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# AS redox reaction MCQ

1. Guys, how do we solve the MCQ (attached) using oxdiation numbers?

I have already found out the oxidation numbers of iodine, it's +1 in HIO, 0 in I2 and +5 in HIO3. But I'm stuck after that. Any help?
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2. The redox needs to balance for each element i.e. the total oxidation change on the left = total oxidation change on the right

m x 1 = (n x 0) + (p x 5) [the 1, 0 and 5 relate to the oxidation states of I on the left and right]

m = 5p

so p = 1 and m = 5

then use these values to get n
3. (Original post by EierVonSatan)
The redox needs to balance for each element i.e. the total oxidation change on the left = total oxidation change on the right

m x 1 = (n x 0) + (p x 5) [the 1, 0 and 5 relate to the oxidation states of I on the left and right]

m = 5p

so p = 1 and m = 5

then use these values to get n
You made it sound so easy, thank you so much!

I didn't know the mole value affects the value of the oxidation numbers.

BTW - Would it be okay if I posted a few more MCQs here if I encounter any other problems? I have my Chemistry MCQ paper tomorrow and while my preparation is good, I might have problems with other MCQs.
4. (Original post by leosco1995)
You made it sound so easy, thank you so much!

I didn't know the mole value affects the value of the oxidation numbers.
It doesn't, it affects the balancing Welcome of course

BTW - Would it be okay if I posted a few more MCQs here if I encounter any other problems? I have my Chemistry MCQ paper tomorrow and while my preparation is good, I might have problems with other MCQs.
Fell free to post as many questions as you like, someone (if not me) will get around to answering them
5. OK, here's another one:

How many different substitution products are possible, in principle, when a mixture of bromine and ethane is allowed to react?

A) 3
B) 5
C) 7
D) 9

I know this involves free radical substitution, but I'm not very good at it, I thought a list of products would be this:

CH3CH2Br
CH3CHBr2
CH3CBr3
CH2BrCBr3
CHBr2CBr3
CBr3CBr3

This is only 6.. but the answer is actually 9, what am I missing?
6. (Original post by leosco1995)
OK, here's another one:

How many different substitution products are possible, in principle, when a mixture of bromine and ethane is allowed to react?

A) 3
B) 5
C) 7
D) 9

I know this involves free radical substitution, but I'm not very good at it, I thought a list of products would be this:

CH3CH2Br
CH3CHBr2
CH3CBr3
CH2BrCBr3
CHBr2CBr3
CBr3CBr3

This is only 6.. but the answer is actually 9, what am I missing?
Theres only one way to get one Br on:

CH3CH2Br

There's two ways to get 2 Br's on:

BrCH2CH2Br and CH3CHBr2

keep going, see if you can find more
7. (Original post by leosco1995)
OK, here's another one:

How many different substitution products are possible, in principle, when a mixture of bromine and ethane is allowed to react?

A) 3
B) 5
C) 7
D) 9

I know this involves free radical substitution, but I'm not very good at it, I thought a list of products would be this:

CH3CH2Br
CH3CHBr2
CH3CBr3
CH2BrCBr3
CHBr2CBr3
CBr3CBr3

This is only 6.. but the answer is actually 9, what am I missing?
you have missed a few isomers
BrCH2CH2Br
BrCH2CHBr2
so on.......
8. Oh, right. Of course..

Another one:

In a car engine, non-metallic element X forms a pollutant oxide Y.

Further oxidiation of Y to Z occurs in the atmosphere. In this further oxidation, 1 mol of Y reacts with 0.5 mol of gaseous oxygen.

What can X be?

1) Carbon
2) Nitrogen
3) Sulfur

2 and 3 are both right, but why can't 1 also be right? Aren't these equations both possible?

C + 1/2 O2 -> CO (in a car engine)
CO + 1/2 O2 -> CO2 (further oxidation)

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