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Net Ionic Reactions Single & Double Replacement. REPLACEMENT REACTIONS - one element replaces another in a compound - more active metals replace less.

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Presentation on theme: "Net Ionic Reactions Single & Double Replacement. REPLACEMENT REACTIONS - one element replaces another in a compound - more active metals replace less."— Presentation transcript:

1 Net Ionic Reactions Single & Double Replacement

2 REPLACEMENT REACTIONS - one element replaces another in a compound - more active metals replace less active metals OR more active nonmetals replace less active nonmetals (see activity chart on other side) A + BC  AC + B KEY: REACTANTS ARE 1 ELEMENT AND 1 COMPOUND - ORDER NOT IMPORTANT - ELEMENT CAN’T BE O 2 A. More active metals replace less active metals from their compounds. Hydrogen can act like a metal. Cu + 2 AgNO 3  Cu(NO 3 ) Ag Zn + H 2 SO 4  ZnSO 4 + H 2 B. More active nonmetals replace less active nonmetals from their compounds. Cl NaI  2 NaCl + I 2 C. Very active metals (only the first 5) can replace one (AND ONLY ONE) of the hydrogens from the stable compound water. 2 Na + 2 H 2 O  2 NaOH + H 2 a metal hydroxide is always one of the products.

3 IONIC REACTIONS - (also called double replacement reactions) ions, usually in solution, exchange respective cations and anions - these reactions occur when ions are removed from solution by the formation of a precipitate or by the formation of a molecular compound. These reactions can be further divided into the following 3 classifications below: KEY: REACTANTS ARE 2 COMPOUNDS - UNLESS BOTH ARE OXIDES AB + CD  AD + CB A. Precipitation Reaction - one of the possible products must be a precipitate or no reaction occurs. AgNO 3 + HCl  AgCl + HNO 3 AgCl is a precipitate. B. Molecular Reaction - the reaction occurs if one of the products is a molecular compound (examples: CO 2, H 2 S, SO 2, H 2 O) H 2 SO 4 + Na 2 O  Na 2 SO 4 + H 2 O H 2 O is molecular CaCO HCl  CO 2 + H 2 O + CaCl 2 CO 2 is molecular C. Acid - Base Reaction - an acid plus a base yields an ionic compound (a salt) and water (a molecular compound). 2 H 3 PO Ca(OH) 2  Ca 3 (PO 4 ) H 2 O these are called HCl + NaOH  NaCl + H 2 O neutralization reactions

4 Directions: For each of the following three reactions, write a balanced equation for the reaction in part (i) and answer the question about the reaction in part (ii). In part (i), coefficients should be in terms of lowest whole numbers. Assume that solutions are aqueous unless otherwise indicated. Represent substances in solutions as ions if the substances are extensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction. You may use the empty space at the bottom of the next page for scratch work, but only equations that are written in the answer boxes provide will be scored.

5 Consider the following reaction: potassium chromate + barium nitrate  barium chromate + potassium nitrate K 2 CrO 4 + Ba(NO 3 ) 2  BaCrO 4 + 2KNO 3 This is an ionic reaction and it occurs in water (like most ionic reactions). The barium chromate is a precipitate. To indicate what is happening in this reaction we sometimes include the physical state of the reactants and products as follows: (aq) means aqueous or dissolved in water K 2 CrO 4 (aq) + Ba(NO 3 ) 2 (aq)  BaCrO 4 (s) + 2KNO 3 (aq)

6 The above equation is called the molecular equation. The complete ionic equation as written below: 2K +1 (aq) + CrO 4 -2 (aq)+ Ba +2 (aq) + 2NO 3 -1 (aq)  BaCrO 4 (s) + 2K +1 (aq)+ 2NO 3 -1 (aq) Notice that all strong electrolytes are written as ions. Also notice that there are K +1 and NO 3 -1 ions on both sides of the reaction. Since these ions are on both sides of the reaction, they are not actually part of the reaction and are called spectator ions.

7 The ions that actually participate in the chemical reaction are the following: CrO 4 -2 (aq) + Ba +2 (aq)  BaCrO 4 (s) This last equation is called the net ionic equation and includes only the participating ions of the reaction.

8 An example forming a molecule instead of a precipitate potassium hydroxide + hydrochloric acid  water + potassium sulfate Molecular equation: 2KOH (aq) + H 2 SO 4 (aq)  2H 2 O (l) + 2KCl (aq) Complete ionic equation: 2 K +1 (aq) + 2OH -1 (aq) + 2H +1 (aq) + 2Cl - (aq)  2H 2 O(l) + 2K +1 (aq) + 2Cl - (aq) Net ionic equation: 2 OH -1 (aq) + 2 H +1 (aq)  2 H 2 O(l)

9 A little lie we told…. Not all acids formed in ionic reactions indicate a reaction …. only WEAK acids When we write ionic or net ionic reactions, all strong electrolytes are written as ions. Reactions that produce one of the 7 strong acids may not occur. Ex: NaCl (aq) + HNO 3 (aq)  NaNO 3 (aq) + HCl(aq) Na + + Cl - + H + + NO 3 -  Na + + NO H + + Cl - No new substances are formed = no reaction

10 Oxidation-Reduction (Redox) Reactions A process where electrons are transferred from one substance to another. One substance is oxidized while another substance is reduced Oxidation loss of electrons an increase in oxidation number Reduction - gain of electrons - a decrease in oxidation number

11 How can you tell when a redox reaction is taking place? 1)Assign oxidation numbers to atoms in substances 2)Compare oxidation numbers before and after reaction to determine if atom has lost or gained electrons

12 Rules For Assigning Oxidation Numbers - A bookkeeping system 1.The oxidation number of an atom in an element is 0. Examples: Na, H 2, Br 2, S 8, Ne Ox. # The oxidation number of a monatomic ion is the same as its charge. Examples: Na +1, Ca +2, Al +3, Cl -1, O -2 Ox #

13 3.The sum of the oxidation numbers of all atoms in a neutral compound is zero. 4.The sum of the oxidation numbers of all atoms in an ion is equal to the charge on the ion.

14 5. In compounds, fluorine is always assigned an oxidation number of -1 6.Hydrogen’s oxidation number will be - +1 when bonded to a nonmetal (HCl) --1 when bonded to a metal (NaH) Examples: NaH CaH 2 HClH 2 S Na—H H—Ca—H H—Cl H—S—H

15 7.Oxygen usually has an oxidation number of -2. Exceptions: in peroxides, oxygen will be -1 and when combined with F, it will be +2 and in O 2 it will be 0 Examples: H 2 O CaO H 2 O 2 H — O — H Ca—O H—O—O—H O 2 -2 OF 2 [O—O] -2 F—O—F

16 8.Halogens usually have an oxidation number of -1. Exception is when chlorine, bromine, iodine are combined with oxygen Examples: NaCl MgI 2 OCl 2 HOBr Na—Cl, I—Mg—I, Cl—O—Cl, H—O—Br ** If none of the above rules help you get started…look for a atom with a known charge and use that charge as its oxidation number CdS: Cd-S +2 -2

17 More examples AlH 3 Al +3 H -1 S 2 O 3 -2 S +2 O -2 Na 2 Cr 2 O 7 Na +1 O –2 Cr +6 CO 2 O -2 C +4

18 Use algebra to determine oxidation numbers of "difficult" atoms. Example:H 2 SO 4 H is +1 * 2 = +2 O is -2 * 4 = x + -8 = 0 S is +6 Example:ClO 4 -1 O is -2 * 4 = x = -1 Cl is +7 Example:NH 4 +1 H is +1* 4 = x = 1 N is -3

19 Tricky Ones: FeSO 4 The SO 4 part has to have a charge of -2 O is -2 S is +6 To make a neutral compound Fe must be +2 Fe 2 (SO 4 ) 3 The SO 4 part has to have a charge of -2 O is -2 S is +6 There are 3 sulfate ions for a total negative charge of -6 We need a total positive charge of +6 Each Fe must be +3

20 2 Fe 2 O 3 (s) + 3 C (s)  4 Fe (s) + 3 CO 2 (g) –2 Fe is reduced, going from +3 to 0 and C is oxidized, going from 0 to +4, the O undergoes no change

21 Reducing agent Causes reduction Loses electrons Undergoes oxidation Oxidation number of atom increases

22 Oxidizing agent Causes oxidation Gains electrons Undergoes reduction Oxidation number of atom decreases

23 Assign oxidation numbers, indicate what is oxidized and reduced, indicate what is the oxidizing agent and reducing agent 1)Ca (s) + 2 H +1  Ca +2 (aq) + H 2 (g) Ca is oxidized – increasing from 0 to +2 H +1 is reduced – decreasing from +1 to 0 Ca is the reducing agent H +1 is the oxidizing agent

24 2)2 Fe +2 (aq) + Cl 2 (aq)  2 Fe +3 (aq) + 2 Cl -1 (aq) Fe +2 is oxidized – increasing from +2 to +3 Cl 2 is reduced – decreasing from 0 to -1 Fe +2 is the reducing agent Cl 2 is the oxidizing agent

25 In general, metals act as reducing agents (are oxidized) and nonmetals act a oxidizing agents (are reduced).

26 Replacement Reactions Are redox reactions! (So are many of the reactions we studied in chapter 3!)

27 Example: It is possible for metals to be oxidized in the presence of a salt (in solution): Fe(s) + Ni(NO 3 ) 2 (aq)  Fe(NO 3 ) 2 (aq) + Ni(s) OR Fe(s) + Ni +2 (aq)  Fe +2 (aq) + Ni(s) In this reaction iron has been oxidized to Fe 2+ while the Ni 2+ has been reduced to Ni.

28 The Activity Series We can list metals in order of decreasing ease of oxidation. This list is the activity series. The metals at the top of the activity series are called active metals and are easily oxidized. The metals at the bottom of the activity series are called noble metals and NOT easily oxidized.

29 A metal in the activity series can only be oxidized by a metal ion below it. (Higher will replace lower) If we place Cu into a solution of Ag + ions, then Cu 2+ ions can be formed because Cu is above Ag in the activity series: Cu(s) + 2AgNO 3 (aq)  Cu(NO 3 ) 2 (aq) + 2Ag(s) or Cu(s) + 2Ag + (aq)  Cu 2+ (aq) + 2Ag(s)

30 Decompositon of acids: F. Acids (hydrogen combined with some negative ion) decompose into water and nonmetallic oxides. H 2 CO 3  H 2 O + CO 2 2 H 3 PO 4  P 2 O H 2 O *This is NOT a redox reaction – the oxidation number on the nonmetal must stay constant. Use this knowledge to predict the formula for the nonmetal oxide.

31 Acid Decomposition H 2 SO 3  S x O y + H 2 O Hydrogen: +1 Oxygen: -2 Sulfur: +4 H 2 SO 3  SO 2 + H 2 O Sulfur: +4 Since oxygen usually Has an oxidation number Of -2. The formula must be SO 2.

32 Composition Reactions: E. Oxides of nonmetals combine with water to form compounds called acids. These acids will usually be made from an “ate” ion. P 2 O H 2 O  2 H 3 PO 4 SO 3 + H 2 O  H 2 SO 4 F. Oxides of metals combine with oxides of nonmetals to yield ionic compounds. These compounds usually contain an “ate” ion. Na 2 O + SO 3  Na 2 SO 4 *These are NOT redox reactions…oxidation numbers stay constant. Use that knowledge to predict the product!

33 SO 2 + H 2 O  H 2 SO xO: -2S: +4 H: +1 In order for the oxidation numbers to add to zero, the formula must be H 2 SO 3

34 P 4 O 10 + CaO  Ca 3 (PO x ) 2O: -2P: +5 H: +1 In order for the oxidation numbers to add to zero, the formula must be Ca 3 (PO 4 ) 2


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