Oxidation and Reduction
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Transcript of Oxidation and Reduction
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Oxidation and Reduction
Redox
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Redox involves two simultaneous reactions◦ An oxidation and a reduction
Oxidation involves a loss of electrons
Reduction involves a gain of electrons
LEO the lion says GER
What exactly is a redox reaction?
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Seeing how in each Redox reaction, there is said to be an oxidation half-reaction and a reduction half-reaction
One molecule will lose electrons (the oxidation) and the electrons will join another atom (reduction)
Redox
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Oxidation half reaction
◦ Fe(s) -> Fe2+(aq) + 2e-
Iron is being oxidized to form the ferrous ion
Reduction half-reaction
◦ Au3+(aq) + 3e- -> Au(s)
The gold ion has been reduced to its ground state
Example
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Agent Electron exchange
Half-reaction The atom is
Reducing Agent Loses electrons Oxidation Oxidized
Oxidizing Agent Gains electrons Reduction Reduced
To help simplify
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Oxidation half-reactionZn(s) -> Zn+2 + 2e-
Cu2+ + 2e- -> Cu(s)
______________________________________________Cu2+ + Zn(s) -> Zn2+ + Cu(s)
By eliminating the common terms on either side, you are left with a simplified final equation
General equation for Redox
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If there is a difference in the number of electrons in the equations, you must use stoichiometrics!
Fe(s) -> Fe2+ (aq) + 2e- x3
Au3+(aq) + 3e- -> Au(s)
x2
3 Fe(s) + 2 Au3+(aq) -> 3 Fe2+
(aq) + 2 Au(s)
General Equation continued
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The reaction of a piece of magnesium (Mg) in hydrochloric acid (HCl) results in the formation of magnesium dichloride (MgCl2). The release of hydrogen (H2) can also be observed. During this reaction, metallic magnesium is oxidized into aqueous Mg2+ ions, while aqueous H+ ions of the acidic solution is reduced to hydrogen gas.
a) What are the half-reactions in this reaction?
b) What is the general equation for oxidation-reduction?
c) Find the oxidizing agent and the reducing agent
Example
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a) Magnesium oxidized◦ Loses electrons
Oxidation reaction◦ Mg (s) -> Mg2+ (aq) + 2e-
Hydrogen reduced◦ Gains electrons
Reduction reaction◦ H+ (aq) + 1e- -> 1/2 H2 (g)
Solution
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b) Add the two half reactions together
Mg (s) -> Mg2+ (aq) + 2e-
2x(H+ (aq) + 1e- -> ½ H2 (g)) 2H+ (aq) +2e- -> H2 (g)
______________________________________________ Mg (s) + 2H+ (aq) -> Mg2+ (aq) + H2 (g)
Solution (cont)
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c) The oxidizing agent is the H+ (aq) because it gains electrons
The reducing agent is the Mg (s) because it loses electons
Remember LEO the lion says GER
Solution
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Do # 1, 2, 4, 5 on the worksheet
We will go over it next class
Have Fun!
Problems
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Oxidation NumberOxidation State
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The oxidation number, also called the oxidation state, indicates the number of electrons an element has lost or gained, in relation to its ground state, during a redox reaction.
All elements in their ground state have an oxidation number of 0
They are considered to be atoms, due to them not having lost any electrons
Oxidation Number
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When atoms are involved in redox reactions, their oxidation numbers vary
Oxidation numbers increase with an oxidation due to a loss in electrons
Oxidation numbers decrease with a reduction due to a gain in electrons
Oxidation number
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To determine the oxidation number of an atom, we must determine whether it is part of an ionic or a covalent compound
Ionic compound◦ A bond between a metal and a non-metal which
share electrons
The oxidation number of each ion is equal to its charge
Oxidation Number
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Example Calcium chloride (CaCl2) is composed of one
Ca2+ ion and two Cl- ions
To distinguish between an ion’s charge and its oxidation number, the convention is different
A charge is written as 2+ while an oxidation number is written as +2
Ionic compounds
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Find the oxidation # for the following atoms
Exercises
Atom Charge Oxidation #Na 1+ +1Sr 2+ +2Ra 2+ +2K 1+ +1Li 1+ +1
Mg 2+ +2Rb 1+ +1
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Substances Charge Oxidation #
Elements in ground state (Li, Mg, Al, Fe, etc.) 0 0Molecules of elements (H2, O2, Cl2, N2, S8, etc.) 0 0Ions of alkali metals (Li+, Na+, K+, etc.) 1+ +1Ions of alkaline earth metals (Ca2+, Mg2+, Be2+, etc.)
2+ +2
Oxidation Numbers for some Substances
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When an atom is part of a molecule or polyatomic ion, convention determines its oxidation number by assigning each pair to the more electronegative atom in the bond, that is, the atom that is more likely to attract electrons to fill its outermost shell.
To determine the oxidation number of a molecule, the molecule can be represented with Lewis notation
Oxidation Number of Covalent Compounds
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If you take a water molecule, you can see that the electrons are shared between the oxygen and the hydrogen
Since oxygen normally has 6 electrons◦ It gets 2 additional ones from hydrogen◦ Oxidation number of -2
Hydrogen atom has lost its electron◦ Oxidation number of +1
So what this means... In English...
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In covalent molecules, the charge is not always the same as the oxidation number
There is no O2- or H+ in water
In covalent compounds, the electrons are not given to the other atom, but rather are shared between the two
Let’s try this again...
Oxidation Number of Covalent Compounds
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What are the oxidation numbers for the atoms in an ammonia (NH3) molecule?
Calculate the Oxidation Number
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1- Since there are no metallic atoms, we know that ammonia is a covalent compound
2- Using the periodic table, we can determine which molecule is more electronegative. In this case, it is Nitrogen. The electrons in the bond are assigned to it. Nitrogen needs 3 electrons to reach its ground state. The oxidation number of Nitrogen is -3
How you do it...
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3- As for Hydrogen, each hydrogen ion has lost its one electron, which gives it an oxidation number of +1
Ans: The oxidation number of nitrogen is -3 and hydrogen is +1
Continued
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The same element can have a different oxidation number depending in which molecule it is found in.
There are a few non-metals that almost always have the same oxidation number◦ Oxygen is always -2, except in peroxide (-1)
Hydrogen is also almost always the same, +1
Overview
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What is the oxidation number for the manganese atom in the permanganate ion (MnO4
-), a very strong oxidizing agent? The permanganate ion’s negative charge indicates that the net charge is 1-. Furthermore, since the oxidation number of oxygen is -2 and there are 4 oxygen atoms in this molecule. How would you be able to figure out the oxidation number for the manganese?
Hint: math!
Example #2
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Mn + 4 O = -1
X + 4 (-2) = -1
X + -8 = -1
X = 7
Solution
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What are the oxidation numbers of all of the atoms in the following redox reaction
Fe2O3 (s) + 2 Al (s) -> Al2O3 (s) + 2 Fe (l)
Example #3
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Seeing how Fe (s) and Al (s) are both solid, their oxidation numbers are 0
For Fe2O3 and Al2O3, oxygen has an oxidation number of -2◦ There are three of them, for a total of -6
The iron and aluminum need an oxidation number of +3 to balance everything out
Solution
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Metals can be classified by their reduction power
Spontaneous reactions
For spontaneous reactions to occur, the stronger reducing agent must be in the solid state and the weaker one must be in its ionic form
Reducing Power of Metals
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Reaction Stronger Reducing Agent
Weaker Reducing Agent
Spontaneous
Solid Aqueous solution
None Aqueous solution Solid
TABLE!
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You have Aluminum (Al) and Copper (Cu)
Which would have to be solid and which would have to be aqueous?
Same situation, but with Gold (Au) and Copper (Cu)
Which would have to be solid and which would have to be aqueous?
Practice
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For the Aluminum (Al) and Copper (Cu) reaction, the Al would need to be solid and the Cu in solution
For the Gold (Au) and Copper (Cu) reaction, the Cu would have to be solid and the Au in solution
Answers
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An electrochemical cell is a device that can spontaneously generate an electrical current◦ i.e. A battery
Electrochemical Cell
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The electrochemical cell is composed of two electrodes, also called half-cells.
They are called half-cells due to the half reactions taking place in the cell
The two electrodes are joined by a wire and are connected by a salt bridge◦ A tube of salt water plugged with a porous
membrane
Electrochemical Cells
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The classic example of an electrochemical cell is a piece of zinc immersed in a solution of zinc sulfate (ZnSO4) and a piece of copper in a solution of copper sulfate (CuSO4)
Since zinc is a better reducing agent than copper, it gives up its electrons to the copper
Positive terminal is called the anode and the negative is the cathode
Zinc is oxidized and the copper is reduced
Example
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The concentration of each half-cell’s ions varies◦ The Zn2+ ions increase◦ The Cu2+ ions decrease
The ions in the salt bridge move towards their respective poles to maintain the number of positive and negative ions◦ Positive towards the Cathode◦ Negative towards the Anode
As the Electrochemical Cell works
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Electrochemical Cells
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During a redox reaction, one of the metals has greater potential energy than the other because it has a greater reducing power.
Without this potential difference between two metals of different types, no chemical reaction would take place
To calculate the potential, we need a reference electrode
In this case, we use the hydrogen electrode
Reduction and Oxidation Potential
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2 H+ + 2 e- H2
This reaction is considered to have a redox potential of 0.00 V
EO = 0.00V
Where EO is redox potential
Hydrogen electrode
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To measure the EO, calculate the potential difference, create an electrochemical cell using the Hydrogen and the electrode which you want to test
The voltage which is read on the voltmeter will be the redox potential of the unknown
How we measure EO
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Redox Potential
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Redox Potential Table
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With this table, we can compare the oxidation or reducing power of substances
And, we can calculate the cell potential
That’s about it for you guys...
What we can do with this...
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The cell potential is equal to the sum of the oxidation potential and the reduction potential of the cell
Calculating the potential of an electrochemical cell made up of two different electrodes can be done without having to use a reference electrode
Cell potential
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First, determine the stronger reducing agent
If we use Silver (Ag) and Magnesium (Mg) as an example, the magnesium is stronger since it is found below silver in the table of redox potentials
This makes it the electrode which will lose electrons and undergo oxidation
How to figure it out...
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To find the cell potential, we must reverse the reduction. In this case, the Mg reaction will be the reduction
(Ag+ + e- -> Ag (s) ) x2 OxidationEO=0.80V
Mg (s) -> Mg 2+ + 2 e- ReductionEO=2.37V
Continued
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Cell potential is calculate using the following equation
EO cell = EO oxidation + EO reduction
EO cell = 0.80V + 2.37V
EO cell = 3.17V
Cell Potential
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The electrochemical cell can also be called a fuel cell
If the EO is above 0, then the reaction is said to occur spontaneously.
If the EO is below 0, then the fuel cell is said to not be spontaneous
What the results mean
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A fuel cell can be represented in this simplified way
Oxidation Reduction Mg|Mg2+ || Ag+|Ag
According to convention, the oxidation reaction is shown on the left
Fuel Cell