Chemistry 2000 (Spring 2014) Problem Set #7: Redox Reactions and Electrochemistry Solutions
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1 Chemistry 2000 (Spring 2014) Problem Set #7: Redox Reactions and Electrochemistry Solutions Answers to Questions in Silberberg (only those w/out answers at the back of the book) 192 An electrochemical process involves electron transfer from one particle to another This is, by definition, a redox reaction 194 ions do not exist in aqueous solution (If added to aqueous solution, would immediately react with H 2 to give two H - ions) 195 The two half-reactions are each multiplied by the appropriate coefficients so that the number of electrons on the product side of the oxidation half-reaction is the same as the number of electrons on the reactant side of the reduction half-reaction 199 (c) (d) (e) (f) True True False In an electrolytic cell, the surroundings do work on the system In a voltaic cell, the system does work on the surroundings True False The cell electrolyte provides a solution capable of conducting current Fe + Mn H + Mn + 5 Fe H 2 This reaction should look familiar You did this titration as part of the analysis in the green crystals lab! 1923 If the two compartments are not connected, there is no way for the electrons produced by the oxidation half-reaction to be consumed by the reduction half-reaction (so neither halfreaction will proceed) If the two compartments were physically connected, there would be no way for the voltaic cell to do work The energy generated by the overall reaction would instead be released as heat 1924 The salt bridge allows both half-cells to maintain a neutral charge (so that the overall reaction does not stop due to charge build-up) It does so by allowing spectator ions (eg Na +, N 3, etc) to travel between the two compartments while preventing the reactive ions from doing so 1935 See lecture notes or pp of text 1952 (a) A reaction is thermodynamically allowed when Q < K A reaction is thermodynamically allowed when ΔG < 0 and E cell > 0 Therefore, ΔG < 0 and E cell > 0 when Q < K If Q < K then Q/K < 1 As noted in part (a), when this is the case, E cell > 0 and the cell can do work
2 Additional Practice Problems 1 Determine the oxidation states for each atom in the following sulfur oxide anions (only one resonance structure is given) (a) peroxodisulfate: S 2 8 dithionite S 2 4 (c) thiosulfate S S S S S S 0 S +4 2 Balance the following redox reactions in acidic aqueous solution (a) 3 U 2 + Te + 4 H + 3 U 4+ + Te H 2 2 PbS H 2 Pb + Pb S H + (c) 4 AsH Ag H 2 As Ag + 24 H + (d) 2 Mn HCN + 5 I + 11 H + 2 Mn + 5 ICN + 8 H 2 (e) H 5 I I + 7 H + 4 I H 2 This is an example of a comproportionation reaction, a reaction in which the same species (I 2 ) is produced by both the oxidation and reduction half reactions (f) 3 U 2 + Cr H + 3 U Cr H 2 3 Balance the following redox reactions in basic aqueous solution (a) 3 Cl - + I - - I Cl - P H H - 2 HP PH 3 (c) 2 Co + 3 Cl + 3 H 2 2 Co(H) Cl (d) 2 As + 6 H 3 2 As H 2 (e) 4 Au + 8 CN H 2 4 [Au(CN) 2 ] + 4H - (Note: while gold is not oxidized by oxygen in aqueous solution, addition of cyanide makes this possible This reaction is heavily used in gold mining operations, especially if the gold concentration of the ores is low The cyanide ponds that are used for this extraction process are serious environmental hazards given the huge quantities of such a toxic substance!)
3 4 The cell described below develops a reversible emf of -097 V at 25 C (assume exact temperature so infinite sig fig) (a) + Pt s H 2g 1 bar H aq ph= 50 V aq mol/l V s The half-reactions are + H 2(g) 2H (aq) 2e - E 0 0V V (aq) 2e - 0 V (s) E V / V verall: H 2(g) V + (aq) 2H (aq) V (s) E 0 0 E V / V with e = 2 The emf generated by the cell under the given conditions is E = -097 V We can calculate E 0 using the Nernst equation: E E 0 RT e F lnq E 0 E RT RT lnq E e F e F ln a H + 2 a H2 a V 097V J K-1 mol K ln C/mol V Since E 0 0 E V, / V 0 E V / V 118 V For the overall reaction from part (a), r e FE C/mol 118 V 227 kj/mol However, we can also write r in terms of the standard free energies of formation of the reactants and products: r 2 f G 0 + H (aq) f G 0 V (s) f G 0 H 2(g) f G 0 V (aq) so f G 0 V (aq) f G 0 V (aq) 227 kj/mol
4 5 Using the data appended to this problem, calculate the equilibrium constant for the reaction below at 25 C (exact temperature) Hg 2aq Hg l +Hg aq Which of the two aqueous ions will be more abundant at equilibrium? (Metallic mercury is insoluble in water, so it precipitates out if formed If necessary, assume that the aqueous layer is in direct contact with metallic mercury initially) Data: Hg aq +2e Hg l Hg 2aq +2e 2Hg l E 0 = +0851V E 0 = V If we turn the first half-reaction around, we have Hg l Hg aq +2e E 0 = 0851V Hg 2aq +2e 2Hg l E 0 = V The overall reaction is Hg 2(aq) Hg (l) Hg (aq) E V with e = 2 We can do the calculation in one or two steps I ll do it in one step here: K exp G 0 r m exp FE 0 e RT RT C/mol 0054 V exp J K -1 mol K 0015 This value of the equilibrium constant implies that the reaction is reactant-favored, ie there will be more Hg 2 than Hg at equilibrium
5 6 Mercury(II) sulfide is only sparingly soluble in water The sulfide ion is a stronger base than hydroxide, so the solubility equilibrium is: HgS (s) H 2 (l) Hg (aq) HS (aq) H (aq) The equilibrium constant for this reaction is at 25 C (exact temperature) (a) Let s start by setting up the equilibrium expression: K a Hg a HS - a H - Since the equilibrium constant is really tiny, this equilibrium will have a negligible effect on the hydroxide concentration in solution Therefore a H We will make one Hg 2= for each HS -, so the activities of these two ions will be equal Thus we have 2 K a a a Hg - Hg K a - H 210 H The concentration is proportional to the activity by a factor of 1 mol/l, so the solubility of HgS is mol/l In other words, in 10 L of water, we should have mol of the Hg and HS - ions But wait! That many moles is (110 3 mol)( mol -1 ) = 9 molecules! In other words, we only expect to find nine molecules of HgS dissolved in 1 L of water This equilibrium constant was clearly not measured by measuring a concentration of dissolved material! We might try something analogous to the AgCl solubility measurement We could make up an Hg HgS electrode (This is clearly not quite the same as an Ag,AgCl electrode since Hg is a liquid, but in principle we could make something like this) The likely half-cell reaction would be HgS (s) H 2 (l) 2e Hg (l) HS (aq) H (aq) Having made this electrode, we could mesure its half-cell potential and calculate the corresponding standard reduction potential Then, combining this standard reduction potential with the standard reduction potential of Hg to Hg (which is already available), we could calculate the equilibrium constant for the solubility equilibrium
6 7 Suppose that we operate an ethanol fuel cell in Lethbridge at 25 C (exact temperature) The partial pressure of 2 is 019 bar and the partial pressure of C 2 is about 0034 bar The fuel cell operates on liquid ethanol (a) The simplest way to do this problem is probably to work out the free energy change, then go from there to an emf We can t do anything without a balanced reaction, of course: C 2 H 5 H (l) 3 2(g) 2C 2(g) 3H 2 (l) r 2 f G 0 C 2(g) 3 f G 0 H 2 (l) f G 0 C 2 H 5 H (l) 2(3944) 3(2371) (1748) kj/mol = kj/mol r G m r G m 0 RT ln Q r RT ln a C 2 a kj/mol kj K -1 mol K ln kj/mol We re going to use the equation r G m = e FE to calculate E ur next problem is to figure out e There are three ways to do this: The hard way: Use the standard method for balancing redox reactions in solution to get the half-reactions We can then read e off from our balanced half-reactions The easier way: Look up the oxygen half-reaction from the table of standard reduction potentials at the back of the book In this case, it doesn t matter if we look up the reaction for acidic or basic reaction conditions because the aqueous intermediates cancel out anyway Note that there are 4 electrons for every 2 For our balanced reaction then, e = 3(4) = 12 The easiest way: In fuel cells that use oxygen as the oxidant, it s always 4 electrons per oxygen molecule, so just remember this ratio E r G m e F J/mol C/mol V If we wanted to make a 240V battery, assuming that the operating voltage is similar to the emf (which, remember, is measured under reversible conditions, and so is generally higher than the operating voltage), then we would need 240/11485 = 209 cells
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