Chem 150 Answer Key Problem Electrochemistry and Thermochemistry
1. Given below is a sketch of a Voltaic Cell. Name the two electrodes:
The copper electrode is the anode.
The silver electrode is the cathode.
The u-shaped glass tube filled with KNO3 connecting the two half cells is called the salt bridge.
Write down the reactions that occur on both electrodes as well as the overall cell reaction.
Cathode reduction:
2[Ag+(aq) + e- → Ag(s)]
E° = +0.800 V (reduction potential)
Anode oxidation:
Cu(s) → Cu2+(aq) + 2e-
E° = +0.340 V (reduction potential)
Overall:
2Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq)
(Note that we do not multiply the cell voltage by 2. The cell voltage is an intrinsic property that does not
depend on the amount of substance.)
Calculate the cell voltage E°cell. Assume both half cells to be in their standard state.
E°cell = E°reduction half cell - E°oxidation half cell = 0.800 V – 0.340 V = 0.460 V
(Methode A)
Finally, draw an arrow in the diagram above to indicate in which direction electrons flow between the
electrodes.
The arrow in the drawing should point to the silver electrode as electrons are being
delivered from the copper anode to the silver cathode.
2. Draw a standard hydrogen electrode and connect it to a half cell consistent of a zinc road immersed in
a 1 M zinc nitrate solution.
Label the electrodes anode and cathode. Show equations for the reaction on both electrodes and give
the overall cell reaction as well as the cell Voltage.
Cathode reduction:
2H+(aq) + 2e- → H2(g)
0 V (as defined)
Anode oxidation:
Zn(s) → Zn2+(aq) + 2e-
+0.763 V
2H+(aq) + Zn(s) → H2(g) + Zn2+(aq)
+ 0.763 V
Overall:
(Methode B)
3. Which is by far the best material to be used in the manufacture of a salt water fishing wheel?
stainless steel
zinc plated steel
chromium plated brass
4. What is anodic protection? Give two examples.
One way of protecting a metal from corrosion is to mechanically attach (there must be eclectic contact)
a metal that is even more prone to oxidation (more negative reduction half potential) to the one that is
to be protected.
Examples are water boilers and aluminum boats. In the former iron in the later aluminum is protected
from corrosion usually by magnesium. Magnesium which is less noble than aluminum (and much less
noble than iron) has a greater negative reduction potential and will oxidize first (anode).
5. Balance the following redox reaction.
3H+ + 3Ag + HNO3 → 3Ag+ + NO + 2H2O
6. Given below is a sketch of a Voltaic Cell. One half cell consists of a platinum wire dipping in to a
solution that is 0.5 M in Cu2+ and 0.1 M in iodide whilst the other half cell is made up of 1 M solution of
sulfate (as sulfuric acid) in which a platinum wire is immersed that is continuously purged with
sulfurdioxide under standard pressure (1 atm).
1 M SO42- (aq)
Cu2+(aq 0.5 M), I- (aq 0.1 M), CuI(s)
Label the two electrodes as anode and cathode.
Write down the overall cell reaction
Calculate the cell voltage.
The two reduction half potentials are as follows:
Cathode: Reduction:
Cu2+(aq) + I-(aq) + e- → CuI(s)
+ 0.86 V
Anode: Oxidation:
SO42- (aq) + 4H+ (aq) + 2e- → SO2 (g) + 2H2O (l)
+ 0.170 V
Since sulfate/sulfur dioxide has the smaller reduction half potential sulfur dioxide will be oxidized and
we reverse the reduction half reaction (and the sign for the voltage). This oxidation happens on the
anode so we label the platinum road on the left half cell as the anode. The other electrode must be the
cathode and the overall cell reaction is:
2Cu2+(aq) +2 I-(aq) + 2e- → 2CuI(s)
+ 0.86 V
SO2 (g) + 2H2O (l) → SO42- (aq) + 4H+ (aq) + 2e-
- 0.17 V
2Cu2+(0.5M)+2 I-(0.1M)+SO2(g,1 atm)+2H2O(l)→2CuI(s)+SO42-(0.011 M)+4H+(0.511 M)
+ 0.69 V
For the concentration of SO42-(0.011 M) and H+(0.511 M) review the pH of 0.5 M sulfuric acid which we
solved in class.
If the cell was under standard conditions the cell voltage would be 0.69 V and the cell reaction
spontaneous in the direction as written.
Note that we do not multiply the voltage of the reduction by 2 (as we would with the enthalpy of a
reaction in Hess law) this is because cell Voltages are intrinsic properties, like densities, that do not
depend on the amount of substance.
Note also that the reaction quotient applies to all products and all reactants but not the water: The
concentration of water does not funnel in to the calculation of Q. It is already part of the – 0.17 V for the
half reaction involving the water. Stoichiometric coefficients in front of reactants or products become
exponents in the reaction quotient. For example:
A + B ⇆ 2C could also be written as A+B ⇆C +C and thus Q =
∏[𝒑𝒓𝒐𝒅𝒖𝒄𝒕]
∏[𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔]
[𝑪][𝑪]
[𝑪]²
= [𝑨][𝑩] = [𝑨][𝑩]
The concentration of solids and pure liquids and gasses under standard pressure is written as 1 e.g. for
water, copper iodide (insoluble) and sulfur dioxide we simply enter 1. For all other species we need to
give the molar concentration including H+ for which need to solve an ICE table like we did in class.
Applying the Nernst equation
Ecell = E°cell ∏[𝒑𝒓𝒐𝒅𝒖𝒄𝒕]
where E°cell = + 0.69 V, n = 2 and Q = ∏[𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔] =
Ecell = +0.69 V – {0.0296 V x log 0.300}
Ecell = + 0.69 V+ 0.015 V
Ecell = + 0.705 V
𝟎.𝟎𝟓𝟗𝟐 𝑽
x
𝒏
log Q
[𝑪𝒖𝑰]²[𝑺𝑶₄²⁻][𝑯⁺]⁴
[𝑪𝒖²⁺]²[𝑰⁻]²[𝑺𝑶₂]
=
[𝟏]²[𝟎.𝟎𝟏𝟏][𝟎.𝟓𝟏𝟏]⁴
[𝟎.𝟓]²[𝟎.𝟏]²[𝟏]²
= 0.300
7.Permanganate [MnO4]- is a powerful oxidizing agent that can oxidize water to oxygen under standard
state. Aqueous solutions of permanganate ions are often employed in a chemistry laboratory. Is a 0.1 M
solution of permanganate stable in water over air? Assume an oxygen pressure of 0.2 atm, an initial
concentration of Mn2+(aq) of 0.1 M and pH of 7 at 25 degrees Celsius.
The relevant reduction half potentials are:
[MnO4]-(ag) + 8H3O+(aq) + 5e- → Mn2+(aq) + 12H2O (l)
1.507 V
O2(g) +4H3O+(aq) + 4e-→ 6H2O (l)-
1.229 V
This is a redox reaction. Let’s set this up like the reaction was running in a cell (not a beaker) by carrying
out the oxidation on an anode and the reduction on a cathode:
-
+
(aq) + 5e- → Mn2+(aq) + 12H2O (l)} x4….. 1.507 V
Cathode:{ [MnO4] (ag) + 8H3O
Anode:
{6H2O (l) → O2(g) +4H3O+(aq) + 4e- } x5
4[MnO4]-(ag) + 12H3O+(aq) + → 4Mn2+(aq) + 18H2O (l) + 5O2(g)
-1.229 V
+ 0.278 V
Under standard conditions, permanganate is reduced to Mn2+(aq) at the cathode and water is oxidized
to oxygen at the anode Ecell° = + 0.278 V. The cell operates at standard temperature but not under
standard concentration:
4[MnO4]-(1 M) + 12H3O+(10-7 M, neutral!) + → 4Mn2+(0.1 M) + 18H2O (liquid) + 5O2(0.2 atm)
So we need to involve the Nernst equation: remember, for aqueous species enter molarities, for gasses
enter partial pressure and for solids and liquids in each phase enter unity (e.g. 1 for water).
Ecell = Ecell° -
𝟎.𝟎𝟓𝟗𝟐 𝑽
𝒏
Ecell = +0.278 V Ecell = +0.278 V Ecell = +0.278 V -
[𝑷𝑶𝒙𝒚𝒈𝒆𝒏]𝟓 [𝑯₂𝑶]𝟏𝟖 [𝑴𝒏²⁺]⁴
log
𝟎.𝟎𝟓𝟗𝟐 𝑽
𝟐𝟎
𝟎.𝟎𝟓𝟗𝟐 𝑽
𝟐𝟎
𝟎.𝟎𝟓𝟗𝟐 𝑽
𝟐𝟎
[𝑴𝒏𝑶₄⁻]𝟒 [𝑯⁺]𝟏𝟐
[𝟎.𝟐]𝟓 [𝟏]𝟏𝟖 [𝟎.𝟏]⁴
log [𝟎.𝟏]𝟒[𝟏𝟎−𝟕]𝟏𝟐
= log3.2x1080
x 80.50
Ecell = +0.278 V – 0.238 V
Ecell = +0.040 V
What does the answer mean? The low hydrogen ion concentration in water at pH 7 (10-7 M) reduces
−𝟕 𝟏𝟐
the cell potential to nearly zero (it funnels in as [𝟏𝟎 ] ) We still have a positive cell potential
though so permanganate solutions slowly deteriorate in water and cannot be stored for long periods
of time but the reaction is usually slow enough that significant oxidation does not occur in a few hours
or days.
8. Draw a diagram for the entropy of a pure substance as a function of temperature (0-500 K). Assume
two phase transitions in this temperature range (melting and boiling).
9. Calculate the enthalpy of formation of Methanol (CH3OH(l)) from standard enthalpies of formation of
CO2 (ΔHf° = -393.5 kJ/mol), H2O (ΔHf° = -285.8 kJ/mol) and the standard enthalpy of combustion for
methanol (-726.56 kJ/mol).
Let’s translate the given enthalpies in chemical equations.
(1)
C(graphite) + 2H2(g) + ½ O2(g) → CH3OH(l)
ΔHf° = ?
(2)
C(graphite) + O2(g) → CO2(g)
ΔHf° = -393.5 kJ/mol
(3)
H2(g) + 0.5O2(g) → H2O(l)
ΔHf° = -285.8 kJ/mol
(4)
CH3OH(l) + 1.5O2(g) → CO2(g) + 2H2O(l) ΔH° = -726.56 kJ/mol
We rearranging and add equations (2)-(4) in such a way as to end up with equation (1). Start by
reversing equation (4) (which reverses the sign on the enthalpy change) :
(-4)
CO2(g) + 2H2O(l) → CH3OH(l) + 1.5O2(g)
ΔH° = +726.56 kJ/mol
(2)
C(graphite) + O2(g) → CO2(g)
ΔHf° = -393.5 kJ/mol
2x(3)
2H2(g) + O2(g) → 2H2O(l)
ΔHf° = 2x(-285.8 kJ/mol)
(1)
C(graphite) + 2H2(g) + ½ O2(g) → CH3OH(l)
ΔHf° = -238.6 kJ/mol
10. Calculate the standard enthalpy of combustion of methane (CH4) from the standard enthalpy of
formation of methane (ΔHf° = -74.8 kJ/mol), CO2 (ΔHf° = -393.5 kJ/mol) and H2O (ΔHf° = -285.8 kJ/mol)
(1)
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
ΔH° = ?
Since the enthalpy of formations of all products and all reactants are known we use Hess law in
this way:
ΔH° = ΣvpΔHf°(products) – ΣvrΔHf°(reactants)
ΔH° = [ΔHf°(CO2) + 2ΔHf°(H2O)] – [ΔHf°(CH4) – 2ΔHf°(O2)]
ΔH° = [-393.5 kJ/mol + 2(- 285.8 kJ/mol)] – [– 74.8 kJ/mol + 2(0 kJ/mol)]
ΔH° = -890.3 kJ/mol
11. Complete the following phrase:
The Gibbs free energy for a spontaneous process is negative.
12. Which of the following (if any) are state functions (depend only on the final and initial state of a
system and not on the path used to get there):
Heat, work, inner energy, enthalpy, entropy, Gibbs free energy
13. Name the three systems below. Describe the differences between them in terms of the flow of heat
and matter.
Open system
closed system
Exchange of heat and matter
Exchange of heat but not matter
isolated system
No exchange of heat or matter
14. This question is related to the experiment Bromination of Acetone where we reacted acetone and
bromine (in the presence of acid) according to the equation bellow:
CH3C(O)CH3 + Br2 → CH2BrC(O)CH3 + HBr
We measured the time it took for the bromine to completely disappear. We ran 4 experiments at room
temperature, one in a cold water bath at 11 ͦC and one in a hot bath at 41 ͦC and observed that the rate
of the reaction depends on the temperature.
A valid question that arose during one of the lab periods was the following. Since this reaction is
spontaneous it is probably exothermic and releases heat. Wouldn’t this heat effect our measurements?
Try to answer the question with the following data I found on our reactants and products. Note that the
HCl we added does not show up in the net stoichiometric reaction. It functions as a catalyst and is not
consumed during the reaction.
species: (gas phase data!)
CH3C(O)CH3(g)
Br2(g)
CH2BrC(O)CH3(g)
HBr(g)
ΔfH ͦ (in kJ/mol)
-217.1
+30.9
-181
-36.3
Assume that ΔH for the gas phase reaction is similar to ΔH for the reaction in solution. This will not
produce a huge error in dilute solution with one exception. HBr will react further in water and ionize
completely according to:
HBr(g) + H2O →H3O+ + Br-
ΔH = -85 kJ/mol
Hint: This heat must be added to the heat of the reaction between the acetone and the bromine!
Assume that the heat capacity of you mixture is that of water (4.184 J/gK) and neglecting the
calorimeter constant for your conical flask calculate the temperature increase of your 50 g solution.
Fair question! Let’s check. It is probably less relevant for the reactions in the hot or cold bath since any
heat released will quickly be absorbed by the water bath. For the conical flasks that are reacting on our
bench tops at room temperature heat would not dissipate that quickly and the reaction mixture could
warm up significantly (e.g. a few degrees would obviously effect the rate). So are we jeopardizing the
room temperature reaction by not controlling the temperature in the flasks?
First we use Hess’s law to find ΔH in the gas phase:
CH3C(O)CH3 + Br2 → CH2BrC(O)CH3 + HBr
ΔH = [(-181 kJ/mol) + (-36.3 kJ/mol)] - [(-217.1 kJ/mol) + (+30* kJ/mol)]
ΔH = -30.2 kJ/mol
For the total heat change we need to add the enthalpy of ionization to the enthalpy of the reaction
which gives: ΔHtotal = -30.2 kJ/mol – 85 kJ/mol = -115 kJ/mol which is mildly exothermic.
Now we are almost there. How much heat was produced? We did not react a whole mole of bromine
but used only 10 mL of a 0.02 M Br2 solution (Check that bromine is the limiting reactant)
n(Br2) = MV = 0.01L x 0.02 mol/L = 2 x 10-4 mol
which gives q = 2 x 10-4 mol x (-115 000 J/mol) = -23 J
How much would the temperature rise in your conical flask that contains approximately 50 g of solution
with a heat capacity assumed to be that of pure water if 23 J of energy are supplied?
q = mcΔT
and
ΔT =
𝑞
𝑚𝑐
=
23 𝐽
50 𝑔 𝑥 4.18 𝐽𝑔⁻¹𝐾⁻¹
= 0.11 K
The temperature in our conical flask will increase but it will increase only slightly. There is not enough
heat produced to sufficiently heat the content of the conical flask and we can continue to run the
experiment on the lab bench. Fair question though and be assured the next time we run this lab I will
have a temperature probe in one of the flasks to see if I can measure a 0.1 degree increase.
Advanced Problem:
D.W. Mitchell, Industrial and Engineering Chemistry, September 1949, p 2027-2031
It is difficult to measure the heat of formation of magnesium nitride from the elements. This reaction
has to be carried out at high temperatures in a furnace. How can we possibly get room temperature
data (e.g. standard state ΔfH ͦ)
Mitchell determined the enthalpy (from here on referred to as heat) of formation of Magnesium nitride
(6) at room temperature and 1 atm by measuring the heat of solution of magnesium nitride in
hydrochloric acid (1) and comparing it to the heat of solution of magnesium in hydrochloric acid (2), the
heat of formation of ammonia (3), The heat of condensation of ammonia (4) and the heat of solution of
ammonia in hydrochloric acid (5)
(1) Mg3N2(s)+ 8HCl (aq) → 3MgCl2(aq) + 2NH4Cl(aq)
ΔH1 = -287,894 cal/mol
(2) Mg(s)+ 2HCl (aq) → MgCl2(aq) + H2(g)
ΔH2 = -111,322 cal/mol
(3) 1.5H2(g) + 0.5N2(g) →NH3 (g)
ΔH3 = -11,036 cal/mol
(4) NH3 (g) → NH3 (l)
ΔH4 = -5,005 cal/mol
(5) NH3 (l) + HCl (aq) → NH4Cl(aq)
ΔH5 = -16,043 cal/mol
(6) 3Mg(s) + N2(g) → Mg3N2(s)
ΔH6 = ?
Calculate the enthalpy (heat) of formation of magnesium nitride (6) in kJ/mol and compare it to the
enthalpy of formation of magnesium oxide (-601.7 kJ/mol).
The Key to this problem is to realize that the products of reaction of magnesium nitride and hydrochloric
acid: Mg3N2(s)+ 8H+ (aq) → 3Mg2+(aq) + 2NH4+(aq)
can also be obtained by dissolving magnesium metal in hydrochloric acid and then dissolving ammonia
in the same solution. Since the heat of formation of ammonia is well known we can use Hess’s law to get
to the unknown ΔH6.
We use the properties of enthalpy to arrange equations 1-5 in such a way that adding them up leads to
equation (6)
(1) 3MgCl2(aq) + 2NH4Cl(aq) → Mg3N2(s)+ 8HCl (aq)
ΔH1 = +287,894 cal/mol
(2) 3Mg(s)+ 6HCl (aq) → 3MgCl2(aq) + 3H2(g)
ΔH2 = -333,966 cal/mol
(3) 3H2(g) + N2(g) →2 NH3 (g)
ΔH3 = -22,072 cal/mol
(4) 2NH3 (g) → 2NH3 (l)
ΔH4 = -10,010 cal/mol
(5) 2NH3 (l) + 2HCl (aq) → 2NH4Cl(aq)
ΔH5 = -32,086 cal/mol
(6) 3Mg(s) + N2(g) → Mg3N2(s)
ΔH6 = -110.240 cal/mol = -461 kJ/mol
How does this compares to the enthalpy of formation of magnesium oxide (-601.7 kJ/mol). At first sight
it compares quite well but careful. Magnesium oxide is a lot more stable! There are three magnesium
ions locked up in Mg3N2 but only one in MgO. Magnesium nitride decomposes at temperatures above
1400-1500 degrees Celsius: Mg3N2(s) → 3Mg(g) + N2 (g) whereas magnesium oxide can be heated to its
boiling point at 3600 degrees Celsius without decomposition MgO(l) → MgO(g).