Redox Reactions & Net Ionic Equations
Oxidizing agent Reducing agent
A ___________ agent undergoes a reduction reaction A ___________ agent undergoes an oxidation reaction
Combustion reaction
A _______________ reaction is a chemical reaction in which a fuel (usually a hydrocarbon) is mixed with an oxidant (usually an oxygen), to form carbon dioxide and water. (CH4 + 2O2 ---> CO2 + 2H2O)
Reducing agent Oxidizing agent
A ________________ agent is the chemical species in a chemical reaction that causes another atom to undergo reduction, while the agent itself is oxidized. A ________________ agent is the chemical species in a chemical reaction that causes another atom to undergo oxidation, while the agent itself is reduced.
Oxidizing Agent Reducing Agent
A ________________ agent usually contains an oxygen or another strongly electronegative element (such as halogens). A __________________ agent usually contains a metal ion or hydrides (H-).
Decomposition reaction
A ________________ reaction is a chemical reaction in which one product breaks down into two or more species. (AB ---> A + B)
Combination reaction
A _________________ reaction is a chemical reaction in which two or more species come together to form a product. (A + B ---> AB)
Redox reaction
A __________________ is a chemical reaction that involves the transfer of electrons from one chemical species to another.
Double-displacement/metathesis reaction
A __________________ reaction is a reaction that involves the switching of counterions between two compounds. (AB + CD -----> CB + AD)
Disproportionation reaction
A __________________ reaction is a redox reaction in which an element undergoes both oxidation and reduction in producing a products. (2H2O2 ---> 2H2O + O2)
1. Balance the equations (both stoichiometry and charge) -1 -1 -1 -1 +1 -1 0 IO3- + I- + H+ -----> I3- + H2O -3 -3 -1 -8 +6 -3 0 IO3- + 8I- + 6H+ -----> 3I3- + 3H2O -3 -3 -1 -2 -1 -2 I3- + S2O3(2-) -----> I- + S4O6(2-) -5 -5 -1 -4 -3 -2 I3- + 2S2O3(2-) -----> 3I- + S4O6(2-) 2. Balance the reactions so that the product of the first matches the reactant of the second. IO3- + 8I- + 6H+ -----> 3I3- + 3H2O 3(I3- + 2S2O3(2-) -----> 3I- + S4O6(2-)) 3I3- + 6S2O3(2-) -----> 9I- + 3S4O6(2-) 3. Determine the moles of IO3- M = mol/L M = 0.01 L = 50 mL = 0.05 L mol = 0.05 * 0.01 = 0.0005 mol 4. Convert moles IO3- to moles S2O3(2-) 0.0005 mol (IO3-) * 6 mol (S2O3(2-))/1 mol (IO3-) = 0.003 mol (S2O3(2-)) 5. Find [S2O3(2-)] M = mol/L mol = 0.003 L = 32 mL = 0.032 L M = 0.003/0.032 = 0.094 M [S2O3(2-)] = 0.094 M
A group of students prepares to standardize a Na2S2O3 solution. 32 mL of the Na2S2O3 solution is titrated into 50 mL of a 0.01 M KIO3 solution to reach the equivalence point. They first titrate the KIO3 solution until it loses its color, then add a starch indicator until the reaction is complete. The reaction proceeds in these following steps: IO3- + I- + H+ -------> I3- + H2O I3- + S2O3(2-) -------> I- + S4O6(2-) Determine the concentration of the sodium thiosulfate solution at the beginning of the experiment.
Find moles of NaI3: M = mol/L M = 0.7 M L = 32.60 mL = 0.0326 L mol = 0.7 * 0.0326 = 0.02282 mol Find mol of Pb(2+): 0.02282 mol NaI3 * (1 mol Pb(2+)/1 mol NaI3) = 0.02282 mol Pb(2+) Find mass of Pb(2+): g/(molar mass) = mol molar mass = 207.2 g mol = 0.02282 mol g = (molar mass) * mol g = 207.2 * 0.02282 = 4.73 g g Pb(2+) = 4.73 g (The 10.00 g is an extraneous value, and does not factor into the solution to the problem)
A sample is assayed for lead by a redox titration with I3- (aq). A 10.00 gram sample is crushed, dissolved in sulfuric acid, and passed over a reducing agent so that all the lead is in the form of Pb(2+). The Pb(2+) (aq) is completely oxidized to Pb(4+) by 32.60 mL of a 0.7 M solution of NaI3. The balanced equation for the reaction is: I3- (aq) + Pb(2+) (aq) ---> Pb(4+) (aq) + 3I- (aq) Calculate the mass of lead in this sample. (Molar mass Pb(2+) = 207.2 g)
Answer: D. Utilize the method described earlier to balance this redox reaction. The balanced half-reactions are: S8 + 24H2O ---> 8SO3(2-) + 48H+ + 32e- NO3- + 4H+ + 3e- ---> NO + 2H2O To get equal numbers of electrons in each half-reaction, the oxidation half-reaction will have to be multiplied by 3, and the reduction half-reaction will have to be multiplied by 32: 3S8 + 72H2O ---> 24SO3(2-) + 144H+ + 96e- 32NO3- + 128H+ + 96e- ---> 32NO + 64H2O This makes the overall reaction: 3S* + 32NO3- + 8H2O ---> 24SO3(2-) + 32NO + 16H+ The sum of the coefficients is therefore: 3 + 32 + 8 + 24 + 32 + 16 = 115.
After balancing the following oxidation-reduction reaction, what is the sum of the stoichiometric coefficients of all the reactants and the products? S8 (s) + NO3- (aq) ---> SO3(2-) (aq) + NO (g) A. 4 B. 50 C. 91 D. 115
Answer: C. First, balance the chemical equation: 4Au + 8NaCN + O2 + 2H2O ---> 4Na[Au(CN)2] + 4 NaOH Now determine the number of moles of NaCN used in the reaction: 0.1 L * 2 M = 0.2 mol NaCN If 0.2 mol NaCN are used in the reaction, then: 0.2 mol NaCN * (4 mol Au/8 mol NaCN) = 0.1 mol Au is oxidized.
An assay is performed to determine the gold content in a supply of crushed ore. One method for pulling gold out of ore is to react it in a concentrated cyanide (CN-) solution. The equation is provided below: Au + NaCN + O2 + H2O ---> Na[Au(CN)2] + NaOH An indicator is used during this reaction, and approximately 100 mL of a 2 M NaCN solution is used to reach the endpoint. How many moles of Au are present in the crushed ore? A. 0.01 mol B. 0.02 mol C. 0.10 mol D. 0.20 mol
Answer: A. The formula for methanol is H3COH, for methanal is HCHO, and for methanoic acid is HCOOH, If we assign oxidation numbers to carbon in each molecule, it starts at -2, then becomes 0, then becomes +2: +1 -2-2+1 +1 0+1-2 +1+2-2-2+1 H3COH -----> HCHO -----> HCOOH +3 In general, it is often easier to think of oxidation as a gain of bonds to oxygen (or a similarly electronegative element) or loss of bonds to hydrogen for organic compounds. Therefore, because the carbon is oxidized as one converts from an alcohol to an aldehyde to a carboxylic acid, the oxidation number must increase.
As methanol is converted to methanal, and then methanoic acid, the oxidation number of the carbon: A. increases B. decreases C. increases, then decreases D. decreases, then increases
The oxidation state of Cl in a compound is typically -1, unless paired with a more electronegative element like O. +2 -1 +4 -1 +4 -1 +2 -1 SnCl2 + PbCl4 -------> SnCl4 + PbCl2 +2 -2 +4 -4 +4 -4 +2 -2 +2 -1 +4 -1 SnCl2 -------> SnCl4 Sn loses electrons, which means it is undergoing an oxidation reaction. Therefore, it is the reducing agent in the redox reaction. +4 -1 +2 -1 PbCl4 -------> PbCl2 Pb gains electrons, which means that it is undergoing a reduction reaction. Therefore, it is the oxidizing agent in the redox reaction.
Assign oxidation numbers to the atoms in the following reaction to determine the oxidizing and reducing agents. SnCl2 + PbCl4 -------> SnCl4 + PbCl2
1. Split the reaction into half reactions +1 -2 -1 ClO- -------> Cl- reduction +3 -1 +6 -2 Cr(OH)4- -------> CrO4(2-) oxidation +3 -4 +6 -8 2. Balance all non- H and O atoms ClO- -------> Cl- Cr(OH)4- -------> CrO4(2-) 3. Balance all O atoms with water ClO- -------> Cl- + H2O Cr(OH)4- -------> CrO4(2-) 4. Balance all H atoms with H+ ions 2H+ + ClO- -------> Cl- + H2O Cr(OH)4- -------> CrO4(2-) + 4H+ 5. Balance charges by adding electrons +1 -1 +2 -1 -1 0 2e- + 2H+ + ClO- -------> Cl- + H2O +2 -1 -2 +4 Cr(OH)4- -------> CrO4(2-) + 4H+ + 3e- 6. Make the electrons in both half reactions equal 3(2e- + 2H+ + ClO- -------> Cl- + H2O) 6e- + 6H+ + 3ClO- -------> 3Cl- + 3H2O 2(Cr(OH)4- -------> CrO4(2-) + 4H+ + 3e-) 2Cr(OH)4- -------> 2CrO4(2-) + 8H+ + 6e- 7. Combine the half reactions and cancel out the electrons and any spectator molecules 3ClO- + 2Cr(OH)4- -------> 3Cl- + 3H2O + 2CrO4(2-) + 2H+ 8. Add OH- ions to both sides to neutralize H+ 3ClO- + 2Cr(OH)4- -------> 3Cl- + 3H2O + 2CrO4(2-) + 2H+ +2OH- +2OH- 2H2O 3ClO- + 2Cr(OH)4- + 2OH- -------> 3Cl- + 5H2O + 2CrO4(2-)
Balance the following redox reaction in a basic solution: ClO- + Cr(OH)4- -------> CrO4(2-) + Cl-
1. Split the reaction into half reactions +6 -2 +3 Cr2O7(2-) -------> Cr(3+) reduction +12 -14 +3 -1 0 Cl- -------> Cl2 oxidation -1 0 2. Balance all non- H and O atoms Cr2O7(2-) -------> 2Cr(3+) 2Cl- -------> Cl2 3. Balance all O atoms with water Cr2O7(2-) -------> 2Cr(3+) + 7H2O 2Cl- -------> Cl2 4. Balance all H atoms with H+ ions 14H+ + Cr2O7(2-) -------> 2Cr(3+) + 7H2O 2Cl- -------> Cl2 5. Balance charges by adding electrons +12 +6 +14 -2 +6 0 6e- + 14H+ + Cr2O7(2-) -------> 2Cr(3+) + 7H2O -2 0 2Cl- -------> Cl2 + 2e- 6. Make the electrons in both half reactions equal 6e- + 14H+ + Cr2O7(2-) -------> 2Cr(3+) + 7H2O 3(2Cl- -------> Cl2 + 2e-) 6Cl- -------> 3Cl2 + 6e- 7. Combine both half reactions, and cancel out electrons on both sides 14H+ + Cr2O7(2-) + 6Cl- -------> 2Cr(3+) + 7H2O + 3Cl2
Balance the following redox reaction in an acidic solution: Cr2O7(2-) + Cl- -------> Cr(3+) + Cl2
1. Split the reaction into half reactions 0 2+ Mg -------> Mg(2+) oxidation +1+5-2 +2-2 HNO3 -------> NO reduction +1+5-6 +2-2 2. Balance the non- H and O atoms Mg -------> Mg(2+) HNO3 -------> NO 3. Balance the O atoms with water Mg -------> Mg(2+) HNO3 -------> NO + 2H2O 4. Balance the H atoms with H+ ions Mg -------> Mg(2+) 3H+ + HNO3 -------> NO + 2H2O 5. Balance the charges by adding electrons 0 +2 Mg -------> Mg(2+) + 2e- +3 0 0 0 3e- + 3H+ + HNO3 -------> NO + 2H2O 6. Make the electrons in both half reactions equal 3(Mg -------> Mg(2+) + 2e-) 3Mg -------> 3Mg(2+) + 6e- 2(3e- + 3H+ + HNO3 -------> NO + 2H2O) 6e- + 6H+ + 2HNO3 -------> 2NO + 4H2O 7. Combine the half reactions 3Mg + 6H+ + 2HNO3 -------> 3Mg(2+) + 2NO + 4H2O
Balance the following redox reaction using the half-reaction method: Mg (s) + HNO3 (aq) -------> Mg(2+) (aq) + NO (g)
1. Separate into two half reactions MnO4- -------> Mn(2+) reduction I- -------> I2 oxidation 2. Balance all the atoms except for H and O in each reaction MnO4- -------> Mn(2+) 2I- -------> I2 3. Balance all the O atoms by introducing water MnO4- -------> Mn(2+) + 4H2O 2I- -------> I2 4. Balance all the H atoms by introducing H+ ions: MnO4- + 8H+ -------> Mn(2+) + 4H2O 2I- -------> I2 5. Balance the charges by adding electrons +7 +2 -1 8+ 2+ 0 5e- + MnO4- + 8H+ -------> Mn(2+) + 4H2O -2 0 2I- -------> I2 + 2e- 6. Make the number of electrons equal in both half reactions: 2(5e- + MnO4- + 8H+ -------> Mn(2+) + 4H2O) 10e- + 2MnO4- + 16H+ -------> 2Mn(2+) + 8H2O 5(2I- -------> I2 + 2e-) 10I- -------> 5I2 + 10e- 7. Add the half reactions together, and cancel out electrons 2MnO4- + 16H+ + 10I- -------> 2Mn(2+) + 8H2O + 5I2
Balance this redox reaction using the half-reaction method: MnO4- + I- -------> I2 + Mn(2+)
Answer: C. The oxidizing agent is the species that is reduced in any given equation. In this problem, six hydrogen atoms with +1 oxidation states in NH3 are reduced to three neutral H2 molecules.
Consider the following equation: 6Na (s) + 2 NH3 (aq) ---> 2Na3N (s) + 3H2 (g) Which species acts as the oxidizing agent? A. Na B. N in NH3 C. H in NH3 D. H2
Answer: A. Step I is a disproportionation reaction because chlorine starts with an oxidation state of 0 in the reactants and ends up with an oxidation state of +1 in HOCl and -1 as Cl-. In the other reactions, no element appears with different oxidation states in two different products. Therefore, only step I is a disproportionation reaction.
Consider the following steps in the reaction between oxalic acid and chlorine: I. Cl2 + H2O ---> HOCl + Cl- + H+ II. H2C2O4 ---> H+ + HC2O4- III. HOCl + HC2O4- ---> H2O + Cl- + 2CO2 Which of these steps, occurring in a aqueous solution, is an example of a disproportionation reaction? A. I only B. III only C. I and III only D. I, II, and III
Answer: D. When assigning oxidation numbers, one starts with elements of known oxidation state first, and determines the oxidation state of the other elements by deduction. As a noble gas, argon, (A.), will always have an oxidation state of 0. As a Group VIIA element, fluorine, (B.), will have an oxidation state of 0 (by itself) or -1 (in a compound). As a Group IIA element, strontium, (C.), will likely have an oxidation state of 0 (by itself) or +2 (in a compound). Like most transition metals, iridium, (D.), can have various oxidation states, ranging from -3 to +8. Therefore, one would have to determine the oxidation states of other atoms in an iridium-containing compound to determine iridium's oxidation number.
During the assigning of oxidation numbers, which of the following elements would most likely be determined last? A. Ar B. F C. Sr D. Ir
+1 -1 0 0 +1 -1 2KI + H2 -------> 2K + 2HI K goes from a +1 oxidation state to a 0 oxidation state, meaning that it underwent reduction. This means that K is the oxidizing agent. H goes from a 0 oxidation state to a +1 oxidation state, meaning that it underwent oxidation. This means that H is the reducing agent.
For the following reaction, identify the oxidation states of each atom, the oxidizing agent, and the reducing agent: 2KI + H2 -------> 2K + 2HI
0 +3 -3 0 +3 -3 Al + BPO4 -------> B + AlPO4 B goes from a +3 oxidation state to a 0 oxidation state, meaning that it underwent reduction, making B the oxidizing agent. Al goes from a 0 oxidation state to a +3 oxidation state, meaning that it underwent oxidation, making Al the reducing agent.
For the following reaction, identify the oxidation states of each atom, the oxidizing agent, and the reducing agent: Al + BPO4 -------> B + AlPO4
Answer: B. First balance the atoms in the equation: Cr2O7(2-) + 14H+ ---> 2Cr(2+) + 7H2O Now, adjust the number of electrons to balance the charge. Currently, the left side has a charge of +12 (-2 from dichromate and +14 from protons). The right side has a charge of +4 (+2 from each chromium cation). To decrease the charge on the left side from +12 to +4, we should add 8 electrons: Cr2O7(2-) + 14H+ + 8e- ---> 2Cr(2+) + 7H2O
How many electrons are involved in the following half-reaction after it is balanced? Cr2O7(2-) + H+ + e- ---> Cr(2+) + H2O A. 2 B. 8 C. 12 D. 16
1. Split the reaction Zn -------> Zn(2+) Cu(2+) -------> Cu 2. Balance charges with electrons Zn -------> Zn(2+) + 2e- oxidation 2e- + Cu(2+) -------> Cu reduction
Identify the oxidation and reduction half-reactions in the following redox reaction: Zn + Cu(2+) -------> Zn(2+) + Cu
Assign oxidation numbers: 0 0 +1 -1 2Na + Cl2 -------> 2NaCl Na loses electrons, which makes it undergo oxidation. Therefore, it is the reducing agent. Cl gains an electron, which makes it undergo reduction. Therefore, it is the oxidizing agent.
Identify the oxidizing and reducing agent in the following redox reaction: 2Na + Cl2 -------> 2NaCl
Answer: B. Recall that oxygen has an oxidation state of -2. Therefore, in tungsten(IV) oxide, (A.), tungsten has an oxidation state of +4. In tungsten(VI) oxide, (B.), it has an oxidation state of +6. In tungsten(III) oxide, (C.), it is +3. In tungsten pentoxide, (D.), it is +5.
If a certain metal has multiple oxidation states, its acidity as an oxide generally increases as the oxidation state increases. Therefore, which of the following tungsten compounds is likely to be the strongest acid? A. WO2 B. WO3 C. W2O3 D. W2O5
Find moles MnO4-: M = mol/L M = 0.02 L = 20 mL = 0.02 L mol = 0.02 * 0.02 = 0.0004 mol Find moles Fe(2+): 1 mol MnO4- : 5 mol Fe(2+) 0.0004 mol MnO4- * (5 mol Fe(2+)/1 mol MnO4-) = 0.002 mol Fe(2+) Find [Fe(2+)]: M = mol/L mol = 0.002 mol L = 10 mL = 0.01 L M = 0.002/0.01 = 0.2 M [Fe(2+)] = 0.2 M
In a redox titration, 20 mL of a 0.02 M solution of K+MnO4- is needed to neutralize 10.0 mL of Fe(2+). What was the concentration of the acidic solution? MnO4- + 5Fe(2+) + 8H+ ---> Mn(2+) + 5Fe(3+) + 4H2O
Answer: C. Start with the atoms that have oxidation states of which you are certain. Potassium is a Group IA metal, and therefore must have an oxidation state of +1. Hydrogen is almost always +1, unless it is paired with a less electronegative element (which is not the case here). Oxygen is generally -2. Because there are four oxygens, they create a total negative charge of -8 which is partially balanced by two hydrogens (+2) and potassium (+1). Therefore, phosphorous has a +5 charge, making it the highest oxidation state.
In the compound KH2PO4, which element has the highest oxidation number? A. K B. H C. P D. O
Find the oxidation states: 0 +1 -2 -1 +1 -1 +1 +5 -2 +1 -2 3Cl2 + 6NaOH ---> 5NaCl + NaClO3 + 3H2O 0 +1 -2 -1 +1 -1 +1 +5 -6 +2 -2 Cl is oxidized in NaClO3 and reduced in 5NaCl, thus it is the element undergoing disproportionation. Cl has an oxidation state of +5 in NaClO3, and -1 oxidation state in NaCl.
In the following reaction which element undergoes disproportionation? What are the element's oxidation states in these products? 3Cl2 (g) + 6NaOH (aq) ---> 5NaCl (aq) + NaClO3 (aq) + 3H2O (l)
Find the oxidation states: +2 -2 +1 0 +4 -2 +1 -2 S2O3(2-) + 2H+ ---> S + SO2 + H2O +4 -6 +1 0 +4 -4 +2 -2 S is oxidized in SO2 and reduced in S, thus it is the element undergoing disproportionation. S has an oxidation state of +4 in SO2, and 0 oxidation state in S.
In the following reaction which element undergoes disproportionation? What are the element's oxidation states in these products? S2O3(2-) (aq) + 2H+ (aq) ---> S (s) + SO2 (g) + H2O (l)
Voltage Endpoint
Indicators in a redox titration changes color when a particular __________ value has been reached, which indicates when the ___________ has been reached.
Answer: A. Hydride ions are composed of a hydrogen nucleus with two electrons, thereby giving it a negative charge and a considerable tendency to donate electrons. LiAlH4 is therefore a strong reducing agent. Strong reducing agents tend to have metals or hydrides; strong oxidizing agents tend to have oxygen or a similarly electronegative element.
Lithium aluminum hydride (LiAlH4) is often used in laboratories because of its tendency to donate hydride ions. Which of the following roles would lithium aluminum hydride likely play in a reaction? A. Strong reducing agent only B. Strong oxidizing agent only C. Both a strong reducing agent and strong oxidizing agent D. Neither a strong reducing agent nor a strong oxidizing agent
Answer: C. What you are shown is a net ionic equation. If two moles of FeSCN are created, two moles of Fe(3+) must be used because the mole ratio is 1:1. Iron sulfate has the formula Fe2(SO4)3 because sulfate has a charge of -2 and iron has a charge of +3 (based on the net ionic equation). Therefore, one mole of iron sulfate is needed to make two moles of iron for the reaction. The molar mass of iron sulfate is: 2 * 55.8 g/mol + 3 * 32.1 g/mol + 12 * 16.0 g/mol = 399.9 g/mol This most closely matches answer (C.). The most common error would be to calculate the amount of iron, which would be 111.6 g (A.).
One way to test for the presence of iron in solution is by adding potassium thiocyanate to the solution. The product when this reagent reacts with iron is FeSCN2+, which creates a dark red color in solution via the following net ionic equation: Fe3+ + SCN- ---> FeSCN(2+) How many grams of iron sulfate would be needed to produce 2 moles of FeSCN(2+)? A. 110 g B. 220 g C. 400 g D. 500 g
Answer: C. Potentiometry refers to carrying out an oxidation-reduction titration with a voltmeter present to get precise readings of the reaction's electromotive force (emf) to determine the endpoint. This is analogous to using a pH meter in an acid-base titration because it uses technology to get precise readings for plotting a titration curve. Indicators, as in (A.) and (B.), can be used in both acid-base and redox titrations, but provide a qualitative (rather than quantitative) analysis of the titration. Oxidizing and reducing agents are used in redox titrations, not acid-base titrations, eliminating (D.).
Potentiometry in an oxidation-reduction titration is analogous to performing an acid-base titration with a(n): A. acidic indicator B. basic indicator C. pH meter D. oxidizing agent
Oxidizing Agents General Chemistry
The following are common ____________ agents in ________________ chemistry: -O2 -H2O2 -H2SO4 -Halogens (F2, Cl2, Br2, I2) -HNO3 -NaClO
Reducing Agents General Chemistry
The following are common _____________ agents in ______________ chemistry: -CO -Pure Metals (Fe, Cu, etc.) -B2H6 -C
Reducing Agents Organic Chemistry
The following are common ______________ agents in __________________ chemistry: -Hydrazine -Lindlar's Catalyst -NaBH4 -Zn(Hg) -LiAlH4
Reducing Agents Biochemistry
The following are common _______________ agents in ________________ chemistry: -NADH -FADH2
Oxidizing Agents Biochemistry
The following are common _______________ agents in ___________________ chemistry: -NAD+ -FADH
Oxidizing Agents Organic Chemistry
The following are common __________________ agents in ______________ chemistry: -CrO3 -Pyridinium Chlorochromate -KMnO4 -NaCr2O7
Answer: D. A strong oxidizing agent will be easily reduced, meaning that it will have a tendency to gain electrons. Atoms usually gain electrons if they are one or two electrons away from filling up a valence shell. (A.) has a full 4s-orbital, meaning that it can only gain an electron if it gains an entire subshell. (B.) has a stable, half-full 3d-orbital, so it is unlikely to pick up electrons unless it can gain five. (C.) has only a single electron in the outer shell, which is more likely lost upon ionization. (D.) would fill up its 4p-orbital by gaining one electron, so it is easily reduced.
The following electronic configurations represent elements in their neutral form. Which element is the strongest oxidizing agent? A. 1s(2)2s(2)2p(6)3s(2)3p(6)4s(2) B. 1s(2)2s(2)2p(6)3s(2)3p(6)4s(2)3d(5) C. 1s(2)2s(2)2p(6)3s(2)3p(6)4s(2)3d(10)4p(1) D. 1s(2)2s(2)2p(6)3s(2)3p(6)4s(2)3d(10)4p(5)
Double-displacement reaction
The only type of reaction that is not considered a redox reaction is _______________________.
Reduction reaction Oxidation reaction
The part of a redox reaction where a chemical species gains an electron is known as the _______________ reaction. The part of the redox reaction where a chemical species loses electrons is known as the _______________ reaction.
0 Charge +1 +2 -1 +1 -1 -2 -1 +2 0 Charge
The rules of oxidation numbers: 1. The oxidation number of a free element (N2, S8, He, etc) is ____. 2. The oxidation number for a monoatomic ion (Na+, Cl-, Fe(3+), etc.) is equal to the __________ of the ion. 3. The oxidation number for each Group 1 element (Li, Na, K, etc.) in a compound is __________. 4. The oxidation number for each Group 2 element (Be, Mg, Ca, etc.) in a compound is _______. 5. The oxidation number for each Group 17 element (F, Cl, Br, etc.) in a compound is _________. 6. The oxidation number of a hydrogen atom is usually ________, but in the presence of less electronegative elements in a compound (such as Group 1 and Group 2 elements) its oxidation number is ________. 7. In most compounds, oxygen has an oxidation number of ________. Exceptions to this rule is if oxygen is part of a peroxide (where each oxygen has an oxidation number of _______), and if it is in a compound with a more electronegative element (such as OF2, where its oxidation number will be _______). 8. The sum of oxidation numbers of all atoms in a neutral compound will equal ______. The sum of oxidation numbers of atoms in a polyatomic ion (such as SO4(2-)) is equal to the __________ of the ion.
Find oxidation numbers -3 +1 +6 -2 0 +3 -2 +1 -2 (NH4)2Cr2O7 (aq) ---> N2 (g) + Cr2O3 (s) + 4H2O (g) +2 +12 -14 +6 -6 Divide into half reactions: 2NH4+ ---> N2 + 8H+ Cr2O7 + 8H+ ---> Cr2O3 + 4H2O Combine half reactions: 2NH4+ + CrO7(2-) ---> N2 + Cr2O3 + 4H2O
What is the net ionic equation of the following reaction? (NH4)2Cr2O7 (aq) ---> N2 (g) + Cr2O3 (s) + 4H2O (g)
Find oxidation states: +1 -1 +1 -1 +1 -1 +1 -1 AgNO3 (aq) + HCl (aq) ---> HNO3 (aq) + AgCl (s) Separate compounds: Ag+ + NO3- + H+ + Cl- ---> H+ + NO3- + AgCl Cancel out spectator ions: Ag+ + Cl- ---> AgCl
What is the net ionic equation of the following reaction? AgNO3 (aq) + HCl (aq) ---> HNO3 (aq) + AgCl (s)
Find oxidation states: -4 +1 0 +4 -2 +1 -2 CH4 (g) + 2O2 (g) ---> CO2 (g) + 2H2O (l) Divide into half reactions: CH4 + 2H2O ---> CO2 + 8H+ + 8e- 2O2 + 8H+ + 8e- ---> 4H2O Combine half reactions: CH4 + 2O2 ---> CO2 + 2H2O
What is the net ionic equation of the following reaction? CH4 (g) + 2O2 (g) ---> CO2 (g) + 2H2O (l)
Find oxidation numbers: 0 0 +1 -1 H2 + F2 ---> 2HF Split into half reactions: H2 ---> 2H+ + 2e- F2 + 2e- ---> 2F- Combine half reactions: H2 + F2 ---> 2H+ + 2F-
What is the net ionic equation of the following reaction? H2 (g) + F2 (g) ---> 2HF (aq)
Zn + CuSO4 ---> Cu + ZnSO4 Zn + Cu(2+) + SO4(2-) ---> Cu + Zn(2+) + SO4(2-) Zn + Cu(2+) ---> Cu + Zn(2+)
What is the net ionic equation of the following reaction? Zn (s) + CuSO4 (aq) ---> Cu (s) + ZnSO4 (aq)
Answer: C. In NaClO (sodium hypochlorite), sodium carries its typical +1 charge, and oxygen carries its typical -2 charge. This means that the chlorine atom must carry a +1 charge in order to balance the overall charge of zero.
What is the oxidation number of chlorine in NaClO? A. -1 B. 0 C. +1 D. +2
Decomposition Reaction
What type of reaction is the following considered? (NH4)2Cr2O7 (aq) ---> N2 (g) + Cr2O3 (s) + 4H2O (g)
Disproportionation Reaction
What type of reaction is the following considered? 2H2O2 (aq) ---catalase---> 2H2O (l) + O2 (g)
Double-Displacement/Metathesis Reaction
What type of reaction is the following considered? AgNO3 (aq) + HCl (aq) ---> HNO3 (aq) + AgCl (s)
Combustion Reaction
What type of reaction is the following considered? CH4 (g) + 2O2 (g) ---> CO2 (g) + 2H2O (l)
Combination Reaction
What type of reaction is the following considered? H2 (g) + F2 (g) ---> 2HF (aq)
Answer: D. A net ionic equation represents each of the aqueous ions comprising the reactants and products as individual ions, instead of combining them as a formula unit. Thus, (A.) is not a net ionic reaction. The term net means that the correct answer does not include any spectator ions (ions that do not participate in the reaction). In this reaction, nitrate (NO3-) remains unchanged. Therefore, (B.) and (C.) are eliminated.
Which of the following is the correct net ionic reaction for the reaction of copper with silver(I) nitrate? A. Cu + AgNO3 ---> Cu(NO3)2 + Ag B. Cu + 2Ag+ + 2NO3- ---> Cu(2+) + 2NO3- + 2Ag C. 2Ag+ + 2NO3- ---> 2NO3- + 2Ag+ D. Cu + 2Ag+ ---> Cu(2+) + 2Ag
This is a double displacement reaction. The solid species does not split. Cu+ + NO3- + Na+ + Cl- -----> CuCl + Na+ + NO3- Cu+ + Cl- -----> CuCl
Write the net ionic equation for the following reaction: CuNO3 (aq) + NaCl (aq) -----> CuCl (s) + NaNO3 (aq)
Balance the equation: 3Mg + 2AlCl3 -----> 2Al + 3MgCl2 3Mg + 2Al(3+) + 2Cl3- -----> 2Al + 3Mg(2+) + 3Cl2- 3Mg + 2Al(3+) -----> 2Al + 3Mg(2+)
Write the net ionic equation for the following reaction: Mg (s) + AlCl3 (aq) -----> Al (s) + MgCl2 (aq)
Redox titrations Acid-Base titrations
_____________ titrations focuses on the transfer of electrons to reach an equivalence point, whereas _________________ titrations focus on the transfer of protons.