Unit 4: Topic 7 - Types of Chemical Reactions Flashcards
Explain what occurs in an acid-base reaction (preferably, Brønsted-Lowry acids and bases), and give an example of such a reaction
In an acid-base reaction, the two reactants are an acid and a base. During
the reaction, the acid will give up one or more of its protons, and the base will accept
these protons. For example, in the reaction H3O+ (aq) + OH- (aq) —> H2O (l) + H2O(l),
the H3O+ is the acid giving up a proton, and the OH- is the base accepting this proton.
HF is a weak acid and NaOH is a strong base. Write out the balanced acid-base equation and come to a conclusion about the products.
If HF is a weak acid and NaOH is a strong base, we should expect that when equal amounts of a weak acid mix with a strong base, the resulting mixture should be more basic than acidic. When we write this reaction out, we have HF (aq) + NaOH (aq) —> H2O(l) + NaF (aq). This might seem strange at first: H2O is neither acidic nor basic, and NaF seems to be a salt. However, since the resulting mixture is basic, we can conclude that NaF (aq) is basic.
Define oxidation and reduction.
Oxidation is the process of an element losing electrons and as a result gaining charge. Reduction is
the opposite: an element gains electrons and loses charge. A mnemonic is oil rig: Oxidation Is Lose (electrons), Reduction Is Gain (electrons)
Explain why combustion of a hydrocarbon (ex. propane) is an example of an oxidation-reduction reaction.
Let us write out the equation for propane: C3H8 + 5O2 —> 3CO2 + 4H2O. On the left hand side, the
reactant O2 implies that oxygen as a reactant has an oxidation number of 0. Similarly, since H is taken
to be +1, the carbon in propane will have an oxidation number of -8/3. However, on the right hand side, oxygen now takes its oxidation number of -2, and in CO2, since oxygen is -2, carbon is +4. Therefore, this reaction is oxidation-reaction because carbon is oxidized (gains charge, loses electrons, oxidized) and oxygen is reduced (loses charge, gains electrons, reduced). Note that carbon’s oxidation number increases from -8/3 to 4, which means it loses 20/3 electrons per carbon: since there are 3 carbons in this reaction, here is a net loss of 20 electrons due to carbon. Oxygen decreases from 0 to -2, so that two electrons are gained per oxygen: since there are 10 oxygens in this reaction, there is a net gain of 20 electrons due to oxygen. Thus electron transfer is conserved as well as mass.
In the (hypothetical) reaction of permanganate ion with chromium (ii) chloride to
form manganese (ii) chloride and chromate ion, determine which reactant is
oxidized and which is reduced, and determine how electrons are transferred.
In permanganate ion MnO4 -, Mn has an oxidation number of +7, whereas in manganese(ii) chloride,
Mn has an oxidation number of +2. Since manganese loses charge, it gains (5) electrons and is reduced. In
chromium(iii) chloride, Cr has oxidation number +2, whereas in chromate ion CrO4 2-, Cr has oxidation number
+6. Chromium gains charge, loses (4) electrons, and is oxidized. Therefore, electrons flow from chromium
to manganese.
State some general rules on how to determine oxidation numbers of atoms.
The oxidation number of an atom in a neutral free element is zero. A free element is considered to be any element in an uncombined state, whether monatomic or polyatomic. For example, the oxidation number of each atom in N2 and P4 are 0.
The oxidation number of a monatomic (composed of one atom) ion is the same as the charge of the ion. For example, the oxidation numbers of K+, Se2−, and Au3+ are +1, −2, and +3, respectively.
The oxidation number of oxygen in most compounds is −2. Exceptions include peroxide, O2(2-) with an oxidation of -1.
The oxidation number of hydrogen in most compounds is +1.
The oxidation number of fluorine in all compounds is −1. Other halogens usually have an oxidation number of −1 in binary compounds, but can have variable oxidation numbers depending on the bonding environment.
In a neutral molecule, the sum of the oxidation numbers of all atoms is zero. For example, in H2O , the oxidation numbers of H and O are +1 and −2 , respectively. Because there are two hydrogen atoms in the formula, the sum of all the oxidation numbers in H2O is 2(+1)+1(−2)=0 .
In a polyatomic ion, the sum of the oxidation numbers of all atoms is equal to the overall charge on the ion. For example, in SO4 −2 , the oxidation numbers of S and O are +6 and −2 , respectively. The sum of all oxidation numbers in the sulfate ion would be 1(+6)+4(−2)=−2 , which is the charge of the ion.
Balance the (hypothetical) reaction of permanganate ion with chromium (ii)
ion to form manganese (ii) ion and chromate ion in acidic solution.
Unbalanced, we have MnO4- + Cr 2+ —> Mn (2+) + CrO4 2-. We want to write half reactions for both
manganese and chromium, and we want to balance electron movement before we account for
conservation of mass. The half reactions are (disregarding the balancing of oxygen),
MnO4- + 5e- —> Mn 2+, and Cr 2+ —> 4e- + CrO4 2-. To make sure the electron transfer is balanced
(i.e. net transfer 0), there should be 20 electrons gained by the manganese and 20 lost by the chromium:
4MnO4- + 5Cr 2+ —> 4Mn(2+) + 5CrO4 2-. This balances the electrons.
Note that the oxygens are not balanced. However, this is where “acidic solution” comes in: we will add
H2O to balance the oxygen, and H+ to then balance the hydrogen. The final solution is
4MnO4- + 5Cr(2+) + 4H2O —> 4Mn(2+) + 5CrO4(2-) + 8H+
Describe what occurs in a precipitation reaction.
In a precipitation reaction, two aqueous compounds will react to form a solid. This solid compound is
then a “precipitate,” and the reactants have participated in a precipitation reaction.
Will mixing sodium chloride and silver nitrate form a precipitate?
Yes: this is simple solubility rules. A double replacement reaction gives
NaCl + AgNO3 —> AgCl + NaNO3, and by solubility rules silver chloride is insoluble.
Mix together 58.44 grams of sodium chloride with 84.93 grams of silver nitrate.
Will there be a precipitate, and how much precipitate should we expect? (Assume
the reaction runs to completion.)
(These are purposely easier numbers) First, we convert grams to moles:
58.44 g NaCl * (1 mol/58.44 g) = 1 mol NaCl, and 84.93g AgNO3 * (1 mol/169.87g) ≈ 0.5 mol AgNO3.
Since silver nitrate is the limiting reactant, we have 0.5 mol AgNO3 * (1 mol AgCl/1mol AgNO3) * (143.32 g AgCl/1mol AgCl) = 71.66g AgCl. This is the amount of precipitate we would expect if the reaction runs to completion.
