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Lecture March 27
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Two separate HWs: Chap. 16 9/10 andanother for Chap. 4/9 and
Chap. 19(I)
One Quiz with Chap. 16 9/10, Chap 4 Sec.4.9 and some Chap. 19 -
DUE 8 AM April 1st
Today - Chapter 19
Half Reactions and GalvanicCells
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Zn(s) + Cu2+(aq) !Zn2+(aq) + Cu(s)
Divide into two half reactions
Zn(s) !Zn2+ + 2e-
Cu2+(aq) + 2e-!Cu(s)
oxidation - loss of electrons
reduction - gain of electrons
Galvanic Cell -a spontaneousreaction generates electrons
OIL RIGOx # from +2 to 0
Ox # from 0 to +2
Half Cell Potentials
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See Appendix 6 - all written as standardreduction potentials
Eo= 0.342 VoltsCu2+(aq) + 2e-!Cu(s)
reductionZn(s) !Zn2+ + 2e-
oxidationEo= 0.762 V
Note: reversedfrom the table
Net cell potential = 0.342 + 0.762 =
1.104 V
Zn(s) + Cu2+(aq) !Zn2+(aq) + Cu(s)
Volts = Joules/Coulombs
Positivemeans thereaction is
spontaneousas written
The electrons must all cancel!!
Eounder standard
conditions
Summary - Standard Cell
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Table A6.1 - Standard ReductionPotentials
Cu2+(aq) + 2e-!Cu(s) 0.3419 V
Zn(s) ! Zn2+(aq) + 2e- 0.7618V
Zn(s) + Cu2+(aq) !Zn2+(aq) + Cu(s)
Reversed
Net Cell Potential = 1.1037 V
reference is the hydrogen electrode:2H+(aq) + 2e- !H2(g) Eo=
0
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Chemical Energy and CellPotential
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Relationship between the Gibbs Free Energy#G (kJ) and the Cell
Potential E (Volts)
#G = - n F En = moles of electrons transferred
F = Faraday constant = 96,500 C/mol electrons
#Go= - n F Eo1 Joule = 1 Coulomb x 1 Volt OR Volt=J/C
Calculating #Gofrom Eo -Exercise 19.4
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Zn(s) + H2O(l)+ Ag2O(s) !Zn(OH)2(s) + 2Ag(s)
Small button batteries used in watches etc.
in basic solution1) Look up the half reactions and find the
Eo
Zn(s) + 2OH-(aq) !Zn(OH)2(s) + 2e-Eo=
+1.249 V
Ag2O(s) + H2O(l) + 2e-!2Ag(s) + 2OH-(aq)
Eo= 0.342 VEobattery = 1.249 + 0.342 =
1.591 V
#19.4 Continued
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3) Use theequation
#Go= - n F Eo
F = 96.5 x 103Coul/mol
2) The number of electrons transferred n = 2
#Go= - (2) (96.5 x 103) (1.591) = - 307 kJ
n is the # of moles of electrons
Joule = (moles of e)x(Coul/mol)x(volts)
Calculating Eofrom #Go
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Zn(s) + Cu2+(aq) !Zn2+(aq) + Cu(s)
From the #Govalues in the appendix
#Go= (-147.1)-(65.5) = -212.6 kJ/mol
#Go= - n F Eo Eo= -#Go/nF
Eo= - (-212.6 x 103) / 96,485 (2) =1.102 V
Eo= 0.7618 + 0.3419 = 1.1037 V from thehalf cell potentials
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E at Non-Standard Conditions- the Nernst Equation
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E = Eo- (RT/nF) lnQ
E = Eo- 2.303(RT/nF) logQ
2.303RT/F = 0.0592 V at 25oC
#G = #Go+ RTlnQ- nFE = -nFEo+ RTlnQ
Eq. 19.9
Eq. 19.10E = Eo- (0.0592/n) logQ
NEED to know n!!!
Example 19.5
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Eo = 1.104 V at STANDARD Conditions[Zn2+] and [Cu2+] = 1.000
M
What is the cell potential E when[Zn2+] = 1.90 M and [Cu2+] =
0.10 M?
E = Eo- (0.0592/n) logQ
n = 2
Q = [Zn2+]/[Cu2+]= 19.0
E = 1.104 - (0.0592/2) log(19.0)
E = 1.066 V
What happens to E as the cell reaction proceeds?
E decreases
Zn(s) + Cu2+(aq) !Zn2+(aq) + Cu(s)
-0.038
Quiz of the Day
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Use the following half cells (written asreduction) to find
#Go(in kJ) for the reaction:
4x(Fe3+(aq) + 1e- !Fe2+(aq))and reverse
O2(g) + 4 H+(aq) + 4 e-!2 H2O(l) Eo= 1.229 V
4 Fe2+(aq) + O2(g) + 4 H+(aq) !4 Fe3+(aq) + 2 H2O(l)
Eo= 0.770 V
Eo= -0.770 V
n = 4 Eocell = 0.459
#Go = -(4)(96.5x103)(0.459) = -177 kJ
