SOLUTIONS TO PRACTICE PROBLEMS

  1. (1.00tonFe)(2000lb/ton)(453.6g/lb)(moleFe/55.85gFe)((moleFe3O4)/3moleFe)(232gFe3O4/moleFe3O4)(100.0g ore/2.50gFe3O4)(lb/453.6g)(ton/2000lb) = 55.3 ton ore
  1. 4FeS2 + 11O2(g) → 8SO2(g) + 2Fe2O3 2SO2(g) + O2(g) → 2SO3(g)

SO3(g) + H2O → H2SO4

  1. (1.00ton soln)(.98ton H2SO4/ton soln)(2000lb/ton)(453.6g/lb)(mole H2SO4/98.1g) (mole SO3/mole H2SO4)(mole SO2/mole SO3)(4mole FeS2/8mole SO2) =

= 4.53 x 103 mole FeS2

(4.53 x 103 mole FeS2)(120.0gFeS2/mole)(100.0g ore/15.0gFeS2)(lb/453.6g) (ton/2000lb) = 4.00 ton ore

b. O2 is required in two steps

(4.53x103mole FeS2)[(11mole O2/4 mole FeS2)+(8mole SO2/4moleFeS2)(moleO2/

2mole SO2)](100.0mole air/21.0mole O2)(.08205Latm/moleK) (298.15K/620.0mmHg)(atm/760.0mmHg) = 2.43x106 L air.

  1. 3Cu2S 16H3O+ +10 NO3-→ 6Cu2+ + 3SO42- + 10NO(g) + 24 H2O

3Cu2+ + 2 Fe → 3Cu + 2Fe3+

  1. (1.00ton ore)(12.0tonCu2S/100.0ton ore)(2000lb/ton)(453.6g/lb)(moleCu2S/ 159.15g)(6mole Cu2+/3mole Cu2S)(2moleFe/3moleCu2+)(55.85gFe/mole)

(lb/453.6g)(ton Fe/2000lb) = 0.0561 ton Fe

  1. (1.00ton ore)(12.0tonCu2S/100.0ton ore)(2000lb/ton)(453.6g/lb)(moleCu2S/ 159.15g)(10moleNO/3moleCu2S)(.08205Latm/moleK)(300.15K/620.0mmHg)

(760.0mmHg/atm) = 6.88x104 L NO(g)

4. 2Ni3S2 + 7O2(g) → 6NiO + 4SO2(g) NiO + C → Ni + CO(g)

a. (kgNi3S2)(1000g/kg)(moleNi3S2/240.2g)(3moleNi/moleNi3S2)(58.7gNi/mole) =

736gNi

  1. (kgNi3S2)(1000g/kg)(moleNi3S2/240.2g)(4moleSO2/2moleNi3S2)(.08205Latm/

moleK)(294.15K/620mmHg)(760mmHg/atm) = 246 L SO2

5. Since Ni → Ni2+ in Ni(NO3)2 and Cu → Cu2+ each has 2 equivalents per mole

(100g alloy)/[(67.0gNi)(mole/58.7gNi)(2eq/moleNi) + (28.0gCu)(mole/63.5gCu) (2eq/moleCu)] = 31.6 g alloy/eq

6. Ni 4CO(g) → Ni(CO)4(g)

(5.60kgCO)(1000g/kg)(moleCO/28.0gCO)(moleNi(CO)4/4moleCO) = 50.0 moleNi(CO)4

(1470gNi)(moleNi/58.7gNi)(molemoleNi(CO)4/moleNi) = 25.0 moleNi(CO)4

Therefore Ni is the limiting reagent, 25.0 moleNi(CO)4 are produced and excess CO = (5600gCO)(moleCO/28.0g) –(25.0 moleNi(CO)4)(4moleCO/moleNi(CO)4) =

100 mole CO left in excess mixed with the 25.0 moleNi(CO)4 thus

(25.0 moleNi(CO)4)/[(25.0 moleNi(CO)4 + 100.0mole CO)=

= 0.200 moleNi(CO)4/mole total

  1. 2 FeTiO3 + 7Cl2(g) + 3C → 2TiCl4 + 2 FeCl3 + 3CO2(g)

TiCl4 + 2Mg → Ti + 2MgCl2

a. (5000gFeTiO3)(moleFeTiO3/151.75g)(moleTiCl4/moleFeTiO3)(moleTi/mole TiCl4)(47.88gTi/mole) = 1.58x103 g Ti

b. (5000gFeTiO3)(moleFeTiO3/151.75g)(3moleCO2/2moleFeTiO3)(.08205Latm/

moleK)(296.15K/550mmHg)(760mmHg/atm) = 1.66x103 L CO2

  1. MnO2 + 2NaCl + 2 H2SO4 → MnSO4 + Cl2(g) + 2H2O + Na2SO4
  2. (1.00gMnO2)(moleMnO2/86.9g)(moleCl2/moleMnO2)=1.15x10-2 moleCl2

(1.00gNaCl)(moleNaCl/58.4g)(moleCl2/2moleNaCl) = 8.56x10-3 mole Cl2

(5.00gH2SO4)(moleH2SO4/98.1gH2SO4)(moleCl2/2moleH2SO4) =

2.55x10-2 mole Cl2

Thus NaCl is the limiting reactant and

(8.56x10-3 mole Cl2)(.08205Latm/moleK)(298.15K/620.0mmHg)(760mmHg/atm) = 0.260L or 260mL Cl2 is formed

b. (1.00L Cl2)(620mmHg)(atm/760mmHg)/[(.08205Latm/moleK)(298.15K)] = 3.33x10-2 mole Cl2

(.0333moleCl2)(2moleH2SO4/moleCl2)(98.1gH2SO4/mole)(g soln/.95gH2SO4)

(ml soln/1.84g soln) = 3.74 mL soln

  1. (.200gCuFeS2/g ore)(moleCuFeS2/183.5g)(moleCu/moleCuFeS2)(63.5gCu/mole) = 0.0692 gCu/g ore

(.160gCu2(OH)2CO3/g ore)(moleCu2(OH)2CO3/221g)(2moleCu/moleCu2(OH)2CO3)(63.5gCu/mole) = .0919gCu/g ore or 9.19% Cu

  1. a. (.0486L Fe3+/.1000L MnO4-)(eqMnO4-/eqFe3+)(2.50eqFe3+/LFe3+) = 1.22 N
  1. (1.22eqMnO4-/L)(mole MnO4-/5eq) = 0.243 M
  2. (.243moleKMnO4/L soln)(158.03gKMnO4/mole)= 38.4 gKMnO4/L soln

1350g soln/L soln – 38.4gKMnO4/L soln 1312gH2O/L soln

(.243moleKMnO4/L soln)(L soln/1312gH2O)(1000g/kgH2O) =

0.183 moleKMnO4/kgH2O or 0.185m

  1. a. (.180gNH3/g soln)(.9295g soln/ml)(1000ml/L soln)(moleNH3/17.0g) = 9.84M

b. (.180gNH3/.820gH2O)(moleNH3/17.0g)(1000g/kgH2O) = 12.9 m

c. total moles = (.180gNH3)(mole/17.0gNH3) + (.820gH2O)(mole/18.0g) = .0561

(.180gNH3)(mole/17.0gNH3)/.0561 total = .189 moleNH3/total moles

  1. 2C + 4H2 + O2 → 2CH3OH

(12.00gO2)(moleO2/32.0g)(2moleCH3OH/moleO2) = 0.75 mole CH3OH

(6.00gC)(moleC/12.0g)(moleCH3OH/moleC) = 0.500 mole CH3OH

So C is the limiting reactant and .500 mole CH3OH form

(.500moleCH3OH)(4moleH2/2moleCH3OH)(.08205Latm/moleK)(303K)/

(685torr)(atm/760torr) = 27.6 L H2

  1. (4.00moleN2)(28.0g/mole) + 19.00 gF2 + (1.204x1024atomAr)(moleAr/6.02x1023 atom)(39.95g/mole) = 211 g total

4.00moleN2 + (19.00 gF2)(mole/38.0g) + (1.204x1024atomAr)(moleAr/6.02x1023

atom) = 6.50 mole total

  1. (211g/6.50mole) = 32.4 g/mole
  2. (625torr)(atm/760torr)/(.08205Latm/moleK)(323K) = 0.0310 mole/L

(0.0310 mole/L)(32.4g/mole) = 1.01 g/L

c. (19.00 gF2)(moleF2/38.0g)/6.50mole total = 0.0769 moleF2/total moles

  1. a. (19.62gH2SO4/.8000L soln)(moleH2SO4/98.10g) = 0.2500 M
  1. (800.0ml soln)(1.0115g soln/ml) – 19.62 gH2SO4 = 789.6 gH2O

(19.62gH2SO4)(mole/98.10g) + (789.6gH2O)(mole/18.0g) = 44.07 moles

(19.62gH2SO4)(mole/98.10g)/44.07 moles total = 4.538x10-3

c. (.1000L)(.2000mole/L)(L soln/.2500mole) =.0800L or 80.0mL soln

d. (.1000L)(.2500mole/L) = .02500 moles

  1. (29.0gCsCl)(moleCsCl/168.35g)(L soln/6.00moleCsCl) = .02871L soln

(29.0+21.0 g soln)/(.02871L soln)(1000ml/L) = 1.75 g/ml

  1. [(.0350L)(2.50mole/L) + (.0500L)(.150mole/L)]/.0850L = 1.12 mole/L
  1. total moles = V(.250mole/L) + (.155L)(2.00mole/L)

total volume = V + .155 L

1.50mole/L = (.250V + .310)/(V + .155) V = .0620L or 62 mL

  1. HCl + NaOH → NaCl + H2O

(.0300L HCl)(.125moleHCl/L)(moleNaOH/moleHCl)(LNaOH/.150moleNaOH) =

=.0250L NaOH or 25.0 mL

  1. KHPhth + base → K+ + Phth2- + baseH+ + H2O

(.8650g sample)(.9995gKHPhth/g sample) = .8646 gKHPhth

(.8646gKHPhth/.04385L base)(moleKHPhth/204.2g)(molebase/moleKHPhth) =

= .09655 M base

  1. (.02365L HCl)(0.1120moleHCl/L)(moleNa2CO3/2moleHCl)(106.0gNa2CO3/

mole) = 0.1404 gNa2CO3

(0.1404 gNa2CO3)/(0.485g sample) = 0.289 gNa2CO3)/g sample or 28.9%

  1. NaOH + HA → NaA + H2O

(.7515gHA/.04320LNaOH)(LNaOH/.0904moleNaOH)(moleNaOH/moleHA) =

192.4 gHA/moleHA

22. OH- + H3O+ → 2H2O (HNO3 reacts with water to form H3O+ and NO3- and

NaOH in water consists of Na+ and Cl- ions)

(.02500LNaOH)(.1500moleOH-/L) = .00375 mole OH-

(.01500LHNO3)(.2000moleH3O+/L) = .00300 moleH3O+

So after reaction .00075 mole OH- remain, all H3O+ has been consumed

(.00075 moleOH-/.04000L total) = .01875 M OH-

(.02500LNaOH)(.1500moleNa+/L)/(.04000L total) = 0.09375 M Na+

(.01500LHNO3)(.2000moleNO3-/L)/(.04000L total) = 0.07500 M NO3-

  1. NH3 + HCl → NH4+ + Cl-

(.02500L NH3)(.1500moleNH3/L) = .00375 mole NH3

(.01000LHCl)(.1200moleHCl/L) = .001200 moleHCl

So after reaction .00375-.00120 = .00255 mole NH3 remain and .001200 mole NH4+ have formed using all the HCl

(.00255moleNH3/.03500L total) = 0.07286 M NH3

(.001200moleNH4+/.03500L) = 0.03429 M NH4+

  1. HCl + NaOH → H2O + Na+ + Cl-

(.0550LHCl)(.200moleHCl/L) – (.0450LNaOH)(.150moleNaOH/L) (moleHCl/moleNaOH) = .00425 mole HCl left (in the form of H+ or H3O+ and Cl-)

So [H3O+] = .00425mole/.1000L total = .0425M

[Cl-] = (.0550L)(.200moleCl-/L)/(.1000L total) = 0.110 M Cl-

  1. 2MnO4- + 10I- + 16 H3O+ → 2Mn2+ + 5I2 + 24H2O

(1.25gKI)(moleKI/166.0g)(2moleKMnO4/10moleKI)(L KMnO4/0.150mole) =

= 0.0100 L or 10.0 mL KMnO4

  1. MnO4- + 5 Fe2+ + 8 H3O+ → Mn2+ + 5 Fe3+ + 12H2O

(.0500L Fe2+)(.100moleFe2+/L)(moleMnO4-/5moleFe2+)(L KMnO4/.150mole MnO4-) = 0.00667L or 6.67 mL KMnO4

  1. 2 S2O32- + I2 → S4O62- + 2 I-

(0.300gI2)(moleI2/253.9g)(2moleS2O32-/moleI2)(LNa2S2O3/.200moleS2O32-) =

= 0.0118 L or 11.8 mL Na2S2O3

  1. MnO2 + HAsO2 + 2 H3O+ → Mn2+ + H3AsO4 + 2H2O

(.0287L)(.0600moleHAsO2/L)(moleMnO2/moleHAsO2)(86.9gMnO2/mole) =

= 0.149 g MnO2

(0.149 g MnO2)/(0.200g sample) =0.749 gMnO2/g sample or 74.9% MnO2

  1. a. -2 +1 +7 -2 +5 -2 +4 -2

N2H4 + MnO4- → NO3- + MnO2

(+14) (-3)

3N2H4 + 14MnO4- → 6NO3- + 14MnO2

3N2H4 + 14MnO4- → 6NO3- + 14MnO2 + 8 OH-

3N2H4 + 14MnO4- → 6NO3- + 14MnO2 + 8 OH- + 2H2O

b. +3 -1 +1 -1 0 0 -1

AuCl63- + H2O2 → O2 + Au + Cl-

(-3) (+2)

2AuCl63- + 3H2O2 → 3O2 + 2Au + 12Cl-

2AuCl63- + 3H2O2 → 3O2 + 2Au + 12Cl- + 6H3O+

2AuCl63- + 3H2O2 + 6H2O → 3O2 + 2Au + 12Cl- + 6H3O+

c. +3 -2 +4 +1 +5 -2 +4 -2 +3

As2S3 + Ce4+ → H3AsO4 + SO2 + Ce3+

(+22) (-1)

As2S3 + 22Ce4+ → 2H3AsO4 + 3SO2 + 22Ce3+

As2S3 + 22Ce4+ → 2H3AsO4 + 3SO2 + 22Ce3+ + 22H3O+

As2S3 + 22Ce4+ + 36H2O → 2H3AsO4 + 3SO2 + 22Ce3+ + 22H3O+

d. +1 -2 +5 -2 +2 +2 -2 0

Cu2S + NO3- → Cu2+ + NO + S

(+4) (-3)

3Cu2S + 4NO3- → 6Cu2+ + 4NO + 3S

3Cu2S + 4NO3- + 16H3O+ → 6Cu2+ + 4NO + 3S

3Cu2S + 4NO3- + 16H3O+ → 6Cu2+ + 4NO + 3S + 24H2O

30. (5.60gN2)(moleN2/28.02g) =0.200 mole N2

(7.20gO2)(moleO2/32.0g) = 0.225 moleO2

a. PN2 = (700torr)(.200moleN2/.425 mole total) = 329torr

b. (12.80 mole total/.425 mole total) = 30.1 g/mole

c. V = nRT/P = (.425mole)(.08205Latm/moleK)(313.15K)/(700/760 atm) =

= 11.9 L

  1. (3.08gN2)(mole/28.0gN2) = 0.110 mole N2

(2.85gF2)(mole/38.0gF2) = .0750 mole F2

(.110 mole N2)(2 mole NF3/mole N2) = .220 mole NF3 but

(.0750 mole F2)(2 mole NF3/3 mole F2) = .0500 mole NF3 so F2 is limiting

.110 mole N2 – (.0500 mole NF3)(mole N2/2 mole NF3) = .0850 mole N2 left

Total moles = .0850 + .0500 = .135 moles

a. V= nRT/P = (.135 mole)(.08205Latm/moleK)(343K)/(1.30atm) = 2.92 L

b. Initially total moles = .110 +.0750 = .185moles

P= (.185 mole)(.08205Latm/moleK)(343K)/(2.92L) = 1.78 atm

  1. (.0850 mole N2)/(.135 mole total) = .630 mole fraction N2
  2. (.0500 mole NF3)/(.135 mole total) = .370 mole fraction NF3

32. Total mass is constant.

(2.00mole H2)(1.008g/mole) + (2.00mole S2)(64.12g/mole) = 132.3 g total

(132.3g)/(.534g/L) = 247.7 L volume

n = PV/RT = (2.00atm)(247.7L)/(.08205Latm/moleK)(1573K) = 3.84 moles

2 H2 + S2 ↔ 2 H2S

2.00-2x 2.00-x 2x so total moles = 2.00-2x+2.00-x+2x = 4.00-x

a. 4.00-x = 3.84 therefore x= 0.16 and moles H2S = 0.34 moles H2S

b. PH2S = (0.34 mole)(.08205Latm/moleK)(1573K)/(247.7L) =0.18 atm

PH2 = (1.66mole))(.08205Latm/moleK)(1573K)/(247.7L) = 0.87 atm

PS2 = (1.84mole)RT/V = 0.95 atm

33. (60.0gCu)(80.0oC-T)(0.0924cal/goC) = (20.0gH2O)(T-10.0oC)(1.00cal/goC)

T = 25.2oC

  1. (95.0g ice)(1435cal/mole)(mole/18.0g ice) = 7.57x104 cal to melt all ice

(155gFe)(215-0oC)(0.106calgFeoC) = 3.53x104 cal to cool Fe to 0.0oC

Therefore the Fe is cooled to 0.0oC melting only part of the ice to water

(3.53x104 cal)(mole/1435cal)(18.0g ice/mole) = 44.3g ice melt

44.3 g water + 50.7g ice at final temperature of 0.0o

35. 2C(gr) + 2O2(g) → 2CO2(g) ∆Ho1

3H2(g) + 3/2 O2(g) → 3H2O(l) ∆Ho2

2CO2(g) + 3H2O(l) → C2H6(g) ∆Ho3

2C(gr) + 3H2(g) → C2H6(g) ∆Hof = ∆Ho1 + ∆Ho2 + ∆Ho3

∆Ho1 = (2moleC)(-94.05kcal/mole) = -188.10kcal

∆Ho2 = (3moleH2)(-68.32kcal/mole) = -204.96 kcal

∆Ho3 = (moleC2H6)(+372.8kcal/mole) = +372.8 kcal

∆Hof = -188.10 -204.96 + 372.8 = -20.3 kcal/moleC2H6(g)

  1. a. HCl and HF both acidic, Cl- is very weak conjugate base of the very strong acid HCl so NaCl is neutral, but F- is the weak conjugate base of the weak acid HF. Thus NaF is basic and has the highest pH.

b.Addition of the weak base NaF to HF raises the pH while the much weaker base NaCl does not significantly raise the pH when added to HCl. Therefore the HF/NaF solution has the highest pH.

  1. NH4+ is the weak conjugate acid of the weak base NH3. Addition of NH4Cl to basic NH3 therefore lowers the pH. So the NH3 solution has the highest pH.
  2. NaCl is neutral because is Cl- such a weak conjugate base. NH4Cl is acidic because NH4+ is a weak acid and Cl- a very weak base. NH4F is slightly acidic because NH4+ is stronger acid than is F- a base. So NaCl has the highest pH.
  3. NaOH is a strong base and HF is a weak acid so equivalence point pH is above 7; NaOH and HCl are both strong so eq. pt. pH is 7; NH3 is a weak base and HCl a strong acid so eq. pyt. pH is acidic. So NaOH/HF has the highest eq. pt. pH.
  1. (.0500L HF)(0.100moleHF/L)(moleNaOH/moleHF)(L NaOH/0.400moleNaOH) = 0.0125 L NaOH at the equivalence point. Total V = 0.0625L

HF + NaOH → NaF + H2O so at eq. pt. solution is NaF solution of concentration

(.0500L HF)(.100moleHF/L)(moleNaF/moleHF)/.0625L = 0.0800 M HF

F- + H2O ↔ HF + OH- K = [HF][OH-]/[F-] x [H3O+]/[H3O+] = Kw/Ka

F- + H2O ↔ HF + OH- K = 1.0x10-14/4.8x10-4 = 2.1x10-11

.080-x x x

(x)(x)/(.080-x) = 2.1x10-11 x= 1.3x10-6 = [OH-]

pOH = -log[OH-] = 5.89 pH = 14.00-pOH = 8.11

  1. B + H3O+ → BH+ + H2O

(.0500L B)(.300mole B/L) = .0150 moleB

(.0250L HCl)(.200moleH3O+/L) = 0.00500 mole H3O+

So after reaction .0150mole B – (.00500moleH3O+)(moleB/moleH3O+) =

= .0100 mole B left and .00500 mole BH+ formed in .0750 L total

B + H2O ↔ BH+ + OH- Kb = [BH+][OH-]/[B]

(.0100/.075) -x (.00500/.0750) + x x

But pH = 8.25, so pOH = 5.75 and [OH-] = 10-pOH = 1.78x10-6 = x

Kb = [BH+][OH-]/[B] = (.0667+1.78x10-6)(1.78x10-6)/(.133-1.78x10-6) =

= 8.9x10-7

  1. a. Greater NaCN is the conjugate base of the weak acid HCN

b. Greater NaCl is the weak conjugate base of the strong acid HCL while NaCN is the stronger base of the weak acid HCN.

c. Greater The more concentrated base has the higher pH

d. Less NH4NO3 contains the fairly strong conjugate acid NH4+ of NH3 and the very weak conjugate base NO3- of the strong acid HNO3 so it is acidic and lowers the pH.

e. Greater due to the basicity of NH3.

f. Less The strong acid HClwins out over an equal amount of the weak base.

g. Same The resulting mixture at the equivalence point is simply 0.500M NH4Cl.

h. Greater The less concentrated solution of the weak conjugate acid has a higher pH.

i. Less HCl is a stronger acid than is HF.

40. a. HSCN + H2O ↔ H3O+ + SCN- Ka = [H3O+][SCN-]/[HSCN] = 0.14

.100-x x x

0.14 = (x)(x)/(.100-x) x= 0.067 pH = -log[H3O+] =-logx = 1.17

b. HSCN + OH- → H2O + SCN-

(.0500L HSCN)(.100mole/L) – (.0150LNaOH)(.200moleNaOH/L)(moleHSCN/moleNaOH) = 0.0020 mole HSCN left in .0650L total

(.0150LNaOH)(.200moleNaOH/L)(moleSCN-/moleNaOH) =

0.0030 mole SCN- formed

HSCN + H2O ↔ H3O+ + SCN-

.0020/.0650)-x x (.00300/.0650)+x

0.14 = (x)(.0462+x)/(.0308-x) x = 0.021 pH = -logx = 1.68

41. a. Electronegativity generally increases left to right and bottom to top in the Periodic Table. Therefore S has the greatest electronegativity among these.

b. Ionization potential increases left to right and bottom to top in the Periodic Table. Ionization of a second electron requires a higher potential than that for the first electron. Therefore K has the small ionization potential among these.

c. Ar has a higher ionization potential than S or Ca as explained above. Removal of an electron from S2- is easer than from S. Ca2+ and Ar contain the same number of electrons but there are 2 more nuclear protons in Ca2+ which therefore attract the electrons more strongly. So Ca2+ has the highest ionization potential among these.

d. Atomic size decreases left to right and bottom to top in the Periodic Table. Anions are larger than the corresponding atoms while cations are smaller. Thus I- has the largest radius among these.

42. a. 2 Hg + 2 Cl- → Hg2Cl2 + 2 e- anode reaction

IO3- + 6 H3O+ + 6 e- → I- + 9 H2O cathode reaction

IO3- + 6 H3O+ + 6 Hg + 6 Cl- → 3Hg2Cl2 + I- + 9 H2O cell reaction

b. Eocell = Eanode + Ecathode = -.355 +1.085 = +.730 volt

c. Ecell = Eocell –(.0591/n)log([I-]/[Cl-]6[IO3-][H3O+]6)

= +.730-(.0591/6)log{(.500)/(.100)6(5.00)(1.0x10-3)6} = +.503 volt

  1. (.600coul/sec)(3.00hr)(3600sec/hr)(eq/96500coul)(moleI-/eq) =

= 0.0112 mole I- produced

[I-] = .500 + (.0112mole/.200L) = 0.556 M

43. HBrO + 7 H2O → BrO3- + 4 e- + 5 H3O+ ∆Go = -(4eq)(96500coul/eq)(-1.59v)

BrO3- + 6 H3O+ + 5 e- → ½ Br2 + 9 H2O ∆Go = -(5eq)(96500coul/eq)(+1.52 v)

HBrO + H3O+ + e- → ½ Br2 + 2 H2O ∆Go = -119660 v-coulomb

Eo = -∆Go/nF = -(-119660 v-coulomb)/(1)(96500coulomb) = +1.24 v.

44. AgBr(s) + e- → Ag(s) + Br- ∆Go = -(1eq)(96500coul/eq)(+.071v)

Ag → Ag+ + e- ∆Go = -(1eq)(96500coul/eq)(-.80v)

AgBr(s) → Ag+ + + Br- ∆Go =.73(96500) = +7.04x104 J

Ksp = exp{-∆Go/RT} = 4.5x10-13

45. 8 OH- + S2O42- → 2 SO42- + 6 e- + 4 H2O ∆Go = -6F(+.99)

2 H2O + 4 e- + 2 SO42- → 2 SO32- + 4 OH- ∆Go = -4F(-.93)

4 OH- + S2O42- → 2 SO32- + 2 H2O+ 2 e-∆Go = -2.22F

Eo = -(-2.22F)/2F = +1.11 volt

46. PbSO4(s) → Pb2+ + SO42- ∆Go =-RTlnKsp = -(8.314)(2998.15)ln(1.6x10-8)

Pb2+ + 2 e- → Pb∆Go = -nFEo = -(2)(96500)(-.126)

PbSO4(s) + 2 e- → Pb + SO42- ∆Go =+4.5x104 J + 2.4x104 J = +6.88x104 J

Eo = -6.88x104/(2)(96500) =-.36 volt

47. a. Cd2+ + 2 e- → CdEo = -.402 v.

2 Cl- → Cl2(g) + 2 e- Eo = -1.3595 v.

Cd2+ + 2 Cl- → Cl2(g) + CdEo = -1.761 v.

E = -1.761 –(.0591/2)log{(1.00atm)/(3.00)(6.00)2} =-1.701 v.

So decomposition potential is +1.701 v.

b. 2 Cl- → Cl2(g) + 2 e- is the anode reaction

c. + pole

48. There are two possible anode reactions:

2 Cl- → Cl2 + 2e- E= E0 = -1.3595v since [Cl-] = 1.00M and Cl2 gas presumably is at 1 atm pressure

2 H2O→O2 + 4 H+ + 4 e- E = -1.229 – (.05917/4)log[H+]4PO2

Assuming pH is 7.00 and oxygen gas is at 1.00 Atm, then

E = -1.229 – (.05917/4) log(1.0x10-7)4(1.00 atm) = -0.815 v.

So oxidation of water is the anode reaction. The cathode reaction is

Ni2+ + 2 e- → Ni E= -.250 –(.05917/2)log(1/0.500) = -0.258v

a.So decomposition potential is ).815+0.258 = 1.074 v.

b. 2 (Ni2+ + 2 e- → Ni) cathode reaction

2 H2O→O2 + 4 H+ + 4 e- anode reaction

2 H2O +2 Ni2+ →O2 + 4 H+ + 2 Ni cell reaction

c.Anode as electrodes must be removed from the anode to produce oxidation.