Oxidation and Reduction Chapters 20 & 21. Oxidation vs Reduction.
Oxidation-Reduction
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Transcript of Oxidation-Reduction
Oxidation-Reduction
Oxidation is the loss of electrons.Reduction is the gain of electrons.The reactions occur together. One does
not occur without the other.The terms are used relative to the
change in the oxidation state or oxidation number of the reactant(s).
Aqueous Reactions:Oxidation - Reduction
In the following reaction, identify what is being oxidized and what is being reduced. What is the total number of electrons involved in the process?
QUESTIONIn a redox reaction, oxidation and reduction must both occur. Which statement provides an accurate premise of redox chemistry?
A.The substance that is oxidized must be the oxidizing agent.B.The substance that is oxidized must gain electrons.C.The substance that is oxidized must have a higher
oxidation number afterwards.D.The substance that is oxidized must combine with oxygen.
Reactivity Tables (usually reducing) show relative reactivities:In the examples from the previous slide, the acid solution (H+) will react with anything below it in the Table but not above. Nickel and Zinc,
….but not Copper.
Select all redox reactions by looking for a change in oxidation number as reactants are converted to products.
I) Ca + 2 H2O → Ca(OH)2 + H2
II) CaO + H2O → Ca(OH)2
III) Ca(OH)2 + H3PO4 → Ca3(PO4)2 + H2O
IV) Cl2 + 2 KBr → Br2 + 2 KCl
A) I and II B) II and III C) I and IV D) III and IV
QUESTION
Cu 2+ Cu (s)
H2 (g) 2 H +
0
0
+2 e-
- 2 e-
Use oxidation numbers to determine what is oxidized and what is reduced.
Number of electrons gained must equal the number of electrons lost.
- 2 e-
+2 e-
Refer to BalancingOxidation-Reduction Reactions
Balancing Redox Equationsin acidic solutions
1) Determine the oxidation numbers of atoms in both reactants and products.
2) Identify and select out those which change oxidation number (“redox” atoms) into separate “half reactions”.
3) Balance the “redox” atoms and charges (electron gain and loss must equal!).
4) In acidic reactions balance oxygen with water then hydrogen from water with acid proton(s).
Fe+2(aq)+ Cr2O7
2-(aq) +H+
(aq) Fe3+(aq)
+ Cr3+(aq) + H2O(l)
Fe 2+(aq)+ Cr2O7
2-(aq) +H+
(aq)
Fe 3+(aq) + Cr 3+
(aq) + H2O(l)
x = ? Cr ; 2x+7(-2) = -2; x = +6
? Cr oxidation number?
Balancing Redox Equationsin acidic solutions
Fe 2+(aq) Fe 3+
(aq)
Cr2O72-
(aq) + Cr 3+(aq)
Cr = (6+)
6 (Fe 2+(aq) -e -Fe3+
(aq))
6 Fe 2+(aq) 6 Fe3+
(aq) + 6 e -
Cr2O72-
(aq) + 6 e - 2 Cr3+(aq)
6 e -
-e -
2
Balancing Redox Equationsin acidic solutions
6 Fe2+(aq) 6 Fe3+
(aq) + 6 e -
Cr2O72-
(aq) + 6 e - 2 Cr3+(aq)
6 Fe2+(aq)+ Cr2O7
2-(aq) + ? 2nd H+
(aq) 6 Fe3+
(aq) + 2 Cr3+(aq)+ ? 1st Oxygen H2O(l)
Oxygen = 7 2nd (Hydrogen) = 14
Balancing Redox Equationsin acidic solutions
Completely Balanced Equation:
6 Fe2+(aq)+ Cr2O7
2-(aq) + 14 H+
(aq)
6 Fe3+(aq) + 2 Cr3+
(aq)+ 7 H2O(l)
Balancing Redox Equationsin acidic solutions
Dichromate ion in acidic medium converts ethanol, C2H5OH, to CO2 according to the unbalanced equation:
Cr2O72−(aq) + C2H5OH(aq) → Cr3+(aq) + CO2(g) + H2O(l)
The coefficient for H+ in the balanced equation using smallest integer coefficients is:
A) 8 B) 10 C) 13 D) 16
QUESTION
Balancing Redox Equationsin basic solutions
1) Determine oxidation numbers of atoms in Reactants and Products
2) Identify and select out those which change oxidation number into separate “half reactions”
3) Balance redox atoms and charges (electron gain and loss must equal!)
4) In basic reactions balance the Oxygen with hydroxide then Hydrogen from hydroxide with water
MnO2 (aq)+ ClO31-
(aq) + OH 1-aq)
MnO41-
(aq)+ Cl 1-(aq) + H2O(l)
Mn4+ (MnO2) Mn7+ (MnO4 ) 1-
Cl+5 (ClO3 ) 1-+ 6 e- Cl 1-
Balancing Redox Equationsin basic solutions
Electronically Balanced Equation:
2 MnO2 (aq)+ ClO31-
(aq) + 6 e - 2 MnO4
1- + Cl 1- + 6 e-
Balancing Redox Equationsin basic solutions
Completely Balanced Equation:
2 MnO2 (aq)+ ClO31-
(aq) + 2 OH 1- (aq)
2 MnO4 (aq)1- + Cl 1-
(aq)+ 1 H2O (l)
9 O in product
Balancing Redox Equationsin basic solutions
QUESTIONOxalate ion can be found in rhubarb and spinach (among other green leafy plants). The following unbalanced equation carried out in a basic solution, shows how MnO4
– could be used to analyze samples for oxalate.
MnO4– + C2O4
2– MnO2 + CO32– (basic solution)
When properly balanced, how many OH– are present?
A.1B.2C.3D.4