BalancingHalf-Reactions in Acid Solution

Example 2 Here, we’ll go through an example of balancing a half-reaction in acid solution.

We’re asked to balance the half-reaction, H2MoO4 gives Mo, in acid solution.

Balance the half-reaction:

H2MoO4 ⇄ Mo (acid solution)

We’ll start by writing H2MoO4 on the left side.

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

And Mo on the right side. We’ve left some room to add other things.

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

We always start by balancing atoms other than oxygen or hydrogen.

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

In this case, it is the element molybdenum, Mo

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance Mo atoms

There is 1 Mo atom on both sides, so Mo is already balanced.

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

1 1

Balance Mo atoms

Our next step is to balance oxygen atoms.

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance O atoms

We see there are 4 oxygen atoms on the left and none on the right.

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance O atoms4

Remember, (click) for every excess oxygen atom on the left, we add one H2O molecule on the right.

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance O atoms4

So we add 4 water molecules to the right side

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance O atoms4

Now, we see there are 4 oxygen atoms on both sides, so oxygen is balanced.

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance O atoms4

1×4= 4

The next step is to balance hydrogen atoms

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance H atoms

We see there are 2 hydrogen atoms on the left

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance H atoms2

And 2 times 4, or 8 hydrogens on the right.

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance H atoms2

2×4 = 8

Because we have 2 H’s on the left and 8 H’s on the right, (click) we need 6 more H’s on the left to balance hydrogens

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance H atoms2

2×4 = 8

We need 6 more H’s on

the left

Remember, for every H needed on the left, we add one H+

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance H atoms2

2×4 = 8

For every H needed, we add one H+

So we add 6 H+’s to the left side.

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance H atoms2

2×4 = 8

Add 6 H+’s to the Left Side

So now we have 2 plus 6

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance H atoms

2 6

Which is equal to 8 hydrogen atoms on the left side…

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance H atoms

2 6

8

And 4 times 2,

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance H atoms

2 6

8

4 2

Which equals 8 hydrogen atoms on the right side. So hydrogens are balanced.

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance H atoms

2 6

8

4 2

8

The last step is to balance charges.

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance Charges

On the left side, we have a total ionic charge of 0 plus positive 6

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance Charges

+6

0

Which is equal to positive 6

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance Charges

+6

0

+6

Looking on the right, Mo and 4H2O both have a zero charge, so the total charge (click) on the right side is zero

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance Charges

+6

0

Remember, to balance charges, we add enough electrons to the more positive side, to make the charges equal.

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance Charges

+6

0Add 6 e– to the more

positive side

Because the charge on the left is +6 and the charge on the right is 0, we must add (click) 6 electrons to the left side

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance Charges

+6

0Add 6 e– to the more

positive side

So we add 6 electrons to the left side.

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance Charges

+6

0Add 6 e– to the

Left side

The total ionic charge on the left side is now 0 + 6 + negative 6

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance Charges

+6

0–60

Which equals zero

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance Charges

+6

0–60

0

So the total ionic charge on each side is zero

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance Charges

Total ionic charge = 0 Total ionic charge = 0

Therefore, the charges are balanced.

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balance Charges

Total ionic charge = 0 Total ionic charge = 0Balanced

And the half-reaction is balanced. At this point you could pause the video and confirm to yourself that all atoms are balanced and total ionic charge is balanced.

Balance the half-reaction: H2MoO4 ⇄ Mo(acid solution)

H2MoO4 + 6H+ + 6e– ⇄ Mo + 4H2O

Balanced