Slutins manual fr Burrws et.al. Chemistry 3 Third editin 16 Electrchemistry Answers t wrked examples WE 16.1 Drawing a cell diagram (n p. 739 in Chemistry 3 ) Draw a cell diagram fr an electrchemical cell with the verall reactin: H2 (g) + Cl2 (g) 2 HCl (aq) Write the tw half-cell reactins and identify the xidatin and reductin reactins. Slutin The half-cell prcess H2 (g) 2 H + (aq) + 2 e represents an xidatin prcess as the H2 (g) lses electrns, and may be written as PtH2 (g)h + (aq) where the vertical lines represent a bundary between slid, aqueus r gas phases and the cmpnents are written with the reduced species first. The halfcell prcess Cl2 (g) + 2 e 2 Cl (aq) is a reductin because the Cl2 (g) gains electrns and may be written PtCl (aq)cl2 (g) By cnventin, the ande, at which xidatin ccurs, is written as the left-hand electrde, and the cathde, at which reductin ccurs, as the right-hand Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin electrde. The tw half cells may then be cmbined using a vertical duble lines t dente a junctin where the junctin ptential has been eliminated, perhaps by the use f a salt bridge. It is als cnventinal t write the cmpnents in the rder that they appear in the verall equatin PtH2 (g) H + (aq) Cl2 (g)cl (aq) Pt WE 16.2 Interpreting a cell diagram (n p. 740 in Chemistry 3 ) What is the verall reactin ccurring in the cell Pt (s) V 2+ (aq), V 3+ (aq) Fe 3+ (aq), Fe 2+ (aq) Pt (s) Identify the tw half cells and write ut their reactins. Cmbine the tw halfcell reactins, remembering the cnventin that xidatin is assumed t ccur in the left-hand cell and reductin in the right-hand cell. Slutin The tw half-cell reactins are Left (xidatin) Right (reductin) V 2+ (aq) V (aq) + e Fe (aq) + e Fe (aq) Then, adding the tw half-cell reactins t get the verall cell reactin. V 2+ (aq) + Fe (aq) V (aq) + Fe (aq) WE 16.3 Predicting the directin f redx reactins (n p. 743 in Chemistry 3 ) Predict whether acidified aqueus ptassium permanganate (KMnO4) culd be used t xidise Fe 2+ t Fe 3+ under standard cnditins. Write equatins fr the pssible half reactins and find their standard reductin ptentials frm Table 16.2. Calculate the cell ptential fr the verall cell ptential, remembering t use the cnventins given in Equatin 16.10. Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin Slutin The apprpriate half-cell reactin fr the Fe 2 /Fe 3+ reductin/xidatin cuple is Fe (aq) + e Fe (aq) fr which E = +0.77 V. The crrespnding xidatin prcess is just the reverse f this prcess Fe (aq) Fe (aq) + e fr which E = 0.77 V. The half cell crrespnding t reductin f permanganate ins is MnO4 (aq) + 8 H + (aq) + 5e Mn 2+ (aq) + 4 H2O (l) fr which E = +1.52 V. Cmbining these tw equatins therefre gives 5 Fe (aq) + MnO4 (aq) + 8 H + (aq) 5 Fe (aq) + Mn 2+ (aq) + 4 H2O (l) with a cell ptential frm Equatin 16.11 f = E R E L = +1.52 V 0.77 V = +0.75 V A psitive value fr the cell ptential indicates that the reactin is spntaneus in the directin written, implying that ptassium permanganate can be used t xidise Fe 2+ (aq) ins t Fe 3+ (aq) ins. WE 16.4 Calculating cell ptentials (n p. 744 in Chemistry 3 ) Use values f standard reductin ptentials frm Table 16.2 t calculate E cell fr the fllwing electrchemical cell Fe (s) Fe 2+ (aq) Fe 3+ (aq), Fe 2+ (aq) Pt (s) Use the cell diagram t write half reactins fr the xidatin reactin and the reductin reactin. Calculate the cell ptential using Equatin 16.11. Slutin The left-hand cell is, by cnventin, the ande and represents an xidatin Fe (s) Fe 2+ (aq) + 2 e whilst the right-hand cell is the cathde and represents a reductin H i g h e r E d u c a t i n Oxfrd University Press, 2017. All rights reserved.
Slutins manual fr Burrws et.al. Chemistry 3 Third editin The verall cell ptential is therefre Fe 3+ (aq) + e Fe 2+ (aq) = E R E L Cnsulting Table 16.2, shws that the standard reductin ptential fr the lefthand Fe/Fe 2+ (aq) cell is 0.44 V and fr the right-hand Fe 2+ (aq)/fe 3+ (aq) cell is +0.77 V. Thus = E R E L = +0.77 V ( 0.44 V) = +1.21 V WE 16.5 Gibbs energy changes frm electrchemistry (n p. 748 in Chemistry 3 ) Calculate the standard Gibbs energy change at 298 K fr the reactin in the Daniell cell Zn (s) Zn 2+ (aq) Cu 2+ (aq) Cu (s) Identify the tw half-cell reactins and determine the standard cell ptential using Equatin 16.11 and the apprpriate values in Table 16.2. Use Equatin 16.14 t calculate the standard Gibbs energy change. Slutin The reductin half reactin fr the left-hand cell is Zn 2+ (aq) + 2 e Zn (s) fr which E L = 0.76 V and fr the right-hand cell is Cu 2+ (aq) + 2 e Cu (s) fr which E L =+0.34 V. By cnventin, the left-hand cell is assumed t be the ande and is therefre an xidatin rather than reductin reactin. Thus, the verall cell ptential is therefre = E R E L = +0.34 V ( 0.76 V) = +1.10 V The standard Gibbs energy f reactin is, frm Equatin 16.14, Δ r G = zfe As written, the tw half-cell reactins are bth tw-electrn prcesses, s that Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin z = 2 and therefre Δ r G = 2 96485 C ml 1 +1.10 V = 212 10 3 CV ml 1 = 212 kj ml 1 because 10 3 CV 1 kj WE 16.6 Calculating a thermdynamic equilibrium cnstant (n p. 749 in Chemistry 3 ) Calculate the thermdynamic equilibrium cnstant at 298 K fr the reactin Cu (s) + 2 Ag + (aq) Cu 2+ (aq) + 2 Ag (s) Determine the standard cell ptential frm Equatin 16.11 by identifying the apprpriate half cells and cnsulting Table 16.2. Use Equatin 16.15 t determine the crrespnding equilibrium cnstant. Slutin The reactin may be brken dwn int tw half-cell reactins. The standard reductin ptential f the first, 2 Ag + (aq) + 2 e 2 Ag (s) is E =+0.80 V. This is a reductin prcess, because the Ag + (aq) ins gain electrns, s frms the cathde. The secnd reactin is Cu (s) Cu 2+ (aq) + 2 e which is the reverse f the reductin half reactin Cu 2+ (aq) + 2 e Cu (s) fr which E =+0.34 V. This is an xidatin prcess, because the Cu (s) lses electrns, and s frms the ande. The verall cell ptential is thus, frm Equatin 16.10, = E cathde E ande Oxfrd University Press, 2017. All rights reserved. = +0.80 V 0.34V = +0.46 V H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin As written, bth reactins are tw-electrn prcesses, s that z = 2. Then, using Equatin 16.15 and substituting in the values, s that ln K = zf RT E 2 96485 C ml 1 cell = 8.3145 J K 1 ml 1 0.46 V = 35.8 298 K K = e 35.8 = 3.62 10 15 The equilibrium therefre lies in favur f the prducts, which is t be expected fr a cell whichhas a psitive cell ptential. WE 16.7 Using the Nernst equatin (n p. 751 in Chemistry 3 ) (a) Write an equatin fr the verall reactin that ccurs in the fllwing cell. Pt (s) H2 (g) H + (aq, 1 ml dm 3 ) Cu 2+ (aq, 1 ml dm 3 ) Cu (s) (b) Calculate the value f E cell. (c) The ptential, E cell, fr the cell belw is measured and fund t be +0.25 V. Pt (s) H2 (g) H + (aq, 1 ml dm 3 ) Cu 2+ (aq, c ml dm 3 ) Cu (s) What is the cncentratin, c, f the Cu 2+ (aq) slutin? Write the chemical equatin fr the cell reactin and hence determine the standard cell ptential. Devise an expressin fr the reactin qutient and use the Nernst equatin, Equatin 16.17, t write an expressin fr the cell ptential in terms f the unknwn cncentratin. Slutin (a) As written, the cell represents the reactin Cu 2+ (aq, 1 ml dm 3 ) + H2 (g) Cu (s) + 2 H + (aq, 1 ml dm 3 ) The cathdic standard half cell, which crrespnds t reductin, is s that E cathde xidatin is Cu 2+ (aq) + 2 e Cu (s) = +0.34 V and the andic half cell, which crrespnds t Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin s that E ande = 0 V. (b) Hence, frm Equatin 16.10, = E cathde H2 (g) 2 H + (aq) + 2 e E ande (c) The Nernst equatin, Equatin 16.16, is = = +0.34 V 0 V = +0.34 V RT ln Q zf As written, the reactin is a tw electrn prcess, s that z = 2. The reactin qutient is, using Equatins 15.2 and 14.19 Q = a 2 Cu(s)a H + (aq) a Cu 2+ (aq)a H2 (g) [H + (aq)] 2 = [Cu 2+ (p (aq)] /c ) p H2 (g) and if it is assumed that the partial pressure f H2 is p H2 (g) = p = 1 bar and the cncentratin f hydrgen ins is then if and [H + (aq)] = c = 1 ml dm 3 Q = c /[Cu 2+ (aq)] = = RT ln Q zf RT zf ln(c / c) ln(c / c) = zf( E cell )/RT ln(c/c ) = zf( E cell )/RT Thus, c = e zf( E )/RT cell c = e 2 96485 C ml 1 (0.25 V 0.34 V)/8.3145 J K 1 ml 1 298 K ml dm 3 Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin = 9.0 10 4 ml dm 3 WE 16.8 Reacting quantities in electrlysis (n p. 759 in Chemistry 3 ) In the electrlysis f Na2SO4, O2 (g) is prduced at the ande. Fr hw lng wuld a current f 10 A need t pass in rder t prduce 0.50 g f O2 (g) at the ande? Calculate the number f mles f O2 (g) prduced, and frm the balanced equatin calculate the number f mles f electrns needed. Use Faraday s cnstant t wrk ut the charge f these electrns, and hence the time needed t pass this much charge with the stated current. Slutin The number f mles f O2 (g) in 0.50 g is given by 0.50 g = 0.0156 ml 31.998 g ml 1 One mle f O2 (g) ccurs with the passage f 4 electrns, accrding t the equatin: 2H2O O2 + 4H + + 4 e Therefre, the number f mles f electrns is 0.0156 ml 4 = 0.0625 ml 1 ml f electrns carries 96485 C f charge, therefre 0.0625 ml is equivalent t: 96485 C ml -1 0.0625 ml = 6030 C. 1 A = 1 C s -1, therefre the time taken t pass this much charge is 6030 C 10 A = 603 s (apprx. 10 min) Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin Answers t bxes Bx 16.1 Ultrapure water and cnductivity (n p. 733 in Chemistry 3 ) (a) Calculate the cncentratins f H + (aq) and OH (aq) in pure water. (b) Frm data in Table 16.1, predict the mlar cnductivity, m, f pure water. (c) Calculate the cnductivity,, f pure water in S cm 1. (d) Assuming that the cnductivity f a sample f water is all due t disslved sdium chlride, calculate the cncentratin needed t give a slutin with = 100 S cm 1. Write an expressin fr the self-inizatin cnstant f water in terms f the cncentratins f the ins. Use Equatin 16.8, the law f independent migratin f ins, t determine the mlar cnductivity using the limiting values frm Table 16.1. Rearrange Equatin 16.7 t determine the cncentratin. Slutin (a) The self-inizatin cnstant K w = [H + (aq)][oh (aq)] = 1 10 14 ml 2 dm 6 But, in water [H + (aq)] = [OH (aq)] s that K w = [H + (aq)] 2 = [OH (aq)] 2 = 1 10 14 ml 2 dm 6 and therefre [H + (aq)] = [OH (aq)] = 1 10 14 ml 2 dm 6 = 1 10 7 ml dm 3 Remembering that 1 m 3 = 10 3 dm 3 then [H + (aq)] = [OH (aq)] = 1 10 4 ml m 3 Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin (b) The cncentratins f the ins are s lw that it is reasnable t use the limiting mlar cnductivities,λ m = Λ m, which are strictly nly valid fr infinite dilutin. Thus, using Equatin 16.8, and nting that the stichimetric cefficients f the ins are + = 1 and = 1, Λ m = ν + λ + + ν λ = (1 35.0 10 3 S m 2 ml 1 ) + (1 19.9 10 3 S m 2 ml 1 ) = 54.9 10 3 S m 2 ml 1 (c) Rearranging Equatin 16.7, Κ = Λ c = 54.9 10 3 S m 2 ml 1 1 10 4 ml m 3 = 5.49 10 6 S m 1 = 5.49 μs m 1 = 0.0549 μs cm 1 (d) Fr NaCl, + = = 1 and assuming that the cncentratins are sufficiently lw that it is reasnable t use the limiting mlar cnductivities, m m = ν + λ + + ν λ = (λ Na + + λ Cl ) = 5.0 10 3 S m 2 ml 1 + 7.6 10 3 S m 2 ml 1 = 12.6 10 3 S m 2 ml 1 Rearranging Equatin 16.7, and nting that 100 μs cm 1 = 10 2 10 6 S cm 1 = 10 4 S cm 1 = 1 10 2 S m 1 then c = Κ/Λ m = 1 10 2 μs cm 1 /12.6 10 3 S m 2 ml 1 = 0.79 ml m 3 = 7.9 10 4 ml dm 3 Bx 16.3 Practical measurement f E values (n p. 745 in Chemistry 3 ) Calculate the cell ptentials, E cell, at 25C fr the electrchemical cells Zn (s) Zn 2+ (aq) reference half cell and reference half cell Cu 2+ (aq) Cu (s) Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin where the reference half cell is: (a) the standard hydrgen electrde; (b) the silver-silver chlride electrde. (c) Wrk ut the difference between the values f E cell fr the tw cells in (a) and the tw cells in (b). Write the full frm f the cells and use Equatin 16.11, tgether with the data fr the standard reductin ptentials in Table 16.2, t determine the cell ptentials. Slutin (a) The ptential f the cell is the difference between the cell ptential f the tw half cells. By cnventin, we dente the left-hand cell as the ande and the right- hand cell as the cathde. Fr the standard hydrgen electrde, E H +,H 2 that, cmbining Equatins 16.10 and 16.11, fr the cell and Zn(s) Zn 2+ (aq) H + (aq), H2(g) Pt (s) = E cathde E ande = E R E L = E H +,H 2 = 0.00 V ( 0.76 V) = +0.76 V Pt(s) H + (aq), H2(g) Cu 2+ (aq) Cu(s) = E cathde E ande = E R E L = E Cu 2+,Cu = +0.34 V 0.00 V = +0.34 V s that the difference between the cell ptentials is 0.42 V. E Zn 2+,Zn E H +,H 2 (b) In the same way, using a silver silver chlride reference, fr which E AgCl,Ag and = +0.22 V Zn(s) Zn 2+ (aq) Ag(s) AgCl(s) Cl (aq) = E cathde E ande = E R E L = E H +,H 2 = 0.22 V ( 0.76 V) = +0.54 V Ag(s) AgCl(s) Cl (aq) Cu 2+ (aq) Cu(s) = E cathde E ande = E R E L = E Cu 2+,Cu = +0.34 V 0.22 V = +0.12 V E Zn 2+,Zn E H +,H 2 = 0 V, s Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin s that the difference between the cell ptentials is, as befre, 0.42 V. Bx 16.4 Crrsin as an electrchemical prcess (n p. 751 in Chemistry 3 ) (a) Suggest why rusting ccurs much mre readily in slutins cntaining disslved electrlytes, fr example in sea-water r in cars during the winter mnths when salt is spread nt rads? (b) Which f the fllwing: sdium, cpper, tin, zinc, graphite culd be used as sacrificial andes fr an irn structure? (a) Cnsider the effect f increasing the number f ins in slutin n the cnductivity and hence the reactivity f the slutin. (b) Use Table 16.2 t write the systems in the rder in which they appear in the electrchemical series and hence determine which f the materials wuld be mre reactive than irn. Slutin (a) The presence f ther ins frm the sdium chlride in either sea water r during the winter mnths n rads increases the number f ins in slutin. The increased inic strength f the slutin increases cnductivity and allws a higher rate f reactin. (b) Cnsidering the standard reductin ptentials in Table 16.2 E Cu 2+ /Cu > E Sn 2+ /Sn > E Fe 2+ /Fe > E Zn 2+ /Zn > E Na + /Na Thus, since bth zinc and sdium are belw irn in the electrchemical series, either culd be used as a sacrificial electrde. In practice, the highly reactive nature f sdium makes its use difficult, s that zinc is mst cmmnly used. Bx 16.5 Bielectrchemistry: nerve cells and in channels (n p. 754 in Chemistry 3 ) (a) The intracellular cncentratin f Cl ins is 7 mml dm 3. If the membrane ptential due t Cl is +76 mv, calculate the extracellular cncentratin f Cl. (b) Calculate the Gibbs energy changes fr the transprt f K + ins and Na + ins Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin against the membrane ptentials. Cmpare these with the energy available frm the hydrlysis f ATP described in Bx 14.8. (a) Rearrange the special frm f the Nernst equatin fr cncentratin cells, Equatin 16.18 and use it t determine the unknwn cncentratin. Remember t use the physilgical temperature, 37 C = 310 K. (b) Use the membrane ptentials fr transprt f the ins acrss the cell membranes in Equatin 16.14. Slutin (a) Fr a cncentratin cell such as this, the standard cell ptential, which is the cell ptential when the cell is at equilibrium and the cncentratins acrss the membrane are equal, is therefre reduces t Equatin 16.18 s that = 0 V. The Nernst equatin, Equatin 16.17, ΔE Cl = RT zf ln [Cl ] utside [Cl ] inside and therefre Thus ln [Cl ] utside [Cl ] inside ln([cl ] utside /c ) = zfδe Cl RT = ln([cl ] utside /c ) (ln[cl ] inside /c ) = + ln([cl ] inside /c ) = 1 96485 C ml 1 +76.0 10 3 V 8.3145 J K 1 ml 1 310 K = 2.11 zfδe Cl RT + ln(7 10 3 ) [Cl ] utside = e 2.11 c = 0.120 ml dm 3 = 120 mml dm 3 (b) Using Equatin 16.4 n p.725, and ΔG K + = zfδe K + = 1 96485 C ml 1 ( 0.090 V) = +8.7 10 3 J ml 1 = +8.7 kj ml 1 Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin ΔG Na + = zfδe Na + = 1 96485 C ml 1 (+0.062 V) = 6.0 10 3 J ml 1 = 6.0 kj ml 1 Bx 16.6 Electrlysis and rechargeable batteries (n p. 759 in Chemistry 3 ) (a) Draw cell diagrams fr (i) the lead-acid battery and (ii) the nickelmetal hydride battery. (b) What is the balanced equatin fr the verall cell reactin fr the nickelmetal hydride battery? Use the principles utlines n p. 734 t write the chemical reactins as electrchemical cell diagrams. Slutin (a) The reactin at the ande, which, by cnventin, is the left-hand cell Pb (s) + SO4 2 (aq) PbSO4 (s) + 2 e and at the cathde, which frms the right-hand cell PbO2 (s) + 4 H + (aq) + SO4 2 (aq) +2 e PbSO4 (s) + 2 H2O (l) s that the cell may be written PbO2 (s) H2SO4 (aq) Pb (s) PbSO4 (s) with the H + (aq) and SO4 2 (aq) ins cmbined as H2SO4 (aq). In the same way, fr the nickel metal hydride system, the reactin at the ande is OH (aq) + MH H2O (l) + M (s) +e and at the cathde NiO(OH) (s) + H2O (l) + e Ni(OH)2 (s) + OH (aq) s that the cmbined cell may be written MH (s), M (s) OH (aq) OH (aq) Ni (s), NiO(OH) (s), Ni(OH)2 (s) Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin (b) Cmbing the tw half reactins fr the nickel nickel metal hydride cell int a single chemical equatin gives NiO(OH) (s) + MH Ni(OH)2 (s) + M (s) Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin Answers t end f chapter questins (Yu will need t use standard reductin ptentials, E, frm Table 16.2 n p.738) 1. (a) Use the data in Table 16.1 t calculate the limiting mlar cnductivities f magnesium sulfate and sdium carbnate in water. (b) Estimate the cnductivity f a slutin f magnesium sulfate with a cncentratin f 1 10-5 ml dm -3. (Sectin 16.2) Frm the chemical frmulae f the cmpunds, yu can wrk ut hw many f each type f in frms when it disslves in water and multiply these by the values f taken frm Table 16.1. Slutin (a) Find the number f ins frmed when the salt disslves. Magnesium sulfate: MgSO4 frms ne Mg 2+ (aq) catin and ne SO4 2 (aq) anin. + =1; - = 1. Sdium carbnate: Na2CO3 frms tw Na + (aq) catins and ne CO3 2 (aq) anin. + = 2; - = 1. Multiply by the inic cnductivities frm Table 16.1 m = + + + - - MgSO4: m = (1 10.6 ms m 2 ml -1 ) + (1 13.9 ms m 2 ml -1 ) m = 24.5 ms m 2 ml -1 Na2CO3 : m = (2 5.0 ms m 2 ml -1 ) + (1 17.0 ms m 2 ml -1 ) m = 27.0 ms m 2 ml -1 (b) Calculate the cnductivity using Equatin 16.7. The slutin is dilute s that it may be assumed that the mlar cnductivity is given by its limiting infinite dilutin value. Λ m s that cnductivity, = m c c Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin SI units are needed s the cncentratin must be cnverted using 1 ml dm -3 = 1 10 3 ml m -3. = m c = 24.5 ms m 2 ml -1 1 10-2 ml m -3 = 24.5 10-2 ms m -1 = 0.245 S m -1 2. The cnductivity f a saturated slutin f silver chlride at 25 C is 1.89 10-4 S m -1. Given that the mlar cnductivities at infinite dilutin f KCl(aq), KNO3(aq) and AgNO3(aq) are 15.0 ms m 2 ml -1, 14.6 ms m 2 ml -1 and 13.4 ms m 2 ml -1, calculate the slubility prduct f AgCl in water at 25 C. (Sectin 16.2) AgCl is sparingly sluble, s that yu can assume that the mlar cnductivity is given by its value at infinite dilutin. AgCl = AgCl. Frm the mlar cnductivity yu can find the cncentratin f Ag + and Cl, then use these t find the slubility prduct. Slutin KCl = K+ + Cl- = 15.0 ms m 2 ml -1 KNO3 = K+ + NO3- = 14.6 ms m 2 ml -1 AgNO3 = Ag+ + NO3- = 13.4 ms m 2 ml -1 Yu need the value fr AgCl. This can be fund by eliminating the values fr K + and NO3 -. (1) + (3) (2): (K+ + Cl-) + (Ag+ + NO3-) (K+ + NO3-) = (Ag+ Cl-) = AgCl AgCl = (15.0 + 13.4 14.6) ms m 2 ml -1 = 13.8 ms m 2 ml -1 Yu can find the cncentratin f the saturated slutin f AgCl frm the cnductivity,. Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin Λ m r c c Λ m 4-1 1.8910 S m c -3 Λ 13.810 S m ml m 2-1 c = 0.0137 ml m -3 = 1.37 10-5 ml dm -3. This is the cncentratin f bth Ag + and Cl. K a a [Ag ]/1 ml dm [Cl ] / 1 ml dm = (1.37 10 ) s + - Ag Cl + -3 - -3-5 2 Ks = (1.37 x 10-5 ) 2 = 1.88 10-10 3. Cnsider the cell Al(s) Al 3+ (aq) Au 3+ (aq) Au(s) (a) At which electrde des reductin ccur? (b) Which electrde is the ande? (c) Which electrde will lse mass in the cell reactin? (d) T which electrde will catins migrate? (e) Which substance is acting as a reducing agent? (f) Write the half reactin fr the gld half cell. (g) Calculate a value fr E cell. (Sectin 16.3) Apply the cell cnventins given. Slutin (a) Reductin, i.e. gain f electrns, ccurs at the cathde which is the right-hand side f the cell diagram, which in this case is the Au 3+ (aq) / Au (s) system. (b) The left-hand electrde which in this example is the Al 3+ (aq) / Al (s) system is, by cnventin, the ande, which is where xidatin ccurs. Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin (c) At the ande, aluminium metal, Al (s) is cnverted t aqueus aluminium (III) ins, Al 3+ (aq), and s lses mass. In cntrast, at the cathde, gld ins frm slutin, Au 3+ (aq), are depsited at the electrde t frm mre gld metal, Au (s). (d) Catins migrate tward the cathde i.e. t the gld electrde. (e) The gld ins, Au 3+ (aq) gain electrns t becme gld metal, Au (s) and s being reduced Au 3+ (aq) + 3 e Au (s) The slid aluminium, Al (s), which lses the electrns is therefre acting as a reducing agent. (f) (g) Using Equatins 16.10 and 16.11 = E cathde Al (s) Al 3+ (aq) + 3 e Au 3+ (aq) + 3 e Au (s) E ande = +1.50 V ( 1.66 V) = +3.16 V = E R E L = E Au 3+,Au E Al 3+,Al 4. Draw cell diagrams fr electrchemical cells that use the fllwing reactins: (a) Cd(s) + Sn 2+ (aq) ; Sn(s) + Cd 2+ (aq) (b) H2 (g) + O2 (g) ; H2O2(aq) (c) Br2 (aq) + Sn 2+ (aq) ; Sn 4+ (aq) + 2Br - (aq) (d) Cu 2+ (aq) + 2Ag(s) + 2Br - (aq) ; Cu(s) + 2AgBr(s) (e) MnO4 - (aq) + 8H + (aq) + 5Fe 2+ (aq) ; Mn 2+ (aq) + 4H2O(l) + 5Fe 3+ (aq) (Sectin 16.3) Write the tw half cell reactins and use Table 16.2 t identify the xidatin and reductin reactins. Remember that the half cell with the mst negative standard reductin ptential will be the xidatin prcess and that with the mst psitive standard reductin ptential the reductin prcess. Write the xidatin Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin reactin, which frms the ande, as the left-hand cell, and the reductin reactins, which is the cathde as the right-hand cell. Slutin (a) The tw half-cell reactins are Cd (s) + Sn 2+ (aq) Sn (s) + Cd 2+ (aq) Sn 2+ (aq) + 2 e Cd (s) Cd 2+ (aq) + 2 e Sn (s) Cnsulting Table 16.2, E Sn 2+ /Sn> E Cd 2+ /Cd s that the cadmium system, Cd (s) Cd 2+ (aq), must be the ande, where xidatin ccurs, and the tin system, Sn 2+ (aq) Sn (s), the cathde, where reductin ccurs. Putting the tw halves tgether, jined by a salt bridge, gives (b) In the same way fr then the tw half cell reactins are s that, because E H + /H 2 Cd (s) Cd 2+ (aq) Sn 2+ (aq) Sn (s) H2 (g) + O2 (g) H2O2 (aq) 2 H2O (l) 2 H + (aq) + 2 e (aq) 2 H + (aq) + 2 e H2 (g) > E H + /H 2 O 2, s that the Pt H2O (l), H + (aq), H2O2 (aq) system frms the ande, and the 2 H + (aq) H2 (g) Pt the cathde. Cmbining then gives (c) Fr s that the tw half cells are and, because E Br2 /Br Pt H + (aq), H2O2 (aq) H2 (g) Pt Br2 (l) + Sn 2+ (aq) Sn 4+ (aq) + 2 Br (aq) Sn 2+ (aq) Sn 4+ (aq) + 2 e Br2 (l) + 2 e 2 Br (aq) > E Sn 2+ /Sn, then the Pt Sn 2+ (aq), Sn 4+ (aq) system must be the left-hand ande, where xidatin ccurs and the Br2 (l), Br (aq)pt system the righthand cathde where reductin ccurs. Thus, putting the tw halves tgether, jined by a salt bridge Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin (d) Fr Pt Sn 2+ (aq), Sn 4+ (aq) Br2 (l), Br (aq) Pt Cu 2+ (aq) + 2 Ag (s) + 2 Br (aq) Cu (s) + 2 AgBr (s) then the tw half cells are with, since E Cu 2+ /Cu 2 Ag (s) + 2 Br (aq) 2 AgBr (s) + 2 e Cu 2+ (aq) + 2 e Cu (s) > E AgBr/Ag, the Ag (s) AgBr (s) Br (aq) cell the left-hand ande and the Ag (s) AgBr (s) Br (aq) cell the right-hand cathde. Thus, cmbining the tw hal;f cells using a salt bridge, Ag (s) AgBr (s) Br (aq) Ag (s) AgBr (s) Br (aq) (e) Fr the cell, MnO4 (aq) + 8 H + (aq) + 5 Fe 2+ (aq) Mn 2+ (aq) + 4 H2O (l) + 5 Fe 3+ (aq) then the half cells are MnO4 (aq) + 8 H + (aq) + 5 e Mn 2+ (aq) + 4 H2O (l) Because E MnO4 /Mn 2+ 5 Fe 2+ (aq) 5 Fe 3+ (aq) + 5 e > E Fe 3+ /Fe 2+, then the ande, where xidatin ccurs is the Pt Fe 2+ (aq), Fe 3+ (aq) system and the cathde, where reductin ccurs is MnO4 (aq), H + (aq), Mn 2+ (aq) Pt. Putting the tw halves tgether, jined by a salt bridge, Pt Fe 2+ (aq), Fe 3+ (aq) MnO4 (aq), H + (aq), Mn 2+ (aq) Pt 5. Fr each f the fllwing electrchemical cells: (i) write the half cell reactins; (ii) calculate E cell. (a) Fe(s) Fe 2+ (aq) Zn 2+ (aq) Zn(s) (b) Pt(s) H2(g), H + (aq) Cl - (aq), AgCl(s) Ag(s) (c) Hg(l) Hg2Cl2(s), Cl - (aq) Cl - (aq), AgCl(s) Ag(s) (d) Pt(s) Fe 2+ (aq), Fe 3+ (aq) Sn 4+ (aq), Sn 2+ (aq) Pt(s) Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin (Sectin 16.3) Identify the half-cell reactins and use Table 16.2 t find the apprpriate standard reductin ptential. Calculate the standard ptential fr the cmplete cell by subtracting the standard reductin ptential fr the left-hand cell frm that fr the right-hand cell accrding t Equatin 16.11. Slutin (a) The tw half cells are Left, xidatin, ande Right, reductin, cathde s that, using Equatin 16.11, (b) The tw half cells are Left, xidatin, ande Right, reductin, cathde s that, using Equatin 16.11, Fe (s) Fe 2+ (aq) + 2 e E L = 0.44 V Zn 2+ (aq) + 2 e Zn (s) E R = 0.76 V = E R E L = 0.76 V ( 0.44 V) = 0.32 V (c) The tw half cells are Left, xidatin, ande Right, reductin, cathde s that, using Equatin 16.11, H2 (g) 2 H (aq) + 2 e E L = 0.000 V 2 AgCl (s) + 2 e 2 Ag (s) + 2 Cl (aq) E R = +0.22 V = E R E L = +0.22 V 0 V = +0.22 V 2 Hg (l) + 2 Cl (aq) Hg2Cl2 (s) + 2 e E L = +0.27 V 2 AgCl (s) + 2 e 2 Ag (s) + 2 Cl (aq) E R = +0.22 V = E R E L = +0.22 V 0.27 V = 0.05 V Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin (d) The tw half cells are Left, xidatin, ande Right, reductin, cathde s that, using Equatin 16.11, 2 Fe 2+ (aq) 2 Fe 3+ (aq) + 2 e E L = +0.77 V Sn 4+ (aq) + 2 e Sn 2+ (aq) E R = +0.15 V = E R E L = +0.15 V 0.77 V = 0.62 V 6. Which f the fllwing is the strngest xidising agent? (a) H202 in acid slutin (b) H2O2 in basic slutin (c) MnO4 - in acid slutin (d) MnO4 - in basic slutin (e) CrO4 2- in acid slutin. (Sectin 16.3) The relative ability f a system as an xidising agent is measured by its E value. The mre psitive the E value, the better the xidising agent. Slutin Examine the E values and find the mst psitive value. (a) H2O2 + 2 H + + 2 e 2 H2O (b) H2O2 + 2 e 2 OH - (c) MnO4 - + 8 H + + 5 e Mn 2+ + 4 H2O (d) MnO4 - + 2 H2O + 3 e MnO2 + 4 OH - (e) CrO4 2- + 8 H + + 3 e Cr 3+ + 4 H2O E = 1.77 V E = 0.88 V E = 1.52 V E = 0.59 V E = 1.20 V Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin Frm these data, the strngest xidising agent is H2O2 in acid. 7. Using standard reductin ptentials, explain why cpper metal des nt disslve in 1 ml dm -3 hydrchlric acid (HCl (aq)) but des disslve in 1 ml dm -3 nitric acid (HNO3(aq)). (Sectin 16.3) Yu need t write the half-reactins that might ccur fr the species invlved. Then cnsider their E values and decide which reactins will take place under standard cnditins. Slutin Fr cpper t disslve, the fllwing reactin has t prceed. Cu (s) Cu 2+ (aq) + 2 e E = -E = 0.34 V In hydrchlric acid, the reactin t cnsume the electrns prduced is: 2H + (aq) + 2 e H 2 (g) + 2H 2 O (l) E = 0.00 V The verall cell ptential is therefre E cell = ( 0.34 V) + 0.00 V = 0.034 V The cell ptential is negative s that the reactin is nt spntaneus. Hwever, in nitric acid, an alternative reactin is pssible NO3 (aq) + 4H + + 3 e NO (g) + 2H2O (l) E = + 0.96 V The verall reactin here is: 3Cu (s) + 2NO3 - (aq) + 8H + + 6 e 3Cu 2+ (aq) + 6 e + 2NO (g) + 4H2O (l) Fr which the verall cell ptential is given by: E cell = ( 0.34 V) + 0.96 V = + 0.62 V Here, the cell ptential is psitive s that the reactin is spntaneus. The reductin f NO3 - t NO gas is sufficient t xidise the cpper metal. Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin 8. Use values f standard reductin ptentials, E in Table 16.2, t decide whether the fllwing reactins ccur spntaneusly. (a) Cr 2+ (aq) + Ni(s) Cr(s) + Ni 2+ (aq) (b) Br2(l) + 2 I (aq) 2 Br (aq) + I2(s) (c) Cl2(s) + Sn 2+ (aq) 2 Cl (aq) + Sn 4+ (aq) (d) Al(s) + Au 3+ (aq) Al 3+ (aq) + Au(s) (Sectin 16.3) Identify the tw half-cell reactins frm the chemical equatin. Use Table 16.2 t find the standard reductin ptentials fr the half cells and use Equatin 16.11 t determine the cell ptential, with a psitive standard cell ptential indicating that the reactin as written will be spntaneus. Slutin (a) The reactin Cr 2+ (aq) + Ni (s) Cr (s) + Ni 2+ (aq) may be written as tw half reactins. The first half reactin is an xidatin prcess, s may be designated as the cnventinal ande, whilst the secnd is a reductin and s frms at the cathde. Ni (s) Ni 2+ (aq) + 2 e E ande Cr 2+ (aq) + 2 e Cr (s) E cathde s that the cell ptential fr the cmbined cell is = E cathde E ande = 0.91 V ( 0.25 V) = 0.66 V Since E cell < 0, the reactin will nt be spntaneus, because, accrding t = 0.25 V = 0.91 V Equatin 16.14, a negative cell ptential implies a psitive standard Gibbs energy change fr the reactin. (b) In the same way, may be written as Br2 (l) + 2 I (aq) 2 Br (aq) + I2 (s) Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin with = E cathde 2 I (aq) I2 (s) + 2 e E ande Br2 (l) + 2 e 2 Br (aq) E cathde E ande = +1.08 V 0.54 V = +0.54 V = +0.54 V = +1.08 V The standard cell ptential is psitive, s the reactin is spntaneus as written. (c) The reactin may be written as with Cl2 (s) + Sn 2+ (aq) 2 Cl (aq) + Sn 4+ (aq) = E cathde Sn 2+ (aq) Sn 4+ (aq) + 2 e E ande Cl2 (g) + 2 e 2 Cl (aq) E cathde E ande = +1.36 V 0.15 V = +1.21 V = +0.15 V = +1.36 V The standard cell ptential is psitive, s the reactin is spntaneus as written. (d) The reactin may be written as with = E cathde Al (s) + Au 3+ (aq) Al 3+ (aq) + Au (s) Al (s) Al 3+ (aq) + 3 e E ande Au 3+ (aq) + 3 e Au (s) E cathde E ande = +1.50 V ( 1.66 V) = +3.16 V = 1.66 V = +1.50 V The standard cell ptential is psitive, s the reactin is spntaneus as written. 9. Silver articles smetimes becme tarnished with a black cating f Ag2S. The tarnish can be remved by placing the silverware in an aluminium pan and cvering it with a slutin f an inert electrlyte such as NaCl. Explain the electrchemical basis fr this prcedure. (Sectin 16.3) Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin Cnsider the relative psitins f the Al(s)/Al 3+ (aq) and Ag2S(s)/Ag + (aq) in the electrchemical series. Slutin The Al(s)/Al 3+ (aq) half cell has E cell = 1.66 V s it is quite highly reducing. It can reduce Ag2S t Ag + (aq), fr which the standard electrde ptential is E cell = 0.64 V and s disslve the tarnish. The silver and aluminium establish an electrchemical cell with the electrlyte slutin as the salt bridge. 10. Use data frm Table 16.2 t calculate the standard Gibbs energy change ΔrG, and the thermdynamic equilibrium cnstant, K, at 298 K fr the reactins: (a) Mg (s) + Zn 2+ (aq) Zn (s) + Mg 2+ (aq) (b) Cl2 (g) + 2 I (aq) I2 (s) + 2 Cl (aq) (c) Cr2O7 2 (aq) + 3 Fe(s) + 14 H3O + (aq) 2 Cr 3+ (aq) + 3 Fe 2+ (aq) + 21 H2O(l) (d) 2 Br (aq) + Cl2 (g) Br2 (l) + 2 Cl (aq) (Sectin 16.4) Identify the half reactins and use Table 16.2 t calculate the cell ptential fr the verall reactin. Then use Equatin 16.14 t determine the standard Gibbs energy change and Equatin 16.15 the equilibrium cnstant at 298 K. Slutin (a) The reactin may be written as Mg (s) + Zn 2+ (aq) Zn (s) + Mg 2+ (aq) Mg (s) Mg 2+ (aq) + 2 e E ande Zn 2+ (aq) + 2 e Zn (s) E cathde = 2.37 V = 0.76 V where the Mg 2+ /Mg half cell is the ande, because the magnesium metal lses electrns and s is xidised. Thus, Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin = E cathde E ande = 0.76 V ( 2.37 V) = +1.61 V Using Equatin 16.14 t calculate the standard Gibbs energy change, and nting that the half-cell reactins are tw-electrn prcesses, s that z = 2, Δ r G = zf = 2 96485 C ml 1 +1.61 V = 310.7 10 3 J ml 1 = 310.7 kj ml 1 The equilibrium cnstant is then, frm Equatin 16.15, s that ln K = zf RT E 2 96485 C ml 1 cell = 8.3145 J K 1 ml 1 1.61 V = 125 298 K (b) In the same way, the reactin may be written as K = e 125 = 2.9 10 54 Cl2 (g) + 2 I (aq) I2 (s) + 2 Cl (aq) Cl2 (g) 2 Cl (aq) + 2 e E ande 2 I (aq) + 2 e I2 (s) E cathde = +1.36 V = +0.54 V where the Cl2/Cl half cell is the ande, because the chlride ins lse electrns and s are xidised. Thus, = E cathde E ande = +0.54 V 1.36 V = 0.82 V Using Equatin 16.14 t calculate the standard Gibbs energy change, and nting that the half-cell reactins are tw-electrn prcesses, s that z = 2, Δ r G = zf = 2 96485 C ml 1 0.82 V = +158.2 10 3 J ml 1 = +158.2 kj ml 1 The equilibrium cnstant is then, frm Equatin 16.15, ln K = zf RT E 2 96485 C ml 1 cell = 8.3145 J K 1 ml 1 0.82 V = 64 298 K s that Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin K = e 64 = 1.8 10 28 The psitin f equilibrium therefre favurs the frmatin f Cl2 (g) and I (aq) ins. (c) The reactin Cr2O7 2 (aq) + 3 Fe(s) + 14 H3O + (aq) 2 Cr 3+ (aq) + 3 Fe 2+ (aq) + 21 H2O may be written as 3 Fe(s) 3 Fe 2+ (aq) + 6 e E ande Cr2O7 2 (aq) + 14 H3O + (aq) + 6e 2 Cr 3+ (aq) + 21 H2O (l) E cathde = 0.44 V = +1.33 V where the Fe 2+ /Fe half cell is the ande, because the irn lses electrns and s is xidised. Thus, = E cathde E ande = +1.33 V ( 0.44 V) = +1.77 V Using Equatin 16.14 t calculate the standard Gibbs energy change, and nting that the half-cell reactins are six-electrn prcesses, s that z = 6, Δ r G = zf = 6 96485 C ml 1 1.77 V = 1024.7 10 3 J ml 1 = 1024.7 kj ml 1 The equilibrium cnstant is then, frm Equatin 16.15, s that ln K = zf RT E 6 96485 C ml 1 cell = 8.3145 J K 1 ml 1 1.77 V = 414 298 K K = e 414 > 1 10 100 The equilibrium cnstant is s large that its value cannt be displayed n a standard calculatr. The thermdynamics therefre favur the prductin f prducts. (d) Fr the reactin then the tw half reactins are 2 Br (aq) + Cl2 (g) Br2 (l) + 2 Cl (aq) Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin Cl2 (g) 2 Cl (aq) + 2 e E ande 2 Br (aq) + 2 e Br2 (s) E cathde = +1.36 V = +1.08 V where the Cl2/Cl half cell is the ande, because the chlride ins lse electrns and s are xidised. Thus, = E cathde E ande = +1.08 V 1.36 V = 0.28 V Using Equatin 16.14 t calculate the standard Gibbs energy change, and nting that the half-cell reactins are tw-electrn prcesses, s that z = 2, Δ r G = zf = 2 96485 C ml 1 0.28 V = +54.03 10 3 J ml 1 = +54.03 kj ml 1 The equilibrium cnstant is then, frm Equatin 16.15, s that ln K = zf RT E 2 96485 C ml 1 cell = 8.3145 J K 1 ml 1 0.28 V = 22 298 K K = e 22 = 3.4 10 10 The psitin f equilibrium therefre favurs the frmatin f Cl2 (g) and Br (aq) ins. 11. Calculate E cell fr C (s)c 2+ (aq) Ni 2+ (aq)ni (s) when: (a) [Ni 2+ ] = 1.0 ml dm 3, [C 2+ ] = 0.10 ml dm 3 ; (b) [Ni 2+ ] = 0.01 ml dm 3, [C 2+ ] = 1.0 ml dm 3 ; (Sectin 16.4) Identify the verall and the tw half reactins and calculate the standard cell ptential using the data in Table 16.2. Write the apprpriate Nernst equatin, using Equatin 16.16, and substitute in the values fr the standard cell ptential and cncentratins. Slutin (a) The verall cell reactin is Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin with half reactins C (s) + Ni 2+ (aq) Ni (s) + C 2+ (aq) C (s) C 2 (aq) + 2 e E ande = 0.28 V Ni 2+ (aq) + 2 e Ni (s) E cathde = 0.25 V with = E cathde E ande = 0.25 V ( 0.28 V) = +0.03 V The Nernst equatin is then = RT ln Q zf where Q is the reactin qutient, Q, which was defined in Equatin 15.2. Fr this cell, because the activity f a pure slid is Q = a Ni(s)a C 2+ (aq) = [C2+ (aq)] a Ni 2+ (aq)a C(s) [Ni 2+ (aq)] a J(s) = 1 and that f a cmpnent in dilute slutin is Thus = RT zf ln [C2+ (aq)] [Ni 2+ (aq)] a J(aq) = [J]/c = +0.03 V 8.3145 J K 1 ml 1 298 K 1.0 ml dm 3 2 96485 C ml 1 ln 0.10 ml dm 3 = 0 V (b) Using the same apprach, = RT zf ln [C2+ (aq)] [Ni 2+ (aq)] = +0.03 V 8.3145 J K 1 ml 1 298 K 0.01 ml dm 3 2 96485 C ml 1 ln 1.00 ml dm 3 = +0.089 V Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin 12. Given the fllwing half cell reactins, calculate the slubility prduct (see Bx 15.1, p.699) f silver brmide at 298 K. (Sectin 16.4) AgBr (s) + e Ag (s) + Br (aq) E = +0.07 V Ag + (aq) + e Ag (s) E = +0.80 V Cmbine the tw half-cell reactins t prduce a chemical equatin fr the disslutin f silver brmide. Calculate the standard cell ptential fr the prcess and use Equatin 16.15 t determine the crrespnding equilibrium cnstant, which is the slubility prduct fr silver brmide. Slutin The disslutin f silver brmide AgBr (s) + H2O(l) Ag + (aq) + Br (aq) may be expressed in the frm f tw half-cell reactins Ag (s) Ag + (aq) + e E ande =+0.80 V AgBr (s) + e Ag (s) + Br (aq) E cathde =+0.07 V where the Ag + /Ag system is designated the ande because the silver is xidised as electrns are remved t frm the silver ins. Thus = E cathde E ande = +0.07 V 0.80 V = 0.73 V Then, using Equatin 16.15, t determine the crrespnding equilibrium cnstant fr the reactin, where because the reactins are ne-electrn prcesses, z = 1, s that ln K = zf RT E 1 96485 C ml 1 cell = 8.3145 J K 1 ml 1 0.73 V = 28 298 K K = e 28 = 4.5 10 13 13. Calculate Ecell and the Gibbs energy change fr the fllwing cells: (a) Ag (s) Ag + (aq, 1.0 ml dm 3 ) Cu 2+ (aq, 1.0 ml dm 3 ) Cu (s) Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin (b) Ag (s) Ag + (aq, 0.1 ml dm 3 ) Cu 2+ (aq, 0.1 ml dm 3 ) Cu (s) (c) Ag (s) Ag + (aq, 1.0 ml dm 3 ) Cu 2+ (aq, 0.1 ml dm 3 ) Cu (s) (Sectin 16.4) Write the chemical equatins fr the verall reactin and fr the tw separate half cells. Use Table 16.2 t determine the standard cell ptential and the Nernst equatin, Equatin 16.16, t calculate the cell ptential fr each f the sets f cnditins. Determine the Gibbs energy change using Equatin 16.17. Slutin (a) The verall reactin may be written 2 Ag (s) + Cu 2+ (aq) 2 Ag + (aq) + Cu (s) and may be expressed in the frm f tw half-cell reactins 2 Ag (s) 2 Ag + (aq) + 2 e E ande =+0.80 V Cu 2+ (aq) + 2 e Cu (s) E cathde =+0.34 V where the Ag + /Ag system is designated the ande because the silver is xidised as electrns are remved t frm the silver ins. Thus = E cathde E ande The Nernst equatin, Equatin 16.16 has the frm = = +0.34 V 0.80 V = 0.46 V RT ln Q zf where Q is the reactin qutient, Q, which was defined in Equatin 15.2. Fr this cell, because the activity f a pure slid is Q = a 2 Cu(s)a Ag + (aq) 2 = [Ag+ (aq)] 2 a Cu 2+ (aq)a Ag(s) [Cu 2+ (aq)] a J(s) = 1 and that f a cmpnent in dilute slutin is a J(aq) = [J]/c Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin Thus = RT zf ln [Ag+ (aq)] 2 [Cu 2+ (aq)] Since the cncentratin f the aqueus slutins is 1.0 ml dm 3 in each case, = The reactin invlves the transfer f 2 electrns, s z = 2, s, using Equatin 16.14 t calculate the Gibbs energy change, Δ r G = zf = 2 96485 C ml 1 0.46 V = +89 10 3 J ml 1 = +89 kj ml 1 (b) If [Ag + (aq)] = [Cu 2+ (aq)] = 0.1 ml dm 3 then and s = RT zf ln [Ag+ (aq)] 2 [Cu 2+ (aq)] = 0.46 V 8.3145 J K 1 ml 1 298 K 2 96485 C ml 1 ln (0.1 ml dm 3 ) 2 0.1 ml dm 3 = 0.43 V Δ r G = zf = 2 96485 C ml 1 0.43 V = +83 10 3 J ml 1 = +83 kj ml 1 (c) If [Ag + (aq)] = [Cu 2+ (aq)] = 0.1 ml dm 3 then and s = RT zf ln [Ag+ (aq)] 2 [Cu 2+ (aq)] = 0.46 V 8.3145 J K 1 ml 1 298 K 2 96485 C ml 1 ln (1.0 ml dm 3 ) 2 0.1 ml dm 3 = 0.49 V Δ r G = zf = 2 96485 C ml 1 0.49 V = +95 10 3 J ml 1 = +95 kj ml 1 Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin 14. Use the fllwing standard reductin ptentials t calculate the equilibrium cnstant fr the frmatin f the Zn(NH3)4 2+ in. (Sectin 16.4) Zn(NH3)4 2+ (aq) + 2 e Zn(s) + 4NH3 (aq) Zn 2+ (aq) + 2 e Zn(s) E = -1.04 V E = -0.76 V Calculate E cell fr the reactin and then use equatin 16.15 t calculate the equilibrium cnstant K. Slutin The reactin f interest is: Zn 2+ (aq) + 4NH3 (aq) Zn(NH3)4 2+ (aq) This can be btained by reversing the first half-reactin and adding it t the secnd. Zn(s) + 4NH3 (aq) + Zn 2+ (aq) + 2 e Zn(s) + Zn(NH3)4 2+ (aq) + 2 e E cell = +1.04 V 0.76 V = +0.28 V Using nf ln K E RT cell -1 K 2 96485 C ml -1-1 ln 0.28 V 8.314 J K ml 298 K ln K = 21.8 K = e 21.8 = 2.9 10 9 15. Calculate the standard electrde ptential fr a cell in which the reactin frms 99.99% prducts at equilibrium under standard cnditins at 298 K. Assume z = 1 fr the reactin. (Sectin 16.4) Use the given prprtins f prducts and reactants t find the value fr K. Then use Equatin.16.15 t find the crrespnding value fr E. H i g h e r E d u c a t i n Oxfrd University Press, 2017. All rights reserved.
Slutins manual fr Burrws et.al. Chemistry 3 Third editin Slutin If the reactin frms 99.99% prducts, there must be 0.01% reactants at equilibrium, s the equilibrium cnstant, K, is given by: K = [prducts] [reactants] = 99.99 0.01 = 9999 Using this value in Equatin.16.15, the standard electrde ptential can be calculated as fllws: nf ln K E RT cell E = (8.314 J ml -1 K -1 x 298 K) ln 9999 / (1 x 96485 C ml -1 ) E = +0.236 V 16. A student measured the emf f the fllwing electrchemical cell at 25 C. Cu(s) CuSO4(aq) (0.050 ml dm -3 ) CuSO4(aq) (0.500 ml dm -3 ) Cu(s) The student cnnected a piece f cpper wire between the electrdes and left the experiment t g fr lunch. Sme time later the student remved the wire and repeated the emf measurement, recrding a value f +0.027 V. (a) Write the reactins which take place at each electrde (b) Calculate the emf recrded during the first measurement (c) Describe briefly what happened in the cell ver lunch (d) Calculate the cncentratin f cpper sulfate in each cell cmpartment after lunch. (Sectin 16.4) Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin Frm the half cell reactins, use the Nernst equatin t calculate emf. Cnsider what happens t the cncentratin f cpper ins when the cells are cnnected, and use the Nernst equatin again t calculate the change in cncentratin f each half cell. Slutin (a) Right hand side: Cu 2+ (aq) + 2 e Cu(s) Left hand side: Cu(s) Cu 2+ (aq) + 2 e (b) Using the Nernst equatin E cell E cell RT ln a a zf a a Cu(s) 2+ Cu (aq), LHS Cu(s) 2+ Cu (aq), RHS E 296485 C ml (1) 0.50-1 -1 8.314 J K ml 298 K (1) 0.05 cell 0 ln -1 Ecell = + 0.030 V (c) If the electrdes are cnnected by a wire, a current will flw as the cell discharges. Cpper metal is depsited n the electrde immersed in the mre cncentrated slutin where the cncentratin f cpper ins decreases. In the ther half-cell, cpper metal disslves and the cncentratin f cpper ins in slutin increases. (d) The stichimetry f the reactin means that the change in cncentratin in each f the half-cells must be the same. Let the change in cncentratin be c ml dm -3. Frm the Nernst equatin, E cell E cell RT ln a a zf a a Cu(s) 2+ Cu (aq), LHS Cu(s) 2+ Cu (aq), RHS RT (0.05 c) 0.027 V 0 ln zf (0.50 c ) (0.05 c) 0.027 V 0.013ln (0.50 c ) (0.05 c) 0.027 exp 0.125 (0.50 c) 0.013 Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin (0.05 c) (0.50 c) 0.125 (0.05 c) 0.125 (0.50 c) (0.05 c) = (0.0625-0.125 c) 1.125 c = 0.0125 c = 0.011 ml dm -3 Hence, Left hand side: [Cu 2+ (aq)] = (0.05 + 0.011) ml dm -3 = 0.06 ml dm -3 Right hand side: [Cu 2+ (aq)] = (0.50-0.011) ml dm -3 = 0.49 ml dm -3 17. Electrlysis f a mlten chrmium salt fr 1.5 h with a 5.00 A current depsited 4.835 g f chrmium metal. Find the charge n the chrmium in in the salt. (Sectin 16.5) Frm the mass f Cr depsited, the number f mles can be calculated. The current and amunt f time can be used t find the ttal amunt f charge passed. The charge n the Cr n+ in can be fund by cmparing these values. Slutin Find the number f mles f Cr 4.835 g 4.835 g f Cr = -1 51.996 g ml = 0.093 ml Find the number f mles f electrns that flw Amunt f charge passed = 5.00 C s -1 1.5 h 60 min h -1 60 s min -1 = 27000 C 1 ml f electrns carries 1 F. 1 F = 96485 C 27000 C 27000 C therefre is equivalent t -1 96485 C ml = 0.280 ml Hence, 0.280 ml f electrns gave rise t 0.093 ml f Cr Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin Each ml f Cr therefre required 0.280 ml 0.093 ml = 3 electrns. Hence, salt must cntain Cr 3+ ins. Cr 3+ + 3 e Cr 18. A Leclanche cell (see Sectin 16.5) prduces a current f 0.002 A. If the battery cntained 2.5 g f MnO2, hw lng will the battery last under these cnditins? (Sectin 16.5) Use the half equatins fr the reactin t calculate the number f mles f electrns prduced during the reactin f 2.5 g MnO2. Wrk ut the charge carried by this many electrns. Slutin The number f mles f MnO2 is given by: 2.5 g = 0.0288 ml 86.937 g ml 1 Accrding t the half equatin fr the reactin: Mn 2+ + 2 e Mn(s) it can be seen that 2 mles f electrns are used per mle f MnO2. Therefre, the amunt f electrns passed in the reactin is: 2 0.0288 ml = 0.0575 ml One mle f electrns carries 96485 C, therefre 0.0575 ml is equivalent t: 0.0575 ml 96485 C ml -1 = 5548 C 1 A = 1 C s -1, therefre the time needed t pass 5548 C f charge is 5545 C 0.002 A = 2.8 106 s (32 days) Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n
Slutins manual fr Burrws et.al. Chemistry 3 Third editin 19. Aluminium is manufactured by the electrlysis f mlten aluminium xide, Al2O3 (l). If an aluminium plant prduces 1000 tnnes f aluminium every 24 hurs, what ttal electric current des it need? (1 tnne = 1000 kg) (Sectin 16.5) Calculate the number f mles f aluminium prduced, and hence the number f mles f electrns needed fr this prcess. Wrk ut the charge passed using Faraday s cnstant. Slutin The number f mles f aluminium is the mass f aluminium in grams divided by its relative mass. Frm the half equatin: 1000 tnnes = 1 10 6 kg = 1 10 9 g 1 10 9 g 26.981 g ml 1 = 3.706 107 ml Al Al 3+ + 3 e Al it can be seen that 3 mles f electrns are needed t prduce ne mle f aluminium. Therefre the number f mles f electrns passed is: 3.706 10 7 ml 3 = 1.112 10 8 ml One mle f electrns carries 96485 C, therefre 1.112 10 8 ml is equivalent t: 1 A = 1 C s -1. Hence: Current = 1.112 10 8 ml 96485 C ml -1 = 1.073 10 13 C 1.073 10 13 C 24 h 60 min h 1 60 s min 1 = 1.24 108 A Oxfrd University Press, 2017. All rights reserved. H i g h e r E d u c a t i n