 Assignment 1

12: 1,11,31,46,69 13: 1, 7, 9 1

12.1

Pt(s)|Cr +2 (aq),Cr +3 (aq)||Cu 2+ (aq)|Cu(s)

Overall cell reaction is

2Cr +2 + Cu +2 = Cu(s) + 2Cr +3

In order to deposit Cu on the Cu electrode, Cu +2 must remove electrons from

that electrode, so electrons in the outer circuit flow from Pt to Cu.

12.11

Ni(s)|Ni 2+ (aq)||Ag + (aq)|Ag(s) E 0 =1.03V.

Cell reaction is

Ni +2Ag + = Ni +2 +2Ag n=2

∆G 0 =-nFE 0 =-2*1.03*96485 J/mole =-198.76kJ/mole (of Ni)

for 1gm of Ag plating out, the number of moles is 0.5/107.9

∆G 0 =-0.5*198.76kJ/107.9=-921J

the maximum work that can be done on the system is ∆G;

the maximum work that the system can do is -∆G.

31. I 2 (s)|I - is connected to an H 3 O + |H 2 (g,1atm) half cell where [H 3 O + ] is

unknown. The measured voltage is 0.841V and the I 2 (s)|I - half cell is the

cathode. What is the pH

Consider the half reaction

I 2 (s) + 2e - = 2 I -

If this electrode is the cathode, the half reaction proceeds as written, and E 0

for this reaction is 0.535 V

The overall reaction is thus

I 2 (s) + H 2 = 2H + +2I -

From the Nernst eq,

E = E 0

- nF

RT lnQ = E 0

- .0592/2 log [H+ ] 2 [I ! ] 2

.841 = .535 + .059pH

pH = 0.306 / .059 = 5.17

Assignment 1

12: 1,11,31,46,69 13: 1, 7, 9 2

46 a) What quantity of charge is fully charged 1.34 V zinc-mercury oxide

watch batter capable of furnishing if the mass of HgO in the battery is 0.50

g

The half cell reactions are

Zn(s)+2OH - (aq) = Zn(OH) 2 + 2e -

HgO(s)+H 2 O(l) + 2e - =Hg(l)+2OH -

The overall reaction is

Zn(s)+HgO(s)+H 2 O(l) =Hg(l)+ Zn(OH) 2

0.5 g of HgO=0.5/(200.6+16)=2.31 mmoles of HgO

for every HgO that disappears, 2e are consumed, so 4.62mmoles of electron

b) What is theoretical amount of work that can be obtained from this

battery

W=∆G 0 =-nFE 0 =2*96485*1.34 coulomb*volts = 2.58 kJ per mole of

HgO. Since there are only 2.31 mmoles, the amt of work is 597 J(max)

69.

Zn(s)|Zn +2 ||Ni +2 |Ni

Zn +2 + 2e - = Zn E 0 =-.763

Ni +2 + 2e - = Ni E 0 =-.23

The spontaneous cell reaction is

Zn + Ni +2 = Ni + Zn +2 E = .763-.23 =.53V

Cell is constructed of 32.68 g Zn (32.68/65.4 =0.5 moles) and 575 mL of 1M

Ni +2 . There is more Ni +2 than Zn, so Zn is the limiting reactant.

There is 0.5 moles of Zn which will produce 1 mole of electrons. They are

removed at the rate of .0715 C/s (1C/s=1A). A mole of electrons is 96485

C, which requires .0715S, so S = 1.35e6 seconds=375 hrs.

The nickel electrode has 0.5 mole of Ni plated out, which is 58.7/2=29.3 g

When the cell is completely discharged,[Ni +2 ]=(.575-.293)/0.575=0.49M

Assignment 1

12: 1,11,31,46,69 13: 1, 7, 9 3

1 draw tangent to plot @200:

Estimated rate is 4*10 -5 mol/L/s

7. in reaction of pyridine C 5 H 5 N with Methyl Iodide the following set of intial rates

was measured

[py](M) CH 3 I(M) Rate (mol/L/sec)

1e-4 1 e-4 7.5e-7

2e-4 2 e-4 3 e -6

2e-4 4 e -4 6 e -6

From the last two entries, [py] is constant, and CH 3 I has doubled. Rate has also

doubled, so rxn in first order wrt CH 3 I. First two entries have CH 3 I and [py] each

doubled. Doubling the CH 3 I will provide a factor of 2 increase in rate, but rate has

increased by 4x. Thus [py] also provides a factor of 2 and rxn is also 1 st order wrt

[py].

Thus rate=-d[py]/dt = k [py][CH 3 I]

k = rate/{[py][CH 3 I]}= 75 Lmol/sec

9. The reaction

SO 2 Cl 2 (g) --> SO 2 (g) + Cl 2 (g)

Is first order with k =2.2e-5 1/s at 320 °C. The partial pressure of SO 2 Cl 2 in a sealed

vessel at 320°C is 1.0 atm. How long will it take for the partial pressure of to fall to

1 atm.

! dS

dt = kS where S denotes P SO2Cl2

integration gives us

Assignment 1

12: 1,11,31,46,69 13: 1, 7, 9 4

S

= exp(!kt) = 0.5

S 0

1

exp(kt) = 2;

t 1/2

= ln2 / k = 3.15x10 4 sec = 8.75hr

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