electrpchemistry Flashcards

(22 cards)

1
Q

why do you want the potential difference to be as large as possible?

A

the larger the potential difference, the more work you get from a particular system

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2
Q

how to electrons flow in spontaneous redox reaction

A

high potential energy to low potential energy
anode to cathode

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3
Q

what is the oxidiation number of Ni in 2NiO(OH)

A

+3

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4
Q

reduction occurs preferentially in the cell with …

A

the more +ve standard reduction potential

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5
Q

Which one of the following statements is true?
A. A reaction can occur spontaneously if DS for the universe is negative.
B. A reaction can occur spontaneously only if DS for the system is positive.
C. A spontaneous reaction with a negative DS for the system must be exothermic.
D. A reaction can occur spontaneously only if DS for the surroundings is positive.

A

C

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6
Q
  1. Bromine boils (Br2() Br2(g)) at 59.0ºC with ΔHvapº = 29.6 kJ mol–1
    . What is the
    value of ΔSvapº?
    A. 11.2 J K–1 mol–1
    B. 1750 J K–1 mol–1
    C. 89.2 J K–1 mol–1
    D. 501 J K–1 mol–1
A

C

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7
Q
  1. Liquid water is injected into an oven at 400 K. What are the signs for DG, DH and DS
    for the transformation that occurs?
    ∆G ∆H ∆S
    A. + – –
    B. + – 0
    C. – + 0
    D. – + +
A

D

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8
Q

SHORT ANSWER QUESTION
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
Calculate the change in the standard Gibbs Free energy. Is this reaction thermodynamically favourable at
298 K? Explain your answer.
Chemical Species
∆Hf (kJ mol–1) ∆S (J K–1 mol–1)
CH4(g) –75 186
O2(g) 0 205.2
CO2(g) –394 214
H2O(g) –242 189

A

DH = -803kJ/mol = 803000J/mol
DS = -4.4J/K
DG = -803kJ/mol - (298K x -4.4J/K)
DG = -801688.8J/mol
DG = -801.6888kJ/mol
Thermodybamicslly favourable

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9
Q

Which of the following is true for a system at equilibrium?
M
a. DS°sys = DS°surr
b. DS°sys = -DS°surr
c. DS°sys = DS°surr = 0
d. DS°univ > 0
e. None of the above is a sufficient condition.

A

b

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10
Q

Iron(III) oxide can be reduced by carbon monoxide.
H
Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)
Use the following thermodynamic data at 298 K to determine the equilibrium constant at this temperature.
Fe2O3(s) CO(g) Fe(s) CO2(g)
DH°f (kJ/mol) -824.2 -110.5 0 -393.5
DG°f (kJ/mol) -742.2 -137.2 0 -394.4
S°(J/K×mol) 87.4 197.7 27.78 213.7

A

1.4*10^5

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11
Q

The reaction of methane with water to form carbon dioxide and hydrogen is non-spontaneous at 298 K.
M At what temperature will this system make the transition from non-spontaneous to spontaneous? The data
refer to 298 K.
CH4(g) + 2H2O(g) CO2(g) + 4H2(g)

CH4(g) H2O(g) CO2(g) H2(g)
DH°f (kJ/mol) -74.87 -241.8 -393.5 0
DG°f (kJ/mol) -50.81 -228.6 -394.4 0
S°(J/K×mol) 186.1 188.8 213.7 130.7

A

955K

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12
Q

a. Explain what is meant by a spontaneous process.
b. Is a spontaneous process necessarily a rapid one? Explain, and provide a real reaction as an
example to illustrate your answer

A

a process that occurs without any external energy input into the system.
no, diamond from graphite, rusting

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13
Q

State the second and third laws of thermodynamics.

A
  1. All spontaneous processes are accompanied by an increase in the total entropy of the universe
  2. the entropy of a perfect crystalline susbtance at abolsute zero is zero.
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14
Q

Given: C2H2(g) ® 2C(graphite) + H2(g)
DG° = -209 kJ
A sample of gaseous C2H2 (acetylene, or ethyne) was stored for one year, yet at the end of this period the
sample remained unchanged and no graphite or hydrogen gas had been formed. Briefly explain why there is
no inconsistency between the sign of DG° and the apparent stability of the sample.

A

Relative to graphite and hydrogen, acetylene is unstable. Its decomposition to form these products is
spontaneous, as the negative sign of DG° suggests. However, the kinetics of the decomposition are
immeasurably slow under normal conditions.

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15
Q

For the reaction of xenon and fluorine gases to form solid XeF4, DH° = -251 kJ and DG° = -121 kJ at
25°C. Calculate DS° for the reaction.

A

-436J/K

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16
Q

Photosynthesis can be represented by the equation
M
6CO2(g) + 6H2O(l) ® C6H12O6(s) + 6O2(g)
a. Calculate DS° for this process, given the following data:
CO2(g) H2O(l) C6H12O6(s) O2(g)
S°: 214 70 212 205
b. Given that DH° for the reaction is 2802 kJ, calculate DG° at 25°C

A

A. -262
b. 2880

17
Q

true or false The free energy of a perfect crystal at absolute zero, is zero.

19
Q
  1. As a fuel, H2(g) produces only non-polluting H2O(g) when it burns: Moreover, it
    combines with O2(g) in a fuel cell to produce electrical energy.
    (a) Calculate DH° and DS° and DG° per mole of H2(g) burned at 298 K.
    (b) Is the spontaneity of this reaction temperature dependent? Explain your answer.
    (c) At what temperature does the reaction become spontaneous?
    Sº (J K–1 mol–1): H2(g), 130.6; O2(g), 205.0; H2O(g), 188.72.
    DHº (kJ mol–1): H2O(g), –241.826
A

a.
D rxnH° = 0 + 0 + (–241.826) = –241.826 kJ
D rxnSº = SSºproducts - SSºreactants = –44.4 J K–1
DGº = –228.6 kJ
[Remember to change J K–1 into kJ K–1]

b. Yes, reaction will become non-spontaneous at higher temperatures

c. Reaction is spontaneous at temperatures less than 5.45 x 103 K.

20
Q

Lead can displace silver(I) ions from solution: Pb(s) + Ag+(aq) ® Pb2+(aq) + Ag(s).
As a consequence, silver is a valuable by-product in the industrial extraction of lead from
its ore. Calculate K and DG° at 298 K for this reaction.
Eº(Pb2+/Pb) = –0.126 V and Eº(Ag+/Ag) = 0.80 V.

A

K = 2.1 x10^31

21
Q

In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a
Zn/Zn2+ half-cell and a H2/H+ hall-cell using the following conditions:
[Zn2+] = 0.010 mol L–1
, [H+] = 2.5 mol L–1 and pH2 = 0.30 atm.
Calculate Ecell at 298 K. Eº(Zn2+/Zn) = –0.76 V and Eº(H+/H2) = 0.00 V

22
Q

Acrylic acid (CH2=CHCOOH) has a pKa of 4.25.
(a) Write an equation for the reaction of acrylic acid with water.
(b) Calculate DG0 for the dissociation of acrylic acid in water at 298 K.
(c) If the [H3O+] = 3.5 ´ 10–3 mol L–1, [CH2=CHCO2
–] = 0.020 mol L–1 and
[CH2=CHCOOH] = 0.05 mol L–1, calculate DG at 298 K.

A

b. 24.2kJ
c. 7.92kJ