Oxidation-Reduction Topic 9.1 1+ 2+ 3+ ...etc. 4+/- 3- 2- 1- 0 Memorize these! NO31- nitrate NO21- nitrite OH1- hydroxide ClO21- chlorite ClO31- chlorate HCO31SO42- hydrogencarbonate (bicarbonate) sulfate SO32- sulfite CO3 2- carbonate PO43- phosphate NH41+ ammonium Introduction • oxidation and reduction can be considered in terms of… 1. oxidation- substance gains oxygen reduction- substance loses oxygen 2. oxidation- the loss of electrons reduction- the gain of electrons 3. oxidation- the oxidation state/# increases reduction- the oxidation state/# decreases O I L - oxidation - is - loss of electrons R I G - reduction - is - gain of electrons Oxidation states Note: oxidation states must be written with the sign in FRONT: +2 not 2+ • the oxidation state (number) is the apparent or theoretical charge of an free element, molecule, or ion • oxidation – a process where the number increases (more positive because loses neg. electrons) • Ex: Mg Mg2+ (aq) + 2e- • reduction – a process where the number decreases (less positive/more negative because gains neg. electrons) • Ex: O + 2 e- O2- (g) Rules for assigning oxidation states 1 The oxidation number of an free element is always 0 O2, H2, Ne, Zn 2 The oxidation number of Hydrogen is usually +1 Metal hydrides are an exception. They are -1 HCl, H2SO4 NaH 3 The oxidation number of Oxygen is usually -2. Peroxides are an exception They are –1. H2O, NO2, etc. -O-Obonding 4 Group 1 metals are always +1 Group 2 metals are always +2 Aluminum is always +3 Li, Na… Mg, Ba.. Al 5 Fluorine is always -1 Other group 17 (halogens) are often -1 HF, OF2 HI, NaCl, KBr 6 Oxidation numbers of monatomic ions follow S2-, Zn2+ the charge of the ion 7 The SUM of oxidation numbers is zero for a neutral compound. LiMnO4… 8 For polyatomic ions, the SUM is their charge. SO42-, NO31- Practice Assigning Oxidation Numbers NO2 O is -2 x 2 = -4 N must equal +4 N2O5 O is -2 x 5 = -10 N must equal +10/2 = +5 HClO3 (+1) + (x) + 3(-2) = 0; x = (+5) HNO3 O is -2 x 3 = -6; H is +1; N must equal +5 Ca(NO3)2 O is -2 x 3 = -6 x 2 = -12; Ca is +2; N must equal +10/2 = +5 KMnO4 O is -2 x 4 = -8; K is +1; Mn must equal +7 Fe(OH)3 O is -2 x 3 = -6; Fe is +3; H is +1 x 3 = +3 K2Cr2O7 O is -2 x 7 = -14; K is +1 x 2 = +2; Cr must equal +12/2 = +6 CO32- x + 3(-2) = -2 x = +4 CN- N is -3; Charge = -1 means that C must be +2 K3Fe(CN)6 N is -3 x 6 = -18; C is +2 x 6 = +12; K is +1 x 3 =+3; Fe must be +3 CH4 H is +1 x 4 = +4; C must be -4 Using Oxidation Numbers • an increase in the oxidation number indicates that an atom has lost electrons and therefore is oxidized • a decrease in the oxidation number indicates that an atom has gained electrons and therefore reduced • Example Zn + CuSO4 ZnSO4 + Cu 0 +2 +6-2 +2+6-2 0 Zn: 0 + 2 Oxidized Cu: +2 0 Reduced S and O are unchanged Exercise For each of the following reactions (not balanced to simplify) find the element oxidized and the element reduced Cl2 Cu + KBr KCl + Br2 + HNO3 Cu(NO3)2 + NO2 + H2O HNO3 + I2 HIO3 + NO2 Exercise For each of the following reactions find the element oxidized and the element reduced Cl2 0 + KBr +1-1 KCl + +1-1 Br loses an electron -- oxidized Cl gains an electron -- reduced K remains unchanged at +1 Br2 0 Exercise For each of the following reactions find the element oxidized and the element reduced Cu + HNO3 Cu(NO3)2 + NO2 + H2O 0 +1+5-2 +2 +5 -2 +4 –2 +1 -2 • Cu increases from 0 to +2. It is oxidized • Only part of the N in nitric acid changes from +5 to +4. It is reduced • The nitrogen that ends up in copper nitrate remains unchanged Exercise For each of the following reactions find the element oxidized and the element reduced HNO3 + +1+5-2 • • • I2 0 HIO3 + +1+5-2 NO2 +4-2 N is reduced from +5 to +4. It is reduced. I is increased from 0 to +5 It is oxidized The hydrogen and oxygen remain unchanged. Agents • all redox reactions have one element oxidized and one element reduced • the compound that supplies the electrons (is oxidized) is the reducing agent • the compound that accepts the electrons (is reduced) is the oxidizing agent • occasionally, the same element may undergo both oxidation and reduction. This is known as an auto-oxidation reduction Review of Stock nomenclature (naming) of transitional metals • transition metals can form more than one type of ion (i.e. lose different amounts of electrons) – Cu1+, Cu 2+ • use roman numerals to indicate the charge – Cu1+ = “Copper I” ; Cu 2+ = “Copper II” • Exceptions: Ag1+, Zn2+, Cd2+, Al3+, Sc3+ • ex. copper (II) oxide (“copper two oxide”) – CuO – oxygen has a -2 charge, so it would only take one Cu2+ to bond with Oxygen. • ex. copper (I) oxide – copper 1+ -we would need two of these to react with oxygen so the formula would be: • Cu2O Examples • lead (II) hydroxide – Pb(OH)2 • cadmium nitrate – cadmium is always +2 – Cd(NO3)2 • MnO2 – manganese (IV) oxide Balancing Redox Reactions (only in Neutral or Acid Solutions) • many chemical reactions involving oxidations and reductions are complex and very difficult to balance by the “guess and check” methods we learned earlier • for complicated reactions, a more systematic approach is required Half-equations • half equations show the changes to individual species in a redox reaction – can use charges or oxidation #’s to do this • Fe2O3 + 2 Al 2 Fe + Al2O3 – Fe3+ + 3 e- Fe ….this is the reduction – Al Al3+ + 3 e- ….this is the oxidation • a wide variety of half equations can be found in the data booklet More… Br2 + 0 • 2e- + Br2 • 2I- 2I- 2Br- + I2 -1 -1 0 2BrI2 + 2e- • 2e- + Br2 + 2I- 2Br- + I2 + 2e- Balancing Redox Reactions in 8 “easy” steps. Page 217 in “the IB text book”. I do NOT like this way any more 1. assign oxidation states for each atom 2. deduce which species is oxidized and which is reduced 3. write half reactions for the oxidation and reduction – take compounds where “action” took place, split them and write them as individual reactions; there will be 2 half reactions 4. balance elements other than O and H 5. balance so # of electrons lost equals the # gained by adding e6. add the two half equations together to write the overall redox reaction and simplify 7. total up the total charges on the reactant and product sides to see if they are the same (not opposite) 8. balance the charges by adding H+ and balance oxygens by adding H2O to the appropriate sides Another way (p. 847 in AP book). I like this way the best but do not yet have examples on this PowerPoint done this way yet, only did in class 1. Divide into half equations 2. Balance 1. 2. 3. 4. elements other than O and H balance O by adding water as needed balance H by adding H+ as needed balance charges by adding e- as needed 3. Make sure electrons lost in one half-reaction are equally gained in the other half-reaction 1. if not, multiply the half reactions by integers as necessary to balance electrons gained and lost 4. Add the two half-reactions back together and simplify/canceling species appearing on both side of the combined reaction as needed Exercise • Deduce and balanced redox equation and identify the oxidizing and reducing agents. Fe2+ + MnO4- Fe3+ + Mn2+ • Step 1. +2 +7 -2 +3 +2 • Step 2. Fe is oxidized (Mn was the reducing agent that caused this) & Mn is reduced (Fe was the oxidizing agent that caused this) Fe2+ Fe3+ + eMnO4- + 5e- Mn2+ • Step 4. 5Fe2+ 5Fe3+ + 5eMnO4- + 5e- Mn2+ • Step 3. Exercise • Step 5. 5Fe2+ + MnO4- 5Fe3+ + Mn2+ • Step 6. Total charge on reactant side = 9+ Total charge on product side = 17+ • Step 7. To balance the equation charges, 8H+ must be added to the reactant side. 5Fe2+ + MnO4- + 8H+ 5Fe3+ + Mn2+ Now need to balance the hydrogens by adding water 5Fe2+ + MnO4- + 8H+ 5Fe3+ + Mn2+ + 4H2O More… • Nitric acid reacts with silver in a redox reaction. Ag(s) + NO3–(aq) → Ag+(aq) + NO(g) • Using oxidation numbers, deduce the complete balanced equation for the reaction showing all the reactants and products. Exercise • Deduce and balanced redox equation and identify the oxidizing and reducing agents. Ag(s) + NO3–(aq) → Ag+(aq) + NO(g) Step 1. 0 +5 -2 +1 +2-2 Step 2. Ag is oxidized & N is reduced Step 3. Ag Ag+ + eNO3- + 3e- NO Step 4. 3Ag 3Ag+ + 3eNO3- + 3e- NO Exercise 3Ag + NO3– → 3Ag+ + NO Total charge on reactant side = 1Total charge on product side = 3+ Step 7. To balance the equation charges, 4H+ must be added to the reactant side. 3Ag + NO3– + 4H+ → 3Ag+ + NO Now need to balance the hydrogens by adding water 3Ag + NO3– + 4H+ → 3Ag+ + NO + 2H2O Step 5. Step 6. Practice #1 Using the “new way” Balance • Cr2O72- + Cl- Cr3+ + Cl2 • 6e- + 14H+ + Cr2O72- Cr3+ + 7H2O • 3[2Cl- Cl2 + 2e-] Answer • 14H+ + Cr2O72- + 6Cl- 2Cr3+ + 7H2O + 3Cl2 #2 Balance and state the oxidizing agent and reducing agent Cu (s) + NO31- (aq) Cu 2+ (aq) + NO2 (aq) 4H+ + Cu (s) + 2NO31- (aq) Cu 2+ (aq) + 2NO2 (aq) + 2H2O #3 Mn 2+ + NaBiO3 Bi 3+ + MnO4 1- + Na 1+ 2Mn 2+ + 5NaBiO3 + 14H+ 5Bi 3+ + 2MnO4 1- + 5Na 1+ + 7H2O #4 NO2- + Cr2O72- Cr 3+ + NO3 13NO2- + Cr2O72- + 8H+ 2Cr 3+ + 3NO3 1- + 4H2O #5 S + HNO3 H2SO3 + N2O 2S + 2HNO3 + H2O 2H2SO3 + N2O #6 1. 2. 3. 4. Balance other than H&O O by H2O H by H+ charges by e- Cr2O72- + CH3OH HCO2H + Cr3+ 16H+ + 2Cr2O72- + 3CH3OH 3HCO2H + 4Cr3+ + 11H2O