Oxidation-Reduction Topic 9.1

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Memorize these! NO 3 1- nitrate NO 2 1- nitrite OH 1- hydroxide ClO 2 1- chlorite ClO 3 1- chlorate HCO 3 1- hydrogencarbonate (bicarbonate) SO 4 2- sulfate SO 3 2- sulfite CO 3 2- carbonate PO 4 3- phosphate NH 4 1+ 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- oxidation I - is L- loss of electrons R- reduction I- is G- gain of electrons

Oxidation states 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  Mg 2+ (aq) + 2e - reduction –a process where the number decreases (less positive/more negative because gains neg. electrons) Ex: O + 2 e -  O 2- (g) Note: oxidation states must be written with the sign in FRONT: +2 not 2+

1The oxidation number of an free element is always 0 O 2, H 2, Ne, Zn 2The oxidation number of Hydrogen is usually +1 Metal hydrides are an exception. They are -1 HCl, H 2 SO 4 NaH 3The oxidation number of Oxygen is usually -2. Peroxides are an exception They are –1. H 2 O, NO 2, etc. -O-O- bonding 4Group 1 metals are always +1 Group 2 metals are always +2 Aluminum is always +3 Li, Na… Mg, Ba.. Al Rules for assigning oxidation states

5Fluorine is always -1 Other group 17 (halogens) are often -1 HF, OF 2 HI, NaCl, KBr 6 Oxidation numbers of monatomic ions follow the charge of the ion S 2-, Zn 2+ 7The SUM of oxidation numbers is zero for a neutral compound. LiMnO 4 … 8For polyatomic ions, the SUM is their charge. SO 4 2-, NO 3 1-

Practice Assigning Oxidation Numbers NO 2 O is -2 x 2 = -4 N must equal +4 N2O5N2O5 O is -2 x 5 = -10 N must equal +10/2 = +5 HClO 3 (+1) + (x) + 3(-2) = 0; x = (+5) HNO 3 O is -2 x 3 = -6; H is +1; N must equal +5 Ca(NO 3 ) 2 O is -2 x 3 = -6 x 2 = -12; Ca is +2; N must equal +10/2 = +5 KMnO 4 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 K 2 Cr 2 O 7 O is -2 x 7 = -14; K is +1 x 2 = +2; Cr must equal +12/2 = +6 CO 3 2- x + 3(-2) = -2 x = +4 CN - N is -3; Charge = -1 means that C must be +2 K 3 Fe(CN) 6 N is -3 x 6 = -18; C is +2 x 6 = +12; K is +1 x 3 =+3; Fe must be +3 CH 4 H is +1 x 4 = +4; C must be -4

Using Oxidation Numbers

Exercise For each of the following reactions (not balanced to simplify) find the element oxidized and the element reduced Cl 2 + KBr  KCl + Br 2 Cu + HNO 3  Cu(NO 3 ) 2 + NO 2 + H 2 O HNO 3 + I 2  HIO 3 + NO 2

Exercise For each of the following reactions find the element oxidized and the element reduced Cl 2 + KBr  KCl + Br Br loses an electron -- oxidized Cl gains an electron -- reduced K remains unchanged at +1

Exercise For each of the following reactions find the element oxidized and the element reduced Cu + HNO 3  Cu(NO 3 ) 2 + NO 2 + H 2 O – 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 HNO 3 + I 2  HIO 3 + NO 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

transition metals can form more than one type of ion (i.e. lose different amounts of electrons) –Cu 1+, Cu 2+ use roman numerals to indicate the charge – Cu 1+ = “Copper I” ; Cu 2+ = “Copper II” Exceptions: Ag 1+, Zn 2+, Cd 2+, Al 3+, Sc 3+ Review of Stock nomenclature (naming) of transitional metals

ex. copper (II) oxide (“copper two oxide”) – CuO – oxygen has a -2 charge, so it would only take one Cu 2+ 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: Cu 2 O

Examples lead (II) hydroxide –Pb(OH) 2 cadmium nitrate –cadmium is always +2 –Cd(NO 3 ) 2 MnO 2 –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 Fe 2 O Al  2 Fe + Al 2 O 3 –Fe e -  Fe ….this is the reduction – Al  Al e- ….this is the oxidation a wide variety of half equations can be found in the data booklet

More… Br 2 + 2I -  2Br - + I e- + Br 2  2Br - 2I -  I 2 + 2e- 2e- + Br 2 + 2I -  2Br - + I 2 + 2e-

Balancing Redox Reactions in 8 “easy” steps. Look on page 847 of the text book for another way to do these steps. 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 e - 6.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 H 2 O to the appropriate sides

Exercise Deduce and balanced redox equation and identify the oxidizing and reducing agents. Fe 2+ + MnO 4 -  Fe 3+ + Mn 2+ Step Step 2. Fe is oxidized (Mn was the reducing agent that caused this) & Mn is reduced (Fe was the oxidizing agent that caused this) Step 3. Fe 2+  Fe 3+ + e- MnO e-  Mn 2+ Step 4. 5Fe 2+  5Fe e- MnO e-  Mn 2+

Exercise Step 5. 5Fe 2+ + MnO 4 -  5Fe 3+ + Mn 2+ 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. 5Fe 2+ + MnO H +  5Fe 3+ + Mn 2+ Now need to balance the hydrogens by adding water 5Fe 2+ + MnO H +  5Fe 3+ + Mn H 2 O

More… Nitric acid reacts with silver in a redox reaction. Ag(s) + NO 3 – (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) + NO 3 – (aq) → Ag + (aq) + NO(g) Step Step 2. Ag is oxidized & N is reduced Step 3. Ag  Ag + + e- NO e-  NO Step 4. 3Ag  3Ag + + 3e- NO e-  NO

Exercise Step 5. 3Ag + NO 3 – → 3Ag + + NO Step 6. Total charge on reactant side = 1- Total charge on product side = 3+ Step 7. To balance the equation charges, 4H + must be added to the reactant side. 3Ag + NO 3 – + 4H + → 3Ag + + NO Now need to balance the hydrogens by adding water 3Ag + NO 3 – + 4H + → 3Ag + + NO + 2H 2 O

Balance Cr 2 O Cl -  Cr 3+ + Cl 2 Answer 14H + + Cr 2 O Cl -  2Cr H 2 O + 3Cl 2 Practice #1

Balance Cu (s) + NO 3 1- (aq)  Cu 2+ (aq) + NO 2 (aq) Answer 4H + + Cu (s) + 2NO 3 1- (aq)  Cu 2+ (aq) + 2NO 2 (aq) + 2H 2 O Practice #2

Balance Mn 2+ + NaBiO 3  Bi 3+ + MnO Na 1+ Answer 2Mn NaBiO H +  5Bi MnO Na H 2 O Practice #3

Balance NO 2- + Cr 2 O 7 2-  Cr 3+ + NO 3 1- Answer 3NO 2- + Cr 2 O H +  2Cr NO H 2 O Practice #4

Balance S + HNO 3  H2SO 3 + N 2 O Answer 2S + 2 HNO 3 + H 2 O  2H 2 SO 3 + N 2 O Practice #5