00:01
In this problem, balance the following redox reaction in acetic solutions.
00:06
So first of all, we will determine the oxidation number.
00:09
So here the oxidation number is plus 2 on sulfur.
00:14
And on cl2, it must be 0.
00:16
Now, if you talk about sulfur on the product side, so it should be plus 6.
00:22
And for cl negative, it should be minus 1 only.
00:28
Right.
00:28
So sulfur is getting oxidized.
00:31
So oxidation number changes from plus 2 to plus 6.
00:38
So this is getting oxidized.
00:40
That is oxidation is done here.
00:43
Right.
00:44
So there is, or you can also say that there is loss of electron.
00:50
Loss of electron.
00:52
Right.
00:53
So chlorine is getting reduced at the same time where oxidation number is getting increased from 0 to, i'm sorry, decreased from 0 to minus 1 and there is gain of electron.
01:05
So now we will write the half reaction.
01:09
Now half reactions will be what? half reactions in accordance with the above equation on as we have already calculated the oxidation number.
01:21
So for oxidation we will write s2 .03 2 minus that gives s .o4 2 minus right.
01:33
And similarly for reduction, what you can do is cl2 is getting reduced to cl negative.
01:43
So now you need to balance the half reactions individually.
01:49
So now this is first part...