Chemistry 122
Mines, Spring 2014
Answer Key, Problem Set 9
1. 18.44(c) (Also indicate the sign on each electrode, and show the flow of ions in the salt bridge.); 2. 18.46 (do this for all cells in
18.44 even though I only asked you to sketch the one in part (c)); 3. 18.48; 4. 18.14 (Note that Tro uses the term “electrode
potential” for what I’ve called a “reduction potential”); 5. NT1; 6. 18.55 & 18.56; 7. NT2; 8. 18.66b (Note: You need to determine
n here in order to complete the calculation!); 9. 18.110 (Add Part (b): Without doing a calculation, state whether the cell potential
+
will increase or decrease if a bit of solid NaOH were added to the solution containing the H (i.e., the electrode in contact with the
Pt(s)) [Assume the T was kept the same.]? Explain without using the Nernst Equation (use “Le Chatelier” type of argument as
discussed in the “Voltaic Cells” experiment reading/handout.); 10. 18.80; 11. NT3; 12. 18.136; 13. 18.89 (NOTE: You must
assume that the concentrations of the ions in each half cell are 1 M [e.g., standard state conditions]) ; 14. 18.91; 15. 18.100 (Add
2+
(b): Would the number of moles of electrons passed be the same if the only difference in the cell were that Cu was plated out as
+
Cu(s) (instead of Ag being plated out as Ag(s)? Would the number of moles of solid produced be the same these two cases?
Explain. Why is it so important to write out the half reaction when doing electrolysis problems such as this?); 16. NT4
-----------------------------------------------1. 18.44(c). Sketch a voltaic cell for each redox reaction.
Label the anode and cathode and indicate the half-reaction
that occurs at each electrode and the species present in each solution. Also indicate the direction of electron
flow. (Also indicate the sign on each electrode, and show the flow of ions in the salt bridge.)
-
+
(c) 2 NO3 (aq) + 8 H
(aq)
2+
+ 3 Cu(s)  2 NO(g) + 4 H2O(l) + 3 Cu
(aq)
Answer:
salt bridge
-
+
(NO3 flows left and K flows right)
e-
anode (-)
(+) cathode
-
NO3
Cu(s)
Pt(s)
+
K
Cu2+(aq)
Cu(s)  Cu2+(aq) + 2 e-
NO(g) [bubbling over electrode;
NO3-(aq)
H
+
(aq)
source, gas inlet and
glassware not shown]
2 NO3-(aq) + 4 H+(aq) + 3 e NO(g) + 2 H2O(l)
Explanation/Reasoning:
The problem implied that the reaction in the cell was the reaction indicated by the balanced
equation, so no knowledge of potentials was used to determine which reaction (forward or
reverse) was the actual cell reaction (it was assumed to be the forward reaction). Half reactions
were obtained from Table 18.1
Since Cu(s)  Cu2+(aq) + 2 e- is an oxidation half reaction, the left side (as I have drawn it—it
could be written with the sides “swapped”) must be the anode. Also, since electrons are being
produced there, they must leave that side and go to the right side. Since this is a voltaic cell, that
means the left side must be negative (since electrons spontaneously flow away from a negative
toward a positive. A Pt electrode was used on the right side because there is no conductive solid
in that half reaction. Negative ions in the salt bridge always flow in the opposite direction as the
(negative) electrons in the wire so that there is “current” looping through the cell. Positive ions
flow in the opposite direction as the negative ones.
2. 18.46.
Calculate the standard cell potential for each of the electrochemical cells in Problem 44. (Do this for all cells in
18.44 even though I only asked you to sketch the one in part (c).)
2+
(a) Ni
(aq) +
+
(b) 2 H
(aq)
-
Mg(s)  Ni(s)
2+
+ Mg
Answers:
(aq)
2+
+ Fe(s)  H2(g) + Fe
+
(c) 2 NO3 (aq) + 8 H
(aq)
(a) 2.14 V; (b) 0.45 V; (c) 0.62 V
(aq)
2+
+ 3 Cu(s)  2 NO(g) + 4 H2O(l) + 3 Cu
Strategy:
PS9-1
(aq)
Answer Key, Problem Set 9
1) Separate the reaction into two half reactions.
2) Look up the standard reduction potential for the reduction half (in a table). You will use this “as
is”.
3) Recognize that the reverse of the oxidation half reaction should match a reduction half reaction
in the Table. The standard oxidation potential is just the opposite of the reduction potential.
4) Sum the standard reduction potential (Ered°) and oxidation potential (Eox°) to get Ecell°.
Execution of Strategy:
(a) Ni2+(aq) + Mg(s)  Ni(s)
+ Mg2+(aq)
Ni2+(aq) + 2 e- → Ni(s)
Mg(s)  Mg
2+
(aq)
Ered°  -0.23 V
+ 2e
-
Eox°  -Ered°(Mg2+)  -(-2.37) V
Ni2+(aq) + Mg(s)  Ni(s)
Ecell°  -0.23 + (+2.37)  2.14 V
(b) 2 H+(aq) + Fe(s)  H2(g) + Fe2+(aq)
2 H+(aq) + 2 e- → H2(g)
Fe(s)  Fe
2+
(aq)
+ 2e
Ered°  0 V
-
Eox°  -Ered°(Fe2+)  -(-0.45) V
Ni2+(aq) + Mg(s)  Ni(s)
Ecell°  0 + (+0.45)  0.45 V
(c) 2 NO3-(aq) + 8 H+(aq) + 3 Cu(s)  2 NO(g) + 4 H2O(l) + 3 Cu2+(aq)
NO3-(aq) + 4 H+(aq) + 3 e- → NO(g) + 2 H2O(l)
Cu(s)
 Cu2+(aq) + 2 e-
Ered°  0.96 V
Eox°  -Ered°(Cu2+)  -(+0.34) V
2 NO3-(aq) + 8 H+(aq) + 3 Cu(s)  2 NO(g) + 4 H2O(l) + 3 Cu2+(aq)
Ecell°  0.96 + (-0.34)  0.62 V**
** NOTE: Since voltage is a “per electron” kind of quantity (energy per electron), reduction (or cell)
potentials do not change when the (half) reaction is multiplied through by a coefficient! In
other words:
2 NO3-(aq) + 8 H+(aq) + 6 e- → 2 NO(g) + 4 H2O(l) ALSO has Ered°  0.96 V (NOT 2(0.96 V)!)
This is an important difference from H, S, or G. Because of this, one need not actually
“bother” to multiply the half reactions above if one only wants to get the Ecell° (as in this
problem). However, it should be pointed out that one must do that multiplication in order to
find n and thus G° (see Problem #8 on this set)
3. 18.48.
Consider the voltaic cell (my picture is not as “pretty” [or accurate] as the text graphic, but the key elements are
there):
(a) Determine the direction of electron
flow and label the anode and the
cathode.
(b) Write a balanced equation for the
overall reaction and calculate Ecell°.
(c) Label each electrode as negative or
positive.
salt bridge containing
NaNO3(aq)
Pb(s)
Pt(s)
Cl2(g) [bubbling over
(d) Indicate the direction of anion and
cation flow in the salt bridge.
1M
Pb2+(aq)
PS9-2
1 M Cl-(aq)
electrode;
source, gas
inlet and
glassware not
shown]
Answer Key, Problem Set 9
Answers:
(d) NO3- flows left and Na+ flows right
(a),(c)
e-
anode (-)
(a)
(+) cathode (a),(c)
-
NO3
Pb(s)
Pt(s)
Na+
Cl2(g) [bubbling over electrode;
source, gas inlet and
glassware not shown]
1M
Pb2+(aq)
Pb(s)  Pb2+(aq) + 2 e-
1 M Cl-(aq)
(b) Pb(s) + Cl2(g)  Pb2+(aq)
+
Cl2(g) + 2 e-  2 Cl-(aq)
2 Cl-(aq) ; Ecell°  1.49 V
Reasoning:
1) NOTE: You cannot ASSUME that the anode will always be on the left! You really have to
do part (b) (or at least consider the potentials) before you can do part (a).
2) To determine the balanced equation, start with the reduction half reactions:
a) On the left, either Pb(s)  Pb2+(aq) + 2 e-
or
Pb2+(aq) + 2 e-  Pb(s) is occurring.
b) On the right, either Cl2(g) + 2 e-  2 Cl-(aq) or 2 Cl-(aq)  Cl2(g) + 2 e- occurs.
3) Then look at the reduction potentials from Table 18.1:
Pb2+(aq) + 2 e-  Pb(s) Ered°  -0.13 V
Cl2(g) + 2 e-  2 Cl-(aq) Ered°  +1.36 V
4) From these values, you can conclude that it is more favorable to reduce Cl2 than Pb2+. This
means that Cl2 is the one that actually gets reduced in the cell (not Pb2+). That means “keep
the second half reaction” and flip the first one. (In other words, Pb is the other reactant with
Cl2). The balanced equation is thus obtained by:
Cl2(g) + 2 e-  2 Cl-(aq) Ered°  +1.36 V
Pb(s)  Pb2+(aq) + 2 e- Eox°  -(-0.13 V)
Pb(s) + Cl2(g)  Pb2+(aq)
+
2 Cl-(aq) ; Ecell° (1.36 + 0.13) 1.49 V
5) Once you know the half reactions (directionality), you can answer part (a). The electrons
must be coming to the side with Cl2 (since it is getting reduced; electrons are on the left side
of the half reaction), and from the side with Pb (it is getting oxidized; electrons are on the
right side of the half reaction).
6) This also allows you to assign the charges on the electrodes, as long as you remember that
in a voltaic cell, electrons “get to go where they want to go”—that is, they flow from negative
to positive (this is because the spontaneous redox reaction is actually occurring in the cell,
dictating the direction of electron flow). Since we already determined that the electrons flow
to the side with Cl2, that electrode must be the positive one.
7) Although not asked here, knowing the half reactions also allows you to state which is the
anode (where oxidation occurs) and the cathode (where reduction occurs).
8) Negative ions in the salt bridge always flow in the opposite direction as electrons in the wire
do (Think of it as a big loop. Negative charges cannot both flow to the same side—the
whole purpose of the salt bridge is to avoid the build-up of charge!)
PS9-3
Answer Key, Problem Set 9
4. 18.14 Does a large positive electrode potential (i.e., “reduction potential”) indicate a strong oxidizing agent or a strong
reducing agent? What about a large negative electrode potential?
Answers: A large positive electrode potential indicates a strong oxidizing agent (which would be on
the left side of the reduction half reaction). This is because a reduction half reaction
involves an oxidizing agent as a reactant (on the left), getting reduced. Since a large
positive E of any kind means “more favorable” (than a smaller one), a large positive
reduction potential indicates a good oxidizing agent.
A large negative reduction potential indicates a strong reducing agent, but on the right side of
the reduction half reaction. Technically, I would prefer to say that a large positive oxidation
potential indicates a strong reducing agent (with the reducing agent being a reactant, on the left
side of the oxidation half reaction). This would be in complete parallel to my response in the
prior paragraph.
5. NT1.
+
+
You are told that metal A is a better reducing agent than metal B. What, if anything, can be said about A and B ?
Explain.
Answer: A+ will be a poorer oxidizing agent than B+. If A is better at giving (than B), then A+ will be
poorer at taking (than B+). The better the “giving power” of a reducing agent, the poorer is the
tendency of the species formed to take electrons back. This is very analogous to the concept
that “the stronger the acid, the weaker is its conjugate base”.
6. 18.55 & 18.56
2+
2+
18.55 Which metal could you use to reduce Mn ions but not Mg ions?
Answer: Al
Reasoning: The standard reduction potential for Mn2+ is -1.18 V, and for Mg2+ it is -2.37 V. Thus, in
order for a metal to reduce Mn2+ under standard conditions, the oxidation potential would have
to be more positive than 1.18 V (so that Ecell° (= Ered° + Eox°) will be positive). That means the
reduction potential (of the cation) would need to be more negative than -1.18 V. Similarly, for
Mg2+, look for a reduction potential (for the cation) more positive than -2.37 V (so that the
cation would be able to oxidize Mg(s), which means that the metal would not be able to reduce
Mg2+). The only cation that has a potential more negative than -1.18 V and more positive than
-2.37 V is Al3+, with a reduction potential of -1.66 V. But be careful to state that it is the metal,
Al, that is used to do the reduction.
2+
2+
18.56 Which metal can be oxidized with an Sn solution but not with an Fe solution?
Answer: Ni and Cd
Reasoning: The standard reduction potential for Sn2+ is -0.14 V, and for Fe2+ it is -0.45 V. Thus, in
order for a metal to be oxidized by Sn2+ under standard conditions, the oxidation potential
would have to be more positive than +0.14 V, which means the reduction potential (of the
cation) would need to be more negative than -0.14 V. Similarly, for Fe2+, look for a reduction
potential more negative than -0.45 V. The only cations that has a potential between these
values Ni2+ and Cd2+, with reduction potentials of -0.23 and -0.40 V. But again, be careful to
state that it is the metals, Ni and Cd, that are being oxidized.
7. NT2.
+
Consider only the following species:
-
+
2+
Na , Cl , Ag , Ag, Zn , Zn, Pb
(a) Which is the strongest oxidizing agent?
(b) Which is the strongest reducing agent?
2-
(c) Which species can be oxidized by SO4 in acid at standard
state conditions?
(d) Which species can be reduced by Al at standard state conditions?
PS9-4
Answers:
(a) Ag+
(b) Zn
(c) Zn and Pb
(d) Ag+ and Zn2+
NOTE: If you have
not included
reasoning on your
paper, add it now
before you look at the
reasoning below.
Deductions will result
without reasoning!
Answer Key, Problem Set 9
Reasoning:
(a) Look for the species on the left of the arrows in Table 18.1 (i.e., on the left side of a reduction
half reaction), and see which has the highest (most positive) reduction potential. This is
because a higher (more positive) reduction potential means “more favorable reduction to occur”
which means the species being reduced (the oxidizing agent) is a “stronger” one:
Na+ (-2.71 V)
Cl- (not on left  not considered an oxidizing agent [extremely poor; It would have to become Cl2-!?])
Ag+ (+0.80 V)
Ag (not on left  not considered an oxidizing agent [extremely poor; would have to become Ag -!?])
Zn2+ (-0.76 V)
Zn (not on left  not considered / negligibly poor)
Pb (not on left  not considered / negligibly poor)
(b): Look for the species on the right of the arrows in Table 18.1 (i.e., on the right in a reduction half
reaction), and see which has the highest (most positive) oxidation potential (i.e., flip the halfreactions in Table 18.1 and change the sign). This is because a higher (more positive) oxidation
potential means “more favorable oxidation to occur” which means the species being oxidized
(the reducing agent) is a “stronger” one:
Na+, Ag+, and Zn2+: not considered reducing agents (not on the right side of any reduction half
reaction  N/A or “extremely poor”)
Cl (-1.36 V)
Ag (-0.80 V)
Zn (-(-0.76 V) = +0.76 V)
Pb (-(-0.13) = +0.13 V)
(c) In order for somebody to be “oxidized by SO42-”, SO42- must accept electrons (i.e., be reduced).
Thus consider the half reaction in which SO42- is reduced (i.e., the reduction half reaction) and
look at its potential. The standard reduction potential for SO42- is +0.20 V. Thus any reducing
agent with an oxidation potential greater than -0.20 V will yield a positive Ecell° and thus will be
able to react with SO42-. So look for the species on the right in Table 18.1 that are below where
SO42- is (remember that for the oxidation potential, you will need to change the sign!). THIS IS
TRICKY! Please take time to fully understand what you are doing and why. Please don’t
memorize the “up here reacts with down there” kind of phrases instead of learning why the result
is what it is. Shortcuts should come only after you have understood the concept on which the
shortcut is based!
From part (b), only Zn (+0.76 V) and Pb (+0.13 V) meet the criteria.
(d) Being reduced by Al means Al is acting as a reducing agent and is getting oxidized. Thus look
at the half reaction in which Al is oxidized (i.e., electrons are being produced). [This oxidation
half reaction will be the opposite of the reduction half reaction for Al3+ in Table 18.1 which has a
value of -1.66 V.] The standard oxidation potential for Al is thus +1.66 V (Ered°(Al3+) = -(-1.66
V) = +1.66 V). Thus any oxidizing agent with a reduction potential greater than -1.66 V will yield
a positive Ecell° and thus will be able to react with Al. So look for the species on the left in Table
17.1 that are above the line where Al is.
From part (a), Ag+ (+0.80 V) and Zn2+ (-0.76 V) meet the criteria. Na+’s reduction potential is
too negative (-2.71 V!)
8. 18.66b
Use tabulated electrode potentials to calculate Grxn for each reaction at 25C.
(b) O2(g) + 2 H2O(l) + 2 Cu(s)  4 OH-(aq) + 2 Cu2+(aq)
Answer: -2 x 104 J or -2 x 101 kJ
PS9-5
Answer Key, Problem Set 9
Strategy:
1) Recognize that the appropriate equation to use is: G° = -nFE°
2) As such, one needs to determine Ecell° and n for the balanced equation.
3) Break the reaction into half reactions.
4) Get the reduction potentials from Table 18.1 and calculate the Ecell° as in Q2 above.
5) Find n by looking at the half reactions. n is the number of electrons transferred ( lost  gained)
Execution of Strategy:
O2(g) + 2 H2O(l) + 2 Cu(s)  4 OH-(aq) + 2 Cu2+(aq) becomes:
O2(g) + 2 H2O(l) + 4 e-  4 OH-(aq) Ered°  +0.40 V
2 Cu(s)  2 Cu2+(aq) + 4 e-
Eox°  -Ered°  -(0.34) V

(recall, Ered’s don’t change with “coefficient”)
Ecell° (0.40 -0.34) 0.06 V and n  4
G°  -nFE°  -(4 mol e-)(96485 C/mol e-)(+0.06 J/C)  -23156 J  -2 x 104 J  -2 x 101 kJ
9. 18.110 The cell potential of this electrochemical cell depends son the gold concentration in the cathode half-cell:
+
3+
Pt(s)|H2(g, 1.0 atm)|H (aq, 1.0 M)||Au (aq, ? M)|Au(s)
3+
(a) What is the concentration of Au in the solution if Ecell is 1.22 V? (b) Without doing a calculation, state whether
+
the cell potential will increase or decrease if a bit of solid NaOH were added to the solution containing the H (i.e.,
the electrode in contact with the Pt(s))? Explain without using the Nernst Equation (use “Le Chatelier” type of
argument as discussed in the “Voltaic Cells” experiment reading/handout.)
Answers: (a) ~6 x 10-15 M;
(b) Ecell would increase, because you would decrease the concentration of a product, H+ (since
OH- would react with it to make H2O). When the concentration of a product is decreased, the
reaction will have a greater tendency to “go” in the forward direction. This is reflected in a
greater (more positive) cell potential. NOTE: You needed to look at the balanced equation for
the cell reaction to see if H+ was a reactant or a product in order to answer this question!
Work / Reasoning:
1) Recognize that the equation to use in this problem is the Nernst Equation:
0.0592
o
log Q
Ecell  Ecell

n
2) As such, you’ll need to find Ecell° and n, and determine the balanced chemical equation (in
order to get an expression for Q)
3) Finding Ecell°, n, and the balanced equation:
2 H+ + 2 e- → H2;
Au
3+
-
+ 3 e → Au;
Ered° = 0.00 V ← More negative, less favorable, so flip it.
Ered° = +1.50 V
Also need to multiply first equation by 3 and second by 2 to get the same number of electrons:
3 H 2 → 6 H + + 6 e -;
3+
2 Au + 6 e
3+
3 H2 + 2 Au
-
→ 2 Au;
E° = 0.00 V
E° = +1.50 V
+
→ 3 Au + 6 H ;
Ecell°  0.00 +
1.50 
1.50 V;
[H ]6
Q = P 3 [Au3  ]2 ; n = 6
H
2
4) Substituting in (recalling that Ecell is given in the problem as 1.22 V)
PS9-6
Answer Key, Problem Set 9
1
1
0.0592 V
0.0592 V
log
 0.28 V  
log
 1.22 V  1.50 V 
2
2
3
3+
6
6
 Au3+ 
1  Au 
6
Ecell

 1028.37.. 
 6 x 10-15 M
10. 18.80
6( 0.28 V)
1
 log
2
0.0592 V
 Au3+ 
1
 Au 
3+
-14
 1 x 10
6
2

 28.37..  log
2
1
 Au3+   28.37..
10

1
 Au3+ 
2
1
 Au3+  
 6.46..x 10-15 M
28.37..
10
(technically, this result doesn’t even have 1 SF! But let’s not dwell on that…
Consider the concentration cell:
salt bridge containing
NaNO3(aq)
Pb(s)
Pb(s)
5.0 x 10-3
M Pb2+(aq)
2.5 M
Pb2+(aq)
(a) Label the anode and cathode.
(b) Indicate the direction of electron flow.
2+
(c) Indicate what happens to the concentration of Pb in each half cell (as time goes by and rxn occurs)
Answers:
Cathode
e-
salt bridge containing
NaNO3(aq)
Pb(s)
Pb(s)
Anode
[Pb2+]
decreases
2.5 M
Pb2+(aq)
5.0 x 10-3
M Pb2+(aq)
[Pb2+]
increases
Reasoning:
Pb2+ will “want” to increase its concentration in the right side and decrease it in the left side
(tendency for “high concentration” to flow to “low concentration”). To make this happen
“chemically” the following reactions should occur:
Left side: Pb2+ + 2 e-  Pb
Right side: Pb  Pb2+ + 2 e-
PS9-7
Answer Key, Problem Set 9
Thus, the left side is the cathode (reduction is occurring), and the right side is the anode
(oxidation).
It also means that electrons flow from right to left.
11. NT3.
True or false questions (correct if false):
(a) The chemical reaction that occurs in a voltaic cell is the spontaneous cell reaction, but the chemical reaction that
occurs in an electrolysis cell is the non-spontaneous cell reaction (the reverse of the spontaneous cell reaction).
TRUE.
(b) The anode is the electrode at which reduction occurs, and thus it is always the negative electrode in a voltaic cell
(a battery). FALSE
Corrected #1: The anode is, by definition, the electrode at which oxidation occurs, and thus it is
always the negative electrode in a voltaic cell (a battery).
OR
Corrected #2: The cathode is, by definition, the electrode at which reduction occurs, and thus it
is always the positive electrode in a voltaic cell (a battery).
(c) The anode is the electrode at which reduction occurs, and thus it is always the negative electrode in an electrolysis
cell. FALSE
Corrected #1: The anode is, by definition, the electrode at which oxidation occurs, and thus it is
always the positive electrode in an electrolysis cell.
OR
Corrected #2: The cathode is, by definition, the electrode at which reduction occurs, and thus it
is always the negative electrode in an electrolysis cell.
Remember that in an electrolysis cell, electrons are “forced” from the negative lead of the
external power source INTO the negative electrode of the cell (i.e., the electrons are “forced”
to go where they would not spontaneously go) and so reduction will occur at the negative
electrode (and oxidation at the positive one).
What is always true in either kind of electrochemical cell is that the anode is where oxidation
occurs and the cathode is where reduction occurs; what changes is whether or not the
anode (or cathode) is positive or negative; that depends on the type of cell you have!
(d) The standard potential of a cell is constant at a given temperature, and does not change with temperature.
FALSE. Corrected: The standard potential of a cell is constant at a given temperature, but it
changes as temperature changes.
We know from the thermodynamics chapter that G°’s vary with T (G° = H° – TS°). But Ecell
changes as G does (G = -nFEcell), so Ecell must also vary with T.
(e) All product-favored oxidation-reduction reactions have a standard cell voltage Ecell with a negative sign.
FALSE. Corrected: All product-favored oxidation-reduction reactions have a standard cell
voltage Ecell with a positive sign.
Reasoning: Product-favored means K > 1. Standard state conditions means Q = 1. So
“product-favored at standard state conditions” means K > Q or “reaction is spontaneous (at
standard state conditions)”. But this means G < 0, which means that Ecell > 0, which
means Ecell is “positive”.
PS9-8
Answer Key, Problem Set 9
12. 18.136
An electrochemical cell has a positive standard cell potential but a negative cell potential. What is true of Q
and K for the cell?
(a) K > 1; Q > K
(b) K < 1; Q > K
(c) K > 1; Q < K;
(d) K < 1; Q < K
Answer (with explanation:
(a). i) A positive Ecell means “spontaneous at standard state conditions” (where Q = 1) and thus
the system must have K > 1 (i.e., Q > 1 at equilibrium).
ii) A negative Ecell means the conditions are such that the reaction not spontaneous in the
forward direction (as the equation was originally written)—in fact, the reaction is
spontaneous in the reverse direction. That means Q must be “too big” to be at equilibrium
(because Q is “products” over “reactants”), which means Q > K
13. 18.89 Consider the electrolytic cell:
salt bridge containing
NaNO3(aq)
(a) Label anode and cathode. Write half
reactions.
(b) Indicate electron flow.
Ni(s)
Cd(s)
(c) Label terminals on battery as positive or
negative, and calculate the minimum
voltage necessary to drive the reaction.
(Assume standard state conditions!)
Ni2+(aq)
Cd2+(aq)
Strategy (one option anyway; there are others):
1) Recognize that in an electrolysis cell, the reaction that is occurring in the cell is the one that is
not spontaneous. That means that in order to proceed with the labeling in this cell (an
electrolysis cell), you must first determine what reaction would be spontaneous without
the “push” from the battery driving the cell. In other words, first pretend that the cell is a
voltaic cell. (Then you will reverse things later.)
2) Treating the cell as if it were a voltaic cell, calculate the Ecell° using reduction potentials (from
Table 18.1) and write the balanced equation corresponding to this Ecell°.
3) Recognize that the reaction you have just determined in (2) above must be reversed in order to
describe the reaction that will be occurring in the actual cell in this problem (an electrolysis cell,
not a voltaic cell). As such, just write the reverse of the balanced equation in (2).
4) Using that equation (broken into half reactions) as your guide, determine the half reactions at
each electrode. From there, you can determine which is the cathode and anode (where
reduction and oxidation are occurring, respectively), and the direction of electron flow.
5) Recognize that the Ecell° you calculated in (2) above is the minimum voltage that would need to
be applied to drive the reaction. (Technically, slightly higher than that, because if it were at that
voltage, the cell would be at equilibrium [no net reaction in either direction].) Why? Let’s take
an example. If the Ecell° in (2) were 1.5 V, then the Ecell° for the reaction in the electrolysis cell is
-1.5 V. That means that +1.5 V is needed to be applied to get the cell potential to be 0 (i.e., at
equilibrium). Any voltage greater than that will make the reaction occur.
Execution of Strategy (and answer, below):
PS9-9
Answer Key, Problem Set 9
Ni2+ + 2 e- → Ni;
Cd
Ni
2+
2+
Ered° = -0.23 V
-
+ 2 e → Cd;
Ered° = -0.40 V ← More negative, less favorable, so flip it.
2+
+ Cd → Ni + Cd ; Ecell° = +0.17 V for spontaneous reaction (if no external voltage were applied)
This is the reaction that would occur without any external voltage “pushing” in the opposite
direction. Thus, in the electrolytic cell noted in this problem, the reaction occurring is
represented by the reverse of the above:
Ni + Cd2+ → Ni2+ + Cd;
Ecell° = -0.17 V (for the reaction occurring in the cell)
Thus, Ni is getting oxidized to Ni2+ in this cell (on the left, anode) and Cd2+ is getting
reduced to Cd.(on the right, cathode).
Thus, electrons must be flowing toward the right (where reduction occurs).
Thus, electrons must be being “pushed” to the right (electrolytic cell), and so the negative
electrode of the battery must be the one on the right side.
**The negative sign of Ecell° here indicates that the reaction won’t happen without an
external voltage “pushing” it. The value of -0.17 V means that the external voltage applied
would have to be +0.17 V just to get the system to be at equilibrium. Anything higher than
that will be needed make the reaction occur as shown.
e-
(+)
Anode
(-)
eCathode
Ni(s)
Ni  Ni2+ + 2e-
Cd(s)
Cd2+(aq)
Ni2+(aq)
Cd2+ + 2e-  Cd
14. 18.91 Answer: K+ + e-  K; 2 Br-  Br2 + 2 e- (Don’t worry about knowing the state
designations)
15. 18.100
(a) 6.8 A 
6.8 C
; 1 min  60 sec
s
 72 min x
(b)
60 s 6.8 C 1 mol e- 1 mol Ag plated 107.87 g
x
x
x
x
 32.8..  33 g Ag plated
min
s
96485 C
1 mol Ag
1 mol e2+
Would the number of moles of electrons passed be the same if the only difference in the cell were that Cu was
+
plated out as Cu(s) (instead of Ag being plated out as Ag(s))? Would the number of moles of solid produced be
the same these two cases? Explain. Why is it so important to write out the half reaction when doing electrolysis
problems such as this?)
Answers:
i) Yes, moles of electrons plated is determined by the current and time, not the chemistry.
PS9-10
Answer Key, Problem Set 9
ii) No. The moles of solid would be half as many, because it takes 2 e- to plate out one Cu
atom, whereas it only takes 1 e- to plate out one Ag. (In other words, the mole ratio in the
equation above would be 1 mol Cu / 2 moles e- ; i.e., “1/2”)
iii) The half reaction tells you the mole ratio of “moles of metal” to “moles of electrons”. That
is how you get the proper mole ratio to do the stoichiometry! 
16. NT4. (i) Balance the equation under the conditions specified
(ii) State the value of n for each balanced redox
equation.
As2O3(s) + NO3-(aq) → H3AsO4(aq) + NO(g) (acidic conditions)
NO2-(aq) + Al(s) → NH3(g) + AlO2-(aq)
(basic conditions)
(a)
(b)
Answers:
(a) (i) 3 As2O3(s) + 7 H2O + 4 NO3-(aq) + 4 H+ → 6 H3AsO4(aq) + 4 NO(g);
(b) (i)
NO2-(aq)
-
2 Al(s) + H2O + OH → NH3(g) + 2
AlO2-(aq);
(ii) n = 12
(ii) n = 6
Work / Procedure:
See handout sheet for the procedure followed here (“Half reaction” method). I did not assign
oxidation numbers. Rather, I let the procedure indicate (at the end) which half was the oxidation half
and which the reduction half.
Note: State designations for H2O(l) and H+(aq) will be omitted in this answer simply for convenience.
As2O3(s) + NO3-(aq) → H3AsO4(aq) + NO(g) (acidic conditions)
(a)
1st Half Reaction:
As2O3(s) → H3AsO4(aq)
(Separate out the species containing As from those with N [see below])
As2O3(s) → 2 H3AsO4(aq)
(Balance atoms other than O and H)
As2O3(s) + 5 H2O → 2 H3AsO4(aq)
(Balance O atoms by adding H2O’s)
As2O3(s) + 5 H2O → 2 H3AsO4(aq) + 4 H+ (Balance H atoms by adding H+’s)
As2O3(s) + 5 H2O → 2 H3AsO4(aq) + 4 H+ + 4 e-
(Balance charge by adding e-’s)
→ Electrons are being produced, so someone is getting oxidized, and this is the oxidation half reaction
-------------------------------------------------
2nd Half Reaction:
NO3-(aq) → NO(g)
NO3-(aq) → NO(g)
NO3-(aq)
NO3-(aq)
(Balance atoms other than O and H—already done!)
→ NO(g) + 2 H2O
(Balance O atoms by adding H2O’s)
+
+ 4 H → NO(g) + 2 H2O
(Balance H atoms by adding H+’s)
NO3-(aq) + 4 H+ + 3 e- → NO(g) + 2 H2O
(Balance charge by adding e-’s)
→ Electrons are being used, so somebody is getting reduced, and this is the reduction half reaction
Bringing half reactions together:
To get the number of electrons to be the same in both reactions (which will be the value of n!),
multiply the first by 3 and the second by 4; then add them together:
3 As2O3(s) + 15 H2O → 6 H3AsO4(aq) + 12 H+ + 12 e4
NO3-(aq)
+
+ 16 H + 12 e
3 As2O3(s) + 15 H2O + 4
-
→ 4 NO(g) + 8 H2O
NO3-(aq)
n is the number of electrons transferred
“per equation unit” of reaction. So n = 12
here.
+ 16 H+ → 6 H3AsO4(aq) + 12 H+ + 4 NO(g) + 8 H2O
PS9-11
Answer Key, Problem Set 9
Checking for “same species” on both sides:
Cancelling out H2O’s and H+’s that appear on both sides (the same number of times) yields:
3 As2O3(s) + 7 H2O + 4 NO3-(aq) + 4 H+ → 6 H3AsO4(aq) + 4 NO(g) ;
n = 12
Check charge: on left, 0 and on right, also zero. Checks!
(b)
NO2 (aq) + Al(s)
→ NH3(g) + AlO2-(aq)
(basic conditions)
Note: All early steps are the same as in part (a), so parenthetical comments will be omitted.
1st Half Reaction:
NO2-(aq) → NH3(g)
NO2-(aq) → NH3(g) + 2 H2O
NO2-(aq) + 7 H+ → NH3(g) + 2 H2O
NO2-(aq) + 7 H+ + 6 e- → NH3(g) + 2 H2O
→ Electrons are being used, so somebody is getting reduced, and this is the reduction half reaction
-------------------------------------------------
2nd Half Reaction:
Al(s) → AlO2-(aq)
Al(s) + 2 H2O → AlO2-(aq)
Al(s) + 2 H2O → AlO2-(aq) + 4 H+
Al(s) + 2 H2O → AlO2-(aq) + 4 H+ + 3 e→ Electrons are being produced, so someone is getting oxidized, and this is the oxidation half reaction
Bringing half reactions together:
To get the number of electrons to be the same in both reactions (which will be the value of n!),
multiply the second by 2; then add them together:
NO2-(aq) + 7 H+ + 6 e- → NH3(g) + 2 H2O
2 Al(s) + 4 H2O → 2 AlO2-(aq) + 8H+ + 6 e-
n is the number of electrons transferred
“per equation unit” of reaction. So n = 6
here.
NO2-(aq) + 7 H+ + 2 Al(s) + 4 H2O → NH3(g) + 2 H2O + 2 AlO2-(aq) + 8 H+
Cancel out species that appear on both sides the same number of times (7 H+’s and 2 H2O’s):
NO2-(aq) 2 Al(s) + 2 H2O → NH3(g) + 2 AlO2-(aq) + H+;
n=6
(acidic solution)
“Converting” to basic conditions:
To “convert” this to basic conditions, add one OH- to both sides for every H+.
NO2-(aq) 2 Al(s) + 2 H2O + OH- → NH3(g) + 2 AlO2-(aq) + H+ + OH-
Make each “OH- + H+” become an H2O molecule (i.e., you are “neutralizing” the H+’s with OH-‘s):
NO2-(aq) 2 Al(s) + 2 H2O + OH- → NH3(g) + 2 AlO2-(aq) + H2O
Finally, cancel out any H2O’s that appear on both sides (if applicable):
NO2-(aq) 2 Al(s) + H2O + OH- → NH3(g) + 2 AlO2-(aq);
Check charge: Left, -1 + -1 = -2; Right, 2 x -1 = -2. Checks!
PS9-12
n=6
(basic solution)