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New Probs.

  • Page ID
    83735
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    Q1

    Determine the oxidation number of each element in the following compounds.

    Rules:

    1. Pure elements have an oxidation number of 0
    2. If the compound is an ionic compound, the oxidation number for each element is the ion’s charge
    3. The oxidation number of hydrogen in a compound is +1
    4. The oxidation number of oxygen in most compounds is –2
      (peroxides are the exception; in peroxides oxygen has an oxidation number of –1)
    5. The sum of the oxidation numbers in a compound is zero.
    6. The sum of the oxidation numbers in a polyatomic ion is equal to the ion charge.
    Hint Oxidation Numbers for each Element
    SnCl4 Rule 2 Sn +4 Cl -1
    Ca3P2 Rule 2 Ca +2 P -3
    SnO Rules 4, 5 Sn +2 O -2
    Ag2S Rule 2 Ag +1 S -2
    HI Rule 3, 5 H +1 I -1
    N2H4 Rule 3, 5 N -2 H +1
    Al2O3 Rule 4, 5 Al +3 O -2
    S8 Rule 1 S 0
    HNO2 Rules 3, 4, 5 H +1 N +3 O -2
    O2 Rule 1 O 0
    H3O+ Rules 3, 4, 6 H +1 O -2
    ClO3- Rules 4, 6 Cl +5 O -2
    S2O22- Rules 4, 6 S +1 O -2
    KMnO4 Rules 4, 5, 6 K +1 Mn +7 O -2
    (NH4)2SO4 Rules 4, 5, 6 N -3 H +1 S +6 O -2

    Q2

    Determine the oxidation number of carbon in each of the following compounds:

    1. methane, CH2
    2. formaldehyde, CH2O
    3. carbon monoxide, CO
    4. carbon dioxide, CO2

    S2

    a) C = -4

    Element Oxidation Number Number of Atoms Total
    H +1 4 +4
    C -4 1 -4

    b) C = 0

    Element Oxidation Number Number of Atoms Total
    H +1 2 +2
    O -2 1 -2
    C 0 1 0

    c) C = -2

    Element Oxidation Number Number of Atoms Total
    O -2 1 -2
    C +2 1 +2

    d) C = +4

    Element Oxidation Number Number of Atoms Total
    O -2 2 -4
    C +4 1 +4

    Q3

    For each of the following reactions, complete the summary table below the equation. If an element does not undergo any change, leave the last two columns blank

    a. \(4 HCl + O_2 → 2 H_2O + 2 Cl_2\)

    element Initial
    Ox. No
    Final
    Ox. No.
    e - gained
    or lost
    Oxidized or reduced Agent
    H +1 +1 0
    Cl -1 0 1 oxidized reducing
    O 0 -2 2 reduced oxidizing

    b. \(4 Al + 3 O_2 → 2 Al_2O_3\)

    element Initial
    Ox. No
    Final
    Ox. No.
    e - gained
    or lost
    Oxidized or reduced Agent
    Al 0 +3 3 oxidized reducing
    O 0 -2 2 reduced oxidizing

    c. \(Fe + SnCl_2 → FeCl_2 (aq) + Sn\)

    element Initial
    Ox. No
    Final
    Ox. No.
    e - gained
    or lost
    Oxidized or reduced Agent
    Fe 0 +2 1 oxidized reducing
    Sn +2 0 2 reduced oxidizing
    Cl -1 -1 0

    d. \(PbO_2 + 4 HI → I_2 + PbI_2 + 2 H_2O\)\

    element Initial
    Ox. No
    Final
    Ox. No.
    e - gained
    or lost
    Oxidized or reduced Agent
    Pb +4 +2 2 reduced oxidizing
    O

    -2

    -2 0
    H +1 +1 0
    I (to I2) -1 0 1 oxidized reducing
    I (to PbI2) -1 -1 0

    Q4

    For each of these reactions, determine whether or not it is a redox reaction. If any are, identify oxidizing and reducing agents in those reactions.

    a. CaBr2 + Pb(NO3)2 → PbBr2 + Ca(NO3)2

    element Initial
    Ox. No
    Final
    Ox. No.
    e - gained
    or lost
    Oxidized or reduced Agent
    Ca - - - - -
    Br - - - - -
    Pb - - - - -
    N - - - - -
    O - - - - -

    Not a redox reaction since no substance undergoes a change in oxidation number.

    b. P4 + 5O2 → P4O10

    element Initial Ox. No Final Ox. No. e - gained or lost Oxidized or reduced Agent
    P 0 +5 5 Oxidized Reducing Agent
    O 0 -2 5 Reduced Oxidizing Agent

    c. SnCl2 + 2 FeCl3 → 2 FeCl2 + SnCl4

    element Initial
    Ox. No
    Final
    Ox. No.
    e - gained
    or lost
    Oxidized or reduced Agent
    Sn +2 +4 2 Oxidized Reducing Agent
    Fe +3 +2 1 Reduced Oxidizing Agent

    Q5

    Two half-cells are connected under standard conditions to make an electrochemical cell. Use the Table of Standard Reduction Potentials (Table P1) to obtain the half-reactions involved. For each:

    1. write the equation for each half-reaction that will occur
    2. label each half-reaction as oxidation or reduction
    3. calculate the voltage of the electrochemical cell
    4. the net overall balanced redox equation.
    5. diagram the cell, clearly indicating the following
    • the electrodes in appropriate electrolytic solutions
    • label each electrode as anode or cathode
    • label each electrode as positive post or negative post
    • diagram the flow of electrons through the external circuit
    • a salt bridge with appropriate electrolytic solution
    • flow of ions from the salt bridge to the two half-cells
    1. iron-iron(II) ion (Fe|Fe2+) and lead-lead(II) ion (Pb|Pb2+)
    2. chromium-chromium(III) ion (Cr|Cr3+) and rubidium-rubidium ion (Rb|Rb+)
    3. copper-copper(I) ion (Cu|Cu+) and aluminum-aluminum ion (Al|Al3+) (NOTE: Be sure to use the Cu1+ half-reaction, not Cu2+)
    4. An electrochemical cell is created using gold and magnesium half-cells.
      • Determine which half-cell will undergo oxidation and which will undergo reduction, identify anode and cathode, and calculate the voltage for the cell. You do not need to diagram the cell.
      • If the mass of the magnesium electrode changes by 5.0 g, what will be the change in mass of the gold electrode, and will its mass increase or decrease?

    Q6

    Using either an activity series or Table of Standard Reduction Potentials (Table P1), predict whether a reaction takes place, and if so, give a balanced reaction equation.

    1. Ag(s) + HCl(aq)
    2. Mg(s) + FeSO4(aq)
    3. Cu(s) + AuCl3(aq)
    4. Sn(s)+ Al2(SO4)3(s)

    S6

    a) By using the activity series, we can determine that no reaction would take place since Ag is lower on the list than H.

    b) By using the activity series, we can determine that a reaction would take place since Mg is higher on the list than Fe.

    c) By using the activity series, we can determine that a reaction would take place since Cu is higher on the list than Au.

    d) By using the activity series, we can determine that no reaction would take place since Sn is lower on the list than Al.

    Q7

    Consider the electrolysis of water. Describe the events at the anode in terms of

    1. the reaction
    2. the pH
    3. gas produced

    S7

    a. At the anode, H2O decomposes to O2 and H3O+ as shown below:

    6H2O(l) → O2(g) + 4H3O+(aq) + 4e-

    b. The water by the electrodes will change pH because of the ions being produced or consumed around it. Around the anode, the pH of the water is acidic.

    c. Oxygen gas is produced at the anode.

    Q8

    In the following decomposition reaction,

    \[2 N_2O_5 → 4 NO-2 + O_2\]

    oxygen gas is produced at the average rate of 9.1 × 10-4 mol · L-1 · s-1. Over the same period, what is the average rate of the following:

    • the production of nitrogen dioxide
    • the loss of nitrogen pentoxide

    S8

    • By looking at the equation, we can see that for every 1 mol of O2 formed, four moles of NO2 are produced. Thus, the rate of production of NO2 is four times that of O2, so:

    rate of NO2 production = 4 × (9.1 × 10-4 mol · L-1 · s-1) = 3.6 × 10-3 mol · L-1 · s-1

    • N2O5 is consumed at twice the rate that oxygen is produced, so

    rate of N2O5 loss = 2× (9.1 × 10-4 mol · L-1 · s-1) = 1.8 × 10-3 mol · L-1 · s-1

    Q9

    Consider the following reaction:

    \[N_2(g) + 3 H-2(g) → 2 NH_3(g)\]

    If the rate of loss of hydrogen gas is 0.03 mol · L-1· s-1, what is the rate of production of ammonia?

    S9

    The Rate of Disappearance and the Rate of Formation are equal. We can see from the reaction above that 2 moles of \NH_3\ are produced for every 2 moles of \H_2\ consumed. Therefore:

    rate of production of ammonia = 2 mol NH3/3 mol H2 x 0.03 mol · L-1 · s-1/1

    = 0.02 mol · L-1 · s-1

    Q10

    Nitrogen monoxide reacts with hydrogen gas to produce nitrogen gas and water vapour. The mechanism is believed to be:

    Step 1: 2 NO → N2O2
    Step 2: N2O2 + H2 → N2O + H2O
    Step 3: N2O + H2 → N2 + H2O

    For this reaction find the following:

    • the overall balanced equation
    • any reaction intermediates

    S10

    In order to find the overall balanced equation, we cross out the substances that appear in equal numbers on both sides of the reaction. Then, we add like items on the same side of the equation:

    Step 1: 2 NO → N2O2
    Step 2: N2O2 + H2N2O + H2O
    Step 3: N2O + H2 → N2 + H2O
    Net Reaction: 2 NO + 2 H2 → N2 + 2 H2O

    To identify the reaction intermediates, look for substances that first appear on the product side of the equation, but then appear in the next step as a reactant. In this example there are two reaction intermediates - N2O2 and N2O.

    Q11

    Give two reasons why most molecular collisions do not lead to a reaction.

    S11

    1. Most collisions are do not have activation energies high enough for a reaction. The colliding particles have kinetic energies that are too low to break or make bonds.

    2. The colliding particles do not have the appropriate orientation to one another.

    Q12

    An important function for managers is to determine the rate-determining steps in their business processes. In a certain fast-food restaurant, it takes 3 minutes to cook the food, 1.5 minutes to wrap the food, and 5 minutes to take the order and make change. How would a good manager assign the work to four employees?

    S12

    The manager would assign two employees to take the order and make change since it is the slowest step (the rate determining step). As for the two other employees, one would be assigned to cook and the other to wrap the food.

    Q13

    Answer the following questions based on the potential energy diagram shown here:

    1. Does the graph represent an endothermic or exothermic reaction?
    2. Label the position of the reactants, products, and activated complex.
    3. Determine the heat of reaction, ΔH, (enthalpy change) for this reaction.
    4. Determine the activation energy, Ea for this reaction.
    5. How much energy is released or absorbed during the reaction?
    6. How much energy is required for this reaction to occur?

    endo_values_2.gif

    S13

    1. Endothermic.

    2.

    Screen Shot 2017-06-15 at 11.45.40 AM.png

    3. ΔH = 100 kJ - 50 kJ = 50 kJ

    4. Activation energy is the minimum energy required for a reaction to happen.

    Therefore, it is the peak of the graph: 250 kJ.

    5. 50 kJ is absorbed.

    6. 250 kJ

    Q14

    Sketch a potential energy curve that is represented by the following values of ΔH and Ea. You may make up appropriate values for the y-axis (potential energy).

    ΔH = -100 kJ and Ea = 20 kJ

    Is this an endothermic or exothermic reaction?

    S14

    1.png

    Reaction is endothermic.

    Q15

    In the next unit we will be discussing reactions that are reversible, and can go in either the forward or reverse directions. For example, hydrogen gas and oxygen gas react to form water, but water can also be broken down into hydrogen and oxygen gas. We typically write a reaction that can be reversed this way, using the double arrow symbol

    \[2 H_2 + O_2 \rightleftharpoons 2 H_2O\]

    This reaction is exothermic in the forward direction:

    \[2 H_2 + O_2 \rightleftharpoons 2 H_2O + \text{ 285 kJ}\]

    but endothermic in the reverse direction:

    \[2 H_2O + \text{285 kJ} \rightleftharpoons 2 H_2 + O_2\]

    Consider a general reversible reaction such as:

    \[A + B \rightleftharpoons C + D\]

    Given the following potential energy diagram for this reaction, determine ΔH and Ea for both the forward and reverse directions. Is the forward reaction endothermic or exothermic?

    practice_3-2.gif

    S15

    ΔHforward = -20 kJ which is the difference in energy level between the reactants and products.
    ΔHreverse = 20 kJ which is the difference in energy from (C + D) and (A + B).

    Ea forward = 60 kJ which is the amount of energy required to go from the initial reactants (A + B) to the the peak of the graph.
    Ea reverse = 80 kJ amount of energy required to go from the products (C + D) to the peak of the graph.

    By looking at the potential energy diagram, we can see that the forward reaction is exothermic since the products, (C + D), are at a lower energy level than reactants (A + B).

    Q16

    Sketch a potential energy diagram for a general reaction

    \[A + B \rightleftharpoons C + D\]

    Given that ΔHreverse = -10 kJ and Ea forward = +40 kJ

    Q17

    Which one of the following reactions would you expect to be fastest at room temperature and why?

    • Pb2+(aq) + 2 Cl-(aq) → PbCl2 (s)
    • Pb(s) + Cl2 (g) → PbCl2 (s)

    S17

    The first reaction, Pb2+(aq) + 2 Cl-(aq) → PbCl2 (s), would be faster, since ions in aqueous solution react very quickly and all the reactants are in the same phase.
    The second reaction, Pb(s) + Cl2 (g) → PbCl2 (s), would be slower, since of the reactants is a solid.

    Q18

    Consider the following reactions. Which do you predict will occur most rapidly at room conditions? Slowest?

    • C2H6 (g) + O2 (g) → 2 CO2 (g) + 3 H2O(g)
    • Fe(s) + O2 (g) → Fe2O3 (s)
    • H2O(l) + CO2 (g) → H2CO3 (g)
    • 2 Fe3+(aq) + Sn2+(aq) → 2 Fe2+(aq) + Sn4+(aq)

    Q19

    Consider the following reaction that occurs between hydrochloric acid, HCl, and zinc metal:

    \[HCl(aq) + Zn(s) \rightarrow H_2 (g) + ZnCl_2 (aq)\]

    Will this reaction occur fastest using a 6 M solution of HCl or a 0.5 M solution of HCl? Explain.

    S19

    The reaction will occur fastest with 6 M HCl, because it is more concentrated than the 0.5 M solution. The more concentrated the solution is (the more moles of HCl present), the more particles there are to react with each other, resulting in more collisions, which in turn, leads to a faster rate of reaction.

    Q20

    Phosgene, COCl2, one of the poison gases used during World War I, is formed from chlorine and carbon monoxide. The mechanism is thought to proceed by:

    step 1: Cl + CO → COCl
    step 2: COCl + Cl2 → COCl2 + Cl
    1. Write the overall reaction equation.
    2. Identify any reaction intermediates.
    3. Identify any catalysts.

    Q21

    We have typically been simplifying our potential energy curves somewhat; for multistep reactions, potential energy curves are more accurately shown with multiple peaks. Each peak represents the activated complex for an individual step. Consider the PE curve for a two-step reaction:

    multistep1.gif

    1. What is ΔH for the overall reaction?
    2. What is ΔH for the first step of the reaction mechanism?
    3. What is ΔH for the second step of the reaction mechanism?
    4. What is ΔH for the overall reverse reaction?
    5. What is Ea for the first step?
    6. What is Ea for the second step?
    7. Which is the rate-determining step: step 1 or step 2? How do you know?
    8. What is Ea for the reverse of step 1?
    9. Is the overall reaction endothermic or exothermic?

    S21

    a) ΔH = -20 kJ
    b) ΔH = 20 kJ
    c) ΔH = -40 kJ
    d) ΔH = 20 kJ
    e) Ea = 80 kJ
    f) Ea = 40 kJ
    g) Step 1 is the rate determining step because it has the largest Ea
    h) Ea ​​​​​​​= ​​​​​​​60 kJ
    i) Exothermic


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