– + 4H+ → 2NO2 + 2H2O reduction half-reaction Cu + 2NO3 – + 4H+ → Cu+2 + 2NO2 +2H2O redox reaction

Oxidation Number Rules The oxidation number of any uncombined element is 0 The oxidation number of a monatomic ion equals the charge on the ion The more electronegative element in a binary compound is assigned the number equal to the charge it would have if it were an ion The oxidation number of fluorine in a compound is always -1

Oxidation Number Rules Oxygen has an oxidation number of -2 unless it is combined with F, in which it is +1 or +2, or it is in a peroxide, in which it is -1 Hydrogen’s oxidation state in most of its compounds is +1 unless it is combined with a metal, in which case it is -1 In compounds, Group 1 and 2 elements and aluminum have oxidation numbers of +1, +2, and +3, respectively

Oxidation Number Rules The sum of the oxidation numbers of all atoms in a neutral compound is 0 The sum of the oxidation numbers of all atoms in a polyatomic ion equals the charge of the ion

Examples: state the oxidation number of every element  Cl2 _________________ SO3 __________________   Fe3+ _________________ H2O2 __________________ NO2 _________________ MgH2 ____________________ OF _________________

Examples: state the oxidation number of every element  Cl2 ______Cl = 0_____ SO3 _S = +6 _ O = – 2   Fe3+ ______Fe = +3___ H2O2 _H = +1 _ O = – 1 NO2 _N = +4 _O = – 2 MgH2 _Mg = +2 _ H = – 1 OF _O = +1 _ F = – 1

Examples: state the oxidation number of every element Li4Pb ____________________ MnBr62– ____________________   AlBr3 ____________________ H3AsO3 ___________________________ PO43– ___________________ Be(BrO3)2 ______________________________ H5P3O10 ___________________________

Examples: state the oxidation number of every element Li4Pb _Li = +1 _Pb = – 4 MnBr62– _Mn = +4 _Br = – 1   AlBr3 _Al = +3 _Br = – 1 H3AsO3 H = +1 _As = + 3 _O = – 2 PO43– _P = +5 _O = – 2 Be(BrO3)2 _Be = +2 _Br = + 5 _O = – 2 H5P3O10 _H = +1 _P = + 5 _O = – 2

Example Problem: Balance the following redox equation. HI + HNO2 → NO + I2 + H2O

Example Problem: Balance the following redox equation. +1 –1 +1 +3 –2 +2 –2 0 +1 –2 HI + HNO2 → NO + I2 + H2O Step 1 – Assign oxidation numbers

Example Problem: Balance the following redox equation. oxidation +1 –1 +1 +3 –2 +2 –2 0 +1 –2 HI + HNO2 → NO + I2 + H2O reduction Step 2 – Identify which elements are changing oxidation state and write the half reactions ox: HI → I2 red: HNO2 → NO

Example Problem: Balance the following redox equation. oxidation +1 –1 +1 +3 –2 +2 –2 0 +1 –2 HI + HNO2 → NO + I2 + H2O reduction Step 3 – Write the half reaction for oxidation. a.) balance the atoms: use H2O to balance O’s, then H+ to balance the H’s b.) balance the charge: add electrons to the right side of the equation ox: 2HI → I2 + 2H+ + 2e– red: HNO2 → NO

Example Problem: Balance the following redox equation. oxidation +1 –1 +1 +3 –2 +2 –2 0 +1 –2 HI + HNO2 → NO + I2 + H2O reduction Step 4 – Write the half reaction for reduction. a.) balance the atoms: use H2O to balance O’s, then H+ to balance the H’s b.) balance the charge: add electrons to the right side of the equation ox: 2HI → I2 + 2H+ + 2e– red: HNO2 + H+ + e– → NO + H2O

Example Problem: Balance the following redox equation. oxidation +1 –1 +1 +3 –2 +2 –2 0 +1 –2 HI + HNO2 → NO + I2 + H2O reduction Step 5 – Multiply the coefficients in the half reactions so that the number of electrons gained = the numbers of electrons lost ox: (2HI → I2 + 2H+ + 2e–) x1 red: (HNO2 + H+ + e– → NO + H2O) x 2

Example Problem: Balance the following redox equation. oxidation +1 –1 +1 +3 –2 +2 –2 0 +1 –2 HI + HNO2 → NO + I2 + H2O reduction Step 6 – Combine the half reactions and cancel out anything common to both sided of the equation. ox: 2HI → I2 + 2H+ + 2e– red: 2HNO2 + 2H+ + 2e– → 2NO + 2H2O overall equation: 2HI + 2HNO2 → 2NO + I2 + 2H2O

Oxidizing and Reducing Agents Reducing agent: is a substance that has the potential to cause another substance to be reduced; they lose electrons (are oxidized) Oxidizing agent: is a substance that has the potential to cause another substance to be oxidized; they gain electrons (are reduced)

Oxidizing and Reducing Agents Relative strengths of oxidizing and reducing agents on page 643 Disproportionation: the process in which a substance acts as both an oxidizing agent and a reducing agent; the substance is self-oxidizing and self-reducing