Redox Equations under Basic Conditions

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Redox Equations under Basic

q

Conditions

B i diti th t h OH d

Basic conditions means that you have more OH- and very

little H+ in your solution. In fact, you have so little H+ that it can’t appear in the equation because it doesn’t exist!

The way you will balance redox reactions under basic

conditions is exactly the same way you balanced equations conditions is exactly the same way you balanced equations under neutral and acidic solution, the only difference is that we will include one additional step to remove the H+ in our

ti d l it ith OH

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Redox Equations under Basic

q

Conditions

The steps to balancing oxidation reduction (redox) reactions under The steps to balancing oxidation-reduction (redox) reactions under basic conditions are:

1. Write separate equations for reduction and oxidation half reactions. 2 For each half reaction

2. For each half reaction

a. First balance all elements except O and H. b. Add H2O to balance O.

c. Add H+ to balance H.

d. Add electrons to balance net charge.

3. If necessary, multiply one or both ½ reaction equations by integers so the total number of electrons used in one reaction is equal to the total number of electrons furnished by the other reaction.

4. Add the two ½ reaction equations together and cancel any common terms.

5 Add OH t BOTH id f th ti t h H+’ i t H O

5. Add OH- to BOTH sides of the equation to change any H+’s into H 2O. 6. Double check that all species and charges balance.

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E ample 1

Example 1.

Let’s try this example Balance the equation:

NO2-(aq) + Al(s) = NH3(g) + AlO2-(aq) nder basic conditions

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NO (

) + Al( )

NH ( ) + AlO (

)

NO

2-

(aq) + Al(s) = NH

3

(g) + AlO

2-

(aq)

½ Reactions: ½ Reactions:

NO2-(aq) = NH3(g) Al(s) = AlO2-(aq)

Balancing: NO2-(aq) = NH3(g) + 2H2O 7H+ + NO 2-(aq) = NH3(g) + 2H2O 6e- + 7H+ + NO 2-(aq) = NH3(g) + 2H2O 2 ( q) 3(g) 2

2H2O + Al(s) = AlO2-(aq)

2H O + Al(s) = AlO -(aq) + 4H+ 2H2O + Al(s) = AlO2 (aq) + 4H

2H2O + Al(s) = AlO2-(aq) + 4H+ + 3e

-l l h d b 2 h 6 l

Multiplying the second equation by 2 so we have 6 electrons:

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-NO (

) + Al( )

NH ( ) + AlO (

)

NO

2

(aq) + Al(s) = NH

3

(g) + AlO

2

(aq)

Adding equations together: Adding equations together:

6e- + 7H+ + NO2-(aq) = NH3(g) + 2H2O + 4H2O + 2 Al(s) = 2 AlO2-(aq) + 8H+ + 6e -6e- + 7H+ + NO2-(aq) + 4H2O + 2 Al(s) =

NH3(g) + 2H2O + 2 AlO2-(aq) + 8H+ + 6e -Removing common terms:

NO2-(aq) +2 H

2O +2Al(s) = NH3(g) +2AlO2-(aq) +1H+

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NO (

) + Al( )

NH ( ) + AlO (

)

NO

2-

(aq) + Al(s) = NH

3

(g) + AlO

2-

(aq)

NO22-(aq) +2 H( q) 22O +2Al(s) = NH( ) 33(g)(g) + 2 AlO22-(aq) + 1 H( q) + Now to our added rule for basic solutions:

5 Add OH- to BOTH sides of the equation to change any H+’s into H 2O

5. Add OH to BOTH sides of the equation to change any H s into H2O To apply this rule find the number of H+’s in the equation.

In this case there is 1 H+ on the right hand side of the equation

In this case there is 1 H on the right hand side of the equation, so we add 1 OH- to BOTH sides of the equation so it remains balanced.

The trick is that on the side with the H+, the H+ and the OH- can

be combined to make H2O, so the H+ disappears from the

2 , pp

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NO (

) + Al( )

NH ( ) + AlO (

)

NO

2

(aq) + Al(s) = NH

3

(g) + AlO

2

(aq)

NO22-(aq) +2 H( q) 22O +2 Al(s) = NH( ) 33(g)(g) + 2 AlO22-(aq) + 1 H( q) +

+ OH- + OH -Rewriting:g NO2-(aq) +2 H2O +2Al(s) + OH- = NH3(g) +2AlO2-(aq) +1 H++ OH -Combining H+ and OH-: NO2-(aq) +2 H

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NO (

) + Al( )

NH ( ) + AlO (

)

NO

2-

(aq) + Al(s) = NH

3

(g) + AlO

2-

(aq)

Checking to remove common terms a second time:g

NO2-(aq) +2H2O +2Al(s) + OH- = NH3(g) +2AlO2-(aq) +1H2O

Becomes:

NO -(aq) + H O +2Al(s) + OH- = NH (g) +2AlO -(aq)

NO2 (aq) + H2O +2Al(s) + OH = NH3(g) +2AlO2 (aq)

Double checking our balance:

NO2-(aq) + H2O +2Al(s) +OH- = NH3(g) +2AlO2-(aq) N 1 =1 O 2 1 1 = 2(2)( ) H 2 1 = 3 Al 2 = 2 Charge Charge -1 -1 = 2(-1)

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NO (

) + Al( )

NH ( ) + AlO (

)

NO

2

(aq) + Al(s) = NH

3

(g) + AlO

2

(aq)

Adding physical forms we get our final answer:

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E ample 2

Example 2.

Let’s try this example Balance the equation: Balance the equation:

Al(s) + MnO4-(aq) = MnO2(s) + Al(OH)4-(aq) under basic conditions.

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Al( ) + M O (

)

M O ( ) + Al(OH) (

)

Al(s) + MnO

4

(aq) = MnO

2

(s) + Al(OH)

4

(aq)

½ Reactions:

Al(s) = Al(OH)4-(aq) MnO4-(aq) = MnO2(s)

Balancing: Balancing:

4 H2O + Al(s) = Al(OH)4-(aq)

4 H2O + Al(s) = Al(OH)4-(aq) + 4H+

4 H O Al( ) Al(OH) ( ) 4H+ 3 4 H2O + Al(s) = Al(OH)4-(aq) + 4H+ + 3e

-M O ( ) M O ( ) + 2 H O MnO4-(aq) = MnO2(s) + 2 H2O

4H+ +MnO4-(aq) = MnO2(s) + 2 H2O 3e- + 4H+ +MnO

4-(aq) = MnO2(s) + 2 H2O

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Al( ) + M O (

)

M O ( ) + Al(OH) (

)

Al(s) + MnO

4-

(aq) = MnO

2

(s) + Al(OH)

4-

(aq)

Both equations have the same # of electrons, q f so we don’t have to multiply anything!

Adding equations together: Adding equations together:

4 H2O + Al(s) = Al(OH)4-(aq) + 4H+ + 3e

-+

3 4H+ M O ( ) M O ( ) 2 H O 3e- + 4H+ +MnO4-(aq) = MnO2(s) + 2 H2O 4 H22O + Al(s) + 3e( ) -+ 4H+ +MnO44-(aq) = ( q)

Al(OH)4-(aq) + 4H+ + 3e- + MnO

2(s) + 2 H2O

Removing common terms: Removing common terms:

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Al( ) + M O (

)

M O ( ) + Al(OH) (

)

Al(s) + MnO

4

(aq) = MnO

2

(s) + Al(OH)

4

(aq)

2 H2O + Al(s) +MnO4-(aq) = Al(OH)

4-(aq) + MnO2(s) 2 ( ) 4 ( q) ( )4 ( q) 2( )

How about that. There aren’t any H+’s so I don’t have to do anything! This happens every once in a while

do anything! This happens every once in a while.

Double checking overall balance:

2 H O Al( ) M O ( ) Al(OH) ( ) M O ( ) 2 H2O + Al(s) +MnO4-(aq) = Al(OH)

4-(aq) + MnO2(s) H 2(2) = 4 O 2 4 = 4 2 Al 1 = 1 Mn 1 1 Charge Charge -1 = -1

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Al( ) + M O (

)

M O ( ) + Al(OH) (

)

Al(s) + MnO

4-

(aq) = MnO

2

(s) + Al(OH)

4-

(aq)

Putting in physical forms to get our final answer:

2 H2O(l) + Al(s) +MnO4-(aq) = Al(OH)

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Practice problems

Practice problems

Here are a couple of problems for you to try.

Balance the following two reactions under BASIC conditions:g CN-(aq) + MnO4-(aq) = CNO-(aq) + MnO2(s)

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Practice problems

Practice problems

CN-(aq) + MnO -(aq) CNO-(aq) + MnO (s) CN-(aq) + MnO4-(aq) = CNO-(aq) + MnO2(s) Answer: 3 CN-(aq) + 2 MnO4-(aq) + 1 H2O(l) =

3 CNO-(aq) + 2 MnO2(s) + 2 OH- (aq)

Cl2(g) = Cl-(aq) + OCl-(aq) Cl2(g) Cl (aq) OCl (aq) Answer:

2 Cl2(g) + 4OH- (aq) = 2 Cl-(aq) + 2 OCl-(aq) + 2 H2O (l) If you didn’t get these answers, look over my notes on the next few pages, otherwise you can exit from this tutorial now.

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CN (

) + M O (

)

CNO (

) + M O ( )

CN

(aq) + MnO

4

(aq) = CNO

(aq) + MnO

2

(s)

½ Reactions: ½ Reactions:

CN-(aq) = CNO-(aq) MnO4-(aq) = MnO2(s)

Balancing:

H22O + CN-(aq) = CNO( q) -(aq)( q)

H2O + CN-(aq) = CNO-(aq) + 2H+

H2O + CN-(aq) = CNO-(aq) + 2H+ +2e -MnO4-(aq) = MnO2(s) + 2H2O

4H+ + MnO

4-(aq) = MnO2(s) + 2H2O

3e- + 4H+ + MnO -(aq) = MnO (s) + 2H O

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CN (

) + M O (

)

CNO (

) + M O ( )

CN

-

(aq) + MnO

4-

(aq) = CNO

-

(aq) + MnO

2

(s)

½ Reactions:

H2O + CN-(aq) = CNO-(aq) + 2H+ +2e -3e- + 4H+ + MnO4-(aq) = MnO2(s) + 2H2O

So we need to multiply the first reaction by 3 and the second by 2!

3x(H2O + CN-(aq) = CNO-(aq) + 2H+ +2e- )

2x( 3e- + 4H+ + MnO4-(aq) = MnO2(s) + 2H2O)

So we have:

3 H2O + 3 CN-(aq) = 3 CNO-(aq) + 6H+ + 6 e

-6e- + 8 H+ + 2 MnO

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CN (

) + M O (

)

CNO (

) + M O ( )

CN

(aq) + MnO

4

(aq) = CNO

(aq) + MnO

2

(s)

Adding together: Adding together:

3 H2O + 3 CN-(aq) = 3 CNO-(aq) + 6H+ + 6 e

-+ 6e- + 8 H+ + 2 MnO

4-(aq) = 2 MnO2(s) + 4 H2O

We get:

3 H2O + 3 CN-(aq) + 6e- + 8 H+ + 2 MnO4-(aq) = 3 CNO-(aq) + 6H+ + 6 e- +2 MnO

2(s) + 4 H2O

Removing common terms: Removing common terms:

3 CN-(aq) + 2 H+ + 2 MnO4-(aq) =

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CN (

) + M O (

)

CNO (

) + M O ( )

CN

-

(aq) + MnO

4-

(aq) = CNO

-

(aq) + MnO

2

(s)

Last equation: Last equation:

3 CN-(aq) + 2 H+ + 2 MnO

4-(aq) =

3 CNO-(aq) +2 MnO

2(s) + 1 H2O

Adding 2OH- to both sides:

3 CN-(aq) + 2 H+ + 2 MnO4-(aq) + 2OH- = 3 CNO-(aq) +2 MnO

2(s) + 1 H2O + 2OH

-Combining the H+ and OH- on the left side to get H2O: Combining the H and OH on the left side to get H2O:

3 CN-(aq) + 2 MnO4-(aq) + 2H2O =

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-CN (

) + M O (

)

CNO (

) + M O ( )

CN

(aq) + MnO

4

(aq) = CNO

(aq) + MnO

2

(s)

Last equation:

3 CN-(aq) + 2 MnO4-(aq) + 2H2O =

3 CNO-(aq) +2 MnO (s) + 1 H O + 2OH

-3 CNO-(aq) +2 MnO

2(s) + 1 H2O + 2OH

-Removing common terms a second time:

3 CN-(aq) + 2 MnO4-(aq) + 1 H2O =

3 CNO-(aq) +2 MnO

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-CN (

) + M O (

)

CNO (

) + M O ( )

CN

-

(aq) + MnO

4-

(aq) = CNO

-

(aq) + MnO

2

(s)

Checking the balance: Checking the balance:

3 CN-(aq) + 2 MnO4-(aq) + 1H2O = 3 CNO-(aq) +2 MnO2(s) + 2OH -C 3 = 3 N 3 = 3 Mn 2 = 2 O 2(4) 1 = 3 2(2) 2(1) H 2 = 2 Charge 3(-1) 2(-1) = 3(-1) 2(-1) P i i h h i l f fi l i

Putting in the physical forms to get our final equation:

3 CN-(aq) + 2 MnO

4-(aq) + 1 H2O(l) =

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Cl ( )

Cl

-

(

) + OCl

-

(

)

Cl

2

(g) = Cl

-

(aq) + OCl

-

(aq)

½ Reactions:

Cl22(g) = Cl(g) -(aq)( q) Cl2(g) = OCl-(aq)

(Note: This is a tricky one the same compound on the left is (Note: This is a tricky one, the same compound on the left is undergoing BOTH oxidation and reductions at the same time. This is called a disproportionation reaction, and you will see

i lik hi i hil )

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Cl ( )

Cl

-

(

) + OCl

-

(

)

Cl

2

(g) = Cl

-

(aq) + OCl

-

(aq)

Balancing: Cl ( ) 2 Cl ( ) Cl2(g) = 2 Cl-(aq) 2e- + Cl2(g) = 2 Cl-(aq) Cl2(g) = 2 OCl-(aq) 2 H2O + Cl2(g) = 2 OCl-(aq) 2 H O + Cl (g) = 2 OCl-(aq) + 4 H+ 2 H2O + Cl2(g) 2 OCl (aq) + 4 H 2 H2O + Cl2(g) = 2 OCl-(aq) + 4 H+ + 2 e

-Both equations have 2 electrons so we can combine without Both equations have 2 electrons so we can combine without a multiplication:

2e- + Cl

2(g) + 2 H2O + Cl2(g) =

2 Cl ( ) + 2 OCl ( ) + 4H+ + 2 2 Cl-(aq) + 2 OCl-(aq) + 4H+ + 2 e

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-Cl ( ) -Cl

-

(

) + OCl

-

(

)

Cl

2

(g) =Cl

-

(aq) + OCl

-

(aq)

Last equation:

2 Cl ( ) 2 H O Cl ( ) 2e- + Cl2(g) + 2 H2O + Cl2(g) =

2 Cl-(aq) + 2 OCl-(aq) + 4H+ + 2 e

-R i d bi Cl

Removing common terms and combing Cl2:

2 Cl2(g) + 2 H2O = 2 Cl-(aq) + 2 OCl-(aq) + 4H+

Addi 4OH b h id f h i

Adding 4OH- to both sides of the equation:

2Cl2(g) + 2H2O + 4OH- = 2 Cl-(aq) +2OCl-(aq) + 4H+ + 4OH

-Combing the H+ and OH- on the right side of the equation:

2Cl2(g) + 2H2O + 4OH- = 2 Cl-(aq) +2OCl-(aq) + 4H2O

Removing common terms:

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Cl ( )

Cl

-

(

) + OCl

-

(

)

Cl

2

(g) = Cl

-

(aq) + OCl

-

(aq)

Last equation:

2 Cl ( ) 4OH 2 Cl ( ) 2 OCl ( ) 2 H O 2 Cl2(g) + 4OH- = 2 Cl-(aq) + 2 OCl-(aq) + 2 H2O

Checking balance:g

2 Cl2(g) + 4OH- = 2 Cl-(aq) + 2 OCl-(aq) + 2 H 2O Cl 2(2) = 2 2 O 4(1) = 2(1) 2(1) O 4(1) 2(1) 2(1) H 4(1) = 2(2) Charge 4(-1) = 2(-1) 2(-1)

Putting in physical forms for final answer:

Figure

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References

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