here, we’ll go through an example of balancing a half-reaction in basic solution

BalancingHalf-Reactions in Basic Solution

Example 1 Here, we’ll go through an example of balancing a half-reaction in basic solution.

We’re asked to balance the half-reaction: H2AlO3 minus gives Al, taking place in basic solution.

Balance the half-reaction:

(basic solution)2 3H AlO Al

We’ll start by adding H2AlO3 minus to the left side

Balance the half-reaction: (basic solution) 2 3H AlO Al

2 3 2

2

2 3 2

H AlO 4H 3e Al + 3H O

4H O 4H 4OH

H AlO H O 3e Al 4OH

And Al to the right side.


2 3 2

2

2 3 2


4H O 4H 4OH

H AlO H O 3e Al 4OH

Even when we’re balancing a half-reaction in basic solution, we still start with the steps for acid solution. We begin by balancing atoms other than hydrogen or oxygen.


2 3

2 3 2

2

24H O 4H 4OH

H

H

AlO H O 3e

AlO 4H 3e Al

Al 4OH

+ 3H O

Balance in Acid Solution

In this case the element is aluminum.


2

2

2 2

2

3

3

Al Al

4H O 4H 4OH

H Al

H O 4H 3e + 3

O H

H O

O 3e Al 4OH

1 1

Balance Aluminum Atoms

We have one Al on each side, so aluminum atoms are already balanced.


2

2

2 2

2

3

3

Al Al

4H O 4H 4OH

H Al

H O 4H 3e + 3

O H

H O

O 3e Al 4OH

1 1

Balance Aluminum Atoms

The next step is to balance oxygen atoms.


2 3

2

2 3

2

2

O

4H O 4H 4OH

H

H

AlO H O 3e A

Al 4H 3e Al + 3H O

l 4OH

Balance Oxygen Atoms

Notice we have 3 oxygen atoms on the left side. Remember we balance oxygen atoms by adding a water molecule on the right for every excess oxygen atom on the left.


3

2

2 3 2

2 2O

4H O 4H 4OH

H

H

AlO H O 3e A

Al 4H 3e Al + 3H O

l 4OH

3

Balance Oxygen Atoms

So we add 3 water molecules to the right side.


3 2

2

2 3 2

2 O 3H O

4H O 4H 4OH

H Al

H Al 4H 3e Al

O H O 3e 4

+

l

A OH

Balance Oxygen Atoms3

Now we have 3 oxygen atoms on both sides, so oxygen is balanced.


3

2

2

2

3 2

2O 3 O

4H O 4H 4OH

H AlO H O 3e A

H Al 4H 3e Al +

l O

H

4 H

Balance Oxygen Atoms3 3

The next step is to balance hydrogen atoms.


2

2

2 3 2

3 2H 3H

4H O 4H 4OH

H AlO H O 3e A

AlO 4H 3e Al +

l O

O

4 H

Balance Hydrogen Atoms

At this point we have 2 hydrogens on the left and 6 hydrogens on the right.


2

2

2 3 2

3 2H 3H

4H O 4H 4OH

H AlO H O 3e A

AlO 4H 3e Al +

l O

O

4 H

Balance Hydrogen Atoms2 6

In order to get 6 hydrogens on the left, we need to (click) add 4 hydrogens to the left side.


2

2

2 3 2

3 2H 3H

4H O 4H 4OH

H AlO H O 3e A

AlO 4H 3e Al +

l O

O

4 H


We need to add 4 H’s to the left side

We do this by (click) adding 4 H+ ions to the left side


32 2

2

2 3 2

H 4H 3H

4H O 4H 4OH

H AlO

AlO 3e

H O 3e Al

Al +

OH

O

4


add 4 H+ to the left side

32 2

2

2 3 2

H 4H 3H

4H O 4H 4OH

H AlO

AlO 3e

H O 3e Al

Al +

OH

O

4

So on the left side, we now have a total of 2 plus 4



2 4 6

32 2

2

2 3 2

H 4H 3H

4H O 4H 4OH

H AlO

AlO 3e

H O 3e Al

Al +

OH

O

4

Which is equal to 6 hydrogens



2 4 6

6

32 2

2

2 3 2

H 4H 3H

4H O 4H 4OH

H AlO

AlO 3e

H O 3e Al

Al +

OH

O

4

And we have 6 hydrogens on both sides, so hydrogen is balanced.



2 4 6

6

2 3

2 3 2

2

24H O 4H 4OH

H

H

AlO H O 3e

AlO 4H 3e Al

Al 4OH

+ 3H O

The next step is to balance the charge.


Balance Charge

2 3

2

2 3

2

2

4

4H O 4H 4OH

H

H AlO

AlO

H 3e Al + 3H

H O 3e Al 4OH

O

Looking on the left side, we see that H2AlO3 minus has an ionic charge of negative 1 and 4H+ has a total ionic charge of positive 4.


Balance Charge

–1+4

2 3

2

2 3

2

2

4

4H O 4H 4OH

H

H AlO

AlO

H 3e Al + 3H

H O 3e Al 4OH

O

So the total ionic charge on the left side is —1 + positive 4,


Balance Charge

–1+4

2 3

2

2 3

2

2

4

4H O 4H 4OH

H

H AlO

AlO

H 3e Al + 3H

H O 3e Al 4OH

O

Which equals positive 3


Balance Charge

–1+4

+3

2 3

2

2 3

2

2

4

4H O 4H 4OH

H

H AlO

AlO

H 3e Al + 3H

H O 3e Al 4OH

O

And on the right side, the total charge is 0 plus 0


Balance Charge

–1+4 00

+3

2 3

2

2 3

2

2

4

4H O 4H 4OH

H

H AlO

AlO

H 3e Al + 3H

H O 3e Al 4OH

O

Which equals zero


Balance Charge

–1+4 00

+3 0

2 3

2

2 3

2

2

4

4H O 4H 4OH

H

H AlO

AlO

H 3e Al + 3H

H O 3e Al 4OH

O

In order to balance charge, we must add enough electrons to the more positive side, to make the charges equal


Balance Charge

+3

0Add 6 e– to the more

positive side

2 3

2

2 3

2

2

4

4H O 4H 4OH

H

H AlO

AlO

H 3e Al + 3H

H O 3e Al 4OH

O

Because the charge on the left side is +3 and the charge on the right side is zero, we must add (click) 3 electrons to the left side.


Balance Charge

+3

0Add 3 e– to the more

positive side

2 3

2

2 3

2

2

3e

4H O 4

H A

H 4OH

H AlO H O 3e Al 4OH

lO 4H Al + 3H O

So we add 3 electrons to the left side.


Balance Charge

+3

0Add 3 e– to the

Left side

2

2 3 2

2 3 24 3e

4H O 4H 4OH

H AlO H O

H AlO H Al + 3

3e Al 4OH

H O

And now the total ionic charge on the left side is –1 plus positive 4 plus – 3


Balance Charge

0–1

+4

–3

2

2 3 2

2 3 24 3e

4H O 4H 4OH

H AlO H O

H AlO H Al + 3

3e Al 4OH

H O

Which is equal to zero


Balance Charge

–1+4

–3

0

0

2

2 3 2

2 3 24 3e

4H O 4H 4OH

H AlO H O

H AlO H Al + 3

3e Al 4OH

H O

So the total charge on the left is equal to the total charge on the right, and charge is balanced.


Balance Charge

–1+4

–3

0

0

2 3 2

2

2 3 2


4H O 4H 4OH

H AlO H O 3e Al 4OH

This half-reaction is now balanced in Acid Solution


Balanced in Acid Solution

Balanced in Acid Solution

2 3 2

2

2 3 2


4H O 4H 4OH

H AlO H O 3e Al 4OH

But the question asks us to balance this in Basic Solution.


Balance in Basic Solution

2 3 2

2

2 3 2

4H

4H O 4H 4OH

H AlO H O 3e

H AlO 3e Al +

Al 4 H

3 O

O

H

For basic solution, we must get rid of the 4 H+. Equations in basic solution cannot have H+ ions in them.



We must get rid of the 4H+

2

2

H O H OH

H OH H O

n n n

n n n

In order to change an equation from acid solution to basic solution, we add either this equation, showing water forming H+ and OH minus,


To Change to Basic Solution

Add

or

2

2H O H

H OH H O

OH

n n

n n n

n

Or it’s reverse, showing H+ and OH minus forming water. We use whichever equation we need to cancel H+ ions out of our equation in acid solution.



Add

or

2

2H O H

H OH H O

OH

n n

n n n

n

n stands for the number of H+ ions that need to be cancelled.



Add

or

n = the number of H+ ions we need to cancel

2 3 2

2

2 3 2

4H

4H O 4H 4OH

H AlO H O 3e

H AlO 3e Al +

Al 4 H

3 O

O

H

Going back to our equation in acid solution, we must cancel out the 4H+ ions on the left side of the equation. So we need to add an equation with 4H+ ions on the RIGHT



We must cancel out

4H+ on the left

side

2 2

2 3 2

3

24 4

4H

H O H OH

H

H AlO

AlO

4

H O 3e

3e Al +

l 4OH

3H O

A

So we add this reaction with water on the left and H+ and OH minus on the right.



n n n

2 2

2 3 2

3

24 4

4H

H O H OH

H

H AlO

AlO

4

H O 3e

3e Al +

l 4OH

3H O

A

Because we need to cancel 4 H+ ions,



n n n

2 2

2 3 2

3

24 4

4H

H O H OH

H

H AlO

AlO

4

H O 3e

3e Al +

l 4OH

3H O

A

n must be equal to 4



n n n

n = 4

2 2

2 3 2

3

24 4

4H

H O H OH

H

H AlO

AlO

4

H O 3e

3e Al +

l 4OH

3H O

A

So 4 is the coefficient we use for all three species in this equation. Make sure they all have the same coefficient!



2 2

2 3 2

3

24 4

4H

H O H OH

H

H AlO

AlO

4

H O 3e

3e Al +

l 4OH

3H O

A

When we add these two equations together, (click) we see the 4H+ are on opposite sides, so they can be cancelled.



2 2

2 3 2

3

24 4

4H

H O H OH

H

H AlO

AlO

4

H O 3e

3e Al +

l 4OH

3H O

A

So we’ll cancel them and we’re left with this.



2

2

2 3

2 3

2

H AlO 4H 3e Al +

4

3H O

4H O

H AlO H O 3e Al 4O

H 4OH

H

Notice we now have (click) 4H2O on the left side, and 3 H2O on the right side.



2

2

2 3

2 3

2

H AlO 4H 3e Al +

4

3H O

4H O

H AlO H O 3e Al 4O

H 4OH

H

We can do something to an equation as long as we do the same thing to BOTH sides. So we can (click) remove 3 H2O molecules from both sides. This is called simplifying the equation.



Remove 3H2O from

BOTH SIDES

2

2

2 3

2 3

2

H AlO 4H 3e Al +

4

3H O

4H O

H AlO H O 3e Al 4O

H 4OH

H

This leaves us with no H2O on the right side of the top equation.



Remove 3H2O from

BOTH SIDES

2

2

2 3

2 3

2

H AlO 4H 3e Al +

4

3H O

4H O

H AlO H O 3e Al 4O

H 4OH

H

And 4 minus 3 equals 1 water on the left of the lower equation.



Remove 3H2O from

BOTH SIDES

1

2

2

2 3

2 3

2

H AlO 4H 3e Al +

4

3H O

4H O

H AlO H O 3e Al 4O

H 4OH

H

We’ll remove the coefficient 1 because we don’t need it.



1

2

2

2 3 2

3 24H 3e Al + 3H O

4H O 4

H AlO

H 4O

H O 3e A

H AlO

l 4OH

H

Now we add what’s left of these two equations, to get our final equation. We’ll start by adding everything on the (click) left side of the arrows. We have 1 (click) H2AlO3 minus



2

2 3 2

2 3

2


4 4H 4OH

H AlO H O 3e Al 4OH

H O

One H2O,



2 3 2

2

2 3 2

H AlO 4H Al + 3H O

4H O 4H 4OH

H AlO H 3e Al

3e

O 4OH

And 3 electrons on the left side. Now we’ll add what we have on the right side



2 3 2

2

2 3 2

H AlO 4H 3e + 3H O

4H O 4H 4OH

H AlO Al 4OHH O 3e

Al

On the right side, we have 1 Al



2 3 2

2

2 3 2


4H O 4H

H AlO H O 3e Al 4 H

4O

O

H

And 4 OH minus



2 3 2

2 3 2

2


4H O 4H 4O

H AlO H O 3e 4

H

Al OH

So now we have our finished balanced half-reaction in basic solution.



Balanced Half-reaction in BASIC Solution

2 3 2H AlO H O 3e Al 4OH

What we should do now is (click) check this to make sure atoms and charge are balanced.


Balanced Half-reaction in Basic Solution

CHECK

2 23AlO O 3e l OH H A 4 H

We see we have a total of 2 + 2, which equals 4 H’s on the left side,



4 H’s

2 3 2H AlO H O 3e Al O4 H

And 4 H’s on the right side. So Hydrogen atoms are balanced.



4 H’s 4 H’s

2 3 2H O H O 3e l OA A 4l H

We have 1 Al on both sides, so aluminum atoms are balanced.



1 Al 1 Al

32 2O OH Al H 3e Al 4OH

We have a total of 3 plus 1, which is 4 O’s on the left side



4 O’s

2 3 2H AlO H O l O3e A 4 H

And 4 O’s on the right side, so oxygen atoms are balanced.



4 O’s 4 O’s

2 3 2H AlO H O Al O3e 4 H

Now we’ll check the charge. We have a total of negative 1 + 0 + negative 3 on the left side,



–1 –30

2 3 2H AlO H O Al O3e 4 H

Negative 1, 0 and negative 3 add up to negative 4, which is the total ionic charge on the left side.



–1 –30

–4

2 3 2H AlO H O 3e Al O4 H

On the right side we have 4 times negative 1, which equals (click) negative 4.



–1 –30

–4

–4


So charge is balanced.



–1 –30

–4

–4


And this is the balanced half-reaction for H2AlO3 minus forming Al in basic solution.



here, we’ll go through an example of balancing a half-reaction in basic solution

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