Electrochemistry L E O

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Barrie Heath
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1 Rmmbr from CHM151 A rdox raction in on in which lctrons ar transfrrd lctrochmistry L O Rduction os lctrons xidation G R ain lctrons duction W can dtrmin which lmnt is oxidizd or rducd by assigning oxidation numbrs. For xampl: H (g) + O (g) H O(g) ox rd Sinc hydrogn wnt from a 0 to +1 oxidation ach hydrogn, lost an lctron Sinc oxygn wnt from a 0 to - rduction ach oxygn, gaind two lctrons Zn(s) + Al + (aq) Zn + (aq) + Al(s) + total charg on this sid In CHM151, w ncountrd quations such as: + total charg on this sid Notic that this is th nt-ionic quation. That is, I rmovd any spctator ions (such as NO - ). Without adding th spctator ion, th chmical quation might look lik it is balancd. But is it? No bcaus th chargs on ach sid ar not balancd. Rmmbr that th whol point of a rdox raction is th transfr of lctrons. W can thrfor brak into two sparat half-ractions Lt s do a littl mor complicatd on btwn Ag and F gas Ag + F Ag + + F - Ag + F Ag + + F- Wll, it looks balancd what s wrong? 0 - Charg imbalanc F F - gain lctrons Al + (aq) Al(s) gain lctrons Al + (aq) + é Al(s) Ag Ag + los lctrons Ag Ag + + é Zn(s) Zn + (aq) los lctrons Zn(s) Zn + (aq) + é Total lctrons transfrrd must b th sam. Al + (aq) + é Al(s) x Total lctrons transfrrd must b th sam. Al + (aq) + 6é Al(s) Zn(s) + Al + (aq) Zn + (aq) + Al(s) Ag Ag + + é x Ag Ag + + é Ag + F Ag + + F - Zn(s) Zn + (aq) + é x Zn(s) Zn + (aq) + 6é Chargs on both sids ar now balancd Cr O - 7 (aq) + HNO (aq) Cr + (aq) + NO - (aq) (acidic) Stp 1: Balanc all lmnts that ar not H or O by using cofficints Cr O 7 Cr + (aq) Stp : Balanc O by using H O HNO (aq) NO - (aq) Stp : Balanc H by using H + Cr O 7 Cr + (aq) Stp 4: Stp 5: Stp 6: Stp 7: If thr is a charg imbalanc, us lctrons to fix th imbalanc Mak sur both half-ractions hav th sam numbr of lctrons Rcombin both half-ractions and mathmatically cancl and spcis prsnt on both sids If th ractions occurs in basic mdia, chang all H + to H O by adding OH -. You must add th sam amount of OH - to both sids of th quation Cr - O 7 (aq) Cr + (aq) + 7H O(l) HNO (aq) + H O(l) NO - (aq) Cr - O 7 (aq) + 14H + (aq) Cr + (aq) + 7H O(l) HNO (aq) + H O(l) NO - (aq) + H + (aq) Cr - O 7 (aq) + 14H + (aq) + 6é Cr + (aq) + 7H O(l) HNO (aq) + H O(l) NO - (aq) + H + (aq) + é (HNO (aq) + H O(l) NO - (aq) + H + (aq) + é) Cr - O 7 (aq) + 514H + (aq) + 6é Cr + (aq) + 47H O(l) HNO (aq) + H O(l) NO HNO (aq) + H O(l) NO - (aq) + 9H + - (aq) + 9H + (aq) + 6é (aq) + 6é Cr - O 7 (aq) + 5H + (aq) HNO (aq) Cr + (aq) + 4H O(l) + NO - (aq) 1

aq) +6H + + 6é Cl - (aq) + H O Cr(OH) (s) + H O CrO 4 + 410H + + 6é 4OH - + ClO - ( aq) + Cr(OH) (s) Cl - (aq) + CrO 4 + 4H O + H O 4OH - + ClO - ( aq) + Cr(OH) (s) Cl - (aq) + CrO 4 + 5H O Cr(OH)

2 Cr(OH) (s) + ClO - ( aq) CrO 4 + Cl - (aq) (basic) Cd(s) + Ni O (s) Cd(OH) (s)+ Ni(OH) (s) (basic) ClO - ( aq) Cl - (aq) Ni O (s) Ni(OH) (s) Cr(OH) (s) CrO 4 Cd(s) Cd(OH) (s) ClO - ( aq) Cl - (aq) Cr(OH) (s) CrO 4 Ni O (s) Ni(OH) (s) ClO - ( aq) Cl - (aq) + H O Cr(OH) (s) +H O CrO 4 ClO - ( aq) + 6H + Cl - (aq) + H O Cr(OH) (s) +H O CrO 4 + 5H + ClO - ( aq) +6H + + 6é Cl - (aq) + H O ClO - ( aq) +6H + + 6é Cl - (aq) + H O Cr(OH) (s) + H O CrO H + + 6é 4OH - + ClO - ( aq) + Cr(OH) (s) Cl - (aq) + CrO 4 + 4H O + H O 4OH - + ClO - ( aq) + Cr(OH) (s) Cl - (aq) + CrO 4 + 5H O Cr(OH) (s) +H O CrO 4 + 5H + + é (Cr(OH) (s) +H O CrO 4 + 5H + + é) Cr(OH) (s) + H O CrO H + + 6é Ni O (s) + H O Ni(OH) (s) Ni O (s) + H O + H + Ni(OH) (s) Ni O (s) + H O + H + + é Ni(OH) (s) Ni O (s) + H O + H + + é Ni(OH) (s) Cd(s) + H O Cd(OH) (s) + H + (aq) + é Cd(s) + H O(l) Cd(OH) (s) Cd(s) + H O Cd(OH) (s) + H + (aq) Cd(s) + H O Cd(OH) (s) + H + (aq) + é Rdox Ractions Anatomy of a Voltaic/Galvanic Cll (NO) Zn(NO) (-) Anod - Zn KNO K + Salt Bridg (+) Cathod Zn + Zn + NO - Zn + + Zn + Although lctrons ar transfrrd, w ar unabl to us ths lctrons. Is thr a way w can harnss ths lctrons?? It is traditional to draw th anod on th lft hand sid, cathod on th othr As th Voltaic/Galvanic Cll runs - - (-) - (+) Anod - KNO K + Zn Salt Bridg Cathod Zn + + Zn + Zn + K + K + >?M <?M It is traditional to draw th anod on th lft hand sid, cathod on th othr Anatomy of a Voltaic/Galvanic Cll - - (-) (+) Anod Cathod - - Ag Ag + NO - Ag + What about a voltaic cll mad from this RDOX raction? How do w us a gas?? Ag(s) + F (g) Ag + (aq) + F - (aq) KNO K + Salt Bridg Pt F - Na + F - Na + 1ATM Platinum (Pt) and graphit (C) ar usd for gass or solutions.

Voltaic/Galvanic Cll Shorthand notation Anothr way w can spcify th spcis in our voltaic cll if to us shorthand notation. Lt s tak our Zn/ cll, for xampl.

3 Voltaic/Galvanic Cll Shorthand notation Anothr way w can spcify th spcis in our voltaic cll if to us shorthand notation. Lt s tak our Zn/ cll, for xampl. With th spcis usd it would bcom: Zn (s) Zn + () () (s) How do w know which dirctions th lctions will travl in a voltaic cll? Ag (s) Ag + () F - () F (1atm) Pt(s) Standard Rduction Potntials All valus ar basd on whthr thy will donat or accpt lctrons to hydrogn. Bcaus of this, hydrogn has a rduction potntial of zro. Notic that all valus ar for raction in th rduction dirction. W can also writ thm in th oxidation dirction (rvrs) if w nd to. Spcis lowr on th list will donat lctrons to spcis highr on th list Strongr oxidizing agnts Hydrogn zro Strongr rducing agnts Positiv valus indicat chmical spcis will accpt lctrons from hydrogn Ngativ valus indicat chmical spcis will donat lctrons to hydrogn Standard Cll Potntial cll rduction + oxidation or cll rd cathod rd anod Rmmbr that oxidation happns at th anod. Standard Cll Potntial Ag + + é Ag Dtrmin th cll for th following raction Ag + F Ag + + F - rd.87 V rd 0.80 V Lt s dtrmin cll btwn and Zn. + - rd 0.4 V Zn Zn rd -0.76V Zn Zn ox 0.76V Which will most likly giv up an lctron and bcom oxidizd? Al(s) + Mn + (aq) Zn(s) + Al + (aq) Al + + é Al rd V Mn + + é Mn rd -1.18V cll 0.4V+ 0.76V 1.10V Rmmbr that this voltag will only tak plac undr standard thrmodynamic conditions (5.0 C, 1atm, aquous solutions 1.0M. Sinc Zn will mor asily giv up its lctron, it will bcom oxidizd. Th raction in th tabl for Zn will actually tak plac in th opposit dirction than writtn, thus ox - rd. Half -ractions + - Zn Zn Cll Potntial and Fr nrgy ΔG - cll mols of lctrons transfrrd Nd to balanc ractions to dtrmin corrct numbr of lctrons + Zn + Zn + Raction is alrady balancd, lctrons transfrrd Standard Cll Potntial (V) Faraday s Constant (96,500 C/mol) C J/V ΔG -(mol - )(96500 J/V mol)(1.10 V) -.1x10 5 J or -1 kj SPONTANOUS Cll Potntial undr ral conditions Th Nrnst quation Sinc clls ar not always undr standard conditions, w nd an quation that w can calculat cll potntial undr diffrnt conditions (highr/lowr molarity, highr/lowr tmpratur) Assum w hav th Zn/ cll with th anod at a concntration of 1.50M and th cathod at 0.50M. (8.14 J/molK)(98.15K) (1.50M) 1.10V ln (mol)(96,500 J/Vmol) (0.50M) 1.09V cll cll lnq Zn(s) + (aq) (s) + Zn + (aq) Q? [Zn ] Q [ ] Al + + é Al Al(s) + Zn + (aq) Zn(s) + Al + (aq) ΔG -(6mol - )(96500 J/V mol)(.00 V) Assum w hav th Al/Zn cll with th anod at a concntration of 0.50M and th cathod at.0m at 70.0 C. Al(s) + Zn + (aq) Zn(s) + Al + (aq) + é -1.56x10 6 J or kj SPONTANOUS (8.14 J/mol K)(4.15K) (0.5M).00V ln (6 mol )(96,500 J/V mol) (.0M) Q? [Al ] Q [Zn ].0V

Concntration Clls Lt s say you want to mak a cll, and you only hav on mtal () and only on ionic solution ((NO) ), cold you mak a cll that would gnrat any voltag?

00V G ( mol)(96,500 J/Vmol)(0.00V) Will this cll work?

knowing - - cll mols of lctrons transfrrd ΔG - cll Faraday s Constant (96,500 C/mol) C J/V G 5. 8x10 Jor - 5.8kJ What would th valu of Q nd to b if you wantd 1.00V? (8.14 J/molK)(98.15K) 1.

4 Concntration Clls Lt s say you want to mak a cll, and you only hav on mtal () and only on ionic solution ((NO) ), cold you mak a cll that would gnrat any voltag? Wll, not in it s currnt condition of 1.00M and 1.00M Lt s chang th concntrations to 0.10M and 1.00M cll (8.14 J/molK)(98.15K) 0.10M 0.00V ln ()(96500 J/V mol) 1.00M cll 0. 00V G ( mol)(96,500 J/Vmol)(0.00V) Will this cll work? cll cll lnq ΔG - Cll Potntial and K Now that w can calculat ΔG for a cll, w should also b abl to calculat th quilibrium constant lik w did in th last chaptr using: Standard Cll Potntial (V) or also knowing - - cll mols of lctrons transfrrd ΔG - cll Faraday s Constant (96,500 C/mol) C J/V G 5. 8x10 Jor - 5.8kJ What would th valu of Q nd to b if you wantd 1.00V? (8.14 J/molK)(98.15K) 1.10V ( mol )(96500 J/V mol) K 5.67x10 Som Shortcuts From th Book ΔG cll (8.14J/V mol)(98.15k) n (96,500 J/V mol) ΔG - cll cll cll ΔG - K cll cll 0.057V n 0.059V logk n lctrolytic Cll A voltaic/galvanic cll with vntually run down to 0V. This is bcaus vntually, on of th ractants will run out. Dirct comparison lctrolytic vs. Voltaic/Galvanic W could rcharg our cll by running th RDOX ractions in rvrs Dad Cll Voltaic/Galvanic Cll lctrolytic Cll An lctrolytic cll is on whr lctrons flow in th nonspontanous dirction. lctrolytic Cll Raction at Anod Raction at Cathod Raction at Cathod Raction at Anod Zn Zn Zn Zn + - Rduction Rduction Zn + Zn + + cll 1.10V Zn + + Zn + cll -1.10V Notic: oxidation always happns at th anod! 4

1 ampr 1 C/s 1 ampr s 1 Coulomb (C) What mass of Zn will b producd if a Zn/ lctrolytic cll is run at.0amps for 5.0hrs?

5 Othr Applications of lctrolytic Clls Faraday s Law (lctrolysis) Th amount of a substanc producd at an lctrod is proportional to th quantity of lctricity (in Coulombs) transfrrd to that lctrod. 1 ampr 1 C/s 1 ampr s 1 Coulomb (C) What mass of Zn will b producd if a Zn/ lctrolytic cll is run at.0amps for 5.0hrs? 96,500 C 1mol - 1 mol molar mass 60 min 60 sc 1C.00 amp 5.00 hr 54,000 C 1hr 1min 1amps 54,000 C 1mol 96,500 C x10 mol 65.9 g Zn 5.60x10-1 mol 1 mol 1mol Zn mol 18.g Zn 5

Electrochemistry. Remember from CHM151 G E R L E O 6/24/2014. A redox reaction in one in which electrons are transferred.

Electrochemistry. Remember from CHM151 G E R L E O 6/24/2014. A redox reaction in one in which electrons are transferred. Electrochemistry Remember from CHM151 A redox reaction in one in which electrons are transferred Reduction Oxidation For example: L E O ose lectrons xidation G E R ain lectrons eduction We can determine