Post on 01-Jan-2016
description
Oxidation-ReductionOxidation-Reduction
Dr. Ron RusayDr. Ron Rusay
Balancing Oxidation-Reduction Reactions
Oxidation-ReductionOxidation-Reduction
OxidationOxidation is the loss of electrons. is the loss of electrons.ReductionReduction is the gain of electrons. is the gain of electrons.The reactions occur together. One does The reactions occur together. One does
not occur without the other.not occur without the other.The terms are used relative to the The terms are used relative to the
change in the change in the oxidation stateoxidation state or or oxidation numberoxidation number of the reactant(s). of the reactant(s).
Aqueous Reactions:Aqueous Reactions:Oxidation - Reduction Oxidation - Reduction
In the following reaction, identify what is being In the following reaction, identify what is being oxidized and what is being reduced. What is oxidized and what is being reduced. What is the total number of electrons involved in the the total number of electrons involved in the process?process?
Oxidation Reduction ReactionsOxidation Reduction Reactions
QUESTIONQUESTIONIn 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.
ANSWERANSWERIn 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.
QUESTIONQUESTION
ANSWERANSWER 2
–
2O2
B) NO
Oxygen almost always has an oxidation state of 2 when part of a compound. The exception is
when it is part of a peroxide. For example, hydrogen peroxide H . Then it has an oxidation state of –1.
What is the oxidation number of chromium in ammonium dichromate?
A) +3 B) +4 C) +5 D) +6
QUESTIONQUESTION
What is the oxidation number of chromium in ammonium dichromate?
A) +3 B) +4 C) +5 D) +6
ANSWERANSWER
(NH4)2Cr2O7
Zinc
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
QUESTIONQUESTION
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
ANSWERANSWER
QUESTIONQUESTION
ANSWERANSWER
C) 2
If an element is a reactant, this usually a redox reaction, since an element usually becomes part of a compound during a chemical reaction.
Cu + 2AgNO3 2Ag + Cu(NO3)2
N2 + 3H2 2NH3
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
QUESTIONQUESTION
ANSWERANSWER
B) the oxidizing agent.
Metals lose electrons, so they are oxidized, making the other reactant an oxidizing agent.
Balancing Redox EquationsBalancing Redox Equationsin acidic solutionsin acidic solutions
1) Determine the oxidation numbers of atoms in both 1) Determine the oxidation numbers of atoms in both reactants and products.reactants and products.
2) Identify and select out those which change 2) Identify and select out those which change oxidation number (oxidation number (““redoxredox”” atoms) into separate atoms) into separate ““half reactionshalf reactions””..
3) Balance the 3) Balance the ““redoxredox”” atoms and charges (electron atoms and charges (electron gain and loss must equal!).gain and loss must equal!).
4) In acidic reactions balance oxygen with water 4) In acidic reactions balance oxygen with water then hydrogen from water with acid proton(s).then hydrogen from water with acid proton(s).
Balancing Redox EquationsBalancing Redox Equations
FeFe+2+2(aq)(aq)+ Cr+ Cr22OO77
2-2-(aq) (aq) +H+H++
(aq)(aq) FeFe3+3+(aq)(aq)
+ Cr+ Cr3+3+(aq) (aq) + H+ H22OO(l) (l)
Fe Fe 2+2+(aq)(aq)+ Cr+ Cr22OO77
2-2-(aq) (aq) +H+H++
(aq)(aq)
Fe Fe 3+3+(aq)(aq) + Cr + Cr 3+3+
(aq) (aq) + H+ H22OO(l)(l)
x x == ?? Cr Cr ; ; 2x+7(-2) 2x+7(-2) = = -2-2; ; x x = = +6+6
? Cr oxidation number?? Cr oxidation number?
Balancing Redox EquationsBalancing Redox Equations
Fe Fe 2+2+(aq) (aq) Fe Fe 3+3+
(aq)(aq)
CrCr22OO772-2-
(aq) (aq) + + Cr Cr 3+3+(aq) (aq)
CrCr = (= (6+6+))
66 (Fe (Fe 2+2+(aq)(aq) -e -e - -FeFe3+3+
(aq)(aq)))
66 Fe Fe 2+2+(aq)(aq) 6 6 FeFe3+3+
(aq) (aq) + + 6 6 ee - -
CrCr22OO772-2-
(aq) (aq) + 6 e+ 6 e - - 2 2 CrCr3+3+(aq)(aq)
6 e6 e - -
-e-e - -
22
Balancing Redox EquationsBalancing Redox Equations
66 Fe Fe2+2+(aq)(aq) 6 6 FeFe3+3+
(aq) (aq) + + 6 6 ee - -
CrCr22OO772-2-
(aq) (aq) + 6 e+ 6 e - - 2 2 CrCr3+3+(aq)(aq)
66 Fe Fe2+2+(aq)(aq)+ Cr+ Cr22OO77
2-2-(aq) (aq) + + ? ? 2nd 2nd HH++
(aq)(aq) 6 6
FeFe3+3+(aq) (aq) + + 2 2 CrCr3+3+
(aq)(aq)+ + ? ? 1st Oxygen 1st Oxygen HH22OO(l) (l)
OxygenOxygen == 7 7 2nd (Hydrogen)2nd (Hydrogen) == 1414
Balancing Redox EquationsBalancing Redox Equations
Completely Balanced Equation:Completely Balanced Equation:
66 Fe Fe2+2+(aq)(aq)+ Cr+ Cr22OO77
2-2-(aq) (aq) + + 14 14 HH++
(aq)(aq)
6 6 FeFe3+3+(aq) (aq) + + 2 2 CrCr3+3+
(aq)(aq)+ + 77 HH22OO(l) (l)
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
QUESTIONQUESTION
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
ANSWERANSWER
Balancing Redox EquationsBalancing Redox Equationsin basic solutionsin basic solutions
1) Determine oxidation numbers of atoms in 1) Determine oxidation numbers of atoms in Reactants and ProductsReactants and Products
2) Identify and select out those which change 2) Identify and select out those which change oxidation number into separate oxidation number into separate ““half reactionshalf reactions””
3) Balance redox atoms and charges (electron gain 3) Balance redox atoms and charges (electron gain and loss must equal!)and loss must equal!)
4) In basic reactions balance the Oxygen with 4) In basic reactions balance the Oxygen with hydroxide then Hydrogen from hydroxide with hydroxide then Hydrogen from hydroxide with waterwater
MnOMnO2 (aq)2 (aq)+ ClO+ ClO331-1-
(aq) (aq) + OH + OH 1-1-aq) aq)
MnOMnO441-1-
(aq) (aq)+ Cl + Cl 1-1-(aq) (aq) + H+ H22OO(l) (l)
MnMn4+4+ (MnO (MnO22) ) MnMn7+7+ (MnO (MnO4 4 )) 1-1-
ClCl+5+5 (ClO(ClO3 3 )) 1-1-+ + 6 6 ee-- Cl Cl 1-1-
Balancing Redox EquationsBalancing Redox Equationsin basic solutionsin basic solutions
Electronically Balanced Equation:Electronically Balanced Equation:
22 MnO MnO2 (aq)2 (aq)+ ClO+ ClO331-1-
(aq) (aq) + + 66 e e - - 2 2
MnOMnO4 4 1-1- + + Cl Cl 1- 1- + + 6 6 ee--
Balancing Redox EquationsBalancing Redox Equationsin basic solutionsin basic solutions
Completely Balanced Equation:Completely Balanced Equation:
22 MnO MnO2 (aq)2 (aq)+ ClO+ ClO331-1-
(aq) (aq) + + 22 OH OH 1- 1- (aq)(aq)
2 2 MnOMnO4 (aq)4 (aq)1-1- + + Cl Cl 1- 1-
(aq)(aq)+ + 1 1 HH22O O (l)(l)
99 O in productO in product
Balancing Redox EquationsBalancing Redox Equationsin basic solutionsin basic solutions
QUESTIONQUESTIONOxalate 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?
•1•2•3•4
ANSWERANSWERD). The redox equation could be balanced according to the steps for an acid solution, but then OH– must be added to neutralize the H+ ions. When done properly 4 OH– ions will be present in the basic equation.