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.