Oxidation - Reduction Reactions

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Oxidation is an increase in oxidation state (a change from +2 to +3 or a change from –2 to –1) ... with atoms of any other element have oxidation numbers of zero.

Oxidation - Reduction Reactions An oxidation-reduction reaction involves a transfer of electrons. In each reaction there will be a substance that loses electrons and one that gains electrons. A molecule is oxidized if it has a loss of electrons. A molecule is reduced if it gains electrons. Oxidation States: Oxidation is an increase in oxidation state (a change from +2 to +3 or a change from –2 to –1) Increase in oxidation state means a change in the positive direction which requires a loss of electrons. Reduction is a decrease in oxidation state (a change from +2 to +1 or a change from –2 to –3) Decrease in oxidation state means a change in the negative direction which requires a gain of electrons. Rules for assigning the oxidation state (also called the oxidation number) 1)

Elements: The oxidation number of an element in its elemental form is zero. Atoms of any element, not combined with atoms of any other element have oxidation numbers of zero. Examples: Na Br2 O2 O3 P4 S8

2)

Monoatomic ions: The oxidation number of a monoatomic ion is the same as its charge. 1+ Examples: K has an oxidation state of +1 2+ Ca has an oxidation state of +2 Cl has an oxidation state of –1 -2 S has an oxidation state of –2

3)

Ions in compounds: (for non-transition state) In their compounds, the oxidation number of group atoms IA is +1, the oxidation number of group IIA atoms is +2, the oxidation number of aluminum is +3. Note, this is applicable to non-transition state elements.

4) Binary compounds of nonmetals and metals: Assign the value for the non-metal first. The oxidation of the nonmetal equals the charge that would be present on the monoatomic ion. Examples: Fe2S3 the oxidation state of sulfur is –2 and that of iron is +3 CrBr3 the oxidation state of bromine is –1 and that of chromium is +3 5) Covalent compounds of nonmetals a) Fluorine is always –1, no exceptions b) Hydrogen is almost always +1. It is –1 if present as a hydride with a metal (Example: NaH) c) Oxygen is usually –2. It is not –2 when it would violate the rule of fluorine or hydrogen. Examples: OF2 O has an oxidation state of +2 H2O2 O has an oxidation state of +2 In general, in a binary compound, the element nearest the upper right hand corner of the periodic table will have a negative oxidation number and the one nearer the lower left hand corner will have a positive number. Assign the value of the element nearest the upper right hand corner of the periodic table first according to rule 4. Example: PCl3 The halide will have a –1 oxidation number, thus phosphorus will have a +3 oxidation number. However apply the rules applicable for F, H, and O first. Example: BrF Fluorine has an oxidation number of –1 and bromine has an oxidation number of +1. ClO3 Oxygen has an oxidation number or –2 and chlorine has an oxidation number of +5 6) The sum of the oxidation states must be zero for an electrically neutral compound. For an ion, the sum of the oxidation states must equal the charge of the ion.

Balancing Oxidation-Reduction Equations: The Half Reaction Method Steps required for balancing reactions taking place in acidic solution: Consider the reaction of dichromate ion and oxalic acid to form chromium (III) ion and carbon dioxide. Write the balanced redox equation for this reaction. 1) Write the skeletal equation for the reaction and identify what is oxidized and what is reduced. 2-

Cr2O7

+ H2C2O4

 Cr

3+

+ CO2

C is oxidized, going from +3 to +4 Cr is reduced, going from +6 to +3 2) Write the initial half reactions for the oxidation and reduction reactions Oxidation: H2C2O4 2Reduction: Cr2O7

 CO2 3+  Cr

3) Balance elements other than H and O. Oxidation: H2C2O4  2CO2 23+ Reduction: Cr2O7  2Cr 4) Balance O by adding H2O Oxidation: H2C2O4  2CO2 23+ Reduction: Cr2O7  2Cr + 7 H2O +

5) Balance H by adding H + Oxidation: H2C2O4  2CO2 + 2H + 23+ Reduction: 14H + Cr2O7  2Cr + 7 H2O -

6) Balance charge by adding e + Oxidation: H2C2O4  2CO2 + 2H + 2e + 23+ Reduction: : 6e + 14H + Cr2O7  2Cr + 7 H2O The half reactions are now balanced. 7) Prepare to combine the half reactions by multiplying by an integer to equate the electrons in each reaction. Oxidation: 3[H2C2O4  2CO2 + 2H + 2e ] + 23+ Reduction: 6e + 14H + Cr2O7  2Cr + 7 H2O +

-

Oxidation: 3H2C2O4  6 CO2 + 6 H + 6 e + 23+ Reduction: 6e + 14H + Cr2O7  2Cr + 7 H2O +

8)

Add half reactions together. Cancel species that exist on both sides. Simplify coefficients. -

+

2-

6e + 14H + Cr2O7 -

Cancel e , reduce H 8H 9)

-

+

2-

(aq)

+ Cr2O7

(aq)

 6 CO2 + 6 H + 6 e +

3H2C2O4

-

2Cr

3+

+ 7 H2O

+

3H2C2O4(aq)

-

 6 CO2(g) + 2Cr

Check that elements and charges are balanced.

3+ (aq)

+ 7 H2O(l)

Balancing Oxidation-Reduction Equations: The Half Reaction Method Steps required for balancing reactions taking place in basic solution: Consider the reaction between permanganate ion and sulfide ion to form manganese sulfide and solid sulfur. 1)

Write the skeletal equation for the reaction and identify what is oxidized and what is reduced. -

MnO4



2-

+S

MnS

+

S

2-

S is oxidized, going from -2 to 0 Mn is reduced, going from +7 to +2 2)

Write the initial half reactions for the oxidation and reduction reactions. Oxidation: S  S Reduction: MnO4  2-

3)

MnS

Balance elements other than H and O Oxidation: S  S 2Reduction: MnO4 + S 2-

4)



MnS



MnS

Balance O by adding water. Oxidation: S  S 2Reduction: MnO4 + S 2-

5)

+

4 H2O

+

-

Balance H by adding H . Then, add that same number of OH to both sides of the half reaction. + Simply by combining H and OH to form H2O. Oxidation: S  S (no hydrogen in reaction so leave as is for this step) + 2+ Reduction: 8H MnO4 + S  MnS + 4H2O (adding H ) 2-

Reduction: 8OH

-

Reduction: 8 H2O 6)

+ 8H

+

-

MnO4 -

+ MnO4

+

S



2-

+

S

 8 OH

2-

8OH -

-

+

+

MnS

MnS

+

+

-

2H2O (adding OH )

4H2O (forming water)

-

Balance charge by adding electrons (e ) Oxidation: S  S + 2e Reduction: 5e + 8 H2O + MnO4 2-

-



2-

+

S

8 OH

-

+

MnS

+

4H2O

At this point, the individual half reactions are balanced. 7)

Prepare to combine half reactions by multiplying by an integer to equate the electrons in each reaction. Oxidation: 5[S  S + 2e ] Reduction: 2[5e + 8 H2O + MnO4 2-

-

Oxidation: 5S  5S + 10e Reduction: 10e + 16 H2O + 2MnO4 2-

8)



2-

+

S

8 OH

-

+

MnS

+

4H2O]

-

+



2-

2S

16 OH

-

+

2MnS

+

8H2O

Add half reactions together. Cancel species that exist on both sides. Simplify coefficients. 2-

-

-

2-

5S + 10e + 16 H2O + 2MnO4 + 2S cancel e , reduce water 2-

7S

-

+ 8 H2O(l) + 2 MnO4 (aq)  5S(s) + 16OH (aq) + 2 MnS(s) -

(aq)

 5S + 10e + 16 OH -

-

+

2MnS

+

8H2O