00:01
Hi everyone.
00:02
Now in this question they asked to balance the redox reaction using basic solution.
00:09
So question number is an o2 minus aqueous plus als gives an h3 gas plus alo2 minus iqs now step one is assign observation number number to each limit.
00:46
So nitrogen oxidation state in this case is plus 3, that is x minus 4 is equal to minus 1.
00:55
Then aluminum is in the free state, so it has a zero oxidation state.
01:00
Then nitrogen is assigned as minus 3 oxidation state, that is x plus 3 is equal to 0.
01:10
Then aluminum is assigned as 3 oxidation state, that is x minus 3 oxidation state, that is minus 4 is equal to minus 1.
01:20
So, you know, oxidation state of nitrogen changes from plus 3 to minus 3.
01:28
So it goes on decreasing and that's why the process is reduction.
01:33
Similarly, oxidation state of adenium changes from 0 to plus 3 and that's why there is an increasing oxidation number is involved.
01:44
But since increasing oxidation number is involved, the process it is called oxidation.
02:00
The process it is called what oxidation.
02:03
So this is oxidation process.
02:09
Then you have to write two separate equation for oxidation and reduction.
02:15
So oxidation is aluminum s 0 changes to air.
02:24
O2 minus acquies now there is no need of this observation state then in a reduction an o2 minus aquevius changes to an h3 gas now you know the actual third rule is balance the atoms other than hydrogen adoption but in this case there is no necessary because you know in this equation hydrogen and oxygions means other items are already balanced like aluminium nitrogen etc now next one is balancing of oxygen now in order to balance oxygen you know there are two oxygen item at rhs in oxidation so add two water molecule on nhs two water molecule on lhs two water molecule on lhs.
03:35
Then in reduction process there are two oxygen atom at lhs so you have to add two atom molecule on rhs now next step is balancing of hydrogen atoms in oxidation you know there are four hydrogen item at lhs so you have to add 4h plus at rhs.
04:16
4h plus at rhs.
04:21
Now in a reduction process you know there are totally 4 plus hydrogen atoms at rhs so you have tried 7h plus at lhs so this is and over to minus plus 7h plus aquius gives an h3 gas plus two water liquid.
04:57
The next since the equation is in basic medium you have to add equal amount of oh minus to both sides.
05:06
You know in first case there are four h2 atoms so you have to add 4 oh minus to lhs and rhs so als plus 2h2 liquid plus 4 oh h minus aqueous gives al o2 minus aqueous plus 4h plus aqueous plus 4 oh h minus acquies similarly there are 7h plus and that's why you have to add 7 o h minus science so an o2 minus aqueous plus 7h plus aqueous xvius plus 7 oh h minus aqueous gives an h3 gas plus two water liquid now you know this 4h plus and 4 h minus they forms a four water molecules similarly 7h plus and 7 ohish minus forms a 7 water molecule then you have to balance the charges by addition of electrons 6th so oxidation you know first is oxidation but in oxidation process or if we see properly you know there is a four negative charge at lhs and on only one negative charge at rhs.
07:06
So you have to add three electrons to the rhs.
07:17
So what will get is that is als plus 4h minus equius...