Homework Problem Chapter 19 Answers. Bi 3+

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1 19.24 Homework Problem Chapter 19 Answers a. The reactions and cell potential are Ni(s) Ni 2+ (aq) + 2e E = 0.23 V 2Cu(aq) + 2e 2Cu(s) E = 0.52 V Ni(s) + 2Cu + (aq) Ni 2+ (aq) + 2Cu(s) E cell = 0.75 V Since the cell potential is positive, the reaction is spontaneous, and the cell is a voltaic cell. b. The current in this cell flows spontaneously. c. The maimum cell potential is E, which is 0.75 V. d. The initial voltage would be 0.75 V. As time went by, the voltage would gradually decrease, until equilibrium was reached, at which time, the voltage would be zero. e. The free energy is G = nfe cell = (2 mol)(96,485 C/mol)(0.75 V) = J = kj f. Over time, G changes, getting larger (less negative) as the cell runs. When equilibrium is reached, G will be zero In balancing oidationreduction reactions in acid, the four steps in the tet will be followed. For part a, each step is shown. For the other parts, only a summary is shown. a. Assign oidation numbers to the skeleton equation (Step 1) Mn 2+ + BiO 3 MnO 4 + Bi 3+ Separate into two incomplete halfreactions (Step 2). Note that manganese is oidized (increases in oidation number), and bismuth is reduced (decreases in oidation number). Mn 2+ MnO 4 BiO 3 Bi 3+ Balance each halfreaction separately. For the oidation halfreaction, add four H 2 O to the left side to balance O atoms (Step 3b), and add eight H + ion to the right side to balance H atoms (step 3c). Finally, add five electrons to the right side to balance the charge (Step 3d). The balanced oidation halfreaction is Mn H 2 O MnO 4 + 8H + + 5e For the reduction halfreaction, add three H 2 O to the right side to balance O atoms (Step 3b), and add si H + ion to the left side to balance H atoms (step 3c). Finally, add two electrons to the left side to balance the charge (Step 3d). The balanced reduction halfreaction is BiO 3 + 6H + + 2e Bi H 2 O

2 Multiply the oidation halfreaction by 2 and the reduction halfreaction by 5, so that, when added, the electrons cancel (Step 4a). 2Mn H 2 O 2MnO H e 5BiO H e 5Bi H 2 O 2Mn BiO H + + 8H 2 O + 10e 2MnO 4 + 5Bi H H 2 O + 10e The equation can be further simplified by canceling siteen H + and eight H 2 O from each side (Step 4b). The net ionic equation is 2Mn BiO H + 2MnO 4 + 5Bi H 2 O b. The two balanced halfreactions are I + 3H 2 O IO 3 + 6H + + 6e (oidation) Cr 2 O H + + 6e 2Cr H 2 O (reduction) Add the two halfreactions together, and cancel the si electrons from each side. Also, cancel si H + and three H 2 O from each side. The balanced equation is Cr 2 O I + 8H + 2Cr 3+ + IO 3 + 4H 2 O c. The two balanced halfreactions are H 2 SO 3 + H 2 O SO H + + 2e (oidation) MnO 4 + 8H + + 5e Mn H 2 O (reduction) Multiply the oidation halfreaction by 5 and the reduction halfreaction by 2, and then add together. Cancel the ten electrons from each side. Also, cancel siteen H + and five H 2 O from each side. The balanced equation is 2MnO 4 + 5H 2 SO 3 2Mn SO H + + 3H 2 O d. The two balanced halfreactions are Fe 2+ Fe 3+ + e Cr + 14H + + 6e 2Cr 3+ 2 O H 2 O (oidation) (reduction) Multiply the oidation halfreaction by si, and then add together. Cancel the si electrons from each side. The balanced equation is Cr 2 O Fe H + 2Cr Fe H 2 O e. The two balanced halfreactions are As + 3H 2 O H 3 AsO 3 + 3H + + 3e (oidation) ClO 3 + 5H + + 4e HClO + 2H 2 O (reduction) Multiply the oidation halfreaction by 4 and the reduction halfreaction by 3, and then add together. Cancel the twelve electrons from each side. Also, cancel twelve H + and si H 2 O from each side. The balanced equation is 4As + 3ClO 3 + 3H + + 6H 2 O 4H 3 AsO 3 + 3HClO

3 In balancing oidationreduction reactions in basic solution, the equation will first be balanced as if the equation were in acidic solution, then the etra two steps in the tet will be followed. a. The two balanced halfreactions are Cr(OH) 4 CrO H + + 3e (oidation) H 2 O 2 + 2H + + 2e 2H 2 O (reduction) Multiply the oidation halfreaction by 2 and the reduction halfreaction by 3, and then add together. Cancel the si electrons from each side. Also, cancel si H + from each side. The balanced equation in acidic solution is 2Cr(OH) 4 + 3H 2 O 2 2CrO H + + 6H 2 O Now add two OH to each side. Simplify by combining the H + and OH to give H 2 O. Then combine the H 2 O on the right side into one term. The balanced equation in basic solution is 2Cr(OH) 4 + 3H 2 O 2 + 2OH 2CrO H 2 O b. The two balanced halfreactions are Br + 3H 2 O BrO 3 + 6H + + 6e (oidation) MnO 4 + 4H + + 3e MnO 2 + 2H 2 O (reduction) Multiply the reduction halfreaction by 2, and then add together. Cancel the si electrons from each side. Also, cancel si H + and three H 2 O from each side. The balanced equation in acidic solution is 2MnO 4 + Br + 2H + 2MnO 2 + BrO 3 + H 2 O Now add two OH to each side. Simplify by combining the H + and OH to give H 2 O. Then cancel one H 2 O on each side. The balanced equation in basic solution is 2MnO 4 + Br + H 2 O 2MnO 2 + BrO 3 + 2OH c. The two balanced halfreactions are Co H 2 O Co(OH) 3 + 3H + + e H 2 O 2 + 2H + + 2e 2H 2 O (oidation) (reduction) Multiply the oidation halfreaction by 2, and then add together. Cancel the two electrons from each side. Also, cancel two H + and two H 2 O from each side. The balanced equation in acidic solution is 2Co 2+ + H 2 O 2 + 4H 2 O 2Co(OH) 3 + 4H + Now add four OH to each side. Simplify by combining the H + and OH to give H 2 O. Then cancel four H 2 O on each side. The balanced equation in basic solution is 2Co 2+ + H 2 O 2 + 4OH 2Co(OH) 3 d. The two balanced halfreactions are Pb(OH) 4 2 PbO 2 + 2H 2 O + 2e (oidation) ClO + 2H + + 2e Cl + H 2 O (reduction)

4 Add the two halfreactions together, and cancel the two electrons from each side. The balanced equation in acidic solution is Pb(OH) ClO + 2H + PbO 2 + Cl + 3H 2 O Now add two OH to each side. Simplify by combining the H + and OH to give H 2 O. Then cancel two H 2 O on each side. The balanced equation in basic solution is Pb(OH) ClO PbO 2 + Cl + 2OH + H 2 O e. The two balanced halfreactions are Zn + 4H 2 O Zn(OH) H + + 2e (oidation) NO 3 + 9H + + 8e NH 3 + 3H 2 O (reduction) Multiply the oidation halfreaction by 4, and then add together. Cancel the eight electrons from each side. Also, cancel nine H + and three H 2 O from each side. The balanced equation in acidic solution is 4Zn + NO H 2 O 4Zn(OH) NH 3 + 7H + Now add seven OH to each side. Simplify by combining the H + and OH to give H 2 O. Then cancel seven H 2 O on each side. The balanced equation in basic solution is Sketch of the cell: 4Zn + NO 3 + 7OH + 6H 2 O NH 3 + 4Zn(OH) 4 2 e Ni (anode) Salt bridge Cu (cathode) + Ni 2+ Cu 2+ Ni Ni e Cu e Cu The electrode halfreactions and the overall cell reaction are Anode: Zn(s) + 2OH (aq) Zn(OH) 2 (s) + 2e Cathode: HgO(s) + H 2 O(l) + 2e Hg(l) + 2OH (aq) Overall: Zn(s) + HgO(s) + H 2 O(l) Zn(OH) 2 (s) + Hg(l)

5 19.50 Because of its less negative E, H + is reduced at the cathode and is written on the right; Al(s) is oidized at the anode and is written first, at the left, in the cell notation. The notation is Al(s) Al 3+ (aq) H + (aq) H 2 (g) Pt The H 2 (g), on the left, is the reducing agent. The Br 2 (l), on the right, is the oidizing agent. The halfreactions and the overall cell reaction are Br 2 (l) + 2e H 2 (g) 2H + (aq) + 2e 2Br (aq) H 2 (g) + Br 2 (l) 2H + (aq) + 2Br (aq) The halfcell reactions are 3O 2 (g) + 12H e 6H 2 O(l) 4Al(s) 4Al 3+ (aq) + 12e n equals 12, and the maimum work for the reaction as written is w ma = nfe cell = (12)( C)(1.15 V) = C V = J For 50.0 g of aluminum, the maimum work is 50.0 g 1 mol Al g Al J 4 mol Al = = J f. The reduction halfreactions and standard potentials are Fe 3+ (aq) + e Fe 2+ (aq) Cr 2 O 2 7 (aq) + 14H + 6e 2Cr 2+ (aq) + 7H 2 O(l) E = 0.77 V E = 1.33 V The stronger oidizing agent is the one involved in the halfreaction with the more positive standard electrode potential, so Cr 2 O 7 2 is the stronger oidizing agent. Thus, Cr 2 O 7 2 will oidize iron(ii) ion in acidic solution under standard conditions. g. The reduction halfreactions and standard electrode potentials are Ni 2+ (aq) + 2e Ni(s) E = 0.23 V Cu 2+ (aq) + 2e Cu(s) E = 0.34 V The stronger reducing agent is the one involved in the halfreaction with the smaller standard electrode potential, so Ni(s) is the stronger reducing agent. Thus, copper metal will not reduce Ni(II) ion spontaneously.

6 Write the equation with the appropriate G f beneath each substance: G f : 5H 2 C 2 O 2 (aq) + 2MnO 4 (aq) + 6H + (aq) 10CO 2 (g) + 8H 2 O(l) + 2Mn 2 (aq) 5(698) 2(447.3) 0 10(394.4) 8(237.1) 2(223)KJ Hence, G = [10( 394.4) + 8( 237.1) + 2( 223) 5( 698) 2( 447.3)] kj = kj = J Obtain n by splitting the reaction into halfreactions. 5H 2 C 2 O 4 (aq) 10CO 2 (g) + 10H + (aq) + 10e 2MnO 4 (aq) + 16H + (aq) + 10e 2Mn 2+ (aq) + 8H 2 O(l) Each halfreaction involves ten electrons, so n = 10. Therefore, G = nfe cell J = C E cell Rearrange and solve for E cell. Recall that J = C V. E cell = J C = = V The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: Ni(s) Ni 2+ (aq) + 2e E = 0.23 V Sn 2+ (aq) + 2e Sn(s) E = 0.14 V Ni(s) + Sn 2+ (aq) Ni 2+ (aq) + Sn(s) E cell = 0.09 V Note that n equals 2. The reaction quotient is Q = 2+ [Ni ] 2+ [Sn ] 1.0 = = The standard cell potential is 0.09 V, so the Nernst equation becomes E cell = E cell n = log Q log ( ) = 0.09 V ( V) = = 0.03 V The conversion of current and time (221 min = s) to coulombs is C = amp sec = 1.51 A s = C The conversion of coulombs to grams of cadmium is C 1 mol Cd C 2 mol e g Cd 1 mol Cd = = 11.7 g Cd

7 a. The halfreactions and the corresponding standard electrode potential values are Cl + 2e 2Cl 2 (g) (aq) MnO + 8H + (aq) + 5e Mn 2+ 4 (aq) (aq) + 4H 2 O(l) E = 1.36 V E = 1.49 V The stronger oidizing agent is the one involved in the halfreaction with the more positive standard electrode potential, so MnO 4 is the stronger oidizing agent. The oidation of chloride ion by permanganate ion is a spontaneous reaction. b. The halfreactions and the corresponding standard electrode potential values are Cr 2 O 2 2 (aq) + 14H + (aq) + 6e 2Cr 3+ (aq) + 7H 2 O(l) Cl 2 (g) + 2e 2Cl (aq) E = 1.33 V E = 1.36 V The stronger oidizing agent is the one involved in the halfreaction with the more positive standard electrode potential, so Cl 2 is the stronger oidizing agent. The oidation of chloride ion by dichromate ion is not a spontaneous reaction. a. The number of faradays is 1.0 mol Fe 3+ 1 mol Fe The number of coulombs is 1.0 F 96,485 C b. The number of faradays is 1.0 mol Fe = 1.0 F = = C 3 mol e 1 mol Fe The number of coulombs is 3.0 F 96,485 C c. The number of faradays is 3+ = 3.0 F = = C 1.0 g Sn 2+ 1 mol Sn 2 mol e g Sn mol Sn = = F The number of coulombs is F 96,485 C = = C

8 d. The number of faradays is 1.0 g Au 3+ 1 mol Au g Au mol e 1 mol Au 3+ = = F The number of coulombs is F 96,485 C = = C a. The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: Fe(s) Fe 2+ (aq) + 2e E = 0.41 V Cu 2+ (aq) + 2e Cu(s) E = 0.34 V Fe(s) + Cu 2+ (aq) Fe 2+ (aq) + Cu(s) E = 0.75 V Use the Nernst equation to calculate the voltage of the cell. E = E n log Q = E n 2+ [Fe ] log 2+ [Cu ] Note that n equals 2, [Fe 2+ ] = 0.15 M, and [Cu 2+ ] = M. E = 0.75 V = = 0.73 V log [0.15] [0.036] b. First, calculate the moles of electrons passing through the cell amp 335 s The moles of Cu deposited are 96,485 C mol e 1 mol Cu 2 mol e = mol e = mol The moles of Cu remaining in the 1.00 L of solution are = = mol Since the volume of the solution is 1.00 L, the molarity of Cu 2+ is M. c. Write the cell reaction with the G f s beneath. Hence, 2Tl(s) + Pb 2+ (aq) 2Tl + (aq) + Pb(s) G f : (32.4) 0 KJ G = [2( 32.4) ( 24.39)] kj = kj = J d. Net, determine the standard cell potential for the cell. Note that n equals 2. G = nfe

9 J = 2( C) E Solve for E to get E = J 4 2( C) = = V The halfreactions and voltages are 2Tl(s) 2Tl + (aq) + 2e E o = E red Pb 2+ (aq) + 2e Pb(s) E red = 0.13 V Cd(s) + Co 2+ (aq) Cd 2+ (aq) + Co(s) E = V The cell potential is E = E red + ( E o ) V = 13 V + E red E red = V V = = 0.34 V e. Write the equation with the appropriate G f beneath each substance. Hence, 2Al(s) + 3Fe 2+ (aq) 2Al 3+ (aq) + 3Fe(s) G f : 0 3 (78.87) 2 (485) 0 KJ G = Σn G f (products) Σm G f (reactants) = [3( 78.87) 2( 485)] kj = kj = J Each halfreaction involves si electrons, so n equals 6. Therefore, G = nfe cell J = C E cell Rearrange and solve for E cell. Recall that J = C V. E cell = J C f. The equilibrium constant is E cell = n log K = = 1.27 V log K = n E = = K = = = g. The value of the equilibrium constant is not affected by the concentration of the ions.

10 The halfcell reactions, the corresponding halfcell potentials, and their sums are displayed below: 2Cr(s) 2Cr 3+ (aq) + 6e E = 0.74 V 3Fe 2+ (aq) + 6e 3Fe(s) E = 0.41 V 2Cr(s) + 3Fe 2+ (aq) 2Cr 3+ (aq) + 3Fe(s) E = 0.33 V Note that n equals 6. Therefore, G = nfe = 6( C)(0.33 V) = J = kj Write the cell reaction with the H f s beneath. Hence, 2Cr(s) + 3Fe 2+ (aq) 2Cr 3+ (aq) + 3Fe(s) H f : 0 2(87.9) 2(143.5) 0 KJ H = Σn H f (products) Σm H f (reactants) Now calculate S. = [2( 143.5) 3( 87.9] kj = 23.3 kj G = H T S kj = 23.3 kj 298 K S Solving for S gives S = 23.3 kj kj 298 K = kj/k = 563 J/K