00:02
Let's write a balanced net ionic equation for the following.
00:06
For a, unbalanced, we have mn -o -4 minus, add c2h5 -o -h aqueous to produce mn2 plus, and c -h -3, c -o -2 -h aqueous.
00:31
Let's break this up in the two half reactions.
00:33
Mn 04 minus to mn2 plus.
00:40
Let's balance the oxygen, balance the hydrogen, and let's balance the charge.
00:54
Five electrons would balance the charge.
00:57
Second half reaction, carbons, two on each side.
01:12
Let's balance the oxygens.
01:13
There's two on the right, so i need an h2o on the left.
01:19
There are two, eight hydrosons on the left, only four on the right.
01:29
So 4h plus and four electrons.
01:37
In order to cancel out the electrons, we'll need to multiply the first half reaction by four, the second half reaction by five.
01:45
Let's rewrite these half reactions.
01:47
It comes four.
02:03
And the second half reaction will be multiplied by five.
02:29
Cancel the electrons, hydrogens will leave me with 12 on the left.
02:37
And oxygen or waters will cancel it and leave me with 11 on the right, putting this all together.
02:59
Here is my balanced equation, and that is done in acidic solution.
03:20
For b, unbalanced, we have h22 chromates to o2, and chromium 3 plus.
03:44
Let's break this up into two half reactions.
03:48
H2o2 to o2 gas.
03:56
Oxygen's balance on each side...