The Foundations of Chemistry

We see that ethanol is oxidized; it is the reducing agent. Cr 2 O 72 ions are reduced; they are

the oxidizing agent.

H

C 2 H 5 OH
Cr 2 O 72 88nCr^3 

C 2 H 4 O
H 2 O

 3


 1

Oxidation Numbers Change/Atom Equalizing Changes Gives

Cr
 6 88nCr
 3 31(3) 3

C 2 88nC 1 
13(
1)
 3

Each change must be multiplied by two because there are 2 Cr in each Cr 2 O 72 and 2 C in

C 2 H 5 OH.

2(3)6 (total decrease) 2(
3)
6 (total increase)

We need 2 Cr and 6 C on each side of the equation to balance the redox part.

H

 3C 2 H 5 OH 
 Cr 2 O 72  88n 2Cr^3 
 
 3C 2 H 4 O 
H 2 O

Now we balance H and O using our chart. There are 10 O on the left and only 4 O on the

right. So we add 6 moreH 2 O molecules on the right.

H

3C 2 H 5 OH
Cr 2 O 72 88n2Cr^3 

3C 2 H 4 O
 7H 2 O

Now there are 26 H on the right and only 19 on the left. So we add 7 moreH
ions on the

left to give the balanced net ionic equation.

8H
(aq)
3C 2 H 5 OH(aq)
Cr 2 O 72 (aq)88n2Cr^3 
(aq)
3C 2 H 4 O(aq)
7H 2 O(
)

You should now work Exercise 54.

Every balanced equation must have both mass balance and charge balance.Once the redox

part of an equation has been balanced, we must next count eitheratoms or charges. After

we balanced the redox part in Example 11-21, we had

H

3C 2 H 5 OH
Cr 2 O 72 88n2Cr^3 

3C 2 H 4 O
H 2 O

The net charge on the left side is (1
 2 ) 1 . On the right, it is 2(3
) 6 
. Because

H
is the only charged species whose coefficient isn’t known,we add 7 moreH
to give a net

charge of 6
on both sides.

8H
 
3C 2 H 5 OH
Cr 2 O 72 88n2Cr^3 

3C 2 H 4 O
H 2 O

Now we have 10 O on the left and only 4 O on the right. We add six moreH 2 O mole-

cules to give the balanced net ionic equation.

8H
(aq)
3C 2 H 5 OH(
)
Cr 2 O 72 (aq)88n2Cr^3 
(aq)
3C 2 H 4 O(
)
 7H 2 O(
)

How can you tell whether to balance atoms or charges first? Look at the equation after

you have balanced the redox part.Decide which is simpler, and do that. In the preceding

equation, it is easier to balance charges than to balance atoms.

424 CHAPTER 11: Reactions in Aqueous Solutions II: Calculations

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 6 
 3  1

8 n

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