Oxidation-Reduction - Oxidation and Reduction Definitions...

Unformatted text preview: Oxidation and Reduction Definitions of oxidation and reduction Oxidation numbers Redox equations Oxidation - reduction Oxidation is loss of electrons Reduction is gain of electrons Oxidation is always accompanied by reduction • The total number of electrons is kept constant Oxidizing agents oxidize and are themselves reduced Reducing agents reduce and are themselves oxidized Follow the electrons Oxidation numbers Oxidation number is the number of electrons gained or lost by the element in making a compound Metals are typically considered more 'cation-like' and would possess positive oxidation numbers, while nonmetals are considered more 'anion-like' and would possess negative oxidation numbers. Predicting oxidation numbers Oxidation number of atoms in element is zero in all cases Oxidation number of element in monatomic ion is equal to the charge sum of the oxidation numbers in a compound is zero sum of oxidation numbers in polyatomic ion is equal to the charge F has oxidation number –1 H has oxidn no. +1; except in metal hydrides where it is –1 Oxygen is usually –2. Except: O is –1 in hydrogen peroxide, and other peroxides O is –1/2 in superoxides KO2 In OF2 O is +2 Position of element in periodic table determines oxidation number G1A is +1 G2A is +2 G3A is +3 (some rare exceptions) G5A are –3 in compounds with metals, H or with NH4+. Exceptions are in compounds to the right; in which case use rules 3 and 4. G6A below O are –2 in binary compounds with metals, H or NH4+. When they are combined with O or with a lighter halogen, use rules 3 and 4. G7A elements are –1 in binary compounds with metals, H or NH4+ or with a heavier halogen. When combined with O or a lighter halogen, use rules 3 and 4. Identifying reagents Those elements that tend to give up electrons (metals) are typically categorized as reducing agents and those that tend to accept electrons (nonmetals) are referred to as oxidizing agents. Iron can reduce Cu2+ to Cu The iron nail reduces the Cu2+ ions and becomes coated with metallic Cu. At the same time, the intensity of the blue color diminishes due to loss of Cu2+ ions from solution. Any element can be both an oxidizer and reducer depending on relative positions in the activity series Fe reduced Cu2+, but Cu can reduce Ag+ (lower activity Fe2+ is reduced by Zn More active metals are strongly reducing Predicting results of displacement reactions In this reaction the element metal A displaces the ion metal B from its compound This will only occur if A lies above B in the activity series A( s ) BX (aq) B ( s ) AX (aq ) Displacement reaction exercises Nuggets of redox processes Where there is oxidation there is always reduction Oxidizing agent Reducing agent Is itself reduced Is itself oxidized Gains electrons Loses electrons Causes oxidation Causes reduction Identify redox by change in oxidation numbers: follow the flow of electrons Reducing agent increases its oxidation number Oxidizing agent decreases its oxidation number In reaction with metals, nonmetals are always oxidizers Reactions of elements are always redox The nonmetal gains electrons, becomes a negative ion The metal loses electrons, becomes a positive ion Identification is harder when there are no elements involved: oxidation numbers must be used Balancing redox equations: systematic methods Oxidation number method – tracking changes in the oxidation numbers Half-reaction method – tracking changes in the flow of electrons Same principles, different emphasis Use is a matter of choice, but familiarity with both is important Oxidation number method What goes up must come down… Sum of the changes in oxidation numbers in any process is zero Six habits of the redox equation balancer Walking through the steps STEP 1: write the unbalanced net ionic equation In an acid solution, permanganate is reduced by bromide ion to give Mn2+ ion and bromine MnO4 (aq) Br (aq) Mn 2 (aq) Br2 (aq) STEP 2: Balance the equation for elements other than O and H* We need to double the bromide ions on the L.H.S. to balance the equation 4 2 MnO (aq) 2 Br (aq) Mn (aq) Br2 (aq) O and H can remain unbalanced because we will top up with water and hydronium ions later * STEP 3: Assign oxidation numbers Use the rules of oxidation numbers Element is zero Monatomic ion: oxidation number is same as charge Oxide +7-2 -1 +2 is 0 4 2 MnO (aq) 2 Br (aq) Mn (aq) Br2 (aq) STEP 4: Identify oxidized and reduced Mn is reduced from +7 to +2 Net gain of 5 electrons Br is oxidized from -1 to 0 Net loss of 1 electron +7 4 -1 +2 0 2 MnO (aq) 2 Br (aq) Mn (aq) Br2 (aq) STEP 5: Balance the oxidized and reduced species For every Mn reduced (decrease in oxidation number of 5), need five Broxidized (increase in oxidation number of 1) Equation becomes 2MnO4 (aq) 5 2 Br (aq) 2Mn2 (aq) 5Br2 (aq) Redox is now complete but material balance is not STEP 6: Material balance with H2O and H+ Strategy: add H2O to the side that lacks for O and add H+ (the reaction is in acid solution) to the other side In basic solution we add OH- and H2O instead of H2O and H+ respectively 2MnO4 (aq) 5 2 Br (aq) 2Mn2 (aq) 5Br2 (aq) 2MnO4 (aq) 10 Br (aq) 16 H (aq) 2Mn2 (aq) 5Br2 (aq) 8H 2O(l ) Test equation for both atoms and charges The Half-Reaction method Any redox process can be written as the sum of two half reactions: one for the oxidation and one for the reduction Six habits of the redox equation balancer STEP 1: the unbalanced equation Dichromate ion reacts with chloride ion to produce chlorine and chromium (III) Cr2O72 (aq) Cl (aq) Cr 3 (aq) Cl2 (aq) STEP 2: identify the oxidized and reduced and write the half reactions Oxidation half-reaction Cl (aq ) Cl2 (aq ) Reduction half-reaction Cr2O72 (aq ) Cr 3 (aq ) STEP 3: Balance the half reactions Oxidation 2Cl (aq) Cl2 (aq) Reduction Cr2O72 (aq) 2Cr 3 (aq) STEP 4: Material balance As with the oxidation number method, add H2O to the side lacking O and add H+ to the other side (for reactions in acid solution) Oxidation reaction – unchanged 2Cl (aq) Cl2 (aq) Reduction reaction 2 14 H (aq ) Cr2O7 (aq ) 2Cr 3 (aq ) 7 H 2O(l ) STEP 5: Balance half-reactions for charge by addition of electrons No explicit calculation of oxidation numbers is required; we balance the charges on both sides of each half-reaction 2Cl (aq) Cl2 (aq) 2e 14 H (aq) Cr2O72 (aq) 6e 2Cr 3 (aq) 7 H 2O(l ) STEP 5 cont: Multiply by factors to balance total electrons Overall change in electrons must be zero Multiply the oxidation half reaction by 3 3 Cl (aq) Cl2 (aq) 2e 14 H (aq) Cr2O72 (aq) 6e 2Cr 3 (aq) 7 H 2O(l ) STEP 6: Add half reactions and eliminate common items 6Cl (aq ) 3Cl2 (aq) 6e + 14 H (aq) Cr2O72 (aq) 6e 2Cr 3 (aq) 7 H 2O(l ) = 14 H (aq) Cr2O72 (aq) 6Cl (aq) 2Cr 3 (aq) 3Cl2 (aq) 7 H 2O(l ) Atoms and charges balance Redox Titrations Acid-base titration is used to determine an unknown concentration (either acid or base) The endpoint is manifested in a color change (of an indicator) or by measuring pH In redox titrations, the concentration of one of the reagents can be measured, provided there is a sharp distinction between the oxidized and reduced states Using the roadmap Oxalic acid is oxidized by MnO4A known quantity of oxalic acid is used to determine the concentration of the MnO4 MnO4- has an intense purple colour, whereas Mn2+ is almost colourless Strategy A known amount of H2C2O4 is used A solution of KMnO4 is titrated till the first purple colour – the endpoint. All the H2C2O4 is oxidized. The equation gives the number of moles of MnO4 The volume of solution yields the concentration ...
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