chapter 19 redox reactions and electrochemistry
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Redox Reactions and ElectrochemistryChapter 19
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2Mg (s) + O2 (g) 2MgO (s)
2Mg 2Mg2+ + 4e-
O2 + 4e- 2O2-
Oxidation half-reaction (lose e-)
Reduction half-reaction (gain e-)
Electrochemical processes are oxidation-reduction reactions in which:
• the energy released by a spontaneous reaction is converted to electricity or
• electrical energy is used to cause a nonspontaneous reaction to occur
0 0 2+ 2-
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Oxidation number
The charge the atom would have in a molecule (or anionic compound) if electrons were completely transferred.
1. Free elements (uncombined state) have an oxidation number of zero.
Na, Be, K, Pb, H2, O2, P4 = 0
2. In monatomic ions, the oxidation number is equal to the charge on the ion.
Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
3. The oxidation number of oxygen is usually –2. In H2O2
and O22- it is –1.
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4. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1.
6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion.
5. Group IA metals are +1, IIA metals are +2 and fluorine is always –1.
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Balancing Redox Equations
1. Write the unbalanced equation for the reaction in ionic form.
The oxidation of Fe2+ to Fe3+ by Cr2O72- in acid solution?
Fe2+ + Cr2O72- Fe3+ + Cr3+
2. Separate the equation into two half-reactions.
Oxidation:
Cr2O72- Cr3+
+6 +3
Reduction:
Fe2+ Fe3++2 +3
3. Balance the atoms other than O and H in each half-reaction.
Cr2O72- 2Cr3+
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Balancing Redox Equations
4. For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms.
Cr2O72- 2Cr3+ + 7H2O
14H+ + Cr2O72- 2Cr3+ + 7H2O
5. Add electrons to one side of each half-reaction to balance the charges on the half-reaction.
Fe2+ Fe3+ + 1e-
6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O
6. If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients.
6Fe2+ 6Fe3+ + 6e-
6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O
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Balancing Redox Equations
7. Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel.
6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O
6Fe2+ 6Fe3+ + 6e-Oxidation:
Reduction:
14H+ + Cr2O72- + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O
8. Verify that the number of atoms and the charges are balanced.
14x1 – 2 + 6 x 2 = 24 = 6 x 3 + 2 x 3
9. For reactions in basic solutions, add OH- to both sides of the equation for every H+ that appears in the final equation.
Example
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19.1
Write a balanced ionic equation to represent the oxidation of iodide ion (I-) by permanganate ion ( ) in basic solution to yield molecular iodine (I2) and manganese(IV) oxide (MnO2).
-4MnO
Example
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19.1
Strategy
We follow the preceding procedure for balancing redox equations. Note that the reaction takes place in a basic medium.
Solution
Step 1: The unbalanced equation is
+ I- MnO2 + I2
-4MnO
Example
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Step 2: The two half-reactions are
Oxidation: I- I2
Reduction: MnO2
19.1
-1 0
+7 +4
Step 3: We balance each half-reaction for number and type of atoms and charges. Oxidation half-reaction: We first balance the I atoms:
2I- I2
-4MnO
Example
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19.1
To balance charges, we add two electrons to the right-hand side of the equation:
2I- I2 + 2e-
Reduction half-reaction: To balance the O atoms, we add two H2O molecules on the right:
MnO2 + 2H2O
To balance the H atoms, we add four H+ ions on the left:
+ 4H+ MnO2 + 2H2O
-4MnO
-4MnO
Example
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19.1
There are three net positive charges on the left, so we add three electrons to the same side to balance the charges:
+ 4H+ + 3e- MnO2 + 2H2O
Step 4: We now add the oxidation and reduction half reactions
to give the overall reaction. In order to equalize the number of electrons, we need to multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 as follows:
3(2I- I2 + 2e-)2( + 4H+ + 3e- MnO2 + 2H2O)________________________________________
6I- + + 8H+ + 6e- 3I2 + 2MnO2 + 4H2O + 6e-
-4MnO
-4MnO
-4MnO
Example
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19.1
The electrons on both sides cancel, and we are left with the balanced net ionic equation:
6I- + 2 + 8H+ 3I2 + 2MnO2 + 4H2O
This is the balanced equation in an acidic medium. However, because the reaction is carried out in a basic medium, for every H+ ion we need to add equal number of OH- ions to both sides of the equation:
6I- + 2 + 8H+ + 8OH- 3I2 + 2MnO2 + 4H2O + 8OH-
-4MnO
-4MnO
Example
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19.1
Finally, combining the H+ and OH- ions to form water, we obtain
6I- + 2 + 4H2O 3I2 + 2MnO2 + 8OH-
Step 5: A final check shows that the equation is balanced in terms of both atoms and charges.
-4MnO