Transcript File

Redox reactions fall into three major categories: synthesis,
decomposition, single replacement
Examples
Synthesis
Decomposition
Single Replacement
N2 + O2  2NO
CaCO3  Ca + CO3
Cu + AgNO3  CuNO3 + Ag
Double Replacement reactions are NOT redox reactions
KI + AgNO3  AgI + KNO3
The elements do not show a change in oxidation number
To identify redox reactions
LOOK for an equation with at least one free element
LOOK for an equation that is showing synthesis,
decomposition, or single replacement
DETERMINE which equation has two elements changing
oxidation numbers
Which equation represents a redox reaction?
1) AgNO3 + LiBr  AgBr + LiNO3
2) KI  K+ + I-1
3) 4Na + O2  2Na2O
4) Na+ + SO4-2  Na2SO4
Practice
Which equation represents an oxidation-reduction
reaction?
1) 2AgNO3 + Cu  Cu(NO3)2 + 2Ag
2) 3O2  2O3
3) H3PO4 + KOH  K3PO4 + H2O
4) H+ + Cl-  HCl
Oxidation: Loss of Electrons is Oxidation LEO
Oxidized substance (aka reducing agent)
substance that is losing its electrons
oxidation number always increases (becomes more
positive)
Reduction: Gain of Electrons is Reduction GER
Reduced substance (aka oxidizing agent)
substance that is gaining electrons
oxidation number always decreases (becomes
more negative)
Half-reaction equation shows either the oxidation portion
or the reduction portion of a redox
must show conservation of atoms, mass, charge
2Na + Cl2  2NaCl
Oxidation-half reaction will show the losing of electrons
2Na0  2Na+ + 2eReduction-half reaction will show the gaining of electrons
Cl20 + 2e-  2Cl-
Equation with electrons on LEFT: Reduction-half reaction
C0 + 4e-  C-4
C0 atom gains 4 electrons to become C-4 ion
C0 oxidation number decreases from 0 to -4
C0 is the reduced substance, and also the oxidizing agent
Equation with electrons on RIGHT: Oxidation-half reaction
Sb+3  Sb+5 + 2eSb+3 loses 2 electrons to become Sb+5
Sb+3 oxidation number increases form +3 to +5
Sb+3 is the oxidized substance, and also the reducing agent
To determine which equation is showing oxidation
LOOK for the equation with: electrons on the right and
oxidation number that is increasing
To determine which equation is showing reduction
LOOK for the equation with: electrons on the left ad
oxidation number that is decreasing
Practice
Which equation is showing oxidation?
1) F20  2F-1 + 2e2) Ca0  Ca+2 + 2e3) Ca+2 +2e-  Ca0
4) 2F + 2e-  F20
Which half-reaction correctly represents reduction?
1) Al (s)  Al+3 (aq) + 3e2) H2 (g) + 2e-  2H+ (aq)
3) I2 (s)  2I- (aq) + 2e4) Cu+2 (aq) + 2e-  Cu0 (s)
To determine which oxidation number change represents
OXIDATION or LOSS of electrons
LOOK for a change that is increasing
REDUCTION or GAIN of electrons
LOOK for a change that is decreasing
To determine which oxidation number change represents:
GREATEST number of electrons LOST or GAINED
LOOK for greatest difference in numbers
LEAST number of electrons LOST or GAINED
LOOK for least difference in numbers
Difference = high oxidation # - low oxidation #
Practice
Which change in oxidation number represents reduction?
1) +1  +2
3) -1  0
2) -2  -1
4) 0  -1
Which change in oxidation number represents the greatest
number of electrons lost?
1) +2 to +1
3) -1 to +1
2) +2 to -1
4) -1 to 0
If the oxidation number of a substance given in the
question goes from low to high
Electrons were LOST, substance is oxidized
If the oxidation number of a substance given in the
question goes from high to low
Electrons were GAINED, substance is reduced
Practice
When Cr+4 changes to Cr+2 there will be
1) 6 electrons gained
3) 2 electrons gained
2) 6 electrons lost
4) 2 electrons lost
Redox equations can be given in two different forms
Fe+3 + Ni0  Fe+ + Ni+2
ionic equation
Mg + H2SO4  MgSO4 + H2 chemical equation
Some common questions asked about redox equations
1) What is the oxidation number change of the substance
2) How many electrons are lost or gained
3) Which substance is being oxidized
4) Which substance is being reduced
5) Which substance is the reducing agent
6) Which substance is the oxidizing agent
7) What is the correct oxidation-half equation
8) What is the correct reduction-half equation
Fe+3 + Ni0  Fe+ + Ni+2
Oxidation half-reaction
Reduction half-reaction
Ni0  Ni+2 + 2eFe+3 + 2e-  Fe+1
Mg + H2SO4  MgSO4 + H2
Oxidation numbers
Mg
0 to +2
H
+1 to 0
oxidation – lose electrons
reduction – gain electrons
Mg  Mg+2 + 2e2(H+1 + e-  H) 2H+ + 2e-  H2
The number of electrons lost must equal the number of
electrons gained
To determine which substance is oxidized, lost electrons, or is a
reducing agent
CHOOSE the substance to the LEFT of arrow with the SMALLER
charge
To determine which substance is reduced, gained electrons, or is an
oxidizing agent
CHOOSE the substance to the RIGHT of arrow with the LARGER
charge
3Mg0 + 2Al+3  3Mg+2 + 2Al0
Mg0 has smaller charge than Al+3
is oxidized
Mg is a reducing agent
Mg loses 2 electrons
Oxidation # increase from 0 to +3
Al+3 has a larger charge than Mg0 Mg
Al is reduced
Al is an oxidizing agent
Al gains 3 electrons
Oxidation # decreases from +3 to 0
Practice
Given the equation
Br2 + 2I-  2Br- + I2
Which substance is the oxidizing agent?
1) Br2
2) I3) Br-
4) I2
Given the equation below
Ca + FeSO4  CaSO4 + Fe
Which of these species is oxidized?
1) Fe+2
2) Fe
3) Ca
4) Ca+2
Oxidation-half reaction: electrons lost are on right and
oxidation number increases
Reduction-half reaction: electrons gained are on left and
oxidation number decreases
Given the redox equation below
Br20 + 2I-  2Br-1 + I20
Which half-reaction represents oxidation that occurs?
1) Br20  2Br- + 2e2) 2I-  I2 + 2e3) Br2 + 2e-  2Br
4) 2I- + 2e-  I2
Given the equation below
Ca + FeSO4  CaSO4 + Fe
Which of these equations represents the reduction that
occurs?
1) Ca+2  Ca0 + 2e3) Ca0 + 2e-  Ca+2
2) Fe0  Fe0 + 2e4) Fe+2 + 2e-  Fe0
Given the equation Ca0 + Fe+2  Ca+2 + Fe0
Write the correct balanced oxidation half-reaction and
reduction half-reaction equations.
Step 1 LOOK at the redox equation given
Write both changes of the elements
Ca0  Ca+2
Fe+2  Fe0
Step 2 Add electrons (add electrons to the side with the higher
oxidation number)
Ca0  Ca+2 + eFe+2 + e-  Fe0
Step 3 Adjust number of electrons to equalize charges
Ca0  Ca+2 + 2eFe+2 + 2e-  Fe0
Step 4 Identify the half-reaction equations
Oxidation
Ca0  Ca+2 + 2eReduction
Fe+2 + 2e-  Fe0
Practice
Given the redox equation below:
Mg (s) + AgCl (aq)  MgCl2 (aq) + Ag (s)
Write balanced oxidation and reduction half reactions.
Half-reaction
Electrons are
Oxidation
number
Example
Oxidation
Lost
Increases
Zn  Zn+2 + 2e-
Gained
Decreases
Si + 4e-  Si-4
Oxidized substance
Reducing agent
Reduction
Reduced substance
Oxidizing agent