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Chemistry 121, Prof. Mines Worksheet to accompany demos on exchange reactions (and net ionic equations that represent them) Write the products of the following exchange reactions (some precipitation reactions and some acid-base reactions), balance, and then write the net ionic equations. *Make sure you consider the correct FORMULAS of the products before you start to balance the equations!* (1) Ca(NO3)2 + Na2CO3 (2) Na2CO3 + HC2H3O2 (3) CaCO3 + H2SO4 (4) NaI + Mg(NO3)2 (5) NaI + AgNO3 1 Chemistry 121, Prof. Mines Solubility, Electrolytes, Net Ionic Equation (Precursor) I. Fill in the table: Substance (or Species) Type (ionic compound, molecular non-acid, molecular-acid, ion) Soluble in water? (write N/A if molecular-nonacid b/c you can’t predict this yet) Strong Electrolyte, Weak Electrolyte, or neither? Na2S H2S HNO3 (NH4)2SO4 BrO2− CH3OH MgCO3 H2SO4 H2SO3 Na2MoO4 AgCl Mg(NO3)2 H3PO4 Li3N II. Complete Ionic Equations Notes: What makes a “complete ionic equation” different from a formula (or “molecular”) equation is that one changes the way that strong electrolytes (soluble compounds that ionize completely when dissolved) are written. Specifically, one writes strong electrolytes as “separated ions”. For example, Na2S would be written as 2 Na+(aq) + S2-(aq) in a complete ionic equation because it is (a soluble ionic compound, and is thus) a strong electrolyte. All other species are written without changing their formulas at all, except that a state designation ((s), (l), (g), (aq), etc.) should be added. Note that the two common classes of “strong electrolyte” are 1) soluble ionic compounds (know the solubility rules!) and 2) strong acids (only 6 common ones). All the above being the case, write how each of the species above (in I) would be written in a complete ionic equation. (Hint: Check for solubility first; then see if it ionizes completely) Na2S H2SO4 H2S H2SO3 HNO3 Na2MoO4 (NH4)2SO4 AgCl BrO2− Mg(NO3)2 CH3OH H3PO4 (soluble) MgCO3 Li3N 2 1. Write net ionic equations: NOTE: On your exam, you will be expected to know the first two solubility rules for ionic compounds (the one about nitrates and the one about Group I and ammonium cations). Other rules will be provided to you. (a) Mg(NO3)2 + + (b) 3 SrBr2 (c) 2 CsF (d) 2 KOH (e) Ni(NO3)2 → MgCO3 2 K3PO4 → Sr3(PO4)2 → Cs2S Na2CO3 H2 S + + H2SO4 + 2 HCl → + 2 NaNO3 + 6 KBr 2 HF 2 H2O → + NiCl2 + K2SO4 + 2 HNO3 2. Write the products of the following exchange reactions (some precipitation reactions and some acid-base reactions), balance, and then write the net ionic equations. *Make sure you consider the correct FORMULAS of the products before you start to balance the equations!* Assume that in each case at least one precipitate (insoluble ionic compound) is formed. (a) K3PO4(aq) + Al(NO3)3(aq) → (b) BaI2(aq) + Cu2SO4(aq) → (c) H3PO4 (aq) + AgNO3(aq) → 3 "Rules" For Assigning Oxidation Numbers Oxidation Numbers can be assigned to any “Atom” in any chemical species using or applying the following rules/principles (See other handout for detailed description of the meaning and purpose of oxidation numbers) 1. Atoms in elements are assigned an oxidation number of ZERO. (Note: Ca2+ is not the same as Ca! Ca2+ is an ion, not an element!) O’s in O2 are assigned ZERO; O’s in O3 are assigned ZERO; Ca in Ca is assigned ZERO. 2. In compounds OR polyatomic ions, O atoms are typically assigned an oxidation number of –2, and H atoms are typically assigned an oxidation number of +1. I will not require you to know the exceptions to this rule (peroxides are a common exception). (Note: Don’t forget about Rule 1 before assigning oxidation numbers to O and H!!!) You can figure out the oxidation numbers of other “atoms” in a species by using Rule number 3: 3. The sum of the oxidation numbers of EACH “ATOM” in a chemical species must equal the net charge on the species. (Don’t forget that if there is more than one atom of the same type in a species, you must sum up the oxidation numbers on ALL of them. BUT that sum is not the oxidation number! Please ask me if you are confused about this.) Example: in NO3-, each O is assigned an oxidation number of –2. Since there are 3 O atoms, the sum total of their oxidation numbers is 3 x –2 = -6. The oxidation number of the single N atom must therefore be the number which when added to –6 yields the overall charge of –1. Thus the oxidation number of N in NO 3- is +5. The oxidation number of O in the above example is –2; it is NOT –6. => For monatomic species, this rule (Rule 3) reduces to the following “trivial” rule: the oxidation number of any monatomic species is equal to the actual charge on the species. E.g., the oxidation number of Ca2+ is +2; for H in H+ the oxidation number is +1 (not zero!!!). Helpful hint: If a compound is ionic, rewrite the formula as the SEPARATED IONS (whose charges you should either know or can figure out). Then assign the oxidation numbers according to rules 2 and 3. Example: Fe2(CO3)3. You need to rewrite as Fe3+ and CO32- to get all ox. numbers. ------------------------------------------------------------------------------------------------------------------------Solubility Rules Comments NOTE: You are responsible for knowing/memorizing the following solubility “rules” (patterns): 1. All “nitrates” are (water) SOLUBLE (i.e., all ionic compounds that have NO 3- as the anion are soluble in water) 2. All ionic compounds having a Group I metal cation (e.g., Li+, Na+, K+, Cs+, etc.) OR NH4+ as the cation are (water) SOLUBLE. The above rules will not be given to you on an exam because I'll assume you've memorized them. However, OTHER solubility rules (similar to the ones in the table in the text, and possibly additional ones, depending on the question asked) will be provided to you, but you will be responsible for knowing how to interpret and apply them (i.e., I will not explain their meaning to you during an exam). 4 What Are ―Oxidation Numbers‖and Why Did Chemists Create Them? This is another bit of text I initially prepared in response to several students’ questions in email about oxidation numbers. I have modified it slightly for this term. There are FOUR "points", some with examples within. I also added a sheet for practicing the ―language‖ of oxidation/reduction. I hope that you find the document and worksheet helpful. Prof. Mines --------------------------1) Since ―to oxidize‖ (something) means "to take electrons from (it)", when a monatomic chemical species ―is oxidized", electrons are taken away from that species. That means that that species must become more "positive". Example 1: If Al becomes Al3+ during some chemical change, we say that Al was oxidized. Electrons were taken from it and it became more positive. Example 2: If Cl- ions are converted into Cl2 molecules then we say that Cl- (or simply ‖Cl‖) was oxidized. Each Cl "atom" started off having a NEGATIVE charge and each Cl atom ended up as a NEUTRAL atom in a chlorine molecule after the change. Thus, Cl became more positive (note: it did not become POSTIVE, just more positive than it was!). The ―take-home‖ lesson(s)? (a) Becoming more positive means "losing electrons" which means "getting oxidized (by something)" or ―undergoing/experiencing oxidation‖. (b) Becoming more negative means "gaining electrons" which means "getting reduced (by something)" or ―undergoing/experiencing reduction”. (c) If all reactants and products were monatomic species (this is not the usual case!), you could easily tell if a reaction were an oxidation-reduction reaction by looking to see if any reactants end up more positive or more negative after chemical change (i.e., compare a (monatomic) reactant to its corresponding product). I will revisit and clarify this point again below when discussing chemical species more complex than in the examples above [generally speaking, you cannot look at overall charge, but I wanted to get the simple idea across first and clarify later]. 2) The idea behind oxidation numbers. In the two examples given above, we were dealing with monatomic ions, which have an actual charge, and so it was relatively straightforward to tell if the charge changed upon chemical reaction. With molecules and polyatomic ions, the individual atoms do not have an overall charge, so the above analysis does not really apply, but clearly oxidation-reduction reactions can involve these kinds of chemical species. So……….even though there are no actual whole-number charges on the atoms in molecular compounds or in polyatomic ions, chemists set up an arbitrary bookkeeping system (the text's own words! see p. 182, bottom) to keep track of electrons in all the atoms of all chemical species (whether they be monatomic, polyatomic, ionic, molecular, whatever!). This is done by assigning "oxidation numbers" to atoms in chemical species. The primary purpose for this is merely to help us assess whether or not the type of chemical change that is being represented is considered to involve oxidation-reduction or not (and if so, to help decide who gets oxidized and who gets reduced when this chemical change occurs). This concept of oxidation number is only a tool that we use to assess oxidation-reduction “behavior”. That is why the system is considered arbitrary. We will see later in this course that there is a second bookkeeping system that chemists use to keep track of electrons that uses different rules!!! Why?! Because that tool is used for a different purpose—to "view" bonding in covalent species, NOT to look at oxidationreduction behavior during chemical change! Please try to understand that "arbitrary" does not mean "unimportant" or "useless"! It means "chosen in order to serve a specific purpose." And don't think that we are being "inconsistent" in a few weeks when I introduce this second way to "keep track of electrons". These are conceptual tools we use; different situations or purposes prompt the use of different tools. 3) The philosophy behind the rules; oxidation numbers are "fictitious charges". NOTE: there are a fixed number of electrons in ANY chemical species. You could figure this out if you had the formula of the substance or ion, right? Essentially what is done when assigning these oxidation numbers is to assign a certain number of electrons to each atom in a species. But you don't actually count the 5 electrons, you just count the fictitious CHARGES. If an atom in a molecule has an oxidation number of zero, that just means that it is assigned (considered from a bookkeeping perspective to ―have‖) the same number of electrons as a neutral, isolated atom has. But let's say that O is assigned an oxidation number of -2 in a chemical species like ―carbon dioxide‖. By doing that, we are saying that each oxygen in a CO2 molecule will be considered to have a charge of -2 (even though we know that it doesn't really have a -2 charge!). That means that it is considered to have two more electrons than a neutral oxygen atom would have. But since there is a FIXED NUMBER OF ELECTRONS in the species (6 + 2 x 8 = 22 in a CO2 molecule), if you assign some extra electrons to the O atoms, the C atom must get short-shrifted! That is, the electrons ―assigned‖ to O were effectively ―taken away‖ from C. Since the two O atoms each get 2 extra electrons, the C atom must be deficient by 4 electrons, and so it would have 4 fewer electrons than a neutral C atom and so it would have a ―fictitious‖ charge (or oxidation number) of +4. Get it? Note that because all we are doing is (mentally) distributing a fixed number of electrons among all the atoms in a species, the (actual) overall charge on the entire species will equal the sum of the oxidation numbers of all the atoms in the species. For example, in our CO2 case, the overall charge of the species, a neutral CO2 molecule, is zero, and the sum of all the oxidation numbers is zero as well: -2 (O) + -2(O) + +4(C) = 0. Assigning the O’s oxidation numbers of –2 ―forced‖ us to take away four of the C atom’s electrons since there were no ―extra‖ electrons to start with (the species was overall neutral). If you DO have extra electrons (i.e., if the species is an anion), then you will adjust the ―distributing‖ (assigning of oxidation numbers) so that the sum of the oxidation numbers adds up to the overall charge on the species. The following example illustrates this. Example: carbonate ion, CO32-. (Note that a carbonate ion has two extra electrons compared to a neutral C atom and three neutral O atoms. Where it ―got‖ these extra electrons from is not known and is of no consequence at this point. It HAS them!) Oxygen is generally given an oxidation number of -2 in compounds or polyatomic ions in which it is found (this is because oxygen tends to attract electrons to itself very well; we’ll discuss this later). So if O is assigned a number of -2, then the O’s end up with a total of 6 ―extra‖ electrons (three O’s x –2). Since there are already two extra electrons in the species overall (the overall charge is –2), only four electrons must come at the expense of the C atom, making the C atom have a (fictitious) charge (oxidation number) of +4. Once you get the hang of this, you don’t even count the electrons anymore; you just ―make the charges add up to the total‖, but I want you to realize that the reason this ―works‖ is because we are just keeping track of (and reassigning) the electrons! In the case of CO32-, you should hopefully soon be able to just ―see‖ that IF O is arbitrarily assigned a value of –2, the C atom MUST have an oxidation number of +4 because the overall charge on the ion is -2 and the three oxygens’ (fictitious) charges add up to –6; the difference is just 4. That is, +4 is what needs to be added to –6 to end up with –2, the actual overall charge. 4) To determine if a reaction represented by an equation is an oxidation-reduction reaction....ASSIGN OXIDATION NUMBERS TO ALL ATOMS IN EACH REACTANT AND PRODUCT. If there is a difference in the oxidation number of ANY ATOM in a reactant species and that same atom in a product species, then the reaction is an oxidation-reduction reaction! If the oxidation number of an atom becomes more positive as a result of chemical reaction (i.e., more positive in the product than in the reactant), then that atom (sometimes we refers to the entire chemical species) is said to be oxidized (see point number 1 above!) when some reaction occurs. We would also say that this atom (species) is the reducing agent or reductant (―one that reduces somebody else‖). If the oxidation number of an atom becomes more negative, then that atom (species) is said to have been reduced. That atom (species) would be referred to as the oxidizing agent or oxidant (―one that oxidizes somebody else‖). See the back page of this handout for practice with the language! (―to reduce‖, ―to be reduced‖, ―a reducing agent‖, ―undergo reduction‖ etc) Example 1: consider the combustion of methane: CH4 + 2 O2 2 H2O + CO2. H is generally assigned a value of +1 in compounds. That makes the oxidation number of C in methane -4 (make the total add up to zero since methane is neutral). The oxidation number of O in oxygen gas is zero (any atoms in an element have oxidation numbers of zero because elements are made of neutral, essentially identical atoms). In water, O is assigned -2 and H +1 (this adds up to zero overall). In carbon dioxide, O is assigned a value of -2, so C ends up having an oxidation number of +4. Thus, we can say the following: 6 (a) This equation represents an oxidation-reduction reaction. (b) C went from -4 in methane to +4 in carbon dioxide. Its oxidation number went "up"; it became more "positive". So C was oxidized. That means C acted as the reductant (reducing agent). (We might also say that "methane" (the chemical species that contained the carbon) was oxidized and that "methane" was the reductant.) (c) O went from 0 in oxygen gas to -2 in water and carbon dioxide. Its oxidation number went "down"; it became more "negative". So O (or oxygen gas) was reduced. So O (or oxygen gas) was the oxidant (oxidizing agent). ---------------------------------------------------------- Note #1: When an equation is written and you are asked ―which is the oxidant and which is the reductant‖, the assumption is made that the chemical change occurs in the forward direction. That means that the reductant and oxidant are the ones who do the giving and taking of electrons that results in the chemical change. The language itself should indicate to you that the oxidant and reductant are REACTANTS, not products (see Final Example, below), but many students seem to have trouble with this kind of question, so please be aware of this possible confusion and ask me about this if necessary. I feel it is important to note that if you consider the reverse reaction to occur in some situation, ―the oxidant‖ and ―the reductant‖ will be different chemical species than those for the ―forward‖ reaction! They will be products of the original reaction, but reactants in the reverse reaction. See ―language practice‖ exercise #2 at the end of this handout. Note #2: although the practicing of the assigning of oxidation numbers is done in a separate problem on the problem set (simply to help you get familiar with the rules), remember that the purpose for knowing them is to use them to determine whether or not a given equation represents an oxidationreduction reaction or not (a separate problem on the problem set). For this latter question, you should assign oxidation numbers to all atoms and be able to identify the oxidizing agent and reducing agent if the reaction is an oxidation-reduction reaction. ---------------------------------------------------------- Final Example (oxidant and reductant are both reactants): Note that Cu is ―copper atom‖, a distinct chemical species from Cu2+ (which is called ―copper(II) ion‖), and S is ―sulfur atom‖, a distinct chemical species from S2- (which is called ―sulfide ion‖). One can imagine a reaction in which a ―copper atom‖ gives two electrons to a ―sulfur atom‖. In the process of doing so (i.e., during this chemical change), the copper atom BECOMES SOMETHING DIFFERENT, ―copper(II) ion‖, and so does the sulfur atom (it becomes ―sulfide ion‖). So who gave the electrons? Copper ATOM. Who took the electrons? Sulfur ATOM. The copper ION and sulfide ION did not exist until the change occurred, so they could not have given nor received anything (they did not exist yet!!!)! So the oxidant in this reaction was ―S‖ and the reductant was ―Cu‖. If the reaction were represented by an equation, it would be (ignoring state designations): 2+ 2Cu + S Cu + S Note that the oxidant, S, and the reductant, Cu, are both reactants (assuming the reaction occurs in the forward direction). 7 “Language” Practice Part I. Fill in the blanks with the appropriate term (use the picture as a guide). [e.g. (Cu) reductant* (Giver) (Ag+) (Cu +) (Ag) ] oxidant* (Taker) "oxidized form" of the reductant "reduced form" of the oxidant The electron was given to the ________________ by the ________________. The electron was taken by the ________________ from the ________________. The reductant ____________ed the oxidant, but in so doing it got ____________ed. The oxidant ____________ ed the reductant, but in so doing it got ____________ed. The oxidant can be considered a(n) "____________izing agent". The reductant can be considered a(n) "_____________ing agent". When the reductant gets oxidized, it is said to undergo ____________tion and its oxidation number becomes more ________tive. When the oxidant gets reduced, it is said to undergo ____________tion and its oxidation number becomes more ________tive. After the oxidant has _______________ed the reductant, it is in its "_______________ed form" (because now it has an electron to give). After the reductant has ______________ed the oxidant, it is in its "_______________ed form" (because now it has "room" to take the electron back). “Language” Practice Part II. Assuming reaction proceeds as indicated by the following equation: 3 Fe2+(aq) + 2 Al(s) 3 Fe(s) + 2 Al3+(aq) Which is the oxidant (oxidizing agent)? Which is the reductant (reducing agent)? Which gets reduced? Which gets oxidized? Which species does the oxidizing? Which species does the reducing? Which species underwent oxidation? (think of who ―undergoes‖ an operation!) __ Which species experienced reduction? IF the REVERSE reaction were to occur, which species would be the oxidant? *Note that we don't typically call a person who participates, a "participater", we call him/her a "participant". The same goes for someone who ―occupies‖ (an "occupant"). Recall the word reactant (―one that reacts‖) as well! 8