physical chemistry ii spring, 2017 third hour exam...physical chemistry ii spring, 2017 third hour...
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PHYSICAL CHEMISTRY II Spring, 2017 Third Hour Exam
Total value of exam: 100 points
Name:__________________________________ (25 points) 1. The following cell can be used to determine the acid dissociation constant, Ka, for the weak
acid HCN: Pt | H2(1 atm) | HCN (0.01 M) | AgCN(s) | Ag | Pt The EMF for the cell is 0.652 volts at 25 °C. E° for the Ag/AgCN electrode is –0.017 volts.
RHE: AgCN(s) + e- <__ __> Ag(s) + CN-(aq)
LHE: H+(aq) + e- <__ __> ½ H2(g) Calculate Ka for HCN and the pH of the HCN solution. *HINTS: Undissociated HCN is neutral (γ = 1). Assume dissociation of HCN is small, but
not zero, i.e., [HCN] ≈ 0.01 M, in Ka =
aH + aCN −
aHCN
.
(25 points) 2. Assuming the DHLL applies, estimate the solubility of CaSO4 (
Ksp = 7.10 x 10−5 ) in aqueous 0.01 M CaCO3 .
*HINT: quadratic formula: x = −b± b2 − 4ac
2a for ax2 + bx + c = 0 .
(25 points) DO EITHER PROBLEM 3 OR 4. 3. Given that the potentials (work per unit charge) of charging a positive ion with and
without an atmosphere are
Ψ+
atm =q+
4πε0Da 1+κ a( ) and Ψ+
no atm =q+
4πε0Da ,
where the work of charging a positive ion is w+ = Ψ+ dq
0
+ ze
∫ ,
and that ΔG = RT lnγ + = N A w+
atm − w+no atm⎡⎣ ⎤⎦ ,
SHOW that lnγ + =
−N A z+e( )2κ
8πε0DRT 1+κ a( )
4. The difference between the electrochemical potential of an ion in two separate but identical phases is given by !µi
α − !µiβ = zi FE , where E is the measured potential difference between
the like phases. By equating chemical potentials across each phase boundary in the cell Cu’ | Zn | Zn2+, Ni2+ | Ni | Cu” SHOW that
µNi + !µZn2+ − µZn − !µNi2+ = −2FE .
(25 points) 5. A solution contains 0.01 M Pd2+ and 0.01 M Pb2+. The following characteristics accompany
reduction of Pd2+ and Pb2+ at the same Pb electrode at 25 °C:
Reaction jo Z α Eoc Pd2+ + e- __> Pd(s) 7.94 x 10-4 A/cm2 +2 1/2 0.532 V
Pb2+ + 2e- __> Pb(s) 3.98 x 10-12 A/cm2 +2 1/2 -0.185 V Using the Butler-Volmer equation
a) on the same graph, sketch jPd2+ and jPb2+ as a function of applied voltage Eapp = Eoc +η
b) determine the total current density jtotal = j
Pd2++ j
Pb2+when the applied voltage is such that
Pd deposition first commences, i.e., Eapp ≈ Eoc(Pd2+ ) = 0.532V .
c) find
Eapp where jtotal = 0 .
USEFUL INFORMATION
R = 8.314 J mol-1 K-1 gas constant
F = 96485.3 coul/equiv the Faraday
volt x coul = joule
E = E°− RT
nFln
aprod
areac
Nernst Equation
ai = i⎡⎣ ⎤⎦γ i activity of an ion i
ΔG = -nFE Gibb free energy
ΔG° = -nFE° = -RT ln Keq standard Gibbs free energy
I = 1
2cizi
2
i
all ions
∑ ionic strength
log γ i = −0.509 zi2 I Debye-Huckel Limiting Law (DHLL)
j = jo e 1−α( )ZFη/ RT − e−αZFη/ RT⎡
⎣⎤⎦ Butler-Volmer equation
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