Unformatted text preview: CHEM 1050 Final Exam Preparation Problems Laboratory Winter 2015 SOLUTIONS 1. Balancing Redox equations There are many possible approaches. The approach we teach has these steps to it: I. Assign oxidation numbers. Identify oxidant and reductant and create the two half-‐reactions for each. II. Balance all atomic species except for H and O. III. Balance the change in oxidation state using electrons. This & the next step are often the difficult steps for students, as some struggle to distinguish between oxidation state and charge. Ensure they recognize the difference. IV. Balance charge using either H+ (if conditions are acidic) or OH– (for basic conditions). V. Balance O by the addition of H2O. VI. Confirm that H is now balanced. If not, a mistake has been made and you need to go back and review the process. VII. Scale the two reactions so they contain the same number of electrons. VIII. Add the two half-‐reactions and cancel any duplication. Check the final equation for atomic and charge balance. (i) Cl–(aq) + MnO4–(aq) → Cl2(g) + MnO2(s) (acidic) Oxidation Reduction -‐1 Cl– → Cl
0 +7 MnO – → MnO
2 2 Cl– → Cl2 MnO4– → MnO2 –
2 Cl → Cl2 + 2e 3e + MnO4– → MnO2 2 Cl– → Cl2 + 2e– 4H+ + 3e– + MnO4– → MnO2 2 Cl– → Cl2 + 2e– 4H+ + 3e– + MnO4– → MnO2 + 2 H2O –
6 Cl → 3 Cl2 + 6e 8H+ + 6e– + 2 MnO4– → 2 MnO2 + 4 H2O 8H+(aq) + 6 Cl–(aq) + 2 MnO4–(aq) → 2 MnO2(s) + 3 Cl2(g) + 4 H2O (l) (ii) NO(g) + Cr2O72–(aq) → HNO2(l) + Cr3+(aq) (acidic) Oxidation Reduction +2 NO → HNO
+3 +6 Cr O 2-‐ → Cr3+ +3 2 2 7
NO → HNO2 Cr2O72-‐ → 2 Cr3+ NO → HNO2 + e– 6e– + Cr2O72-‐ → 2 Cr3+ NO → HNO2 + e– + H+ 14H+ + 6e– + Cr2O72-‐ → 2 Cr3+ H2O + NO → HNO2 + e– + H+ 14H+ + 6e– + Cr2O72-‐ → 2 Cr3++ 7 H2O – +
6H2O + 6NO → 6HNO2 + 6e + 6H 14H+ + 6e– + Cr2O72-‐ → 2 Cr3++ 7 H2O 8H+(aq) + 6 NO(g) + Cr2O72-‐(aq) → 2 Cr3+(aq) + 6 HNO2(l) + H2O(l) (iii) Fe(OH)3(s) + OCl–(aq) → FeO42–(aq) + Cl–(aq) (basic) Oxidation Reduction +3 Fe(OH) → FeO 2– +6 +1 OCl– → Cl– -‐1 3
Fe(OH)3 → FeO42– OCl– → Cl– Fe(OH)3 → FeO42–+ 3e– 2e– + OCl– → Cl– 5 OH– + Fe(OH)3 → FeO42–+ 3e– 2e– + OCl– → Cl– + 2 OH– 5 OH– + Fe(OH)3 → FeO42–+ 3e– + 4 H2O H2O + 2e– + OCl– → Cl– + 2 OH– –
10 OH + 2 Fe(OH)3 → 2 FeO4 + 6e + 8 H2O 3 H2O + 6e– + 3 OCl– → 3 Cl– + 6 OH– 2 Fe(OH)3(s) + 3 OCl–(aq) + 4 OH–(aq) → 2 FeO42–(aq)+ 3 Cl–(aq) + 5 H2O(l) (iv) CN–(aq) + MnO4–(aq) → MnO2(s) + CNO–(aq) (basic) Oxidation Reduction -‐1 CN–→ CNO– +1 +7 MnO – → MnO
2 CN–→ CNO– MnO4– → MnO2 –
CN → CNO + 2e 3e + MnO4– → MnO2 2 OH– + CN–→ CNO– + 2e– 3e– + MnO4– → MnO2 + 4 OH– –
2 OH + CN → CNO + 2e + H2O 2 H2O + 3e– + MnO4– → MnO2 + 4 OH– 6 OH– + 3 CN–→ 3 CNO– + 6e– + 3 H2O 4 H2O + 6e– + 2 MnO4– → 2 MnO2 + 8 OH– 2 MnO4–(aq) + 3 CN–(aq) + H2O(l) → 2 MnO2(s) + 3 CNO–(aq) + 2 OH–(aq) Note: The cyanide moiety remains together during this reaction. It is sufficient to note that (CN) changes from -‐1 to +1 and 2 electrons are lost. 2. Strategy: Identify which species is oxidized and which is reduced. Look up their standard reduction potentials in the table. Determine E˚cell = E˚SRP, cathode – E˚SRP, anode. Alternatively, students use a Hess’s Law approach, where the oxidized species reduction reaction is reversed and the sign of the SRP is changed to produce an SOP (standard oxidation potential). In this case, E˚cell = E˚SRP,cathode + E˚SOP, anode. We see that both methods produce the same result, but students commonly mix the two and switch the sign twice. Help them with this problem. If E˚cell is positive, the reaction will favour the formation of more products; a negative E˚cell will favour more reactants. Another source of confusion may be mix up E˚ with ∆G˚, where a negative value means it was favours products. Make sure they understand the correct signs. (i) Sn(s) + Pb2+(aq) → Sn2+(aq) + Pb(s) Sn is being oxidized. Pb2+ is being reduced. E˚SRP(Sn2+/Sn) = -‐0.137 V E˚SRP(Pb2+/Pb) = -‐0.125 V E˚cell = E˚SRP, cathode – E˚SRP, anode = -‐0.125 V – (-‐0.137 V) = + 0.012 V Products favoured. (ii) Cu2+(aq) + 2 I–(aq) → Cu(s) + I2(s) I– is being oxidized. Cu2+ is being reduced. E˚SRP(I2/I–) = +0.535 V E˚SRP(Cu2+/Cu) = + 0.340 V E˚cell = E˚SRP, cathode – E˚SRP, anode = +0.340 V – (+0.535 V) = – 0.195V Reactants favoured. (iii) 4 NO3–(aq) + 4 H+(aq) → 3 O2(g) + 4 NO(g) + 2 H2O(l) O is being oxidized. N is being reduced. E˚SRP(H2O / O2) = + 1.229 V E˚SRP(NO3–/NO) = + 0.956 V E˚cell = E˚SRP, cathode – E˚SRP, anode = +0.956 V – (+1.229 V) = – 0.273 V Reactants favoured. (iv) O3(g) + Cl–(aq) → OCl–(aq) + O2(g) Cl is being oxidized. O is being reduced. E˚SRP(O3/ O2) = + 2.075 V E˚SRP(Cl–/ OCl–) = + 0.890 V E˚cell = E˚SRP, cathode – E˚SRP, anode = + 2.750 V – (0.890 V) = + 1.860 V Products favoured. 3. Similar to question 2 above, except students need to translate the word description into a chemical equation. (i)
Mg(s) dissolves in a solution of Pb2+ ions which in turn precipitate as Pb(s). Reactants, Mg and Pb2+. By inspection of the SRP table: Mg is oxidized to Mg2+ and Pb2+ is reduced to Pb(s). E˚SRP(Mg2+/Mg) = -‐ 2.356 V E˚SRP(Pb2+/Pb) = -‐0.125 V E˚cell = E˚SRP, cathode – E˚SRP, anode = – 0.125 V – (– 2.356 V) = + 2.231V Yes, the reaction will proceed as described. (ii)
Sn(s) will react with and dissolve in 1 M HCl. The reactants are Sn, Cl–, and H+. List possible reactions and select the two to occur. Sn is being oxidized. H is being reduced. E˚SRP(Sn2+/Sn) = -‐0.137 V E˚SRP(H+/H2) = 0.000 V E˚cell = E˚SRP, cathode – E˚SRP, anode = 0.000 V – (– 0.137 V) = + 0.137 V Yes, the reaction will proceed as described. (iii) MnO4–(aq) will oxidize H2O2(aq) to O2(g) in acidic solution. The reactants are permanganate and peroxide. E˚SRP(MnO4–/ MnO2) = + 1.700 V E˚SRP(O2/ H2O2) = + 0.695 V E˚cell = E˚SRP, cathode – E˚SRP, anode = + 1.700 V – (0.695 V) = + 1.005 V Yes, the reaction will proceed as described. (iv) I2(s) will displace Br–(aq) to produce Br2(l). The reactants are I2 and Br–. E˚SRP(I2/ I–) = + 0.535 V E˚SRP(Br2/ Br–) = + 1.065 V E˚cell = E˚SRP, cathode – E˚SRP, anode = + 0.535 V – (1.065 V) = – 0.530 V NO, the reaction will NOT proceed as described. 4. Students are asked to translate the conventional shorthand cell notation into the respective reactions. The notation declares an oxidation process on the left and a reduction process on the right. Each half-‐cell has its components listed in a reactant-‐product order. This does not claim that the resultant reaction will occur spontaneously. One can always refer to the SRP table to clarify the reactions. Don’t forget to scale them for electron cancellation before forming the overall reaction. (i) Al(s) | Al3+(aq) || Sn2+(aq) | Sn(s) Oxidation: Al(s) → Al3+(aq) + 3 e– Reduction: Sn2+(aq) + 2 e– → Sn(s) Overall: 2 Al(s) + 3 Sn2+(aq) → 2 Al3+(aq) + 3 Sn(s) (ii) Pt(s) | Fe2+(aq), Fe3+(aq) || Ag+(aq) | Ag(s) Pt is an inert electrode here and is not part of the reaction. Oxidation: Fe2+(aq) → Fe3+(aq) + e– Reduction: Ag+(aq) + e– → Ag(s) Overall: Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s) 5. This question has two objectives: to explore the mathematical relationship between E˚, ∆G, and K and to see how the stoichiometry of the reaction affects these variables. I2(aq) + Cd(s) → 2 I–(aq) + Cd2+(s) (i) Calculate E˚, ∆G˚, and K for this reaction Find the two half-‐reactions in the table, and calculate E˚. E˚overall = E˚(I2/I–) – E˚( Cd2+/Cd) = + 0.535 V – (– 0.403 V) = + 0.938 V ∆G˚ = -‐nF E˚ = –(2)(96485.34 C/mol) (0.938 V) = – 181.0 kJ/mol K = exp(-‐∆G˚/RT) =exp(–(– 181,000 J/mol)/(8.3145 J/K mol)(298 K)) = e73.05 = 5.32 x 1031 What if we rewrote the reaction as 0.5 I2(aq) + 0.5 Cd(s) → I–(aq) + 0.5 Cd2+(s) (ii) Calculate E˚, ∆G˚, and K for this reaction. E˚overall = E˚(I2/I–) – E˚( Cd2+/Cd) = + 0.535 V – (– 0.403 V) = + 0.938 V ∆G˚ = -‐nF E˚ = –(1)(96485.34 C/mol) (0.938 V) = – 90.5 kJ/mol K = exp(-‐∆G˚/RT) =exp(–(– 90,500 J/mol)/(8.3145 J/K mol)(298 K)) = e36.53 = 7.29x 1015 Note that E˚ is the same for both reactions because it represents a Gibbs energy per electron measurement. Both the Gibbs energy and the equilibrium constant depend upon how we write the chemical equation. In this case, the second reaction is just half of the first, so we would expect that to be only half as much Gibbs energy driving it. 6. What is the value for K and ∆G˚ for the reaction at 298 K between Ag(s) and Zn2+(aq)? This is just like question 5, but a word problem that must be translated into a chemical equation. Ag(s) + Zn2+(aq) → 2 Ag+(aq) + Zn(s) E˚overall = E˚( Zn2+/Zn) – E˚( Ag+/Ag) = – 0.763 V – (–0.800 V) = – 1.563 V ∆G˚ = -‐nF E˚ = –(2)(96485.34 C/mol) (–1.563 V) = + 301.6 kJ/mol K = exp(-‐∆G˚/RT) =exp(–(301,600 J/mol)/(8.3145 J/K mol)(298 K)) = e–121.7 = 1.37 x 10–53 7. Consider the voltaic cell 2 Ag+(aq) + Cd(s) → 2 Ag(s) + Cd2+(aq) (i) What is E˚cell for this cell? Ag is being reduced and Cd is being oxidized. Therefore, E˚cell = E˚(Ag+/Ag) – E˚(Cd2+/Cd) = 0.800 – (–0.403) = +1.203 V (ii) If [Cd2+] = 2.00 M and [Ag+] = 0.250 M, what is Ecell? Use the Nernst equation. We will use the natural logarithm, so the term RT/F = 0.02568 V. By inspection, we see 2 electrons are being transferred so n = 2. !"## = ˚!"## − !!
= 1.203 − = 1.16 !
0.25! (iii) If Ecell = 1.25 V and [Cd2+] = 0.100 M, what is [Ag+]? Use the Nernst equation to solve for [Ag+]. !!
!"## = ˚!"## − = 1.203 − = 1.25 !
1.25 − 1.203 = − 2
2 1.203 − 1.25
= = −3.660 0.02568
= !!.!!" = 0.02572 ! !
! = 0.100
= 1.97 0.02572 8. For each of these solutions, indicate what occurs at each electrode during electrolysis. (i) CuI2(aq) (ii) CdCl2(l) (iii) MgF2(aq) Follow this approach in solving these problems (1) Note that for aqueous solutions, the oxidation and reduction of water must be considered. (2) Decide the identity of the ions when dissolved in water. (3) Find the standard reduction potentials for the species involved. Don’t forget the oxidation and reduction of water, when present. O2(g) + 4 H+(aq) + 4 e– → 2 H2O(l) (reverse it to position water as a reactant) 2 H2O(l) → O2(g) + 4 H+(aq) + 4 e– E˚ = – 1.229 V (possible anode) 2 H2O(l) + 2e– → H2(g) + 2 OH–(aq) (acceptable as is; water already reactant) 2 H2O(l) + 2e– → H2(g) + 2 OH–(aq) E˚ = – 0.8277 V (possible cathode) (4) Species found on the reactant side of a half-‐reaction equation, are candidates for being reduced at the cathode. (5) Species found on the product side are candidates for being oxidized at the anode. (6) Group all potential cathode half-‐reactions together and identify the one with the least negative potential as this reaction will occur at the cathode. (7) Group all potential anode half-‐reactions together. Reverse each reaction and change the sign of its standard potential to match the reaction direction based on which reactants are present Identify the one with the least negative potential, as this reaction occurs first at the anode. Since water is involved in all three questions, the relevant reactions are given here: (i) CuCl2 Possible reactants: Cu2+, Cl–, and H2O. The relevant copper reaction is Cu2+(aq) + 2e– → Cu(s) E˚ = +0.340 V (cathode candidate) Since chloride is found on the product side in the SRP table, we must switch the reaction around, making it an oxidation, and so it becomes a candidate for the anode reaction. 2 Cl–(aq) → Cl2(l) + 2 e– E˚ = –1.3583 V (anode candidate) The two water reactions are 2 H2O(l) → O2(g) + 4 H+(aq) + 4 e– E˚ = – 1.229 V (anode candidate) 2 H2O(l) + 2e– → H2(g) + 2 OH–(aq) E˚ = – 0.8277 V (cathode candidate) These are all of the possible reactions. At the cathode, we must consider the reduction of water (-‐
0.8277 V) and the reduction of copper ions (+0.340 V). The copper reduction is the least negative (since it is positive), so it is the reaction that occurs at the cathode. This is where students are often confused, since the copper reaction is not negative. What we are aiming for is a final cell potential that is the least negative, and positive half-‐cell potential will certainly contribute to that goal. At the anode, we consider the oxidation of water (-‐1.229 V) and the oxidation of chloride ions (-‐
1.3583 V). Clearly water is the least negative so water oxidation occurs at the anode. Hence, copper reduction occurs, plating out copper on the cathode while bubbles of oxygen gas appear at the anode from the oxidation of water. The overall cell potential is E˚ = +0.340 – 1.229 = – 0.889 V (note that the water potential’s sign was already changed, so we just add the two reactions together as in Hess’s Law). We must apply at least 0.889 V to cause the reaction to begin to occur. (ii) CdCl2(l) When the phase is given as a liquid, that implies the ionic compound has been heated to melting and the ions Cd2+ and Cl– ions are present in the melt. No water is present, so we do not consider those reactions. The cadmium reaction is Cd2+(aq) + 2 e– → Cd(s) E˚ = – 0.403 V. (cathode candidate) The chloride reaction is 2 Cl–(aq) → Cl2(l) + 2 e– E˚ = –1.3583 V (anode candidate) These are the only possible reactions, so Cd metal is deposited at the cathode and chlorine gas is formed at the anode. (iii) MgF2(aq) Mg2+ and F– ions are in solution. The magnesium reaction is Mg2+(aq) + 2 e– → Mg(s) E˚ = -‐2.356 V. (cathode candidate) For the F– ions, the relevant reaction is 2 F–(aq) → F2(g) + 2 e– E˚ = -‐2.866 V (anode candidate) Water reactions must be considered as it is an aqueous solution. 2 H2O(l) → O2(g) + 4 H+(aq) + 4 e– E˚ = – 1.229 V (anode candidate) 2 H2O(l) + 2e– → H2(g) + 2 OH–(aq) E˚ = – 0.8277 V (cathode candidate) Both water oxidation at the anode and water reduction at the cathode are the favoured reactions, so H2 gas bubbles up at the anode and O2 gas bubbles up at the cathode. 9. A current of 2.50 A is passed through a solution of Cu(NO3)2 for 2.00 hr. What mass of copper is deposited at the cathode? An ampere is a coulomb/s, so convert all time units into seconds. Reaction is Cu2+(aq) + 2e– → Cu(s) 3600 = = 2.50 2.00 ℎ ×
= 18,000 ℎ
, = ∴ 18,000 = =
= 0.187 96485.32 Based on the reaction equation, it takes two electrons for each atom of copper, n(Cu) = 0.187/2 = 0.0933 mol Cu is deposited. The molar mass of Cu is 63.546 g/mol, so the mass of deposited copper is = × = 0.0933 ×63.546 = 5.93 KINETICS 10. Experimental data in the table are for the hypothetical reaction A → 2 B Time (s) [A] mol/L [B] mol/L 0.00 1.000 0.000 10.0 0.833 0.334 20.0 0.714 0.572 30.0 0.625 0.750 40.0 0.555 0.890 1) Draw a graph of [A] vs. time. Calculate the average rate of change of [A] in each 10.0 s time interval. Why is the rate changing from one interval to the next? Average rate = ∆[A]/∆t = (0.833 – 1.000)/(10.0 – 0.0) = -‐0.0167 M/s The remaining values are: 20 s: -‐0.0119 M/s; 30 s: -‐0.0089 M/s; 40s: -‐0.0070 M/s Since the rate decreases and changes with changing concentration, the reaction is not zero-‐
order. The blue, decreasing line is [A] vs. time.
Concentration (M) 1 0.8 0.6 0.4 0.2 0 0 10 20 30 40 50 Time (s) 2) If [B] begins at 0.00 M, sketch [B] on the same graph with [A]. How does the rate of change of [B] compare with that of [A] in a given time interval? From the stoichiometry, each time one A disappears, 2 B particles must appear. At each time point, find the amount of A that has been lost since the beginning and double it to be the amount of B at that time point. Since B is increasing, its rate is positive and must be double that of A’s loss. The rate of change of B at the time points are 10 s: 0.0334 M/s; 20 s: 0.0238 M/s; 30 s: 0.0178 M/s; 40s: 0.0140 M/s [B] is the red, ascending curve in the graph above. 3) Sketch a new graph that plots the rate of change of [A] and [B] against time (average rate against the average time (5.00, 15.00, 25.00, …) in each interval. Watch out for the sign of the change. 0.04 0.03 0.02 d[A]/dt 0.01 d[B]/dt 0 -‐0.01 0 10 20 30 40 -‐0.02 Note how both rates are approaching 0. The rate of [A]’s change is negative(it is a reactant) while that for [B] is positive (it being a product). Now graph the rate of the reaction on the same graph. Compare how the reaction rate compares to the rates of change of the reaction participants. It is important to distinguish between the rate of change of one of the reaction participants and the rate of the reaction itself. This is contained in the definition that scales the reaction rate by the inverse stoichiometric coefficient of each participant and includes a negative sign for reactants. In this way, a reaction rate is always positive, though the rate of change of reactants is negative. In this case, the reaction rate will have the same magnitude as d[A]/dt, but with a positive sign since the stoichiometric coefficient is 1. Rate of Change 4) 0.04 d[A]/dt 0.03 d[B]/dt 0.02 0.01 0 -‐0.01 -‐0.02 0 5 10 15 20 Time (s) 25 30 35 40 11. The transfer of an oxygen atom from NO2 to CO has been studied at 540 K. CO(g) + NO2(g) → CO2(g) + NO(g) These data were collected. Experiment Initial Rate Initial Concentration (mol L-‐1) -‐1
# (mol L hr ) [CO] [NO2] -‐4
1 5.1 x 10 0.35 x 10 3.4 x 10-‐8 2 5.1 x 10-‐4 0.70 x 10-‐4 1.7 x 10-‐8 -‐4
3 5.1 x 10 0.18 x 10 6.8 x 10-‐8 4 1.0 x 10-‐3 0.35 x 10-‐4 6.8 x 10-‐8 5 1.5 x 10-‐3 0.35 x 10-‐4 10.2 x 10-‐8 Using these data Write down the rate law for this reaction, specifying the reaction order with respect to all reactants. To determine the rate law, find two experiments in which the concentration of one reactant has been held constant and see how that affects the rate. For example, look at Expts 1 and 4. [CO] is held constant but the [NO2] is doubled. By comparing the two rates, we see that doubling NO2 doubles the rate. This indicates that the rate law has [NO2] raised to the first power. Now compare Expts 3 and 4 where [NO2] is constant but [CO] is doubled. The rate is also doubled, so [CO] appears to the first power also. The rate law, then is Rate = k [CO] [NO2] All of the other data can be examined to confirm this relationship. (i) (ii) Calculate the rate constant. The rate constant can be found from any of the five experiments, entering the concentrations and the rate (or averaged from all five experiments). Ensure that the units of the rate constant are such that the rate units are M/hr. Using experiment #1, we find Rate = 5.1 x 10-‐4 M/hr = k [CO] [NO2] = k (0.35 x 10-‐4 M) (0.34 x 10-‐8 M) k = 5.1 x 10-‐4 M/hr/(0.35 x 10-‐4 M) (0.34 x 10-‐8 M) = 4.29 x 109 M-‐1 hr-‐1 12. The initial concentration of a reactant in a first-‐order reaction is 0.64 mol L-‐1. The half-‐life of the reaction is 30.0 s. (i) Calculate the concentration of the reactant 60.0 s after the initiation of the reaction. The half-‐life is the time it takes the reactant to drop to one-‐half its starting concentration. Since 60.0 s is two half-‐lives, the concentration will drop by ½ x ½ = ¼ of the original value, or 0.64 x ¼ = 0.16 M (ii) How long would it take for the concentration of the reactant to drop to one-‐eighth of
its initial value? One-‐eighth would occur when ½ x ½ x ½ = 1/8, or three half-‐lives have transpired, or, in this case, 90.0 s. (iii) How long would it take for the concentration of the reactant to drop to 0.040 mol L-‐1? You could use the integrated rate law, find the value for k from the given half-‐life time, and then find t for this value of the concentration. But as soon as you determine the ratio of 0.04/0.64 = 1/16, one should realize that this is just 1/8 x ½ = 1/16 or just one more half-‐
life beyond part (ii). So after 4 half-‐lives, or 120 s, one would reach this concentration. 13. For the reaction of Pt(NH3)2Cl2 with water Pt(NH3)2Cl2 + H2O → Pt(NH3)2(H2O)Cl+ + Cl– the rate law was given as rate = k[Pt(NH3)2Cl2] with k = 0.090 hr-‐1. Calculate the initial rate of reaction when the concentration of Pt(NH3)2Cl2 is (i) 0.010 M Rate = k [Pt(NH3)2Cl2] Rate = 0.090 hr-‐1 x 0.010 M = 9.0 x 10-‐4 M hr-‐1 (ii) 0.020 M Rate = 0.090 hr-...
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