Oxidation-Reduction

© Copyright 2002-2008 R.J. Rusay© Copyright 2002-2008 R.J. Rusay

Oxidation-ReductionOxidation-Reduction

Dr. Ron RusayDr. Ron Rusay

Spring 2008Spring 2008


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).

Oxidation Reduction ReactionsOxidation Reduction Reactions

QuickTime™ and aSorenson Video decompressorare needed to see this picture.



Oxidation StateOxidation State(Oxidation Number)(Oxidation Number)


QUESTIONQUESTIONIn which of the following does nitrogen have an oxidation state of +4? 1) HNO3 2) NO2 3) N2O 4) NH4Cl 5) NaNO2

ANSWERANSWER 2

-

–

2O2

2) NO

Section 4.9 Oxidation Reduction Reactions(p. 154)

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.

Zinc

QUESTIONQUESTIONHow many of the following are oxidation-reduction reactions? NaOH + HCl → + NaCl H2O + 2Cu AgNO3 → 2 + (Ag Cu NO3)2 ( )Mg OH 2 → + MgO H2O N2 + 3H2 → 2NH3 1) 0 2) 1 3) 2 4) 3 5) 4

ANSWERANSWER

-

3) 2


If an element is found on the reactant’s side, this is almost always a redox reaction, since an element usually becomes part of a compound during a chemical reaction.

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

QUESTIONQUESTIONIn the reaction 2Cs(s) + Cl2(g) → 2 ( ), CsCl s Cl2 is 1) .the reducing agent 2) .the oxidizing agent 3) oxid .ized 4) .the electron donor 5) two of these

ANSWERANSWER

-

2) 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 (“redox” atoms) into separate oxidation number (“redox” atoms) into separate “half reactions”.“half reactions”.

3) Balance the “redox” atoms and charges (electron 3) Balance the “redox” 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).

1


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)

1

x x == ?? Cr Cr ; ; 2x+7(-2) 2x+7(-2) = = -2-2; ; x x = = +6+6



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+))

1

66 (Fe (Fe 2+2+(aq)(aq) -e -e - - ---> Fe---> Fe3+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



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)

1

OxygenOxygen == 7 7 2nd (Hydrogen)2nd (Hydrogen) == 1414



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)

1


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 “half reactions”oxidation number into separate “half 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

1


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) ---> Mn) ---> Mn7+7+ (MnO (MnO4 4 )) 1-1-

ClCl+5+5 (ClO(ClO3 3 )) 1-1-+ + 6 6 ee-- ---> Cl ---> Cl 1-1-



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--



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


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. 12. 23. 34. 4

ANSWERANSWERChoice 4 is correct. 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.

Section 4.10: Balancing Oxidation–Reduction Equations

Documents

Oxidation-Reduction