balancing oxidation-reduction equations - mister chemistry

Balancing Oxidation-Reduction Equations

Co2+ (aq) + Al(s) e)

Using the activity series provided, write the net ionic chemical equations for the following reactions. If no reaction occurs, simply write NR

2+ 0

best reducing agents

best oxidizing agents

Co2+ (aq) + Al(s)

Using the activity series provided, write the net ionic chemical equations for the following reactions. If no reaction occurs, simply write NR

2+ 0

Co (s) + Al3+ (aq)0 3+

oxidation-reduction reactions

Co2+(aq) + Al(s) + Al3+(aq)Co (s)

Al 3+(aq) + 3e- Al(s) oxidation

Co(s)+ 2e- Co2+(aq)reduction

the number of electrons lost has to be equal to the number of electrons gained.



Al 3+(aq) + 3e- Al(s) oxidation

Co(s)+ 2e- Co2+(aq)reduction

the of electrons lost has to be equal to the number of electrons gained.

2( )

3( )



2Al 3+(aq) + 6e-

3Co(s)+ 6e-

Al(s) oxidation

Co2+(aq)reduction

the of electrons lost has to be equal to the number of electrons gained.

2

3

2( )

3( )

3Co (s) + 2Al3+ (aq) 3Co2+ (aq) + 2Al(s)

2Al 3+(aq) + 6e- Al(s) oxidation

3Co(s)+ 6e- Co2+(aq)reduction

2

3


Adding the half reactions to gather gives us the overall redox reaction.

Example

is this a redox reaction?

KMnO4 + FeSO4

*All species are (aq)

K2SO4 MnSO4Fe2(SO4 )3+ +

7+ 2+

3+ 2+

Example

KMnO4 + FeSO4



Balance the following equation ?

Half-Reaction Method in Acid

1. Write the unbalanced equation in ionic form.

2. Separate the equation into two half-reactions.

3. Balance each half reaction (except for O and H).

5. Balance the charges by adding electrons.

4. In acid solution, balance O by adding H2O and H by adding H+

6. Add the half reactions

7. Check to make sure atoms and charges are balanced

Example

Balance the following equation for the reaction in acid solution. *

KMnO4 + FeSO4



Example


MnO4- + Fe2+ Mn2+ + Fe3+



MnO4- + Fe2+ Mn2+ + Fe3+

2. Separate the equation into two half-reactions

Fe2+ Fe3+

MnO4- Mn2+



Fe2+ Fe3+

MnO4- Mn2+8H++ 4H2O+


Fe2+ Fe3+ + 1e-

MnO4- Mn2+8H++ 4H2O++ 5e-

5 x

1 x

You need the same number of electrons on both sides of the equation.

(

( )

)


5Fe2+ 5Fe3+ + 5e-

MnO4- Mn2+8H++ 4H2O++ 5e-

5Fe2+ 5Fe3+ + 5e-

MnO4- Mn2+8H++ 4H2O++ 5e-

MnO4-

Mn2+

8H++

4H2O ++

+ 5Fe2+

5Fe3+


This is the balanced equation

Example

Balance the following equation for the reaction in basic solution. *


HS- + NO3- S + NO2

-

Half-Reaction Method in Base

1. Use the half-reaction method as specified for acidic solutions to obtain the final balanced equation as if H+ ions were present.

2. Add the number of OH- ions to both sides of the equation to turn the remaining H+ ions to H2O

3. Eliminate waters that appear on both sides of the equation.

Example



HS- S + H+

2H+ + + H2O

+ 2e-

2e- +

HS- + NO3- S + NO2

-

NO3- NO2

-

Example


HS- S

NO3- NO2

-H+ + + H2O

HS- + NO3- S + NO2

-

NO3-H+ +HS- + NO2

- + H2O+ SOH- + + OH-

Example


HS- S

NO3- NO2

-H+ + + H2O

HS- + NO3- S + NO2

-

NO3-HS- + NO2

- + H2O+ S + OH-H2O +

Example


HS- + NO3- S + NO2

-

NO3-HS- + NO2

- + S + OH-

Cr2O72- (aq) + I- (aq) a)

Complete and balance the following equations, and identify the oxidizing and reducing agents:

Cr3+ (aq) + IO3- (aq)

I2 (s) + OCl- (aq) b) IO3- (aq) + Cl- (aq)

As2O3 (s) + NO3- (aq) c) H3 AsO4 (aq) + N2O3 (aq)

acidic solution

acidic solution

acidic solution

MnO4- (s) + Br- (aq) d) MnO2 (s) + BrO3- (aq)

BrO3- (s) + N2H4 (g) e) Br- (aq) + N2 (g)

NO2- (aq) + Al (s) f) NH4+ (aq) + AlO2- (g)

basic solution

acidic solution

basic solution


2. Separate the equation into two half-reactions.


6. Balance the charge in the half reaction by adding electrons.

4. In acid solution, balance O by adding H2O

8. Add the half reactions (canceling anything that is redundant)

9 Check to make sure atoms and charges are balanced

Half-Reaction Method in Acid

5. In acid solution, balance H by adding H+

7. multiply by half reactions by an integer to make the electrons lost equal to the electrons gained

Half-Reaction Method in base

2. Eliminate waters that appear on both sides of the equation.

1. Use the acid method, then add the number of OH- ions to both sides of the equation to turn the remaining H+ ions to H2O

Cr2O72- (aq) + I- (aq)

a)

Cr3+ (aq) + IO3- (aq)

I- IO3-

Cr2O72- Cr3+

+ 6H+ + 6e-3H2O +

+ 7H2O 214H+ + 6 e- +

8H+ + I- + Cr2O72-

IO3- + 2Cr3+ + 4H2O

4


acidic solution

2H+ + 2 e- + 5 551010

5 ( )

I2 IO3-

OCl- Cl-

+ 12H+ + 10e-6H2O +

+ H2O

H2O + I2 + 5OCl-

2IO3- + 5Cl- + 2H+

I2 (s) + OCl- (aq)

b)

IO3- (aq) + Cl- (aq)

2


acidic solution

As2O3 H3 AsO4

NO3- N2O3

+ 4H+ + 4e-5H2O +

+ 3H2O 6H+ + 4 e- +

2

As2O3 (s) + NO3- (aq)

c)

H3 AsO4 (aq) + N2O3 (aq)

2

2H+ + 2H2O + 2NO3-

2H3 AsO4 + N2O3

+ As2O3


acidic solution

+ 2OH-

2OH- +

+ 6e-

3 e- + 2 ( )

MnO4- (s) + Br- (aq)

d)

MnO2 (s) + BrO3- (aq)

MnO4- MnO2

Br- BrO3-

+ 2H2O 4H+ +

+ 6H+ 3H2O +

2H+ + Br- + 2MnO4-

BrO3- + 2 MnO2 + H2O

6 8 2 2 4

H2O + Br- + 2MnO4-

BrO3- + 2 MnO2 + 2OH-


basic solution

BrO3- (s) + N2H4 (g)

e)

Br- (aq) + N2 (g)


acidic solution

N2H4 N2 + 4H+ + 4e-

BrO3- Br- + 3H2O 6H+ + 6 e- +

3 ( )

2 ( )

12123 3

12 12 2 2 6

3N2H4 + 2BrO3- 2Br- + 3N2 + 6H2O

NO2- (aq) + Al (s)

f)

NH4+ (aq) + AlO2- (g)


basic solution

NO2- NH4+

Al AlO2-

+ 2H2O 8H+ + 6 e- +

2H2O + + 4H+ + 3e- 2 ( )4 2 2 8 6

2BrO3- + NH4+2H2O + + 2Al AlO2-

2 ( )

5 ( )

+ 4H2O 8H+ + 5 e- +

+ 2e-

MnO4- (s) + Br- (aq) Mn2+(aq) + Br2 (aq)

MnO4- Mn2+

Br- Br2

16H+ + 10 Br- + 2 MnO4-

5 Br2 + 2 Mn2+ + 8 H2O


acid solution

2

10 16 2 2 8

10 5 10

NH4NO3 (s) + Zn (s) ZnO (s) + 2H2O (g) + N2 (g)

problems 34-37 refer to the chemical reaction below

which atom is acting as the oxidizing agent?

What is the oxidation state of zinc ion at the end of the reaction?

Which element is reduced during the reaction reaction?

Example


HNO3(aq) + H3PO3(aq) NO(g ) + H3PO4aq) + H2O(l)


2. Separate the equation into two half-reactions

NO3- NO

H3PO3 H3PO4

H+ (aq) NO3- (aq) + H3PO3(aq) NO(g )

+ H3PO4aq) + H2O(l)



NO3- NO

H3PO3 H3PO4

+ 2H2O4H+ +

H2O + + 2H+


NO3- NO +

H3PO3 H3PO4

2H2O4H+ +

H2O + + 2H+

3e- +

+ 2e-


NO3- NO +

H3PO3 H3PO4

2H2O4H+ +

H2O + + 2H+

3e- +

+ 2e-

2 x

3 x

(

( )

)


2NO3- 2NO +

3H3PO3 3H3PO4

4H2O8H+ +

3H2O + + 6H+

6e- +

+ 6e-


2NO3- 2NO +

3H3PO3 3H3PO4

4H2O2H+ +

3H2O + + 6H+

6e- +

+ 6e-


2NO3- 2NO +

3H3PO3 3H3PO4

H2O2H+ +

3H2O + + 6H+

6e- +

+ 6e-

2NO3-2H+ + 3H3PO3+

2NO + H2O + 3H3PO4


2NO3 +-2H+ + 3H3PO3 2NO + H2O + 3H3PO4

This is the balanced equation

Example


Fe(s) + HCl(aq) HFeCl4(aq) + H2 (aq )

H+ H2

Fe HFeCl44Cl- +H+ +

22e- +

+ 3e-( )2

( )3

Example


Fe(s) + HCl(aq) HFeCl4(aq) + H2 (aq )

H+ 3H2

2Fe 2HFeCl48Cl- +2H+ +

66e- +

+ 6e-

Example

H+ 3H2

2Fe 2HFeCl48Cl- +2H+ +

66e- +

+ 6e-

+ 3H22Fe 2HFeCl48Cl- +8H+ +

+ 3H22Fe 2HFeCl48HCl +or

balancing oxidation-reduction equations - mister chemistry

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