Electrochemistry Redox Half-Reactions

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1 Electrchemistry Electrchemistry deals with the relatiship betwee chemical chage ad electricity Electrchemical s (tw types) Galvaic s use a sptaeus ( G < 0) reacti t prduce electricity (batteries) Electrlytic s use a surce f electricity t drive a -sptaeus ( G > 0) reacti (electrlysis) 21.1 Redx Half-Reactis Redx reactis ivlve e - trasfer Oxidati lss f e - (xidati state ) Reducti gai f e - (xidati state ) Half-reactis fcus xidati ad reducti separately Example: Ca(s) Cl 2 (g) CaCl 2 (s) CaCl 2 (s) csists f Ca 2 ad Cl - is Ca(s) Ca 2 (s) 2e - (lss f 2e -, xidati) Cl 2 (g) 2e - 2Cl - (s) (gai f 2e -, reducti) Ca(s) Cl 2 (g) 2e - Ca 2 (s) 2e - 2Cl - (s) Addig the half-reactis gives the verall reacti Ca is xidized (Ca is the reducig aget) Cl 2 is reduced (Cl 2 is the xidizig aget) Geeralized expressis fr half reactis: Red Ox e - r Ox e - Red Ox/Red frm a redx cuple (Ex: Ca 2 /Ca; Cl 2 /Cl - ) Balacig Redx Reactis Half-reacti methd divides the verall reacti it tw half-reactis Balacig i acidic slutis 1. Idetify the redx cuples ad write the halfreactis 2. Balace each half-reacti separately: 1 st, balace all elemets ther tha O ad H 2 d, balace O by addig H 2 O 3 rd, balace H by addig H 4 th, balace the charge by addig e - 3. Multiply the half-reactis by itegers t equal the # f e - i them 4. Add the half-reactis ad cacel the e - Example: Balace the fllwig skelet equati i acidic sluti: V 3 Ce 4 VO 2 Ce 3 1. Redx cuples: VO 2 /V 3 ad Ce 4 /Ce 3 Half-reactis: V 3 VO 2 ad Ce 4 Ce 3 2. V 3 VO 2 (V is balaced) V 3 2H 2 O VO 2 (balace O) V 3 2H 2 O VO 2 4H (balace H) V 3 2H 2 O VO 2 4H 2e - (balace charge) Ce 4 Ce 3 (Ce, O ad H are balaced) Ce 4 1e - Ce 3 (balace charge) 3. Multiply the 2 d half-reacti by 2 t get 2e - 2Ce 4 2e - 2Ce 3 4. Add the half-reactis V 3 2H 2 O VO 2 4H 2e - 2Ce 4 2e - 2Ce 3 V 3 2H 2 O 2Ce 4 2e - VO 2 4H 2e - 2Ce 3 V 3 2Ce 4 2H 2 O VO 2 2Ce 3 4H Nte: If H 3 O is required i the equati istead f H, add as may water mlecules bth sides as the # f H is V 3 2Ce 4 6H 2 O VO 2 2Ce 3 4H 3 O Balacig i basic slutis The same fur steps are used plus a fifth step: 5. Add OH - bth sides f the equati i rder t eutralize the H, ad cacel the water mlecules if ecessary Example: Balace the fllwig skelet equati i basic sluti: CrO BrO 4- CrO 4 BrO 3-1. Redx cuples: CrO 4 /CrO ad BrO 4- /BrO 3 - Half-reactis: CrO CrO 4 ad BrO 4- BrO 3-2. CrO CrO 4 (Cr is balaced) CrO 2H 2 O CrO 4 (balace O) CrO 2H 2 O CrO 4 4H (balace H) CrO 2H 2 O CrO 4 4H 3e - (balace charge) BrO 4- BrO - 3 (Br is balaced) BrO 4- BrO 3- H 2 O (balace O) BrO 4-2H BrO 3- H 2 O (balace H) BrO 4-2H 2e - BrO 3- H 2 O (balace charge)

2 3. Multiply the 1 st half-reacti by 2 ad the 2 d by 3 t get 6e - i bth half-reactis 2CrO 4H 2 O 2CrO 4 8H 6e - 3BrO 4-6H 6e - 3BrO 3-3H 2 O 4. Add the half-reactis 2CrO 4H 2 O 3BrO 4-6H 6e - 2CrO 4 8H 6e - 3BrO 3-3H 2 O 2CrO 3BrO 4- H 2 O 2CrO 4 3BrO 3-2H 5. Add 2OH - bth sides f the equati 2H 2CrO 3BrO 4- H 2 O 2OH - 2 O 2CrO 4 3BrO 3-2H 2OH - 2CrO 3BrO 4-2OH - 2CrO 4 3BrO 3- H 2 O 21.2 Galvaic (Vltaic) Cells Prduce electricity frm a sptaeus chemical reacti Example: Z metal reacts sptaeusly with Cu 2 slutis t yield metallic Cu ad Z 2 is Z(s) Cu 2 Z 2 Cu(s) (SO 4 cuter is) The tw half-reactis are: Z(s) Z 2 2e - (xidati) Cu 2 2e - Cu(s) (reducti) The tw half-reactis ca be physically separated by placig them i separate ctaiers (half-s) Half-s where the half-reactis ccur Ade half- where xidati ccurs Cathde half- where reducti ccurs Electrdes i ctact with the electrlyte slutis ad the exteral electrical circuit Ade (xidati) Cathde (reducti) I vltaic s, the ade is (-) ad the cathde is () The e - s flw frm the ade tward the cathde Salt bridge cmpletes the electrical circuit ad maitais electrical eutrality f the half-s (prus material saked i a ccetrated electrlyte sluti) Ais i the salt bridge flw tward the ade Catis i the salt bridge flw tward the cathde By cveti, the ade half- appears the left Galvaic Cell Ntati Half- tati Differet phases are separated by vertical lies Species i the same phase are separated by cmmas Types f electrdes Active electrdes ivlved i the electrde half-reacti (mst metal electrdes) Example: Z 2 /Z metal electrde Z(s) Z 2 2e - (as xidati) Ntati: Z(s) Z 2 Iactive (iert) electrdes t ivlved i the electrde half-reacti (iert slid cductrs; serve as a ctact betwee the sluti ad the exteral el. circuit) Example: Pt electrde i Fe 3 /Fe 2 sl. Fe 3 e - Fe 2 (as reducti) Ntati: Fe 3, Fe 2 Pt(s) Electrdes ivlvig metals ad their slightly sluble salts Example: Ag/AgCl electrde AgCl(s) e - Ag(s) Cl - (as reducti) Ntati: Cl - AgCl(s) Ag(s)

3 Electrdes ivlvig gases a gas is bubbled ver a iert electrde Example: H 2 gas ver Pt electrde H 2 (g) 2H 2e - (as xidati) Ntati: Pt(s) H 2 (g) H Example: A cmbiati f the Z(s) Z 2 ad Fe 3, Fe 2 Pt(s) half-s leads t: Cell tati The ade half- is writte the left f the cathde half- The electrdes appear the far left (ade) ad far right (cathde) f the tati Salt bridges are represeted by duble vertical lies Z(s) Z 2 2e - (ade, xidati) Fe 3 e - Fe 2 ( 2) (cathde, reducti) Z(s) 2Fe 3 Z 2 2Fe 2 Z(s) Z 2 Fe 3, Fe 2 Pt(s) Example: A cmbiati f the Pt(s) H 2 (g) H ad Cl - AgCl(s) Ag(s) half-s leads t: Nte: The reactats i the verall reacti are i differet phases ( physical ctact) eed f a salt bridge H 2 (g) 2H 2e - (ade, xidati) AgCl(s) e - Ag(s) Cl - ( 2) (cathde, reducti) 2AgCl(s) H 2 (g) 2Ag(s) 2H 2Cl - Pt(s) H 2 (g) H,Cl - AgCl(s) Ag(s) Example: Write the reacti ad the tati fr a csistig f a graphite cathde immersed i a acidic sluti f MO 4- ad M 2 ad a graphite ade immersed i a sluti f S 4 ad S 2. Write the half reactis (a list f the mst cmm half-reactis is give i Appedix D) S 2 S 4 2e - 5 (xidati) MO 4-8H 5e - M 2 4H 2 O(l) 2 (reducti) 5S 2 2MO 4-16H 10e - 5S 4 10e - 2M 2 8H 2 O(l) The graphite (C) electrdes are iactive C(s) S 2,S 4 H,MO 4-,M 2 C(s) Why D Galvaic Cells Wrk? Csider a made f tw active metal electrdes, M 1 ad M 2, ad their is. If the circuit is pe, the tw metals are i equilibrium with their is 1) M 1 M 1 e - 2) M 2 M 2 e - The prduced electrs accumulate i the metal electrdes ad prduce electrical ptetials If M 1 has a greater tedecy t give ut its electrs, the 1 st equilibrium is shifted further t the right ad the ptetial f M 1 is mre egative Whe the circuit is clsed, electrs flw frm the mre egative M 1 (ade) tward the less egative M 2 (cathde) 21.3 Cell Ptetials Electrmtive frce (emf) drives the electrs i the el. circuit emf is the differece betwee the electrical ptetials f the tw electrdes (vltage) Cell ptetial (E ) E = emf Uits vlts (V) (1 V = 1 J/C sice the electrical wrk is equal t the applied vltage times the charge mvig betwee the electrdes) Stadard ptetial (E ) the ptetial at stadard-state cditis (gases 1 atm, slutis 1 M, liquids & slids pure)

4 E is measured with a vltmeter If the () termial f the vltmeter is cected t the () electrde (cathde), the vltmeter shws a psitive readig E characterizes the verall reacti If E > 0, the reacti is sptaeus If E < 0, the reacti is -sptaeus If E = 0, the reacti is at equilibrium Example: Z(s) Z 2 (1M) Cu 2 (1M) Cu(s) 1.10 V Z(s) Cu 2 Z 2 Cu(s) E = 1.10 V > 0 sptaeus reacti Electrde ptetials (E) characterize the idividual electrdes (half-reactis) The ptetial is the differece betwee the electrde ptetials f the cathde ad ade E = E cathde E ade Stadard electrde ptetials (E ) electrde ptetials at the stadard-state E = E cathde E ade E values are reprted fr the half-reacti writte as reducti (stadard reducti ptetials) listed i Appedix D Abslute values fr E ad E ca t be measured A referece electrde (half-) is eeded The ptetials f all electrdes are measured relative t the referece electrde Stadard hydrge electrde used as a referece electrde E ref = 0 V (assumed) H (1M) H 2 (g, 1atm) Pt(s) 2H (1M) 2e - H 2 (g, 1atm) T fid the ptetial f ay electrde, a is cstructed betwee the ukw electrde ad the referece electrde The ptetial is directly related t the ukw electrde ptetial If the ukw electrde is the cathde f the E = E uk E ref E uk = E E ref = E 0= E > 0 If the ukw electrde is the ade f the E = E ref E uk E uk = E ref E = 0 E = E < 0 Example: Pt(s) H 2 (g, 1atm) H (1M),Cl - (1M) AgCl(s) Ag(s) H /H 2 ade Ag/AgCl cathde E = E Ag/AgCl E ref = E Ag/AgCl E Ag/AgCl = 0.22 V Determiati f Electrde Ptetials Electrde ptetials ca be determied by measuremets versus the stadard H-electrde r ther electrdes with kw ptetials Example: E = 0.46 V fr the reacti: Cu(s) 2Ag Cu 2 2Ag(s) If E = 0.34 V fr the Cu 2 /Cu redx cuple, what is E fr the Ag /Ag redx cuple? Split it half-reactis: Cu(s) Cu 2 2e - E Cu = 0.34 V (ade, x) Ag e - Ag(s) E Ag =??? V (cathde, red) E = E Ag E Cu = E Ag (0.34) = 0.46 E Ag = 0.46 (0.34) = 0.80 V Usig Cell Ptetials i Calculatis Cell ptetials are additive If tw reactis are added, their ptetials are added t Cell ptetials are itesive prperties remai idepedet f the system size If a reacti (r a half-reacti) is multiplied by a umber, its ptetial remais the same Example: Cu(s) 2Ag Cu 2 2Ag(s) ( 3) E = 0.46 V 3Ag(s) Au 3 3Ag Au(s) ( 2) E = 0.70 V 3Cu(s) 2Au 3 3Cu 2 2Au(s) E = = 1.16 V

5 Stregths f Oxidizig ad Reducig Agets E values are always tabulated fr reducti Ox e - Red (E ) Ox is a xidizig aget; Red is a reducig aget E is a measure fr the tedecy f the half-reacti t uderg reducti Higher (mre psitive) E meas Greater tedecy fr reducti Lwer tedecy fr xidati Higher (mre psitive) E meas Strger xidizig aget (Ox) Ox is reduced Weaker reducig aget (Red) Red is xidized Electrchemical series a arragemet f the redx cuples i rder f decreasig reducti ptetials (E ) Appedix D The mst psitive E s are at the tp f the table The mst egative E s are at the bttm f the table The strgest xidizig agets (Ox) are at the tp f the table as reactats The strgest reducig agets (Red) are at the bttm f the table as prducts Every redx reacti is a sum f tw half-reactis, e ccurrig as xidati ad ather as reducti Red 1 Ox 1 e - Ox 2 e - Red 2 Red 1 Ox 2 Ox 1 Red 2 I a sptaeus redx reacti, the strger xidizig ad reducig agets react t prduce the weaker xidizig ad reducig agets Strger Red 1 Strger Ox 2 Weaker Ox 1 Weaker Red 2 Example: Give the fllwig half-reactis: Cl 2 (g) 2e - 2Cl - E = 1.36 V O 2 (g) 4H 4e - 2H 2 O(l) E = 1.23 V Fe 3 e - Fe 2 E = 0.77 V Fe 2 2e - Fe(s) E = 0.44 V a) Rak the xidizig ad reducig agets by stregth Ox agets the left; Red agets the right Oxidizig (Tp) Cl 2 > O 2 > Fe 3 > Fe 2 (Bttm) Reducig (Bttm) Fe > Fe 2 > H 2 O > Cl - (Tp) Appedix D b) Ca Cl 2 xidize H 2 O t O 2 i acidic sluti? Cl 2 /Cl - has higher E (Cl 2 /Cl - is abve O 2,H /H 2 O) Cl 2 is a strger xidizig aget tha O 2 Cl 2 ca xidize H 2 O t O 2 at stadard cditis c) Write the sptaeus reacti betwee the Cl 2 /Cl - ad Fe 3 /Fe 2 redx cuples ad calculate its E Cl 2 /Cl - has the higher reducti ptetial (E ) Cl 2 /Cl - uderges reducti Fe 3 /Fe 2 uderges xidati (reverse equati) Cl 2 (g) 2e - 2Cl - (reducti) E = 1.36 V Fe 2 Fe 3 e - 2 (xidati) E = 0.77 V Cl 2 (g) 2e - 2Fe 2 2Cl - 2Fe 3 2e - E = E cath E ad = 1.36 (0.77) = 0.59 V d) Is the reacti f disprprtiati (simultaeus xidati ad reducti) f Fe 2 t Fe 3 ad Fe(s) sptaeus at stadard cditis? Need the sig f E Fe 2 /Fe(s) uderges reducti Fe 3 /Fe 2 uderges xidati (reverse equati) Fe 2 2e - Fe(s) (reducti) E = 0.44 V Fe 2 Fe 3 e - 2 (xidati) E = 0.77 V 3Fe 2 2e - Fe(s) 2Fe 3 2e - E = E cath E ad = 0.44 (0.77) = 1.21 V E < 0 the reacti is -sptaeus at stadard cditis Relative Reactivity f Metals The activity series f metals raks metals based their ability t displace H 2 frm acids r water r displace each ther s is i sluti Metals that ca displace H 2 frm acids The reducti f H frm acids t H 2 is give by the stadard hydrge half-reacti 2H 2e - H 2 (g) E = 0 V I rder fr this half-reacti t prceed as writte, the metal must have lwer reducti ptetial (the metal must be belw H 2 /H i Appedix D) If E metal < 0, the metal ca displace H 2 frm acids If E metal > 0, the metal cat displace H 2

6 Example: Ca Fe ad Cu be disslved i HCl(aq)? Fe 2 /Fe is belw ad Cu 2 /Cu is abve H 2 /H 2H 2e - H 2 (g) (reducti) E = 0.00 V Fe(s) Fe 2 2e - (xidati) E = 0.44 V 2H 2e - Fe(s) H 2 (g) Fe 2 2e - E = E cath E ad = 0.00 ( 0.44) = 0.44 V E > 0 sptaeus (Fe disslves i HCl) 2H 2e - H 2 (g) (reducti) E = 0.00 V Cu(s) Cu 2 2e - (xidati) E = 0.34 V 2H 2e - Cu(s) H 2 (g) Cu 2 2e - E = E cath E ad = 0.00 (0.34) = 0.34 V E < 0 -sptaeus (Cu des t disslve) Metals that ca displace H 2 frm water The reducti f H 2 O t H 2 is give by: 2H 2 O(l) 2e - H 2 (g) 2OH - E = V The value f E is fr ph = 7 (stadard state) Metals that are belw H 2 O/H 2,OH - i Appedix D ca displace H 2 frm water at stadard cditis Metals that have E metal < ca displace H 2 frm water at ph = 7 Example: Ptassium, K, disslves readily i water 2H 2 O(l) 2e - H 2 (g) 2OH - (reducti) E = 0.42 V K(s) K e - 2 (xidati) E = 2.93 V 2H 2 O(l) 2e - 2K(s) H 2 (g) 2OH - 2K 2e - E = 0.42 ( 2.93) = 2.51 V > 0 (sptaeus) 21.4 Free Eergy ad Electrical Wrk Relatiship Betwee E ad G r Electrical wrk (w) w = (charge trasferred) (vltage) # ml e - trasferred F charge f 1 ml e - (charge trasferred) = F (vltage) = E w = FE (w < 0 sice the system des wrk) G is the maximum wrk the system ca d, s G= w max If the prcess is carried ut reversibly (w = w max ) G r = FE ad G r = FE F = C/ml Faraday cstat G r, G r, E, ad E are all depedet T (superscripts, T, are mitted fr simplicity) G r, G r are extesive prperties E, E, E, ad E are itesive prperties If a redx equati is multiplied by a umber, G is als multiplied, but E is t Example: Usig E values frm appedix D, calculate G r at 298 K fr the reacti: 2Cr 3 2Br - 2Cr 2 Br 2 (l) Fid the redx cuples i Appedix D (298 K): Cr 3 e - Cr 2 E = V Br 2 (l) 2e - 2Br - E = 1.06 V Ivert the 2 d half-reacti t match the verall eq. Cr 3 e - Cr 2 2 (reducti) E = 0.41 V 2Br - Br 2 (l) 2e - (xidati) E = 1.06 V 2Cr 3 2e - 2Br - 2Cr 2 Br 2 (l) 2e - Calculate E E = E cath E ad = 0.41 (1.06) = 1.47 V Calculate G r ( = 2 # ml e - i verall eq.) G r = FE = (2 ml) (96485 C/ml) (-1.47 V) G r = C V = J = 284 kj G r > 0 ad E < 0 the reacti is sptaeus at stadard cditis The reverse reacti is sptaeus at stadard cditis Relatiship Betwee E ad K Frm G r = FE ad G r = RT l K FE = RT l K E = RT l K F At 298 K, RT/F = V E = l K RT FE K = e E K = e l K = lg K ad = E E = lg K K = 10.

7 Example: Usig E values frm appedix D, calculate K at 298 K fr the reacti: 2Cr 2 Br 2 (l) 2Cr 3 2Br - This is the reverse f the reacti i the previus example (E = V frm previus example) E reverse = - E frward E = -(-1.47 V) = 1.47 V ad = 2 E K = e = e 0 = E > 0 the reacti is sptaeus at stadard cditis K >> 1 the prducts are favred at equilibrium 49 Iterrelatiship betwee G r, E, ad K Example: Calculate K sp f PbSO 4 at 298 K. PbSO 4 (s) Pb 2 SO 4 K sp =? Nt a redx reacti, but it ca be represeted as a sum f tw redx half-reactis PbSO 4 (s) 2e - Pb(s) SO 4 (reducti) E = V Pb(s) Pb 2 2e - (xidati) E = V PbSO 4 (s) 2e - Pb(s) Pb(s) SO 4 Pb 2 2e - E = E cath E ad = 0.36 ( 0.13) = V K sp = e E 2 ( 0.23) = e = E < 0 the dissluti f PbSO 4 is sptaeus at 298 K (K sp << 1 ) The Effect f Ccetrati E TheNerst equati gives the variati f the ptetial with cmpsiti G r = G r RT l Q G r = FE ad G r = FE FE = FE RT l Q E At 298 K E = E RT lq F = E lq E = E lgq The Nerst equati applies als t half-reactis (E ad E are used istead f E ad E ) Example: Calculate the electrde ptetial f the Cu 2 /Cu redx cuple at 298 K, if the ccetrati f Cu 2 is M. Cu 2 2e - Cu(s) E = 0.34 V Use the Nerst eq. t get E Q = 1/[Cu 2 ] = 1/0.025 ad = E = E lq = l E = = V E i Relati t Q ad K Frm E = E (RT/F)lQ: If Q < 1 (mre reactats), l Q < 0, ad E > E If Q > 1 (mre prducts), l Q > 0, ad E < E If Q = 1 (stadard state), l Q = 0, ad E = E Cmbiig E = E (RT/F)lQ with E = (RT/F)lK leads t: E = (RT/F)lK (RT/F)lQ If Q < K, E > 0 frward reacti is sptaeus If Q > K, E < 0 reverse reacti is sptaeus If Q = K, E = 0 reacti is at equilibrium

8 Ccetrati Cells Ccetrati ctais the same redx cuple i bth the ade ad cathde half-s The ade ad cathde are the same E = E cathde E ade = 0 The ccetratis f the cmpets are differet i the tw half-s E = 0 (RT/F)lQ 0 Example: Cu 2 /Cu ccetrati (E Cu= 0.34 V) Cu 2 (1.0 M) 2e - Cu(s) (cathde,reducti) Cu(s) Cu 2 (0.1 M) 2e - (ade,xidati) Cu 2 (1.0 M) 2e - Cu(s) Cu(s) Cu 2 (0.1 M) 2e - E = E Cu E Cu = 0.34 (0.34) = 0 V E = E lgq = 0 lg E = 0 ( 1) = = V 2 The ctiues t wrk util [Cu 2 ] is equalized i the tw half-s ad E decreases t zer I-selective electrdes have ptetials that are directly related t the ccetrati f specific is such as H (ph), K, F -, Cl -, Br -, Example: Calculate the ph f a sluti i which the ptetial f the H H 2 (g,1atm) Pt electrde is E = V. 2H 2e - H 2 (g,1atm) E = 0.00 V P Q = [H ] E = E H 2 2 1atm = 2 [H ] = lgq = 0 lg 2 [H ] E = lg = lg [H ] 2 2 [H ] 2 ( 2) E = lg [H ] = ph 2 E 0.15 ph = = = The ptetial f a i-selective electrde is measured by cmbiig it i a with a referece electrde havig a well kw ad cstat electrde ptetial such as the calmel (Hg/Hg 2 Cl 2 /Cl - ) r the Ag/AgCl/Cl - electrdes Optially, the i-selective electrde ca be perated i a ccetrati with the same electrde immersed i a sluti with kw ccetrati as a referece 21.5 Galvaic Cells as Batteries Primary s ca t be recharged The battery dies whe the reactats are exhausted Example: The alkalie battery a dry Cathde (reducti): 2MO 2 (s) H 2 O 2e - 2M 2 O 3 (s) 2OH - Ade (xidati): Z(s) 2OH - ZO(s) H 2 O 2e - Overall: 2MO 2 (s) Z(s) 2M 2 O 3 (s) ZO(s) Sice all reactats ad prducts are slids (dry ) Q = 1 E = E RT l 1 = E 1. 5 V F Secdary s ca be recharged The is peridically cverted t a electrlytic i rder t cvert sme f the prducts back t reactats Example: The lead-acid battery E = (-0.356) = V E = E l 2 2 [H2 SO4] As H 2 SO 4 is csumed, E drps The eeds t be recharged

9 Fuel s use cmbusti reactis The must be ctiuusly prvided with fuel ad xyge (flw s) Example: The hydrge fuel E = E cath - E ad E = 1.23 (0.00) 1.2 V PH 2 Q = P P E H 2 = E O 1 / 2 O l Q 2 PH 2 ad PO 2 leads t E 21.6 Crrsi Uwated xidati f metals i the evirmet If the metal (M) is i ctact with water Cathde, reducti: 2H 2 O(l) 2e - H 2 (g) 2OH - E = V (at ph = 7) Ade, xidati: E = V M(s) M e - E < V Ay metal with E < V ca be xidized by H 2 O Cathde, reducti: O 2 (g) 4H 4e - 2H 2 O(l) E = 1.23 V (at ph = 7) Ade, xidati: E= 0.82 V M(s) M e - E < 0.82 V Ay metal with E < 0.82 V ca be xidized by H 2 O i the presece f O 2 At ph < 7 (acid rai, etc.), the reducti ptetials f H 2 O ad O 2 are eve higher Easier xidati f the metal Example: Rustig f ir (Fe E = V) Cathde, reducti: O 2 (g) 4H 4e - 2H 2 O(l) E= 0.82 V (ph = 7) Ade, xidati: Fe(s) Fe 2 2e - ( 2) E = V Overall: 2Fe(s) O 2 (g) 4H 2Fe 2 2H 2 O(l) E = E cath E ad = 0.82 (-0.44 ) = 1.26 V E > 0 sptaeus reacti Rust Further xidati: 2Fe 2 ½O 2 (g) (2)H 2 O(l) Fe 2 O 3 H 2 O(s) 4H Overall rustig prcess: 2Fe(s) O 2 (g) 4H 2Fe 2 2H 2 O(l) 2Fe 2 ½O 2 (g) (2)H 2 O(l) Fe 2 O 3 H 2 O(s) 4H 2Fe(s) 3/2O 2 (g) H 2 O(l) Fe 2 O 3 H 2 O(s) Crrsi is ehaced by acidic cditis ( [H ]) ad by salty slutis (imprved cductivity) Crrsi prtecti Adic prtecti prexidati f the metal by frmati f a thi layer f prtective metal xide Cathdic prtecti cectig the metal t a mre strgly reducig metal with lwer E value called sacrificial ade Example: Prtectig Fe cstructi elemets by cectig them t blcks f Mg r Al (sacrificial ades) E Mg = V E Fe = V Example: Galvaizati f Fe by catig it with Z E Z = V 21.7 Electrlytic Cells ad Electrlysis Electrlytic s use exteral electrical surce t drive a -sptaeus reacti

10 Electrlytic s act i reverse (-sptaeus) directi cmpared t galvaic s E < 0 ad G > 0 (-sptaeus reacti) The ade is psitive ad the cathde is egative There are sme similarities betwee electrlytic ad galvaic s Oxidati is always the ade ad reducti is always the cathde Electrs always flw frm ade tward cathde Electrlysis the passage f electrical curret thrugh a electrlyte by applyig exteral vltage (the prcess i electrlytic s) Electrlysis causes a -sptaeus reacti (fte a splittig f a substace t its elemets) The applied vltage must be greater tha the ptetial f the reverse sptaeus reacti The electrlyte ca be a mlte salt r a aqueus electrlyte sluti Salt bridges are fte t ecessary Durig electrlysis the catis are attracted t the cathde (egative) ad the ais are attracted t the ade (psitive) Predictig the Prducts f Electrlysis The catis () are attracted t the cathde (-) ad the ais (-) are attracted t the ade () Electrlysis f mlte salts used fr idustrial islati f the mst active elemets (Na, Li, Mg, Al, ; F 2, Cl 2, Br 2, ) The cati is reduced at the cathde The ai is xidized at the ade Example: Islati f Na ad Cl 2 by electrlysis f mlte NaCl Na (l) e - Na(l) ( 2) cathde, reducti 2Cl - (l) Cl 2 (g) 2e - ade, xidati 2Na (l) 2Cl - (l) 2Na(l) Cl 2 (g) I the Dws fr prducti f Na, CaCl 2 is added t reduce the meltig pit f NaCl Electrlysis f mixed mlte salts The cati with higher E value (the strger xidizig aget) is reduced at the cathde The ai with lwer E value (the strger reducig aget) is xidized at the ade Nte: E values i appedix D are fr aqueus is ad ca be used ly as apprximate guidace. Istead, EN values ca be used t estimate the strger xidizig ad reducig agets. Example: Predict the prducts f the electrlysis f a mlte mixture f NaCl ad AlF 3 Pssible cathde half-reactis (reducti) 1) Reducti f Na ad 2) Reducti f Al 3 Al 3 is the strger xidizig aget because Al is mre EN tha Na, s Al 3 gais electrs easier Cathde half-reacti: Al 3 (l) 3e - Al(l) Pssible ade half-reactis (xidati) 1) Oxidati f F - ad 2) Oxidati f Cl - Cl - is the strger reducig aget because Cl is less EN tha F, s Cl - lses electrs easier Ade half-reacti: 2Cl - (l) Cl 2 (g) 2e - Al 3 (l) 3e - Al(l) ( 2) cathde, reducti 2Cl - (l) Cl 2 (g) 2e - ( 3) ade, xidati 2Al 3 (l) 6e - 6Cl - (l) 2Al(l) 2Cl 2 (g) 6e - The prducts are Al(l) ad Cl 2 (g)

11 Electrlysis f water Pure water is hard t electrlyze (lw cductivity), s a small amut f a -reactive salt (NaNO 3 ) is added (ca be eglected) H 2 O is reduced at the cathde: 2H 2 O(l) 2e - H 2 (g) 2OH - E = V (at ph = 7) E = V H 2 O is xidized at the ade: 2H 2 O(l) O 2 (g) 4H 4e - E = 1.23 V (at ph = 7) E= 0.82 V 4H 2 O(l) 4e - 2H 2 O(l) 2H 2 (g) 4OH - O 2 (g) 4H 4e - Overall: 2H 2 O(l) 2H 2 (g) O 2 (g) E = E cath E ad = (0.82 ) = V E < 0 -sptaeus reacti T drive the reacti, the exteral vltage must be greater tha 1.24 V Overvltage the extra vltage (i the case f water ver 1.24) eeded t drive the reacti Fr H 2 O mst iert electrdes, the vervltage is 0.4 t 0.6 V per electrde Cathde, reducti: E = V Ade, xidati: E= V Ttal: E = -1.0 (1.4 ) -2.4 V Electrlysis f aqueus slutis Pssible cathde half-reactis (reducti) 1. Reducti f H 2 O 2. Reducti f catis i the sluti Pssible ade half-reactis (xidati) 1. Oxidati f H 2 O 2. Oxidati f active metal electrdes 3. Oxidati f ais i the sluti The half-reacti with the higher E value (havig the strger xidizig aget) ccurs the cathde The half-reacti with the lwer E value (havig the strger reducig aget) ccurs the ade Example: Predict the prducts f the electrlysis f a mixture f 1M NaCl(aq) ad 1M KNO 3 (aq) with iert electrdes at ph = 7. Pssible cathde half-reactis (reducti) 1. Reducti f H 2 O 2H 2 O(l) 2e - H 2 (g) 2OH - E = V with vervltage E -1.0 V 2. Reducti f catis i the sluti Na e - Na(s) E = V K e - K(s) E = V Half-reacti (1) has the highest (mst psitive) E value H 2 O is reduced t H 2 the cathde Pssible ade half-reactis (xidati) 1. Oxidati f H 2 O 2H 2 O(l) O 2 (g) 4H 4e - E= 0.82 V with vervltage E 1.4 V 2. Oxidati f active metal electrdes e 3. Oxidati f ais i the sluti 2Cl - Cl 2 (g) 2e - E = 1.36 V Nte: NO 3- ca t be further xidized (t a prduct i ay half-reacti i appedix D) Half-reacti (3) has the lwest E value Cl - is xidized t Cl 2 the ade Overall: 2H 2 O(l) 2Cl - H 2 (g) 2OH - Cl 2 (g) Nte: Withut the vervltage, H 2 O wuld be xidized t O 2 at the ade The Chlr-alkali fr prducti f Cl 2 ad NaOH

12 Electrlysis f aqueus slutis is fte used fr prducti r purificati f less active elemets Catis f less active metals (Cu, Ag, Au, Pt, ) are reduced the cathde Ais f less active metals (I 2, Br 2, ) are xidized the ade (icludig Cl 2 due t the vervltage f water) Catis f mre active metals (Na, K, Mg, Ca, ) are t reduced (H 2 O is reduced t H 2 istead) ca t be prduced by electrlysis f aqueus sl. Ais f mre active metals (F - ) ad xais f elemets i their highest xidati state (NO 3-, CO 3, SO 4, ) are t xidized The Stichimetry f Electrlysis Faraday s law the amut f substace prduced each electrde is directly prprtial t the amut f charge trasferred thrugh the I el. curret t time f electrlysis Q charge trasferred # ml e - trasferred I = Q/t Q = I t I t = Q = F = Q/F F is related t the amut f substace thrugh the stichimetry f the half-reacti Allws the determiati f the amut f substace prduced by measurig I ad t Example: Durig electrrefiig f Cu, hw much time is eeded t prduce 250 g Cu the cathde if the curret is kept at 11 A? I = 11 A = 11 C/s t? Half-reacti: Cu 2 2e - Cu(s) = 250 g 1ml Cu g Cu - 2 ml e = 7.9 ml 1ml Cu I t F = t = F I 7.9 ml C/ml 4 t = = s = 19 hr 11C/s Elctrrefiig f Cu