Download Worksheet to accompany demos on exchange reactions

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


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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) →
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"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).
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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
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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:
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(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).
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“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! 
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