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Pre-AP Chemistry - Simple Rules for Electron Exchange
First Rule
Metals lose electrons when they become ions. Nonmetals gain electrons when they become ions.
Second Rule
Number of electrons gained by the reduced species must equal number of electrons lost by the oxidized
species.
Third Rule
Polyatomic ions that are not broken up during chemical reactions are usually spectator ions. Spectator ions can
be removed from net ionic equations in order to make electron bookkeeping easier
Simple rules for assigning oxidation numbers
Tracking electron gain and loss for simple reactions like metals becoming ionized is easy. But in most chemical
reactions, it is impossible to simply look at the reactants and products and track the electron exchange. In those
cases, we must do some electron bookkeeping. Oxidation numbers are the basis for electron bookkeeping.
Tracking how they change as each chemical species goes from reactants to products helps us keep track of which
species loses and which species gains electrons. You will note that oxidation numbers (or “oxidation states” for
groups of like atoms) are similar, but not identical to, formal ionic charges.
Rule #1: The oxidation number for a pure or uncombined element is always 0.
Rule #2: The oxidation state for a monoatomic ion is the same as its formal ionic charge.
Rule #3: In a compound or polyatomic ion, oxygen is always -2. There are some exceptions, but we deal with
them in AP Chemistry.
Rule #4: When paired with a negative ion (mono- or polyatomic) hydrogen is always +1.
Examples: H2S, any acid (HCl, HNO3, etc.)
Rule #5: In ionic compounds made of non-metals (like SO 2 or H2S) the element with the largest electronegativity
is considered “minus” and is given a numerical value equal to its formal ionic charge.
Rule #6: For any ionic compound, the sum of the oxidation numbers of all elemental constituents must equal 0.
Example: Al2(SO4)3: total oxidation numbers for Al, S, and O, multiplied by the number of atoms, = 0.
Rule #7: For any polyatomic ion, the sum of the oxidation numbers of all elemental constituents must equal the
charge of the ion.
Example: MnO4– : total oxidation numbers for Mn and O, multiplied by number of atoms, must = -1
Strategy:
1) Identify pure uncombined elements first.
2) Identify all ionic compounds, then identify the positive ions (metals or ammonioum) and negative ions (non-metals
or polyatomic ions) in each ionic coupound.
3) Identify all monoatomic ions and use rule #2 to assign oxidation numbers.
4) Identify all polyatomic ions and use rule #7 to determine oxidation numbers of each constituent element. It is often
easiest to start with oxygen, since rule #3 tells us it is always -2.
NOTE: Even though the formal ionic charge of all non-metals is always negative, when combined in polyatomic ions
non-metals can have positive oxidation numbers.
5) Use rule number 6 to combine the results of steps 3 through 5 and determine the oxidation numbers for all ionic
compounds.
Examples:
H3PO4 :
1) No uncombined elements.
2) H is combined with polyatomic ion PO4-3.
3) By rule #4, H is +1.
4) Look at PO4-3 ion. Rule #3 tells us that O is -2. There are 4 O's, so O's total oxidation state is -8. By rule #7, the
constituents of PO4-3 must combine to equal -3. So, -3 = (-8) + x; therefore x = 5. So the oxidation number of P must
be +5.
5) Combine everything according to number of atoms in the formula:
3(hydrogen) + 1(phosphorus) + 4(oxygen) → 3(+1) + 1(+5) + 4(-2) = 0 so rule #6 is satisfied.
So H is +1, P is +5, O is -2
Al2(SO4)3:
1) No uncombined elements.
2) Al is combined with polyatomic ion SO4-2.
3) By rule #2, Al is +3.
4) Look at SO4 ion. Rule #3 tells us that O is -2. There are 4 O's, so O's total oxidation state is -8. By rule #7, the
consitutents of must combine to equal -2. So, -2 = (-8) + x; therefore, x = 6. So the oxidation state of
S must be +6.
5) Combine everything according to number of atoms in the formula:
2(aluminum) + 3(sulfur) + 12(oxygen) → 2(+3) + 3(+6) + 12(-2) = 0 so rule #6 is satisfied
So Al is +3, P is +5, and O is -2
MnO4– :
This is a polyatomic ion, so we start with step 4: Rule #3 tells us that O is -2. There are 4 O's, so O's total oxidation
state is -8. By rule #7, the consitutents of must combine to equal -1 (the charge of the ion). So, -1 = (-8) + x;
therefore, x = 7. So the oxidation state of Mn must be +7.
Now step 5: Combine everything according to number of atoms in the formula:
1(manganese) + 4(oxygen) → 1(+7) + 4(-2) = -1 so rule #6 is satisfied
So Mn is +7 and O is -2