Section 13.2 p. 568 - 582
Using the Reductions Table
• Why is this table called a “reduction half
reaction” table?
– all reactions listed have electrons in them and all
reactions show a gain of electrons.
• What would the table be called if it was
written backwards?
– a table showing a loss of electrons would be an
oxidation half-reaction table.
• Note that as you read the half reactions from
left to right, they all gain electrons. That is
why the reactions are called reduction half
reactions. Within each reduction half reaction,
one major species gains electrons. This
species is the oxidizing agent. Conversely, any
species that loses electrons is the reducing
agent.
• Given:
Pb(s) + 2 Ag+(aq)
• The reduction half reaction is:
Ag+(aq)
+ e–


2 Ag(s) + Pb2+(aq)
Ag(s) (G.E.R.)
• The oxidation half reaction is:
Pb(s)  Pb2+(aq) + 2 e –
(L.E.O.)
• The species gaining electrons is: Ag+(aq)
(oxidizing agent)
• The species losing electrons is: Pb(s)
(reducing agent)
• The oxidizing agents are found on the left side
and are listed in order from the most reactive or
strongest at the upper left-hand corner to the
least reactive or weakest near the lower left.
• Similarly, the most reactive or strongest reducing
agents are found near the lower right side of the
table and get weaker as they move upward.
• When a reduction half reaction (showing a gain
of electrons) is added to an oxidation half
reaction (showing a loss of electrons) we get a
REDOX (reduction-oxidation) reaction (net ionic
equation).
Determining Whether a Reaction is
Spontaneous or Non-Spontaneous
• A strip of zinc will react when placed in a solution of
copper (II) nitrate but a strip of copper will not react in
a solution of zinc nitrate. How does this work?
• Consider the following lab results:
• Which species would gain electrons (are OA)
in a reduction half reaction?
– all the metallic ions
• Which species would lose electrons (are RA) in
an oxidation half reaction?
– all the solid metals
• Which ion reacts the most? the least?
– D3+(aq) (most)
- E+(aq) (least)
• Which solid metal reacts the most? the least?
– E(s) (most)
- D(s) (least)
• Rank the OA from the most reactive to the
least reactive.
– D3+(aq) , A+(aq) , F2+(aq) , B2+(aq) , C3+(aq) , E+(aq)
• Using reduction half reactions, make a
“Reductions Table” similar to the one in your data
books.
D3+(aq) + 3 e –  D(s)
A+(aq) +
e–
 A(s)
F2+(aq) + 2 e –  F(s)
B2+(aq) + 2 e –  B(s)
C3+(aq) + 3 e –  C(s)
E+(aq)
+
e –  E(s)
• So, how can you determine if a reaction will
take place?
In any redox reaction, an OA must react with a
RA. (a species from the left side must react
with a species from the right side)
• If the OA is positioned higher on the table than the
RA with which it reacts, the reaction will proceed
naturally, therefore, be spontaneous.
• If the OA is positioned lower on the table than the
RA with which it reacts, the reaction will not proceed
naturally, therefore, be non-spontaneous.
p. 573
Activity: Practice Questions #11 – 23, p. 573 574
Creating Net Ionic (Redox) Equations Using
The Reductions Table
• Example: A strip of tin is placed in a solution of
silver nitrate.
Creating Net Ionic (Redox) Equations Using
The Reductions Table
• Example: A strip of tin is placed in a solution
of silver nitrate.
Step 1: Identify all species present.
Sn(s)  Ag+(aq)  NO3–(aq)  H2O(l)
Step 2: Identify which species are oxidizing
agents (OA) and which are reducing agents
(RA). The OA are found on the left side of the
table and the reducing agents are found on
the right side of the table.
Sn(s)  Ag+(aq)  NO3–(aq)  H2O(l)
RA OA
“normal” OA/RA
not found on tables
• Step 3: Identify the strongest OA and the
strongest RA. The top left side of the table
contains the strongest OA and the bottom
right contains the strongest RA.
• Sn(s)  Ag+(aq)  NO3–(aq)  H2O(l)
SRA SOA normal OA/RA
• Step 4: Write down the reduction half
reaction (the OA gaining e-).
Ag+(aq) + e –  Ag(s)
• Step 5: Write down the oxidation half reaction
(the RA losing e-).
Sn(s)  Sn2+(aq) + 2 e –
• Step 6: Add the two half reactions.
Ag+(aq) + e –  Ag(s)
Sn(s)  Sn2+(aq) + 2 e –
• Step 7: Balance the electrons.
2 (Ag+(aq) + e–  Ag(s) )
Sn(s)  Sn2+(aq) + 2 e–
• Step 8: Add the two equations to yield the
redox reaction.
2 (Ag+(aq) + e–  Ag(s) )
Sn(s)  Sn2+(aq) + 2 e –
2 Ag+(aq) + Sn(s)  2 Ag(s) + Sn2+(aq)
Writing Net Redox Reactions
• Assumption: The SOA present will react with the SRA present.
1) A 5¢ piece is left in chlorinated laundry bleach.
SOA
OA
Ni(s) / Cl2(aq) / H2O(l)
SRA
RA
RED:
Cl2(aq) + 2 e –  2 Cl –(aq)
OX:
Ni(s)  Ni 2+(aq) + 2 e –
REDOX: Cl2(aq) + Ni(s)  2 Cl –(aq) + Ni 2+(aq)
2) A magnesium ribbon is immersed in an iron (III) chloride
solution.
SOA
OA
Mg(s) / Fe 3+(aq) / Cl –(aq) / H2O(l)
SRA
RA
RA
RED:
2(Fe 3+(aq) + 1 e - → Fe 2+(aq) )
OX:
Mg(s) → Mg2+(aq) + 2 e REDOX: 2Fe 3+(aq) + Mg(s) → 2 Fe 2+(aq) + Mg2+(aq)
3) Tin (II) bromide solution is poured into acidified sodium dichromate
solution.
OA
OA
OA
SOA
OA
Sn2+(aq) /Br –(aq) / H+(aq) / Na+(aq) / Cr2O7 2–(aq) / H2O(l)
SRA
RA
RA
RED: Cr2O7 2–(aq) + 14 H +(aq) + 6 e –  2 Cr 3+(aq) + 7 H2O(l)
OX:
3 (Sn 2+(aq)  Sn 4+(aq) + 2 e –)
REDOX: Cr2O7 2–(aq) + 14 H +(aq) + 3 Sn 2+(aq) 
2 Cr 3+(aq) + 7 H2O(l) + 3 Sn 4+(aq)
4) Hydrochloric acid is poured into an aluminum pop can.
H+(aq) / Cl –(aq) / Al(s) / H2O(l)
RED:
OX:
REDOX:
5) Acidified potassium permanganate is accidentally spilled onto a "chrome"
faucet.
OA OA
SOA
OA
H+(aq) / K+(aq) / MnO4–(aq) / Cr(s) / H2O(l)
SRA RA
RED: 2(MnO4–(aq) + 8 H +(aq) + 5 e – Mn 2+(aq) + 4 H2O(l))
OX:
5 (Cr(s)  Cr 2+(aq) + 2 e –)
REDOX: 2 MnO4–(aq) + 16 H +(aq) + 5 Cr(s)  2 Mn 2+(aq) + 8 H2O(l)
+ 5 Cr 2+(aq)
6) A mixture of sodium hydroxide and sodium sulfite are mixed with a
stirrer than churns air into the solution.
OA
SOA
Na +(aq) / OH –(aq) / SO32–(aq) / O2(g) / H2O(l)
RA
SRA
RED:
OX:
O2(g) + 2 H2O(l) + 4 e –  4 OH –(aq)
2 (SO32–(aq) + 2 OH –(aq)  SO42–(aq) + H2O(l) + 2 e –)
REDOX: O2(g) + 2 SO32–(aq)  2 SO42–(aq)
7) A steel underground gasoline storage tank is surrounded by
moist acidic soil.
OA
SOA
Fe(s) / H2O(l) / H +(aq) / O2(g)
SRA RA
RED: O2(g) + 4 H +(aq) + 4 e –  2 H2O(l)
OX: 2 ( Fe(s)  Fe 2+(aq) + 2 e – )
REDOX: O2(g) + 4 H +(aq) + 2 Fe(s)  2 H2O(l) + 2 Fe 2+(aq)
8) Pennies are dropped into concentrated nitric acid.
SOA
OA
Cu(s) / H +(aq) / NO3–(aq) / H2O(l)
SRA
RA
RED: 2NO3–(aq) + 4 H +(aq) + 2e –  N2O4(g) + 2H2O(l)
OX:
Cu(s)  Cu 2+(aq) + 2 e –
REDOX: 2 NO3–(aq) + 4 H +(aq) + Cu(s)  N2O4(g) + 2 H2O(l) + Cu 2+(aq)
Disproportionation
Disproportionation is a reaction in which a species
is both oxidized and reduced.
Example: What happens if two iron(II) ions in a
solution collide?
SOA
OA
Fe2+(aq) / H2O(aq)
SRA
RA
Red:
Oxid:
Redox:
Fe2+ (aq) + 2e- → Fe(s)
2[Fe2+ (aq) → Fe3+ (aq) + e- ]
3Fe2+ (aq) → Fe(s) + 2 Fe3+ (aq)
Activity: Practice Questions #25 – 30 p. 579
Predicting Redox Reactions by Constructing Half-Reactions
Steps to Predict Balanced Redox Equations
1. Use the information provided to start two halfreaction equations.
2. Balance each half-reaction equation (Using the rules
from 13.1)
3. Multiply each half-reaction equation by simple whole
numbers to balance the electrons lost and gained.
4. Add the two half-reaction equations, cancelling the
electrons and anything els that is exactly the same on
both sides of the equation.
Activity:
Practice Questions #31 – 33, p. 581
Section Questions #1 – 18, p. 582