Today What is the oxidatio umber of N i KNO3? Review for our Quiz! Thermo ad Electrochemistry What happes whe the coditios are ot stadard Nerst Equatio! A.!! 0! B.!! +1! C.!! -1! D.!! +3! E.!! +5 K is +1, O is -2 molecule is o charge 1(+1) + 3(-2) = -5 N must be +5 Balace this half reactio HNO2 NO3 - How may protos are i the balace 1/2 reactio? (keepig the coeficiet for NO3 - as 1) Give followig stadard reductio potetial, which do you thik would make the best reducig aget? Cl2(g) + 2e - 2Cl - E = 1.36 V I2(s) + 2e - 2I - E = 0.54 V Br2(l) + 2e - 2Br - E = 1.06 V! A.!! 0! B.!! 1 o the left! C.!! 1 o the right! D.!! 2 o the left! E.!! 3 o the right HNO2 NO3 - HNO2 + H2O NO3 - HNO2 + H2O NO3 - + 3H + HNO2 + H2O NO3 - + 3H + + 2e -! A.!! Cl -! B.!! Cl2! C.!! I2! D.!! I -! E.!! Br2 reducig agets are oxidized hardest to reduce is easiest to oxidize Lowest potetial Need to pick the reduced species (it ca be oxidized)
You reduce H + to H2 i a electrochemical cell. Your cell has a curret of 1 Amp for 10 miutes What is the total charge that is passed through the cell? You reduce H + to H2 i a electrochemical cell. Your cell has a curret of 1 Amp for 10 miutes How may moles of electros pass through the cell?! A.!! 1 C! B.!! 10 C! C.!! 600 C 1 A x (10 mi) x (60 s mi -1 ) = 600 C! A.!! 600 C / F! B.!! 600 C x F! C.!! 1 A x F! D.!! 6000 C F is C mol - Therefore the umber of moles of electros is q/f You reduce H + to H2 i a electrochemical cell. The umber of moles of electros that pass through the cell is 6.2 x 10-3. How may moles of H2 are formed? 2H + + 2e - -----> H2(g) You reduce H + to H2 i a electrochemical cell. Your cell has a curret of 1 Amp for 10 miutes. How may moles of H2 are formed? 2H + + 2e - -----> H2(g)! A.!! 6.2 x 10-3! B.!! 3.1 x 10-3! C.!! 1.2 x 10-2 For every mole of H2 you eed two moles of electros! A.!! 6.2 x 10-3! B.!! 3.1 x 10-3! C.!! 1.2 x 10-2 For every mole of H2 you eed two moles of electros
I the followig stadard cell, what is E cell? Z(s) Z 2+ H + H2 (g) Z 2+ + 2e - ---> Z(s) 2H + + 2e - ---> H2 E = -0.76 V E = 0.0 V I the followig stadard Ecell, what is the sig of the cathode? Z(s) Z 2+ H + H2 (g) Z 2+ + 2e - ---> Z(s) 2H + + 2e - ---> H2 E = -0.76 V E = 0.0 V! A.!! 0.0 V! B.!! +0.76 V! C.!! -0.76 V aode o the left Z Z 2+ aode H + H2 cathode E cell = E cathode - E aode = 0 - (-0.76) = +0.76 V Voltaic Cell! A.!! + E cell = E cathode - E aode = 0 - (-0.76) = +0.76 V! B.!! - Voltaic Cell therefore cathode +! C.!! either E cell = 0 For a battery which of the followig is correct? We'll look at stadard cocetratios! A.!! E > 0,!G > 0, K > 1! B.!! E > 0,!G < 0, K > 1 Battery = voltaic Spotaeous X 1.1 V! C.!! E > 0,!G < 0, K < 1! D.!! E < 0,!G > 0, K > 1! E.!! E < 0,!G < 0, K > 1 E > 0!G < 0 K>1 volt meter! F.!! E < 0,!G < 0, K < 1 hit its ot F. 1 M Z 2+ (aq) ad 1 M Cu 2+ (aq) (ote this is ridiculously cocetrated)
What about other cocetratios? Relatioship betwee E ad "G X????? volt meter "G is eergy E is electrical potetial Electric work (eergy) is -charge x potetial work = -q x E "G = workmax "G = - q x Emax 10-3 M Z 2+ (aq) ad 10-1 M Cu 2+ (aq)??? From ow o well ow the Potetial we calculate are the theoretical maximum Real world ever actually that good Relatioship betwee E ad "G "G = - q x E What is the charge q? q = X F is umber of moles of electros F is the charge of oe mole of electros F = 96,485 C (Faraday's Costat) "G = - FE Other cocetratios ad equilibrium Let s remember equlibrium! assume 25 C "G = "G + RTlQ at equilibrium "G = 0 so "G = -RTlK -FE = -FE + RTlQ E = E - RT F lq log!
What about other cocetratios? X????? volt meter 10-3 M Z 2+ (aq) ad 10-1 M Cu 2+ (aq)??? 1 M Z 2+ (aq) ad 1 M Cu 2+ (aq) stadard Z(s) + Cu 2+ (aq) Q = [Z2+ ] [Cu 2+ ] = 1 1 Z 2+ (aq) + Cu(s) = 1 E = 1.10 V - 0.0591 2 log(1) = 1.10V 10-3 M Z 2+ (aq) ad 10-1 M Cu 2+ (aq)??? Z(s) + Cu 2+ (aq) Z 2+ (aq) + Cu(s) Q = [Z2+ ] [Cu 2+ ] = (10-3 ) (10-1 ) = 10-2 E = 1.10 V - 0.0591 2 log(10-2 ) = 1.16V Curret will flow util E = 0 Equilibrium E = + 0.0591 logk = E 0.0591 logk
.1 M Ag + (aq) ad 1 M Ag + (aq) Same reactio! E = 0 V Q = [Ag+ ]aode = [Ag + ]cathode.1 1 =.1 Cocetratio Differeces will lead to potetial differece E = 0 V - 0.0591 1 log(.1) = 0.0591V each factor of te will be aother 0.0591 V F = 96,485 C q = I x t If E < 0, the the reactio ca be force i the o-spotaeous directio by applyig a potetial greater tha E to the cell