3/12/16
Chem 30 Unit 1: REDOX Building A Redox Table
Given experimental evidence, it is possible to make ones own redox table. This is an
important skill to develop: you will need to use it in a lab later this week and it almost
always gets tested on the big D.
In this hand out, you will see 3 worked problems. I’ve used one method for the first two
problems, and a second method for the last one. Use whichever one you find most
helpful. After the worked problems, you have some homework problems to work on.
Question:
The following reactions were carried out in a laboratory. Use the information
given below to construct a small reduction/oxidation table.
Test 1: Cobalt (II) chloride and zinc metal react to form cobalt solid and zinc
chloride.
Test 2: Magnesium nitrate and zinc metal are mixed, but there is no evidence of
reaction.
Solution:
Step 1: Let’s look at Test 1. Write out a net ionic reaction.
To write a net ionic equation,
start by writing a normal
balanced equation.
CoCl2(aq) + Zn(s)  Co(s) + ZnCl2(aq)
2+
Co
(aq)
+ 2Cl
-
(aq)
+ Zn(s)Co(s) + Zn
Co2+(aq) + Zn(s)  Co(s) + Zn2+(aq)
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2+
(aq)
+
2Cl-(aq)
Then break all of the (aq)
compounds into ions. Cancel
out the spectator ions (i.e. the
ones that do not exchange
electrons).
This is your net ionic
equation.
Step 2: Identify the reducing agent (reducing agents are oxidized, so they will lose
electrons) and the oxidizing agent (oxidizing agents are reduced, so they will gain
electrons).
Co2+(aq) + Zn(s)  Co(s) + Zn2+(aq)
OA
RA
Cobalt goes from ion to
element. It gains electrons,
therefore it is an OA.
Zinc goes from a element to
an ion. It gains electrons,
therefore it is a RA.
Step 3: Recall that for a reaction to take place spontaneously, the OA must be above the
RA. Since this reaction took place in the lab spontaneously, place cobalt higher than zinc
in the table, like below.
Recall:
2+
Co
(aq)
+
OA
Zn(s)
+
= spontaneous
RA
Step 4: Repeat steps 1-3 with Test 2.
Mg(NO3)2(aq) + Zn(s)  Mg(s) + Zn(NO3)2(aq)
Mg2+(aq) + Zn(s)  Mg(s) + Zn2+(aq)
RA
OA
Recall:
Zn(s)
RA
+
Mg2+(aq)
+
= non –spontaneous
OA
Step 5: You now see that cobalt is above zinc and that zinc is above magnesium. You can
now complete the table by writing each half-reaction in the appropriate order.
Co2+(aq) + 2e-  Co(s)
Zn2+(aq) + 2e-  Zn(s)
Mg2+(aq) + 2e-  Mg(s)
You can check your work by referencing the redox table in the data booklet. Hey, you
could even copy it directly…but if you did, you’d never learn anything about redox. So
on second thought, don’t do that.
Here’s another worked problem. This one has more tests in it, but the steps are the same.
In a laboratory, three reactions were recorded. Use these reactions to establish a
reduction/oxidation table for the four metals.
Test 1: Radium metal reacts with a solution of 1.0 mol/L aluminum chloride. Solid
aluminum is observed, plated to the radium.
Test 2: Aluminum metal reacts with a solution of 1.0 mol/L cobalt (III) chloride. Co2+ ions
were identified as the products.
Test 3: Neither radium metal or aluminum metal will react in a solution of 1.0 mol/L
KNO3(aq).
Step 1: Write out the net ionic reactions. Step 2: Label RA and OA. Step 3: decide
whether reaction is spontaneous and use the spontaneity rule to order the metals.
Ra(s) + Al3+ (aq)  Ra2+ (aq) + Al(s)
RA
OA
Al3+(aq)
+
Note: balancing equations
in this step is optional and
should not affect your
answer.
Ra(s)
Al(s) + Co3+(aq)  Al3+(aq) + Co2+(aq)
RA OA
Co3+(aq)
+
In this reaction, cobalt
goes from 3+ to 2+. But it
still gains electrons, so
you can still think of it as
an oxidizing agent.
Al(s)
From the first two tests, we know aluminum is above radium, and that cobalt is above
aluminum. The last test tells us that potassium will not react with any metals given. I.e.,
potassium can not take electrons from either metal to turn them into ions. It is therefore
the weakest OA and must go on the bottom of the list.
Step 5: Write out the table.
Co3+(aq) + 3e-  Co(s)
Al3+(aq) + 3e-  Al(s)
Ra2+(aq) + 2e-  Ra(s)
K+(aq) + e-  K(s)
Note: you can check your
list by using the
spontaneity rule for each
test described in the
question.
You can also build a redox table by remembering the trends present in these tables. The
strongest oxidizing agents are always on the top left. A strong oxidizing agent is one that
is really good at being reduced, or at gaining electrons. You can now place the metal
which gains electrons best on the top of the table and fill down from there.
From the following hypothetical net reactions, establish the redox table.
Test 1: R2+(aq) + X(s)  X2+(aq) + R(s)
Test 2: Q2+(aq) + X(s)  no reaction
Test 3: X2+(aq) + M(s)  M2+(aq) + X(s)
Test 4: Q2+(aq) + M(s)  no reaction
Solution:
Test 1: R2+(aq) is able to take two electrons from X(s). Therefore, R is a better oxidizing
agent than X. Place R above X in the final table.
Test 2: Q2+(aq) could not take electrons from X(s). Therefore, Q must be below X on the
table.
Test 3: X2+(aq) is able to take two electrons from M(s). Therefore, X must be above M on
the table.
Test 4: Q2+(aq) could not take electrons from M(s). Therefore, M must be above Q on the
table.
R2+(aq) + 2e-  R(s)
X2+(aq) + 2e-  X(s)
M2+(aq) + 2e-  M(s)
Q2+(aq) + 2e-  Q(s)
Practice Problems:
1. The results from six lab tests are recorded below. Write the net ionic reactions for each
test and use them to build a redox table. The nitrate ion, NO3-, is a spectator ion.
Test 1: A strip of cobalt, Co(s), was placed in a solution of 1.0 mol/L Pb(NO3)2(aq). The
Co(s) started to dissolve to Co2+(aq) and crystals of lead started to form.
Test 2: A bar of lead, Pb(s), began to dissolve in 1.0 mol/L Cu(NO3)2(aq), forcing copper
out of solution.
Test 3: A rod of copper, Cu(s), was placed in a solution of 1.0 mol/L Co(NO3)2(aq). No
reaction occurred.
Test 4: A rod of copper, Cu(s), was placed in a 1.0 mol/L solution of iodine, I2(l). Copper
began to dissolve.
Test 5: No reaction occurred when a strip of silver, Ag(s), was placed in a 1.0 mol/L
solution of I2(l).
Test 6: Silver crystals formed when a cobalt strip, Co(s), was immersed in a 1.0 mol/L
AgNO3(aq).
2. For the reduction table established in question 1, name:
a) the best oxidizing agent
b) the worst oxidizing agent
c) the best reducing agent
d) the worst reducing agent
3. Using the hypothetical reactions below, establish a redox table.
Test 1: X3+(aq) + D(s)  D3+(aq) + X(s)
Test 2: X3+(aq) + T(s)  no reaction
Test 3: T3+(aq) + Z(s)  Z3+(aq) + T(s)
Test 4: D3+(aq) + Z(s)  Z3+(aq) + D(s)
4. Use the standard reduction table (in data booklet) to answer the following.
a) Will cobalt, Co(s) dissolve in 1.0 mol hydrochloric acid?
b) Can zinc, Zn(s) reduce the copper (II) ion, Cu2+(aq)?
c) Can an aqueous solution of chlorine, Cl2(g), oxidize zinc, Zn(s)?
5. For each of the chemical changes below, state whether the initial substance was
oxidized or reduced.
a) Ba(s) is changed to Ba2+(aq).
b) Br2(l) is changed to 2Br-(aq)
c) Fe2+(aq) is changed to Fe3+(aq)
d) Sn4+(aq) is changed to Sn2+(aq)
e) Cr3+(aq) is changed to Cr(s)
f) Te2-(aq) is changed to Te(s)
6. When chlorine gas is passed over hot cesium, the two elements combine vigorously to
produce a white solid. The exothermic reaction releases energy as a brilliant flash of
light.
a) What is the compound product?
b) Write the net ionic reaction.
c) Write the two half-reactions.
d) Which is the oxidizing agent and which is the reducing agent?
7. Five spontaneous reactions are given below. Use these reactions to build a reduction
table of the five elements involved.
a) Br2(l) + 2Fe2+(aq)  2Br-(aq) + 2Fe3+(aq)
b) 2Fe3+(aq) + 2I-(aq)  I2(aq) + 2Fe2+(aq)
c) Cl2(g) + 2Br-(aq)  2Cl-(aq) + Br2(l)
d) Hg(l) + Br2(l)  2Br-(aq) + Hg2+(aq)
e) Hg2+(aq) + 2Fe2+(aq)  2Fe3+(aq) + Hg(l)
8. For the reduction table established in #7, which substance is the best oxidizing agent
and which substance is the best reducing agent?