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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