Chemistry Final Exam (Answered)

Chemistry Final Exam (Answered) Oxidation Rules Rule 1: The oxidation number of an element in its free state = 0 (Also true for diatomic elements) -Mg=0,H2=0, C=0 Rule 2: The oxidation number of a monatomic (one-atom) ion is the same as the charge on the ion -Mg^2+=+2, Na^+=+1, O^2-=-2 Rule 3: The sum of all oxidation numbers in a neutral compound is zero. The sum of all oxidation numbers in a polyatomic (many-atom) ion is equal to the charge on the ion. Rule 4: The oxidation number of an alkali metal (IA family) in a compound is +1; the oxidation number of an alkaline earth metal (IIA family) in a compound is +2. Rule 5: The oxidation number of oxygen in a compound is usually -2. If, however, the oxygen is in a class of compounds called peroxides (H2O2), then the oxygen has an oxidation number of -1. If the oxygen is bonded to fluorine, the number is +1. Rule 6: The oxidation state of hydrogen in a compound is usually +1. Rule 7: The oxidation number of fluorine is always -1. Chlorine, bromine, and iodine usually have an oxidation number of -1, unless they're in combination with an oxygen or fluorine. -In HF=-1, H=0 because F=-1 A) What is the oxidation state of an individual sulfur atom in MgSO4? B) What is the oxidation state of an individual nitrogen atom in NH2OH? C) What is the oxidation state of an individual phosphorus atom in PO33−? D) What is the oxidation state of each individual carbon atom in C2O42−? A) +6 B) -1 C) +3 D) +3 Which compound contains phosphorous in its highest oxidation state? PO43- PI3 P4 P2O2 PO4^3- P+4(-2)=-3 PI3 (P=3) P4 (P=0) 2Na (s) + Cl2 (g) = 2NaCl (s) What is the oxidation, reduction, oxidizing agent, and reducing agent What are the half reactions? Oxidation: Na= Na^+1 + 1e^- Reduction: Cl + 2e^- = 2Cl Na is oxidized by Cl2 (Cl2 is the oxidizing agent) Cl is reduced by Na (Na is the reducing agent or reductant) What is the oxidation and reduction half reaction for: Zn (s) + Cu2+ (aq) = Zn2+ (aq) + Cu (s) Oxidation 1⁄2 -reaction (Anode): Zn (s) = Zn2+ (aq) + 2 e- Reduction 1⁄2 -reaction (Cathode): 2 e- + Cu2+ (aq) = Cu (s) Electrochemistry Flow goes from anode to cathode Anode=oxidation (Anode half cell = oxidation occurs) Cathode=Reduction (Cathode half cell= reduction occurs) Cations go from anode to cathode Anions go from cathode to anode The "salt bridge" is a reservoir for cations/anions to replenish electrons and close the electrical circuit Electrons/electricity always headed towards CATHODE Eocell (Cell potential) = Eordn (cathode) - Eordn (anode) *Half-reactions are always written as reductions Substances with more NEGATIVE reduction potentials reduce substances with more positive reduction potentials Substances with very NEGATIVE reduction potentials are good reducing agents (More negative= more likely to be a reduction) Substances with very POSITIVE reduction potentials are good oxidizing agents Nernst Equation Ecell= E°cell - (RT/nF) (lnQ) At 25 DegC, Eo - (0.0592/n) logQ Among the following reactions, identify which ones are redox reactions (and the oxidizing agent if they are redox reactions) A) NO2- (aq) + H+ (aq) = HNO2 (aq) B) 4Al(s) + 3O2 (g) = 2Al2O3 (s) C) 3Cl2 (g) + 2Fe(s) 2FeCl3 (s) A) No redox B) Oxidizing agent = O2 C) Oxidizing agent = Cl2 What is the emf for a galvanic cell that employs the following balanced overall cell reaction: 2 Al (s) + 3 I2 (s) = 2 Al3+ (aq) + 6 I- (aq) I2 (s) + 2e^- = 2I (aq) = 0.54 Al3+ (aq) + 3e^- = Al (s) = -1.66 What is the emf for a cell employing the reaction below if [Al3+] = 0.004M and [I- ] = 0.010M at 298K? 2Al(s) + 3I2 (s) → 2Al3+ (aq) + 6I- (aq) 2.2 V Eordn are independent of stoichiometric factors ! V = J/C (energy per charge transferred) = const. 2.37V Q= [I- ]6 [Al3+]2= 0.0106 × 0.0042 = 1.6 × 10^-17 Eo = (RT/nF) lnK n=6 E = 2.20 V- .0592/6 * log(1.6 × 10^-17) Balancing Redox Equations in Basic Solution: Cu(s) + 4 H+ (aq) + 2 NO3- (aq) → Cu2+ (aq) + 2 NO2 (aq) + 2 H2O (l) 2H2O (l) + Cu + 2NO3^- = Cu^2+ + 2NO2 + 4OH^- In a basic solution, H+ need to be replaced by OH- in the final equation! Take H2O on right, move to left Take 4H+ on left, add 4OH- on right Calculate Oxidation values for: 1. FeO3 2. O3 3. MgCl 4. LiOH 5. H2O2 6. NH4+ 7. HSO4- 8. ClO2- 9. Cr2O7^2- 10. K2O2 1. Fe (+6), O (3*-2) 2. O (3*0=0) 3. Mg (+2), Cl (-2) 4. Li (+1), O (-2), H (+1) - Start with Li value 5. H2 (21), O (2-1) 6. N (-3), H (4*1) = 1 7. H (1), S (6), O (4*-2)= -1 8. Cl (3), O (2*-2)= -1 9. Cr (26), O (7-2) = -2 10. K (21), O (21) - Start with K (rule 4) What is the value of ΔGo (in kJ/mol) for the following reaction 25.0 oC? Al (s) + Ag+ (aq) = Ag (s) + Al3+ (aq) Eo = + 2.46 V - 712 kJ/mol ΔGo = - 3 x 96,500 J/V/mol x (+2.46 V) Chapter 20 Equations ΔGo = - RT lnK Nernst: Eo = (RT/nF) lnK Spontaneity of redox reactions: ΔGo = - n F Eo n = # electrons transferred F = Faraday's constant Spontaneity of Redox Reactions DeltaGo=-nFEo n = # electrons transferred F = Faraday's constant Oxidation-Reduction Reactions (Redox Reactions) -Reactions that involve the transfer of electrons from one species to another. -In general, one element will lose electrons (oxidation), with the result that it will increase in oxidation number, and another element will gain electrons (reduction), thereby decreasing in oxidation number. An oxidizing agent is an element or compound in a redox reaction that oxidizes another species and itself gets reduced and is therefore the electron acceptor in the reaction. A reducing agent is an element or compound in a redox reaction that reduces another species and itself gets oxidized and is therefore the electron donor in the reaction. -Oxidation means an increase in oxidation state and a loss of electrons and involves a reducing agent. -Reduction means a decrease in oxidation state and a gain of electrons and involves an oxidizing agent. FeO+CO→Fe+CO2 A) Which element is oxidized in this reaction? B) Which substance is the oxidizing agent in this reaction? Cr2O72−+3HNO2+5H+→2Cr3++3NO3−+4H2O C) Which element is reduced in this reaction? D) Which substance is the reducing agent in this reaction? A) Carbon is oxidized B) FeO is the oxidizing agent C) Cr is reduced D) HNO2 is the reducing agent. Determine the oxidation number for the indicated element in each of the following substances: A) S in SO2 B) C in COCl2 C) Mn in MnO−4 D) Br in HBrO E) As in As4 F) O in K2O2 A) 4 B) 4 C) 7 D) 1 E) 0 F) -1 (Rule 4 takes precedence over rule 5) For the given reactions, classify the reactants as the reducing agent, oxidizing agent, or neither: N2 + 3H2 → 2NH3 2S + 3O2 → 2SO3 3S^2− + 14H^+ + Cr2O72− → 3S +2Cr^3+ + 7H2O Oxidizing Agents: N2, O2, Cr2O72- Reducing Agents: H2, S, S2- Neither: H+ In the process of oxidizing I− to I2, SO42− is reduced to SO2. How many moles of SO2 are produced in the formation of one mole of I2? 1 mol The iodine half-reaction generates two electrons. The sulfur half-reaction consumes two electrons. So we can determine that I2 and SO2 are produced in a 1:1 mole ratio without ever writing the full equation. (2 electrons lost from 2I- to I2) The iodate anion (IO3^-) reacts with SO3^- to form iodide (I-) and X according to the reaction IO3^- +3SO3^- = 1^- + X The IO3- ions disappear at rate of 810^-5 Ms^-1. The unknown compound appears at rate of 2.410^-4 Ms^-1. What is the coefficient? 3 rate[A] = -delta[A]/delta t -delta[IO3-]/delta t = 1* Delta [x]/?delta t 810^-5 = 1/? 2.4 * 10^-4 Galvanic Cells -A galvanic cell (or voltaic cell) produces electricity using a spontaneous redox reaction -The components of this reaction are separated by a salt bridge and connected with a wire, forcing the electrons to travel across the wire, creating electricity. -The salt bridge is a U-shaped glass tube that is filled with a gel-like substance