Review of Oxidation States/Numbers and Balancing Redox Reactions Chemistry 1050/1051 Winter 2014 Oxidation States/Oxidation Numbers: A bookkeeping device used to tell whether a neutral atom has gained or lost electrons when it forms a compound. They are not real charges except for simple ions. They do give a qualitative indication of the electronic environment of an atom in a molecule which results from the bonds it forms with other atoms. A positive value indicates an electron deficient environment, a negative value, an electron rich environment. Background Atoms in molecules and ions are assigned oxidation numbers by assuming total transfer of all shared bonding electrons to the more electronegative atom. The resulting net charge on each atom is its oxidation number. Cl is more electronegative than H. H x o o Cl o o o X H = +1 X Cl = 1 Rules for Assigning Oxidation Numbers (X) Note: The oxidation number of most elements can vary considerably from one substance to another. The oxidation number of such elements can be determined in a given compound or ion using rule #8 if, as is usually the case, the oxidation number of the other elements present can be positively determined from the other rules. 1. In free (uncombined) elements, each atom has an oxidation number of zero. e.g. all atoms in Fe, O 2, P 4, S 8, Ca, Li have oxidation number = 0 2. For an ion composed of only one atom, the oxidation number of the atom is equal to the charge on the ion. e.g. Fe 3+ (X Fe = +3), S 2 G (X S =!2) 3. The oxidation number of fluorine in all its compounds with other elements is equal to!1. This is because each fluorine atom makes only one single bond and fluorine is the most electronegative element. e.g. in CF 4, OF 2, SF 6, XeF 2, X F =!1 4. The oxidation number of hydrogen (H) in most compounds is equal to +1. e.g. in H 2 O, NH 3, C 2 H 5 OH, X H = +1 exception: metal/hydrogen compounds where it is equal to!1 e.g. in NaH, LiAlH 4, X H =!1 5. The oxidation number of oxygen in most compounds is equal to!2. It is the second most electronegative element and almost always forms two bonds to less electronegative atoms. e.g. in H 2 O, SO 2, H 2 C 2 O 4, KMnO 4, X O =!2 1
exceptions: 1. The oxidation number of oxygen can be +1 or +2 only in binary compounds oxygen forms with fluorine. Fluorine is more electronegative than oxygen. e.g. OF 2 : X O = +2 O 2 F 2 : X O = +1 (see rules # 3 and # 8) 2. The oxidation number of oxygen is!1 (a) in hydrogen peroxide (H 2 O 2 ) which contains an oxygen-to-oxygen bond. The Lewis structure of H 2 O 2 is H O O H (b) in Alkali metal and Alkaline Earth metal peroxides, which contain the peroxide ion (O 2 2 G). e.g. Na 2 O 2, BaO 2 3. The oxidation number of oxygen is!½ in the binary compounds formed when the more reactive Alkali metals and Alkaline Earth metals react with O 2 (g). Such compounds contain the superoxide ion (O 2 G) e.g. K(s) + O 2 (g) KO 2 (s) 6. In binary ionic compounds, the oxidation numbers of the atoms are equal to ionic charges. e.g. CrBr 3 X Br =!1, X Cr = +3 7. In binary compounds of non-metals, the oxidation number of the more electronegative atom is identical to the charge on its common anion. e.g. in CI 4, X I =!1 (common anion of I is IG). The more electronegative atom is almost always the second atom in the formula. 8. The algebraic sum of the oxidation numbers of the elements in a neutral compound must equal zero. In a polyatomic ion the sum must equal the ionic charge. e.g. identify the oxidation numbers of all elements in each species given below: HNO 3 X H + X N + 3X O = 0 SO 2 4 G X S + 4X O =!2 +1 + X N + 3(!2) = 0 X S + 4(!2) =!2 X N = +5 X S = +6 PH 3 X P + 3X H = 0 VO 2+ X V + X O = +2 X P + 3(+1) = 0 X V + (!2) = +2 X P =!3 X V = +4 O 2 G 2X O =!1 CI 4 X c + 4X I = 0 X O =! ½ X c + 4(!1) = 0 X c = +4 9. In all compounds, the oxidation number of Alkali metals is equal to +1, of Alkaline earth metals is equal to +2 and, of Aluminum is equal to +3. The oxidation number of Na in Na 2 O 2 or K in KO 2 must be +1. Hence it is used to find the oxidation number of oxygen which can vary. See the examples below. e.g. in Na 2 O 2, X Na = +1 : 2X Na + 2X O = 0 2(+1) + 2X O = 0 X O =!1 in KO 2, X K = +1 : X K + 2X O = 0 +1 + 2X O = 0 2X O =!1 X O =!½ 2
Oxidation/Reduction (Redox) Reactions. 1. They are electron transfer reactions. 2. They involve two complementary processes, oxidation and reduction. Definitions: Oxidation involves loss of electrons by a reactant. Reduction involves gain of electrons by a reactant. NOTE: OIL-RIG is a good mnemonic for distinguishing between oxidation and reduction. OIL = oxidation is loss RIG = reduction is gain Examples of redox reactions 1. All single displacement reactions e.g. Cu(s) + 2 AgNO 3 (aq) 2 Ag(s) + Cu(NO 3 ) 2 (aq) 2. All reactions involving combining 2 pure elements 3. All combustion reactions e.g. 2 Na(s) + Cl 2 (g) 2 NaCl(s) e.g. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(l) 4. All reactions where a free element is a reactant or a product. Oxidizing Agent (Oxidant) e.g. C 2 H 4 (g) + H 2 (g) C 2 H 6 (g) Definition: An electron acceptor. It causes another substance to undergo loss of electrons, i.e. to become oxidized. An oxidizing agent gains electrons from the oxidized species and hence undergoes reduction. Common strong oxidizing agents are the gases, F 2, O 2 and Cl 2 and acidified solutions of MnO 4 G, Cr 2 O 7 2 G and NO 3 G. Reducing Agent (Reductant) Definition: An electron donor. It causes another substance to gain electrons and hence become reduced. A reducing agent loses electrons to the reduced species and hence undergoes oxidation. Note: Metals cannot gain electrons in chemical reactions. They can only act as reducing agents. Examples: Alkali metals and alkaline earth metals are strong reducing agents. Zinc is a moderate reducing agent. C(s), CO(g) and H 2 (g) are good reducing agents at high temperature. Definitions of oxidation and reduction based on oxidation numbers. Oxidation: An element is oxidized if it s oxidation number increases (becomes more positive) in a chemical reaction. 3
Reduction: An element is reduced if it s oxidation number decreases (becomes more negative) in a chemical reaction. Balancing redox reactions: The half!reaction method Note: for reactions in solution, write an ionic equation if strong electrolytes are present. Strong electrolytes are ionic compounds, strong acids and strong bases. Represent strong electrolytes in the ionic equation with the formulas of their ions, all other substances including aqueous weak electrolytes with their molecular formula. Step 1. Identify the unbalanced oxidation and reduction ½ reactions, assigning oxidation numbers if needed. The oxidation number of each element is written above its symbol in the example below. H 2 C 2 O 4 (aq) + KMnO 4 (aq) + HCl(aq) MnCl 2 (aq) + CO 2 (g) + KCl(aq) + H 2 O(l) Strong electrolytes: KMnO 4 (aq), HCl(aq), MnCl 2 (aq), KCl(aq) Weak electrolytes: H 2 C 2 O 4 (aq) non!electrolytes: CO 2 (g), H 2 O(l) +1 +3-2 1+ +7-2 +1-1 - H 2C 2O 4(aq) + K+(aq) + MnO H + 4 (aq) + (aq) + Cl-(aq) Mn(+7 to +2) reduced +2 Mn2+(aq) + -1 +4-2 +1-1 +1-2 Cl- (aq) + CO2(g) + K+(aq) + Cl-(aq) + H2O(l) C (+3 to +4) oxidized note: The presence of HCl(aq) means acidic medium. Step 2. Balance each ½ reaction for atoms and charge using rules for acidic/neutral or basic medium. A set of rules for balancing atoms and charge in acidic/neutral medium 1. Balance all atoms, other than O and H 2. Balance O by adding H 2 O in sufficient numbers to deficient side 3. Balance H by adding H + in sufficient numbers to deficient side 4. Balance charge by adding electrons in sufficient numbers to more positive side Reduction MnO 4 G Mn 2+ MnO 4 G is oxidizing agent balance Mn already balanced balance O MnO 4 G Mn 2+ + 4 H 2 O balance H MnO 4 G+ 8 H + Mn 2+ + 4 H 2 O balance charge LHS RHS + 7 + 2 result: add 5 eg to LHS [MnO 4 G+ 8 H + + 5 eg Mn 2+ + 4 H 2 O 4
Oxidation H 2 C 2 O 4 CO 2 H 2 C 2 O 4 is reducing agent balance C H 2 C 2 O 4 2 CO 2 balance O already balanced balance H H 2 C 2 O 4 2 CO 2 + 2 H + balance charge LHS RHS 0 + 2 result: add 2 eg to RHS H 2 C 2 O 4 2 CO 2 + 2 H + + 2 eg Step 3. Combine ½ reactions balancing eg lost and gained. Cancell H 2 O s, H + s (or OHG s) from both sides as needed. [MnO 4 G + 8 H + + 5 eg 2 Mn 2+ + 4 H 2 O] x 2 [H 2 C 2 O 4 2 CO 2 + 2 H + + 2 eg] x 5 Now add the two half reactions: 2 MnO 4 G + 16 H + + 10 eg 2 Mn 2+ + 8 H 2 O 5 H 2 C 2 O 4 10 CO 2 + 10 H + + 10 eg 2 MnO 4 G + 6 H + + 5 H 2 C 2 O 4 2 Mn 2+ + 10 CO 2 (g) + 8 H 2 O Note: Restore and balance spectators ions if you wrote an ionic equation in Step 1. 2 KMnO 4 + 5 H 2 C 2 O 4 + 6 HCl 2 MnCl 2 + 10 CO 2 + 2 KCl + 8 H 2 O A set of rules for balancing atoms and charge in basic medium Follow rules for balancing half reactions in acidic/neutral medium and Either neutralize H + by adding sufficient OHG to both sides in each half reaction (gives correct half reactions) or in the combined equation (simplest). Each H + is converted to H 2 O. Cancel H 2 O s as needed. 0!2 +1 +4!2!1 S(s) + OClG(aq) SO 3 2 G(aq) + ClG(aq) basic medium oxidation: [S + 3 H 2 O SO 2 3 G + 6 H + + 4 eg] x 1 reduction: [ OClG + 2 H + + 2 eg ClG + H 2 O] x 2 S + 3 H 2 O SO 3 2 G + 6 H + + 4 eg 2 OClG + 4 H + + 4 eg 2 ClG + 2 H 2 O S + 2 OClG + H 2 O SO 3 2 G + 2 H + + 2 ClG S + 2 OClG + H 2 O + 2 OHG S + 2 OClG + H 2 O + 2 OHG S + 2 OClG + 2 OHG SO 3 2 G + 2 H + + 2 ClG + 2 OHG SO 3 2 G + 2 H 2 O + 2 ClG SO 3 2 G + H 2 O + 2 ClG (one H 2 O cancelled from both sides) 5
Special Cases 1. more than one oxidized species O ( 1 to 2) reduced 3+ 2 +1 1 +5 2 +6 2 +1 2 As 2 S 3 (s) + H 2 O 2 (aq) AsO 4 3 (aq) + SO4 2 (aq) + H2 O(l) acidic medium As (+3 to +5) oxidized S ( 2 to +6) oxidized unbalanced half reactions oxidation: As 2 S 3 (s) AsO 3 4 G(aq) + SO 2 4 G(aq) reduction: H 2 O 2 (aq) H 2 O(l) 2. disproportionation: A redox reaction in which the same element is both oxidized and reduced. 0!1 +5 Cl 2 ClG + ClO 3 G basic medium Cl 2 must be the reactant in both half reactions unbalanced half reactions: oxidation: Cl 2 ClO 3 G reduction: Cl 2 ClG 3. reverse disproportionation: A redox reaction in which two reactants, each containing the same element but in different oxidation states form a product in which that element has an oxidation state intermediate between the oxidation states of the two reactants.!2 +4 +2 HSG + HSO 3 G unbalanced half reactions: oxidation: HSG S 2 O 2 3 G reduction: HSO 3 G S 2 O 2 3 G S 2 O 3 2 G (must be the product in both half reactions) Note: Practice balancing redox reactions from the text. Developed by Dr. Chris Flinn 6