Department of Chemistry University of Texas at Austin

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1 Electrochemical Cells II Supplemetal Worksheet All the electrochemical cells o this worksheet are the same oes o the first Electrochemical Cells worksheet. To make the work o this worksheet easier, refer to the work you did o the previous Electrochemical Cells worksheet. Experimetal Observatios of Electrochemical Cells 1. Cosider the voltaic cell that cotais stadard Co 2+ /Co ad Au 3+ /Au electrodes. The followig experimetal observatios were oted: (1) Metallic gold plates out o oe electrode ad the gold io cocetratio decreases aroud that electrode, ad (2) The mass of the cobalt electrode decreases ad the cobalt (II) io cocetratio icreases aroud that electrode. a. Recall the diagram, overall balaced redox reactio, ad the stadard cell potetial for this cell? KEY <Co(s) Co 2+ (aq) Au 3+ (aq) Au(s)> E cell = E cathode - E aode = 1.50 V (-0.28 V) = 1.78 V b. What is the stadard reactio free eergy for this cell? ΔGr = -FE = - ( = kj mol rx 6 mol e ) 1 mol rx (96,485 C 1 mol e ) (1.78 J C ) = J mol rx

2 c. Calculate the equilibrium costat for the overall redox reactio of this cell at 25 C. Because we are lookig for K at 25 C, we ca use the equatio logk = E So, K = 10 E = K = = 5.14* This is a very large K, this reflects the spotaeous ature of this reactio. d. Calculate the emf at 25 C of this cell i which the cocetratio of Co 2+ ios is 0.30 mol/l ad that of the Au 3+ ios is mol/l. Here we use the Nerst Equatio. Sice we are fidig the emf at 25 C we ca use this versio of the Nerst Equatio: Ecell = E cell log Q = E cell log [Co2+ ] [Au 3+ ] 2 = 1.78 V log [0.30]3 6 [0.0010] 2 = 1.74 V e. Assume you use this cell as a battery to power oe of your electric devices that draws 1 ma of curret. If you ra the battery at a costat 1 ma for 3000 hours, how may grams of Co would be cosumed? To figure this problem out, we will do a series of coversios: curret ad time charge moles of e- moles of Co grams of Co curret ad time charge: charge = curret * time = C/s * (3000 hrs *60 mi/hr * 60 s/mi) = C charge moles of e-: C * 1 mol e 96,485 C moles of e- moles of Co: mol e- * moles of Co grams of Co: mol Co * 3.30 g Co would be cosumed = mol e- 1 mol Co 2 mol e 58.9 g Co 1 mol Co = mol Co = 3.30 g Co

3 2. Cosider the electrolysis of molte calcium chloride with iert electrodes. The followig experimetal observatios were oted: (1) Bubbles of pale gree chlorie gas are produced at oe electrode, ad (2) Silvery white molte metallic calcium is produced at the other electrode. a. Recall the diagram, overall balaced redox reactio, ad the stadard cell potetial for this cell? <Pt(s) Cl - (aq) Cl2(g) Ca 2+ (aq) Ca(s) Pt(s)> E cell = E cathode - E aode = V (1.36 V) = V b. What is the stadard reactio free eergy for this cell? ΔGr 2 mol e = -FE = - ( ) 1 mol rx (96,485 C ) ( 4.12 J ) = kj 1 mol e C mol rx c. Calculate the equilibrium costat for the overall redox reactio of this cell at 25 C. Because we are lookig for K at 25 C, we ca use the equatio logk = E So, K = 10 E = K = 10 2 ( 4.12) = 3.76* This is a very small K, this reflects the o-spotaeous ature of this reactio

4 d. How may hours are required to plate g of metallic calcium from 1.00 M CaCl2 (aq) by usig a curret of 3.00 A? To figure this problem out, we will do a series of coversios: grams of Ca moles of Ca moles of e- charge charge ad curret time grams of Ca moles of Ca: 12 g Ca * 1 mol Ca moles of Ca moles of e-: mol Ca * 40.1 g Ca 2 mol e = mol Ca 1 mol Ca moles of e- charge: mol e- * 96,485 C 1 mol e = C charge ad curret time: time = charge = C curret 3 C/s = mol e- = s * 1 hr = 5.4 hr 3600 s It would require 5.4 hours to plate g of metallic calcium from 1.00 M CaCl2 (aq) by usig a curret of 3.00 A. e. Determie the volume (i liters, at STP) of chlorie gas that ca be produced i this cell by usig a curret of 7.30 ma for 2.11 hours. To figure this problem out, we will do a series of coversios: curret ad time charge moles of e- moles of Co liters of Cl2 curret ad time charge: charge = curret * time = C/s * (2.11 hrs *60 mi/hr * 60 s/mi) = C charge moles of e-: C * 1 mol e 96,485 C = mol e- moles of e- moles of Cl2: mol e- * 1 mol Cl 2 = mol Cl2 2 mol e moles of Cl2 liters of Cl2: mol Cl2 * 22.4 L Cl 2at STP = L Cl2 = 1 mol Cl 2 at STP 6.44 ml Cl ml Cl2 would be produced

5 Short Had Notatio of Electrochemical Cells 3. Cosider the followig cell <Ni(s) Ni 2+ (aq) Ag + (aq) Ag(s)> a. Recall the diagram, overall balaced redox reactio, ad the stadard cell potetial for this cell? E cell = E cathode - E aode = 0.80 V (-0.23 V) = 1.03 V b. What is the stadard reactio free eergy for this cell? ΔGr 2 mol e = -FE = - ( ) 1 mol rx (96,485 C ) (1.03 J ) = kj 1 mol e C mol rx c. Calculate the equilibrium costat for the overall redox reactio of this cell at 25 C. Because we are lookig for K at 25 C, we ca use the equatio logk = E So, K = 10 E = K = = 7.18*10 34 This is a very large K, this reflects the spotaeous ature of this reactio.

6 d. What is the cocetratio of Ni 2+ ios if the emf at 25 C of this cell is V ad the cocetratio of Ag + ios is mol/l? Here we use the Nerst Equatio. Sice we are fidig the emf at 25 C we ca use this versio of the Nerst Equatio: Ecell = E cell log Q Ecell - E cell = log [Ni2+ ] [Ag + ] (E cell - Ecell) = log [Ni2+ ] 10 [Ag + ] (E cell Ecell) = [Ni2+ ] [Ag + ] 2 [Ni 2+ ] = [Ag + ] (E cell Ecell) = [ mol/l] (1.03V 1.1V) = 1.2 * 10-4 M e. How may hours would it take for this galvaic cell to plate g of metallic silver from 1.00 M solutios of Ni 2+ (aq) ad Ag + (aq) assumig it produces a costat curret of 2.25 A? To figure this problem out, we will do a series of coversios: grams of Ag moles of Ag moles of e- charge charge ad curret time grams of Ag moles of Ag: 30.0 g Ag * moles of Ag moles of e-: mol Ag * 1 mol Ag g Ag 1 mol e 1 mol Ca = mol Ag moles of e- charge: mol e- * 96,485 C 1 mol e = C charge ad curret time: time = charge = C curret 2.25 C/s = mol e- 1 hr = s * = 3.3 hr 3600 s It would require 3.3 hours to plate g of metallic silver from 1.00 M solutios of Ni 2+ (aq) ad Ag + (aq) assumig it produces a costat curret of 2.25 A.

7 4. Cosider the followig cell <Pt(s) Ce 3+ (aq) Ce 4+ (aq) Cu 2+ (aq) Cu(s)> a. Recall the diagram, overall balaced redox reactio, ad the stadard cell potetial for this cell? E cell = E cathode - E aode = 0.34 V (1.70 V) = V b. What is the stadard reactio free eergy for this cell? ΔGr 2 mol e = -FE = - ( ) 1 mol rx (96,485 C ) ( 1.36 J ) = kj 1 mol e C mol rx c. Calculate the equilibrium costat for the overall redox reactio of this cell at 25 C. Because we are lookig for K at 25 C, we ca use the equatio logk = E So, K = 10 E = K = 10 2 ( 1.36) = 9.47*10-47 This is a very small K, this reflects the o-spotaeous ature of this reactio

8 d. What is the cocetratio of Ce 4+ ios if the emf at 25 C of this cell is 1.20 V, the cocetratio of Cu 2+ ios is 0.60 mol/l ad the cocetratio of Ce 3+ ios is 0.30 mol/l? Here we use the Nerst Equatio. Sice we are fidig the emf at 25 C we ca use this versio of the Nerst Equatio: Ecell = E cell log Q Ecell - E cell = [Ce log 4+ ] 2 [Cu 2+ ] [Ce 3+ ] (E cell - Ecell) = log [Ce 4+ ] 2 10 [Cu 2+ ] [Ce 3+ ] (E cell Ecell) [Ce = 4+ ] 2 [Cu 2+ ] [Ce 3+ ] 2 [Ce 4+ ] 2 = [Cu 2+ ] [Ce 3+ ] 2 10 [Ce 4+ ] = [Cu 2+ ] [Ce 3+ ] 2 10 (E cell Ecell) (E cell Ecell) = [0.60 mol/l] [0.30 mol/l] ( 1.36V ( 1.20 V)) = 4.56 * 10-4 M e. Determie the mass (i grams) of metal copper that ca be produced i this cell by usig a curret of 5.0 A for 2.7 days g Cu would be produced (see problem 1e for a similar explaatio for this problem)

9 Electrochemical Cell Diagrams 5. Cosider the followig cell: a. Recall the diagram, overall balaced redox reactio, ad the stadard cell potetial for this cell? <Cu(s) Cu 2+ (aq) Z 2+ (aq) Z(s)> E cell = E cathode - E aode = V (0.34 V) = V b. What is the stadard reactio free eergy for this cell? ΔGr = -FE = kj mol rx

10 c. Calculate the equilibrium costat for the overall redox reactio of this cell at 25 C. K = 5.96*10-38 This is a very small K, this reflects the o-spotaeous ature of this reactio. d. What is the potetial for this cell if the Cu 2+ cocetratio is 5.8 x 10-3 M ad the Z 2+ cocetratio is 1.3 x 10-1 M? Here we use the Nerst Equatio to fid that E = E logq = log [Z2+ ] 2 [Cu 2+ ] = log (0.13) 2 (0.0058) = -1.14V e. Determie the mass (i grams) of zic metal that ca be produced i this cell by usig a curret of 6.0 A for 1.5 days g Z would be produced (see problem 1e for a similar explaatio for this problem)

11 6. Cosider the followig cell: a. Recall the diagram, overall balaced redox reactio, ad the stadard cell potetial for this cell? <Z(s) Z 2+ (aq) Cu 2+ (aq) Cu(s)> E cell = E cathode - E aode = 0.34 V (-0.76 V) = 1.10 V b. What is the stadard reactio free eergy for this cell? ΔGr = -FE = kj mol rx

12 c. Calculate the equilibrium costat for the overall redox reactio of this cell at 25 C. K = 1.1*10 37 This is a very large K, this reflects the spotaeous ature of this reactio. d. Calculate the emf at 25 C of this cell i which the cocetratio of Cu 2+ ios is mol/l ad that of the Z 2+ ios is mol/l. Here we use the Nerst Equatio. Sice we are fidig the emf at 25 C we ca use this versio of the Nerst Equatio: Ecell = E cell log Q = E cell log [Z2+ ] [Cu 2+ ] = 1.15 V = 1.10 V log [0.0020] 2 [0.1250] e. How may hours would it take for this galvaic cell to plate g of metallic copper from 1.00 M solutios of Cu 2+ (aq) ad Z 2+ (aq) assumig it produces a costat curret of 5.0 A? It would require 2.53 hours to plate g of metallic copper from 1.00 M solutios of Cu 2+ (aq) ad Z 2+ (aq) assumig it produces a costat curret of 5.0 A.

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