Question

A question on redox equations please?

Write down the overall redox equations for the following:
(a) Acidified potassium dichromate oxidising sulphur dioxide to the sulphate ion.
oxidation 1/2:
reduction 1/2:
overall ionic equation:

Answer 1

Which one undergoes oxidation - K2Cr2O7 or SO2?

Answer 2

I think the SO2... I'm not sure. I've never done any questions formatted like this...

Answer 3

Yup.
So, for oxidation, you'll have to write an equation for SO2.
Now, SO2 will form SO4^-2
\[SO_2 \rightarrow SO_4^{2-}\]Can you balance the equation?

Answer 4

Would it be...

Answer 5

Yes?

Answer 6

SO2 + H2O --> SO4(2-) + 2H(+)

Answer 7

Oh wait the oxygens aren't balanced

Answer 8

SO2 + 2H2O --> SO4(2-) + 4H(+) + 2e(-)

Answer 9

Is that correct @Callisto ? :S

Answer 10

Yes. Sorry.. My chrome had a fight with OS for a while :(

For reduction:
\[Cr_2O_7^{2-} \rightarrow Cr^{3+}\]Can you balance it?

Answer 11

Thanks, haha no problem :)

How do you know that the potassium dichromate reduces to the chromic ion?

Answer 12

*chromate ion

Well... there are some basic equations that I have to remember. Cr2O7 is one of them.

Answer 13

Oops, chromate. Oh okay, so it's something you memorise...

Answer 14

*Cr2O7 ^(2-)

Answer 15

Yup!

Answer 16

Dang! Ok I'm gonna try balancing it

Answer 17

In my exam, the electroreactivity series is not given, the equations of the redox reactions are not given. All I have to do is to memorize them.. in order..

Answer 18

Oh man that's so annoying!

Answer 19

Balance Cr on the right first. Then, balance the number of O

Answer 20

Ok thank you

Answer 21

Cr2O7(2-) + 14H(+) --> 2Cr(3+) + 7H2O

Answer 22

I'm confused about balancing the charges by adding electrons...

Answer 23

What's the total charge on the left?

Answer 24

Umm would it be +12?

Answer 25

Yes.
What about for the right?

Answer 26

+3

Answer 27

Not really.
There are TWO chromium(III) ions there

Sorry!! Cr^(3+) should be called as chromium(III) ion.
Chromate ion is actually CrO4^(2-)!!

Answer 28

OHH of course

Answer 29

so it's +6

Answer 30

Do you add 6 electrons to the reactant side?

Answer 31

Indeed!

Answer 32

Can you write the whole equation now?

Answer 33

Oh thank you so much for being so patient, you're a legend!!

Answer 34

Thanks! But I'm not :)

Answer 35

Cr2O7(2-) + 14H(+) + 6e(-) + 3SO2 +6H2O --> 2Cr(3+) + 7H2O + 3SO4(2-) + 12H(+) + 6e(-)

and so the electrons cancel out

Answer 36

so it simplifies down to...

Answer 37

Cr2O7(2-) + 2H(+) + 3SO2 --> 2Cr(3+) + H2O + 3SO4(2-) I think

Answer 38

And yes you are a really great teacher! :P

Answer 39

Yup. It's correct! :)

Answer 40

Ahhhhhhh thanks so much!!

Answer 41

Do you have to write the ionic equation or the equation?

Answer 42

The ionic equation thank goodness

Answer 43

Then that's what you need :)
Well done!!~

Answer 44

Thank you!!

Answer 45

Welcome :)