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Topic 9: Redox Processes. SWBAT: Define the processes of reduction & oxidation Define and identify reducing and oxidizing agents Deduce the oxidation.

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Presentation on theme: "Topic 9: Redox Processes. SWBAT: Define the processes of reduction & oxidation Define and identify reducing and oxidizing agents Deduce the oxidation."— Presentation transcript:

1 Topic 9: Redox Processes

2 SWBAT: Define the processes of reduction & oxidation Define and identify reducing and oxidizing agents Deduce the oxidation state of an atom in an ion or compound Agenda Redox reactions Oxidation states Oxidizing & reducing agents

3 Mrs Teter OUT

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6 Reduction Oxidation (Redox) Reactions Reduction & oxidation occur whenever electrons are transferred from one reactant to another (photosynthesis, batteries, voltaic cells, electrochemical cells, electrolysis)

7 Reduction Oxidation (Redox) Reactions Transferring electrons from one substance to another leads to a flow of electrons, in other words an electric current Chemical reactions can be used to generate electricity (voltaic cell or battery) By reversing the process and using electric current to drive reactions, stable compounds can be broken down into their elements (electrolysis)

8 Reduction Oxidation (Redox) Reactions Oxidation is the loss of electrons Reduction is the gain of electrons  OIL RIG (oxidation is loss, reduction is gain)  Leo the lion says gerrrrrr (loss of electrons is oxidation, gain of electrons is reduction)

9 Reduction Oxidation (Redox) Reactions How do we know when a redox reaction has happened? When magnesium is burned in air, it gives a bright white flame with the production of a white powder, magnesium oxide: 2Mg(s) + O 2 (g)  2MgO(s) Compare the charges of the reactants to the products: 2Mg  2Mg 2+ Because there was a transfer of e - s from one O 2  2O 2- reactant to another, it was a redox rxn

10 Reduction Oxidation (Redox) Reactions 2Mg(s) + O 2 (g)  2MgO(s) Compare the charges of the reactants to the products: 2Mg  2Mg 2+ Because there was a transfer of e - s from one O 2  2O 2- reactant to another, it was a redox These eventually become half-equations or half-reactions: 2Mg  2Mg 2+ + 4e - O 2 + 4e -  2O 2-

11 Oxidation Numbers Oxidation numbers (or oxidation states) are a measure of the electron control or possession it has relative to the atom in the pure element This can be used with covalent bonds where electrons are shared and not transferred to identify if a redox reaction has happened Oxidation state is written sign 1 st and value 2 nd (unlike charge which is value 1 st and sign 2 nd )

12 Strategy for assigning oxidation states 1.Atoms in the free (uncombined) element have an oxidation state of zero ex. Mg, O 2, N 2, Ar 2.In simple ions, the oxidation state is the same as the charge on the ions ex. Mg 2+ O 2- N 3- Mg 2+ oxid#=+2 O 2- oxid#=-2 N 3- oxid#=-3 3.The oxidation states of all the atoms in a neutral (uncharged) compound must add up to zero H 2 SO 4 sum of oxidation states=0

13 Strategy for assigning oxidation states 4. The oxidation states of all the atoms in a polyatomic ion must add up to the charge on the ion SO 4 2- the sum of oxidation states=-2 5.The usual oxidation state for an element is the same as the charge on its most common ion Hydrogen is usually +1, O is usually -2 6.Most main group non-metals (at the bottom of grp 14) & transition elements have oxidation states that vary in different compounds depending on the conditions & other elements present

14 Strategy for assigning oxidation states Element Usual Oxidation State ExceptionsExplanation Li, Na, K+1 Mg, Ca+2 F no exceptions because F is the most electronegative element O-2 peroxides such as H 2 O 2, where it is -1; OF 2, where it is +2 H+1 metal hydrides such as NaH, where it is -1 H is more electronegative than Na and so gains electron control Cl when it is combined with O or F Cl is less electronegative than O and F, and so loses electron control

15 Assigning Oxidation States H 2 SO 4 We know the oxidation states have to add to 0 because this is a neutral compound H is +1, oxygen is -2 2*+1 + S+ 4*-2=0 so S = +6 SO 3 2- Here the oxidation states have to add up to -2 Oxygen is -2 S + 3*-2=-2 S = +4

16 Assigning Oxidation States Work out the oxidation # of chlorine in the following species: a)Cl 2 O b)HCl c)ClO 4 - d)ClF 3 e)HClO 3 Assign oxidation states to the metal ion in a)[Co(NO 3 ) 6 ] 3+ b)[CuCl 4 ] 2-

17 Interpreting Oxidation States An element such as sulfur can have a wide range of oxidation states in different compounds H 2 S S SCl 2 SO 2, SO 3 2- SO 3, H 2 SO 4 -2 0 +2 +4 +4 +6 +6 The higher the positive #, the more the atom has lost control over e - s (the more oxidized it is) The greater the negative #, the more it has gained e - control (the more reduced it is)

18 Interpreting Oxidation States Any change in oxidation state during a reaction is an indication that redox processes are occurring An increase in oxidation # represents oxidation A decreases in oxidation # represents reduction 2H 2 (g) + O 2 (g)  2H 2 O(l) oxid #s 0 0 +1 -2 Hydrogen has been oxidized (oxidation state increased from 0 to +1) Oxygen has been reduced (oxidation state decreased from 0 to -2)

19 Interpreting Oxidation States Use oxidation states to deduce which species is oxidized and which is reduced in the following reactions: a)Br 2 + SO 2 + 2H 2 O  H 2 SO 4 + 2HBr b)6OH - + 3Cl 2  5Cl - + ClO 3 - + 3H 2 O c)Cr 2 O 7 2- + H 2 O ↔ 2CrO 4 2- + 2H + d)Ca(s) + Sn 2+ (aq)  Ca 2+ (aq) + Sn(s) e)4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(l)

20 Oxidizing and Reducing agents An oxidizing agent is a substance that oxidizes something else. Are they themselves reduced. Takes electrons away from something A reducing agent is a substance that reduces something else. They themselves are oxidized. Gives electrons to something 2Br - + Cl 2  2Cl - + Br 2 Cl 2 is the oxidizing agent because it oxidizes the Br- to Br 2. The Br- is the reducing agent because it causes the Cl 2

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