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In this video, we want to balance this redox reaction in base.
00:04
The way i do it, i balance an acid and then change the acid to the base.
00:08
So we're going to go with that.
00:09
So, all right, we have mno4 minus to mno2.
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So one of the reactions is going to involve manganese.
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And what is the change in oxidation number? so oxygen is minus 2, so minus 8 because there's four of them.
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To get to minus 1, we need the manganese to be plus 7.
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Here, minus 2 times 2 is minus 4.
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So this is a plus four.
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So that means this is a reduction because the manganese goes from plus seven to plus four decreases by three.
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All right.
00:39
So then we can go ahead and write the reaction and we balance all atoms other than oxygen and hydrogen.
00:47
So manganese is balanced.
00:48
And then from plus seven to plus four, so we'll say plus three electrons here.
00:53
Now we need minus one, minus three.
00:56
So there's a minus four charge here and zero charge here.
00:59
So we'll add four.
01:00
H pluses on this side to balance out the charge.
01:03
And that means that we'll need to add two waters to balance out the hydrogens.
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That also makes the four oxygen on each side.
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So this is a balanced reduction reaction.
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Now, the other reaction in question is going to be oxidation, because they're common pairs.
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So we'll have s203 -2 -1 -2 -minus becomes s -o -4 -2 -minus.
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Let's figure out oxidation numbers.
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So we have three times two, negative six.
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To get to negative two, we need plus four, but there's two solvers, so it's plus two each.
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Overall, a plus four.
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And then we have minus eight, two times four.
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So this is going to be a plus six.
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Now we need to balance out all atoms other than oxygen and hydrogen.
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So if there's two solvers here, that means that this is a 2.
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This was a plus 4, and this will be a plus 12.
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So counting all of the atoms...