REDOX REACTIONS

(reduction/oxidation)

REDOX REACTIONS

 Involve the exchange of electrons in a chemical reaction  Electrons are lost by one substance and gained by another substance  One substance goes through oxidation, while the other substance goes through reduction

REDUCTION

 gain of electrons  the gain of hydrogen  the loss of oxygen  a decrease in oxidation number for the substance being reduced

OXIDATION

 loss of electrons  the gain of oxygen  the loss of hydrogen  an increase in oxidation number for the substance being oxidized

REDUCING AGENT  experiences oxidation  is an electron donor

OXIDIZING AGENT  experiences reduction  is an electron acceptor

OXIDATION STATES     Also known as oxidation numbers Allows understanding of what is oxidized and what is reduced Imaginary charges that atoms would have if shared electrons were divided equally in a covalent bond Or real charges that monatomic ions have in an ionic bond

ASSIGNING OXIDATION STATES  Written directly above a symbol with the sign and then the number…unlike charges

ASSIGNING OXIDATION STATES  The oxidation state of an atom of an element in its natural state is zero.

Na(s)

0 0

Br 2 (l)

0

Cl 2 (g)

0

C(s)

ASSIGNING OXIDATION STATES  The oxidation state of a monatomic ion is equal to its charge.

+1

Na 1+ (aq)

+2

Fe 2+ (aq)

+1 -1

NaCl(g)

+3

Al 3+ (aq)

ASSIGNING OXIDATION STATES  Oxygen is assigned an oxidation number of -2 in compounds; an exception is found in the peroxide ion, O number of -1.

2 2 , where each oxygen is assigned an oxidation

+1 -2

Na 2 O

+3 -2

Fe 2 O 3

ASSIGNING OXIDATION STATES  Hydrogen is assigned an oxidation number of +1 in its covalent compounds with nonmetals. In compounds with metals, the oxidation number of hydrogen is -1.

+1 -2

H 2 O

+1 -1

HCl

ASSIGNING OXIDATION STATES  The sum of the oxidation states in a compound must be zero, as a compound’s charge is zero.

+1 -2

Na 2 O

+1 -2

H 2 O

+3 -2

Fe 2 O 3

+1 -1

HCl

ASSIGNING OXIDATION STATES  The sum of the oxidation states in a polyatomic ion must be equal to that ion’s charge.

-2 +1

OH 1-

+2 -3

CN 1-

+4 -2

SO 3 2-

+7 -2

ClO 4 1-

ASSIGNING OXIDATION STATES  Non-integer oxidation states do exist and indicate the average division of electrons among the elements.

+8/3 -2

Fe 3 O 4

ASSIGNING OXIDATION STATES

+2 -2

CO

+4 -2

CO 2

+2 -2

NO

+4 -2

NO 2

+3 -2

NO 2 1-

0

N 2

OXIDATION STATES METHOD OF BALANCING    Assign oxidation numbers to every atom in the reaction Connect the atoms involved in oxidation with a line above the reaction.

Connect the atoms involved in reduction with a line beneath the reaction.

OXIDATION STATES METHOD OF BALANCING     Write the change in oxidation number on each line with parentheses around it.

Determine the least common multiple of the changes.

Use multipliers to achieve the LCM.

The multipliers guide you in determining your coefficients.

OXIDATION STATES METHOD OF BALANCING

1 (+2) +2 -2 +2 -2

PbO(s) + CO(g) 

0 +4 -2

Pb(s) + CO 2 (g)

1 (-2)

OXIDATION STATES METHOD OF BALANCING

1 (+2)

2

+4 -1

CeCl 4

+2 -1

+ SnCl 2  2

+3 -1

CeCl 3

+4 -1

+ SnCl 4

2 (-1)

OXIDATION STATES METHOD OF BALANCING

3 (+1) +1-1

3

+2 -1

2

+1 +5 -2

+ KNO 3  3

+3 -1

FeCl 3

+2 -2 +1 -2 +1 -1

+ NO + H 2 O + KCl

1 (-3)

OXIDATION STATES METHOD OF BALANCING 4 HCl+ FeCl 2 +KNO 3  3 FeCl 3 2 2 O+KCl

HALF-REACTION METHOD OF BALANCING  Assign oxidation numbers to every atom in the reaction  Identify what is oxidized and what is reduced. Anything whose oxidation number does not change is a spectator.

HALF-REACTION METHOD OF BALANCING   Write a ½ reaction for the reduction reaction without the spectators.

If an atom being reduced is part of a solid, a polyatomic ion or part of a molecular compound, you will keep that ion or cpd together when you bring it down for the ½ rxn.

HALF-REACTION METHOD OF BALANCING   Balance the ½ rxn by first balancing the non-H, non-O atoms.

Then balance the H’s and O’s using the following guide:   Use H 1+ and H 2 O if the rxn occurs in an acidic medium.

Use OH 1 and H 2 O if the rxn occurs in a basic medium.

HALF-REACTION METHOD OF BALANCING  Balance the ½ rxn electrically (charge-wise) by adding electrons (e ) to the left side since a reduction rxn goes GER

HALF-REACTION METHOD OF BALANCING   Write a ½ reaction for the oxidation reaction without the spectators.

If an atom being oxidized is part of a solid, a polyatomic ion, or a molecular compound, you will keep that solid, ion, or cpd together when you bring it down for the ½ rxn.

HALF-REACTION METHOD OF BALANCING   Balance the ½ rxn by first balancing the non-H, non-O atoms.

Then balance the H’s and O’s using the following guide:   Use H 1+ and H 2 O if the rxn occurs in an acidic medium.

Use OH 1 and H 2 O if the rxn occurs in a basic medium.

HALF-REACTION METHOD OF BALANCING  Balance the ½ rxn electrically (charge-wise) by adding electrons (e ) to the right side since an oxidation rxn is LEO

HALF-REACTION METHOD OF BALANCING  Normalize the electrons in each ½ rxn by finding the LCM of the two numbers of electrons and distributing the multipliers through each entire ½ rxn.

HALF-REACTION METHOD OF BALANCING  Add the two ½ rxns and cancel any like atoms, ions, or cpds that appear on both sides.

 Add the spectators back into the rxn and balance the rest by inspection.

½ RXN METHOD OF BALANCING

+2 -2 +2 -2

PbO(s) + CO(g) 

0 +4 -2

Pb(s) + CO 2 (g)

½ RXN METHOD OF BALANCING PbO(s)  Pb(s) + H 2 O

½ RXN METHOD OF BALANCING CO(g)  CO 2 (g) + 2H 1+ + 2e -

½ RXN METHOD OF BALANCING 1 PbO(s) + 2e  Pb(s) + H 2 O 1 CO(g)  CO 2 (g) + 2H 1+ + 2e -

½ RXN METHOD OF BALANCING 1 PbO(s) + 2e  Pb(s) + H 2 O 1 CO(g)  CO 2 (g) + 2H 1+ + 2e PbO(s) + 2H + + 2e + CO(g) + H 2 O  Pb(s) + H 2 O + CO 2 + 2H + + 2e -

½ RXN METHOD OF BALANCING PbO(s) + CO(g)  Pb(s) + CO 2 (g) Ta-da!

½ RXN METHOD OF BALANCING

+4 -1

CeCl 4

+2 -1

+ SnCl 2 

+3 -1

CeCl 3

+4 -1

+ SnCl 4

½ RXN METHOD OF BALANCING Ce 4+ + 1e  Ce 3+

½ RXN METHOD OF BALANCING Sn 2+  Sn 4+ + 2e -

½ RXN METHOD OF BALANCING 2 Ce 4+ + 1e 1 Sn 2+  Ce 3+  Sn 4+ + 2e -

½ RXN METHOD OF BALANCING 2 Ce 4+ + 1e 1 Sn 2+   Ce 3+ Sn 4+ + 2e 2Ce 4+ + Sn 2+ + 2e  2Ce 3+ + Sn 4+ + 2e -

½ RXN METHOD OF BALANCING 2CeCl 4 + SnCl 2  2CeCl 3 + SnCl 4 Ta-da!

½ RXN METHOD OF BALANCING HCl + FeCl 2 + KNO 3  FeCl 3 + NO + H 2 O + KCl