Oxidation reduction reactions honors
-
Upload
currituck-county-high-school -
Category
Education
-
view
187 -
download
1
description
Transcript of Oxidation reduction reactions honors
![Page 1: Oxidation reduction reactions honors](https://reader036.fdocuments.in/reader036/viewer/2022082413/5583fa16d8b42aa82c8b4a3e/html5/thumbnails/1.jpg)
OXIDATION-REDUCTION REACTIONS
Settle in, this is going to take a while…
![Page 2: Oxidation reduction reactions honors](https://reader036.fdocuments.in/reader036/viewer/2022082413/5583fa16d8b42aa82c8b4a3e/html5/thumbnails/2.jpg)
What is redox?
Reaction where there is a transfer of electrons between reactants
Oxidation involves the loss of electrons (OIL) Oxidation number/state of the element
increases Oxidized element is the reducing agent
Reduction involves the gain of electrons (RIG) Oxidation number/state of the element
decreases Reduced element is the oxidizing agent
![Page 3: Oxidation reduction reactions honors](https://reader036.fdocuments.in/reader036/viewer/2022082413/5583fa16d8b42aa82c8b4a3e/html5/thumbnails/3.jpg)
Example
Complete Reaction:Mg + Zn(NO3)2 Mg(NO3)2 + Zn
Net-ionic Reaction:Mg + Zn2+ Mg2+ + Zn
The magnesium metal was oxidized by the zinc and the zinc was reduced by the magnesium.
![Page 4: Oxidation reduction reactions honors](https://reader036.fdocuments.in/reader036/viewer/2022082413/5583fa16d8b42aa82c8b4a3e/html5/thumbnails/4.jpg)
Do what?!?!
The oxidation state of the magnesium changed from 0 to +2 Oxidation state increased = oxidation Because magnesium gave its electrons
away, it is the reducing agent The oxidation state of zinc changed from
+2 to 0 Oxidation state decreased = reduction Because zinc took the electrons, it is the
oxidizing agent
![Page 5: Oxidation reduction reactions honors](https://reader036.fdocuments.in/reader036/viewer/2022082413/5583fa16d8b42aa82c8b4a3e/html5/thumbnails/5.jpg)
How do you know oxidation states? The oxidation number for any pure
element is zero. Group 1 metals form +1 ions, group 2
metals form +2 ions, group 13 metals form +3 ions.
Transition metals can be all kinds of oxidation numbers (ranging from +1 to +7)
Transition metal oxidation states can be determined based on the nonmetal(s) it’s bonded to…
![Page 6: Oxidation reduction reactions honors](https://reader036.fdocuments.in/reader036/viewer/2022082413/5583fa16d8b42aa82c8b4a3e/html5/thumbnails/6.jpg)
Nonmetal oxidation states
Fluoride is ALWAYS -1, the other halides are usually -1.
Oxide is usually -2, except when it’s in the peroxide ion (-1) or bonded to fluorine (+2)
Hydrogen is +1, unless it is the hydride ion (-1)
![Page 7: Oxidation reduction reactions honors](https://reader036.fdocuments.in/reader036/viewer/2022082413/5583fa16d8b42aa82c8b4a3e/html5/thumbnails/7.jpg)
Putting it all together
The total charge on a compound is zero, so all oxidation numbers must cancel out.
The total charge of elements in a polyatomic ion must add to the charge on the ion
![Page 8: Oxidation reduction reactions honors](https://reader036.fdocuments.in/reader036/viewer/2022082413/5583fa16d8b42aa82c8b4a3e/html5/thumbnails/8.jpg)
Practice
What is the oxidation number of each element in the following compounds?
1. Zn(NO3)2
2. H2SO4
3. KMnO4
4. N2O4
5. PCl3
![Page 9: Oxidation reduction reactions honors](https://reader036.fdocuments.in/reader036/viewer/2022082413/5583fa16d8b42aa82c8b4a3e/html5/thumbnails/9.jpg)
What’s the point?
When an element gains electrons, another element must accept those electrons (Newton’s 3rd law).
If you separate the reaction into half-reactions, you can exploit this electron transfer to generate electricity.
The study of this is electrochemistry, but more on that later…
![Page 10: Oxidation reduction reactions honors](https://reader036.fdocuments.in/reader036/viewer/2022082413/5583fa16d8b42aa82c8b4a3e/html5/thumbnails/10.jpg)
Half-Reactions?
You can separate a redox reaction into the reduction reaction and the oxidation reaction.
First you have to identify which element is oxidized and which is reduced.
So let’s practice identification first:
![Page 11: Oxidation reduction reactions honors](https://reader036.fdocuments.in/reader036/viewer/2022082413/5583fa16d8b42aa82c8b4a3e/html5/thumbnails/11.jpg)
Practice
Determine the oxidation states of all elements in the following reactions and then identify which element is oxidized and which is reduced.
N2 + 3H2 2NH3
2MnO2 + Zn + 2H2O 2MnO(OH) + Zn(OH)2
AgNO3 + Cu Cu(NO3)2 + Ag
![Page 12: Oxidation reduction reactions honors](https://reader036.fdocuments.in/reader036/viewer/2022082413/5583fa16d8b42aa82c8b4a3e/html5/thumbnails/12.jpg)
N2 + 3H2 2NH3
![Page 13: Oxidation reduction reactions honors](https://reader036.fdocuments.in/reader036/viewer/2022082413/5583fa16d8b42aa82c8b4a3e/html5/thumbnails/13.jpg)
2MnO2 + Zn + 2H2O 2MnO(OH) + Zn(OH)2
![Page 14: Oxidation reduction reactions honors](https://reader036.fdocuments.in/reader036/viewer/2022082413/5583fa16d8b42aa82c8b4a3e/html5/thumbnails/14.jpg)
AgNO3 + Cu Cu(NO3)2 + Ag
![Page 15: Oxidation reduction reactions honors](https://reader036.fdocuments.in/reader036/viewer/2022082413/5583fa16d8b42aa82c8b4a3e/html5/thumbnails/15.jpg)
Separating reactions
Once the oxidized and reduced elements have been identified, separate the reactions.
Use net ionic reactions instead of complete reactions
2AgNO3 + Cu Cu(NO3)2 + 2Ag
2Ag+1 + Cu Cu2+ + 2Ag
![Page 16: Oxidation reduction reactions honors](https://reader036.fdocuments.in/reader036/viewer/2022082413/5583fa16d8b42aa82c8b4a3e/html5/thumbnails/16.jpg)
2Ag+1 + Cu Cu2+ + 2Ag
The silver is reduced, so that is the reduction reaction:
2Ag+1 2Ag The masses are balanced, but the
charges are not, so add the electrons being transferred:
2Ag+1 + 2e- 2Ag Notice that the reduction half reaction
has electrons as reactants
![Page 17: Oxidation reduction reactions honors](https://reader036.fdocuments.in/reader036/viewer/2022082413/5583fa16d8b42aa82c8b4a3e/html5/thumbnails/17.jpg)
2Ag+1 + Cu Cu2+ + 2Ag
The copper is oxidized, so that is the oxidation reaction:
Cu Cu2+
The masses are balanced, but the charges are not, so add the electrons being transferred:
Cu Cu2+ + 2e-
Notice that the oxidation half reaction has electrons as products