00:01
Hello students, in this question we have to balance the given redox reaction.
00:04
So, at first we are going to split this equation into two halves.
00:08
So, let us consider the first half mno4 - is converted to mno2.
00:15
So, in order to balance the oxygen we are adding water molecule to this side.
00:19
In order to balance the hydrogen we are adding 4 proton to this side.
00:24
So, in order to balance the charge we are adding electron to the more positive side that is to the reactant side.
00:33
So, we are adding 3 electron to this and this would give mno2 plus 2h2o.
00:41
Coming to the second half of the reaction that is ch3oh is converted to hcooh.
00:49
In order to balance the oxygen we are adding water to this side h2o plus ch3oh will gives us hcooh plus in order to balance the hydrogen we are adding 4h plus to this side and in order to balance the positive charge we are adding 4 electron to this side.
01:13
So, in order to cancel out the electron we are multiplying this equation with the 4 and this equation with the 3.
01:19
So, that this equation will become 3h2o plus 3ch3oh will gives us 3hcooh plus 12h plus plus 12 electron and this equation will become 16h plus plus 4mno4 2 minus plus 12 electron will gives us 4mno2 plus 8h2o.
01:59
Now, we are going to cancel out the similar terms 12 electron 12 electron cancels each other this cancel and remaining will be 4h plus and this cancel and remaining will be 5h2o and hence the final balance equation can be written as follow 3ch3oh plus 4mno4 minus plus 4h plus will gives us 3hcooh plus 4mno2 plus 5h2o...