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: H 2 MoO 4 ⇄ Mo (acid solution)

We’ll start by writing H2MoO4 on the left side. Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O

And Mo on the right side. We’ve left some room to add other things. Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O

We always start by balancing atoms other than oxygen or hydrogen. Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O

In this case, it is the element molybdenum, Mo Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance Mo atoms

There is 1 Mo atom on both sides, so Mo is already balanced. Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O 11 Balance Mo atoms

Our next step is to balance oxygen atoms. Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance O atoms

We see there are 4 oxygen atoms on the left and none on the right. Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance O atoms 4

Remember, (click) for every excess oxygen atom on the left, we add one H2O molecule on the right. Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance O atoms 4

So we add 4 water molecules to the right side Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance O atoms 4

Now, we see there are 4 oxygen atoms on both sides, so oxygen is balanced. Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance O atoms 4 1×4 = 4

The next step is to balance hydrogen atoms Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance H atoms

We see there are 2 hydrogen atoms on the left Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance H atoms 2

And 2 times 4, or 8 hydrogens on the right. Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance H atoms 2 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: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance H atoms 2 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: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance H atoms 2 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: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance H atoms 2 2×4 = 8 Add 6 H + ’s to the Left Side

So now we have 2 plus 6 Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance H atoms 26

Which is equal to 8 hydrogen atoms on the left side… Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance H atoms 26 8

And 4 times 2, Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance H atoms

Which equals 8 hydrogen atoms on the right side. So hydrogens are balanced. Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance H atoms

The last step is to balance charges. Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance Charges

On the left side, we have a total ionic charge of 0 plus positive 6 Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance Charges +60

Which is equal to positive 6 Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance Charges +60

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: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance Charges +60

Remember, to balance charges, we add enough electrons to the more positive side, to make the charges equal. Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance Charges +60 Add 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: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance Charges +60 Add 6 e – to the more positive side

So we add 6 electrons to the left side. Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance Charges +60 Add 6 e – to the Left side

The total ionic charge on the left side is now negative 6 Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance Charges +6 0 –60

Which equals zero Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance Charges +6 0 –60 0

So the total ionic charge on each side is zero Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance Charges Total ionic charge = 0

Therefore, the charges are balanced. Balance the half-reaction: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balance Charges Total ionic charge = 0 Balanced

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: H 2 MoO 4 ⇄ Mo (acid solution) H 2 MoO 4 + 6H + + 6e – ⇄ Mo + 4H 2 O Balanced