CHEM 1105 REDOX REACTIONS
1. Definition of Oxidation and Reduction The old definition of oxidation was "addition of oxygen". Consider the following reactions: 4 H + 2- (a) 2Na + O2 Na2O (2Na + O ) 4 H + - (b) Na + Cl2 NaCl (Na + Cl ) By the old definition, reaction (a) is an oxidation. But in both cases Na is converted into Na+. Hence, in reaction (b) Na is also oxidized. In both cases, Na lost electrons in forming Na+ ions. Oxidation and reduction occur simultaneously; when one species is oxidized, the other is reduced. Hence we call these redox reactions. In reaction (a), Na is oxidized by losing electrons and O is reduced by gaining electrons to form O2- ions. Similarly, in reaction (b) Cl is reduced. New definition: Oxidation is the loss of electrons and reduction is the gain of electrons. LEO GER is a convenient way of remembering this (Loss of Electrons is Oxidation; Gain of Electrons is Reduction). [Another memory aid is OIL RIG]. In reaction (b), we can say that Na was oxidized by Cl and Cl was reduced by Na. Hence, we call Cl the oxidizing agent and Na the reducing agent. In general, the species oxidized is the reducing agent and the species reduced is the oxidizing agent.
2. Oxidation Numbers In reaction (b), Na and Cl were changed to NaCl in a redox reaction. In NaCl, we have Na+ ions and Cl-ions and we say that Na+ has an oxidation number of +1 and Cl- has an oxidation number of -1. We can consider the free elements (in this case, Na and Cl) to have an oxidation number of 0. We can hence define oxidation as an increase in oxidation numner (Na: from 0 to +1) and reduction as a decrease in oxidation number (Cl: from 0 to -1) In more complicated species (compounds and polyatomic ions), we can make rules for calculating an oxidation number of an element in the ion or compound, but we must remember that in those cases the use of oxidation numbers is only a convenient bookkeeping device.
Oxidation Number Rules 1. The sum of all oxidation numbers of elements in a compound is zero. 2. The oxidation number of any free element is zero. 3. The oxidation number of a monatomic ion is the charge on the ion. 4. The sum of all oxidation numbers of elements in a polyatomic ion is the charge on the ion. 5. The oxidation number of any Group 1 metal in combination is +1. 6. The oxidation number of any Group 2 metal in combination is +2. 7. The oxidation number of any halogen (Group 7) in combination with one other element other than oxygen is -1. 8. The oxidation number of hydrogen in combination is +1 unless the H is present as H-1 (hydride, as in NaH) in which case the oxidation number is -1. 9. The oxidation number of oxygen in combination is -2 unless the O is present as peroxide in which case the oxidation number is -1. 10. The oxidation number of fluorine is always -1.
Exercises Calculate the oxidation number of the underlined element in each case.
2- + HCl H2SO4 CuCl2 S2O3 O3 NH4 2- HNO3 Cr2O7 K2XeF6 N2O3 Mg(ClO4)2 -2-
3. Balancing Redox Reaction Equations in Aqueous Acid
The Half-Reaction (Ion-Electron) Method
This method divides the reaction into two half-reactions; one involving the species being oxidized and the other involving the species being reduced.
(1) Use oxidation numbers to identify which species is oxidized and which is reduced. (2) Write the half-reactions. For each half-reaction, balance the element undergoing oxidation + number change, then the O (by adding H2O) and finally the H (by adding H ). Then balance the charge by adding electrons. (3) Multiply one or both half-reactions by the smallest integer(s) so that they have the same number of electrons on opposite sides. (4) Add the multiplied half-reactions and cancel the charges and any other species that occur on both sides. (5) Check that the resulting equation is balanced for mass (atoms) and charge.
Examples
Simple; no other atoms needed 3+ - H Fe + Cl Fe + Cl2 Solution
(1)Fe3+ is being reduced to Fe (ox. no. decreases from 3 to 0) - Cl is being oxidized to Cl2 (ox. no. increases from -1 to 0)
(2)Fe3+ H Fe Fe3+ + 3e- H Fe (balanced) (i)
- H Cl Cl2 - H 2Cl Cl2 - H - 2Cl Cl2 + 2e (balanced) (ii)
(3) (i) x 2 gives: 2Fe3+ + 6e- H 2Fe (iii) - H - (ii) x 3 gives: 6Cl 3Cl2 + 6e (iv)
3+ - H (4) add (iii) + (iv): 2Fe + 6Cl 2Fe + 3Cl2
(5) Check: LHS 2Fe; RHS 2Fe; LHS 6Cl; RHS 6Cl; LHS charge = 0; RHS charge = 0 EQUATION IS BALANCED
+ More complex; oxygen (from H2O) and H (from H ) needed
- - H 2+ MnO4 + Cl Mn + Cl2
Solution
- 2+ (1) Mn in MnO4 is being reduced Mn (ox. no. decreases from 7 to 2) - Cl is being oxidized to Cl2 (ox. no. increases from -1 to 0) -3-
- H 2+ (2) MnO4 Mn - H 2+ MnO4 Mn + 4H2O + - H 2+ 8H + MnO4 Mn + 4H2O + - - H 2+ 8H + MnO4 + 5e Mn + 4H2O (balanced) (i)
- H Cl Cl2 - H 2Cl Cl2 - H - 2Cl Cl2 + 2e (balanced) (ii)
+ - - H 2+ (3) (i) x 2 gives: 16H + 2MnO4 + 10e 2Mn + 8H2O (iii) - H - (ii) x 5 gives: 10Cl 5Cl2 + 10e (iv)
+ - - H 2+ (4) add (iii) + (iv): 16H + 2MnO4 + 10Cl 2Mn + 5Cl2 + 8H2O
(5) Check: LHS 2Mn; RHS 2Mn; LHS 10Cl; RHS 10Cl; LHS 16H; RHS 16H; LHS 8 O; RHS 8 O LHS charge = 4; RHS charge = 4 EQUATION IS BALANCED
2- H - 2- (a) S2O3 + I2 I + S4O6 H (b) HNO3 + H2S NO + S
H 2+ (c) Cu + HNO3 Cu + NO
- H - - (d) I2 + ClO3 IO3 + Cl
2- H 3+ (e) Cr2O7 + CH3OH Cr + CH2O
Disproportionation
A disproportionation reaction is a redox reaction in which one substance is both oxidized and 0 - reduced. In the reaction below, Cl (in Cl2) is being reduced to Cl (in HCl) and is being oxidized to Cl with an ox. no. of 1 in HOCl. Each half-reaction starts with the same reactant.
H Cl2 + H2O HOCl + HCl
H - (f) P4 PH3 + H2PO2
2- H - (g) MnO4 MnO4 + MnO2