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2001 - Multiple Choice (Q35) - Written Paper (Q14(a)) SectionA part 2) 35 and Section B) 14a.

xlynsey

Posted 27 May 2006 - 03:44 PM

First of all, Section A part 2 Q 35...
I can understand how B is correct, but not E?? How do you know that they are oxidised?

And, 14a, How do you know which one is reduction and when you worked that out how do you go about writing the equation anyway??

If anyone could help me that would be great!

Allana

Posted 27 May 2006 - 03:57 PM

QUOTE(xlynsey @ May 27 2006, 04:44 PM) View Post

First of all, Section A part 2 Q 35...
I can understand how B is correct, but not E?? How do you know that they are oxidised?

And, 14a, How do you know which one is reduction and when you worked that out how do you go about writing the equation anyway??

If anyone could help me that would be great!


For Section A Part 2, isn't the answers B+D?

Well, statement D appears to be true : In redox reactions in solution, the sulphite ion acts as a reducing agent.

Look at the equation for the reduction of the sulphate ion into sulphite ions , taken from the date booklet. As you see, in a reduction reaction, the sulphite ion is produced with water. But, we can spin this equation around so that it is added to water - and so is therefore oxidised. As the sulphite ion is oxidised in solution, it therefore causes something to be reduced, and so is a reducing agent.

There is a more complex theoretical way to work it out, but why do that when it's in the data book smile.gif



xlynsey

Posted 27 May 2006 - 04:05 PM

Ok thanks i understand now smile.gif its complicated though!

Does anyone know about 14a?

Allana

Posted 27 May 2006 - 04:16 PM

For 14a) it tells you that palladium is formed in a redox reaction. All you have to do is remember that whenever a pure element is formed from existing within a compound, then this is an example of Reduction. You can also work this out by elimination. Notice how the CO isn't reduced, it is Oxidised into CO2 due to the addition of Oxygen. We can cancel that out now. Now look at the compound which contains chlorine and water, and the other 2 products are 2HCL and H2O. This is merely a displacement reaction, so this can be cancelled out too.

That leaves you with the palladium being reduced:)

Now to write it, you will notice you can't simply take it from the data book. However, look at the section of the formula for the compound: PdCL2

Now you know that the charge on a Chlrone atom is 1-, and as it's CL2 then this makes the charge 2-. The charges of the Pd and Cl have to balance, so the charge on the Pd must be 2+, giving 2 electrons.

Pd2+ + 2e- ---> Pd

Just a quick note. My answers to things are quite lengthy but that's only because I like explaining in detail to help people understand - the actual answer isn't complicated so don't be put off! smile.gif

d00pyd

Posted 27 May 2006 - 04:17 PM

For 14 the clue is in the question - it says palladium is formed. The Pd is reduced because it was a 2 + ion and is now PD atoms. We know this by looking at PdCL2. As Cl is CL - it must be a + ion, and morevoer must be a 2+ ion. Then we can write the reduction. As it is 2+ there will be 2 electrons gained.

xlynsey

Posted 27 May 2006 - 04:18 PM

Thanks v much - would never have got that!


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