0:00
Hello.
00:01
So this question it is based upon the oxidation reduction reaction.
00:10
So in the first part it is asked about to balance the equation.
00:16
So balanced equation is that is mn -o -4 negative.
00:23
Two molecules are there plus 5h c -o -o -h plus 6 -h positive which produce 2 m -n plus.
00:33
2 plus 8h2 plus 5co2.
00:40
So this is the balanced equation.
00:43
Next we are asked to find out the oxidation number of carbon in h -c -o -o -h.
00:53
So we know that carbon for this we need to find out.
00:58
So oxidation number of hydrogen it is 1 plus x.
01:04
Let us suppose the oxidation number is x for carbon plus oxidation state for oxygen is minus 2 plus minus 2 plus 1 is equal to 0 because this compound is carrying no charge not a positive neither positive nor negative so therefore on solving this we will have that is x is equal to that is plus 2 oxidation state now let us find out the oxidation number of carbon in co2.
01:43
So carbon again this x plus minus 2 it is and as there are two atoms therefore multiply by 2 is equal to 0 as it doesn't contain any charge so it will be x minus 4 is equal to 0 and hence the charge will be x is equal to plus 4.
02:03
Next is the charge oxidation number of mn of magnesium in mno4 negative.
02:17
So, x for mn plus oxygen oxidation state is minus 2, multiply by as there are four atoms, to multiply by 4 is equal to this compound it is containing minus charge, that is minus 1.
02:35
So therefore x minus 8 is equal to minus 1 and hence on solving this we will have x is equal to plus 7 oxidation state.
02:48
Next is the oxidation number of oxidation number of mn plus 2.
03:00
It is equal to plus 2 as it is already mentioned.
03:06
Now the half reaction for oxidation.
03:12
As we can see here that h -c -o -o -h, here the carbon having the plus 4 oxidation state and 2h plus plus co2 plus 2 electron loss...