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Chemistry Specialist: Alicia Hart Chem 2A Workshop Oxidation State (Oxidation Number) Simply put, it is the degree to which an element is oxidized in a chemical compound. We assign a number to the element according to how many e- s the element appears to lose, gain or contribute to a bond in a chemical compound. This is said to be a hypothetical charge because the oxidation state of the atom is not always consistent with its actual charge. Below are the rules that govern the assignment of oxidation numbers. Assigning Oxidation Numbers 1. An atom in its elementary (uncombined) state is assigned an oxidation number of zero (e.g. O2). 2. The oxidation number of a monatomic ion corresponds to its charge (e.g. Fe3+ Ox. # is +3). 3. The algebraic sum of the elemental oxidation states in a neutral molecule is zero (e.g. CO2, C is +4 and O is -2; since there are two O’s [2(-2) = -4], the Ox. #’s sum to zero). In contrast, the algebraic sum of the elemental oxidation states in a polyatomic ion is = to the net charge on the molecule (e.g. CO32-, since O is -2, C must be +4 because [3(-2) + x= -2]). When in combination with other atoms: 4. Typically, group 1 elements (i.e. alkali metals) are assigned an oxidation state of +1 (Ag is also assigned +1) and group 2 elements (i.e. alkaline earth metals) are assigned an oxidation state of +2 (Zn, Cd also assigned +2) 5. Halogens are assigned an oxidation number of -1. EXCEPTION: Oxidation number varies when combined with oxygen (F always assigned -1). 6. Hydrogen is assigned an oxidation number of +1. EXCEPTION: In metal hydrides, hydrogen is assigned an oxidation number of -1 (e.g. NaH). 7. Oxygen is assigned an oxidation number of -2. EXCEPTION: Oxygen is assigned an oxidation state of -1 when forming peroxides (e.g. H2O2). * Note: the more electronegative atom is assigned the (-) oxidation state. OIL RIG Oxidation Is Loss Reduction Is Gain (in e- s) Reduction: occurs when the oxidation number of an atom becomes more negative. Oxidation: occurs when the oxidation number of an atom becomes more positive. Reducing agent: this is the chemical species that causes the reduction. In other words, the reducing agent is the compound that contains the element being oxidized. Oxidizing agent: this is the chemical species that causes the oxidation. In other words, the oxidizing agent is the compound that contains the element being reduced. Balancing Redox Reactions Acidic Solution: 1. 2. 3. 4. 5. 6. Break up the chemical reaction into two half-reactions (red-ox) Balance all atoms other than H or O Balance O’s by adding H2O to the side with less O Balance H’s by adding H+ to the side with less H Balance charge by adding e- s to appropriate side Combine the half-reactions canceling any like terms (1 H+ and 1 –OH = 1 H2O) and balance the overall charge by multiplying each half-reaction by the appropriate integer. 7. Check your work and then you’re done! Example: SO2(aq) + Cr2O72-(aq) SO42-(aq) + Cr3+(aq) * remember: reaction carried out in acidic conditions 1. SO2 SO42Cr2O72- Cr3+ S from +4 to +6 (oxidation) Cr from +6 to +3 (reduction) 2. Cr2O72- 2Cr3+ 3. SO2 + 2 H2O SO42Cr2O72- 2 Cr3+ + 7 H2O 4. SO2 + 2 H2O SO42- + 4 H+ Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O 5. SO2 + 2 H2O SO42- + 4 H+ + 2 eCr2O72- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O 6. 3 (SO2 + 2 H2O SO42- + 4 H+ + 2 e-) 2+ Cr2O7 + 14 H + 6 e 2 Cr3+ + 7 H2O ____ 3 SO2 + Cr2O72- + 2 H+ 3 SO42- + 2 Cr3+ + H2O Basic Solution: Nearly identical procedure, but we need to add a step between steps 4 and 5. We shall label this step 4B: after balancing the H+, add the same amount of –OH to both sides of each of the half-reactions. Example: MnO4-(aq) + H2O2(aq) MnO2(s) + O2(g) * remember: reaction is carried out in basic conditions 1. MnO4- MnO2 H 2O2 O2 2. Not necessary; all atoms other than H and O are balanced. 3. MnO4- MnO2 + 2 H2O Mn from +7 to +4 (reduction) O from -1 to 0 (oxidation) 4A. MnO4- + 4 H+ MnO2 + 2 H2O H 2O2 O2 + 2 H+ 4B. MnO4- + 4 H+ + 4 -OH MnO2 + 2 H2O + 4 -OH H2O2 + 2 -OH O2 + 2 H+ + 2 -OH 5. MnO4- + 4 H+ + 4 -OH + 3 e- MnO2 + 2 H2O + 4 -OH H2O2 + 2 -OH O2 + 2 H+ + 2 -OH + 2 e- 6. 2 (MnO4- + 4 H+ + 4 -OH + 3 e- MnO2 + 2 H2O + 4 -OH) 3 (H2O2 + 2 -OH O2 + 2 H+ + 2 -OH + 2 e-) 2 MnO4- + 3 H2O2 2 MnO2 + 3 O2 + 2 -OH + 2 H2O Jim and Rolf’s alternative method: Balancing Redox Reactions A six step method: HElOHEN H El O Half rxn's: Elements: Oxygen: break up into balance the Elements balance O Half (1/2) rxn's except O and H with H2O Example: Balance the following reaction in acid: 1) H: 1/2 rxn's Cr2O7-2 Þ H 2O 2 Þ Cr+3 O2 red (+6 Þ +3) ox (-1 Þ 0) 2) El: elements Cr2O7-2 Þ H 2O 2 Þ 2 Cr+3 O2 red ox 3) O: oxygen Cr2O7-2 Þ H 2O 2 Þ 2 Cr+3 + 7 H2O red O2 ox H Hydrogen: balance H with H+ E N Electrons: Number of electrons: add the e- to multiply so the Number balance charge of e- is the same in both Cr2O7-2 + H2O2 Þ Cr+3 + O2 4) Cr2O7-2 + 14 H+ Þ 2 Cr+3 + 7 H2O red H 2O 2 Þ O2 + 2 H+ ox 5) E: electrons Be sure to use the net (total) charge on each sides. Multiply coefficients by charges to get the net. Cr2O7-2 + 14 H+ + 6e- Þ 2 Cr+3 + 7 H2O red H 2O 2 Þ 2 e- + O2 + 2 H+ ox 6) N: number Cr2O7-2 + 14 H+ + 6e- Þ 2 Cr+3 + 7 H2O red 3 H 2O 2 Þ 6 e- + 3 O2 + 6 H+ ox 7) Now add them up. Cr2O7-2 + 8 H+ + 3 H2O2 Þ 2 Cr+3 + 7 H2O + 3 O2 8) Always check that in your answer all electrons cancel and atoms and charges balance! Also check to see if you can divide by a common divisor (see if you have the lowest possible coefficients) and if you can collect terms. Balancing in Base The procedure is the same as with acid except that after you add the H+ to balance the hydrogens, add the same number of OH- to both sides and form water on one side. Example: Balance the following 1/2 rxn in base: BrO3- Þ BrO4- 1) H: (The 1/2 rxn is given in this case) BrO3Þ BrO42) El: elements BrO3Þ BrO4Bromines are already balanced. 3) O: oxygen BrO3- + H2O Þ BrO44) H: hydrogen (but because its really in base, after adding H+, then add OH- to both sides.) BrO3- + H2O + 2 OH- Þ BrO4- + 2H+ + 2 OHIf you put H+ and OH- together, they will form water. BrO3- + H2O + 2 OH- Þ BrO4- + (2H+ + 2 OH- Þ 2 H2O) Cancel the waters and continue. BrO3- + 2 OH- Þ BrO4- + H2O 5) E: electrons BrO3- + 2 OH- Þ BrO4- + H2O + 2 e-