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