Chemistryyy (3 Viewers)

[drsabz101]

let me know if its right or wrong ^^

 

[Shuuya]

oxidation reaction is correct but you made a mistake in balancing the reduction reaction:

4H+ + 2e- ---> 2H2

your equation has an overall charge of 2- on the left side and 0 on the right side

It should be:
2H+ + 2e- ---> H2 ( so just add 2e- instead of starting with e-)

 

[drsabz101]

yes thankyou, I realised that after I posted it haha

 

[drsabz101]

Anymore practice questions?
Is this also hsc content?

 

[Shuuya]

yes thankyou, I realised that after I posted it haha

Don't worry haha, I make those kinds of mistakes all the time

 

[Shuuya]

Anymore practice questions?

Here's a harder one: I2O5 + 3CO ---> I2 + 3CO2

(use the fact that oxygen always has an oxidation state of 2- to find the oxidation states of C and I)

 

[drsabz101]

Okay , I'll do it now

 

[drsabz101]

can I have a hint about the ionic equation?

 

[Shuuya]

can I have a hint about the ionic equation?

Sorry this is probably not the best example because it doesn't have an ionic equation (because the reactants aren't ionic compounds)

So to find out which species reduces/oxidises you have to compare oxidation states.

so for I2O5:

I2O5 is neutral, so it has an overall charge of 0. The oxidation state of O is always 2-, so we can work out the oxidation state of I using this information. Let the oxidation state of I be x:

2x + 5(-2) = 0
2x =10
x = 5 i.e. the oxidation state of I is 5 in this compound

On the products side, I2 has an oxidation state of 0, and therefore I also has an oxidation state of 0.

So Iodine goes from oxidation state of 5 to 0, and since a lowering in oxidation number signifies reduction, Iodine is the oxidising agent.

Similarly, we have to do the same for C in CO (which has an overall charge of 0). Let x be the oxidation state of C

On the reactants side:

x+-2 =0
x=2

On the products side:
x+2(-2)=0
x=4

Therefore the oxidation state of Carbon goes from 2-->4, and an increase in oxidation state signifies oxidation, meaning that C is the reducing agent.

Sorry about that dodgy example, here's another one that's more like the ones that we've been doing:

Zinc + Iron Sulfate

 

[drsabz101]

interesting. Is this a yr 11 standard question or actual yr 12 chem?

 

[Shuuya]

interesting. Is this a yr 11 standard question or actual yr 12 chem?

Its more yr 12, sorry I forgot that you weren't up to this stuff yet redox reactions are in the first topic in year 12, so its good that you're learning about them early on

 

[drsabz101]

Haha that's okay, I like challenging questions . I am trying to practice these things, so I don't suffer when start chem. I have a really bad chem teacher

 

[drsabz101]

Wait, I just realised you added another example.
is that a metal+acid= metallic acid + hydrogen gas form

 

[Shuuya]

Wait, I just realised you added another example.
is that a metal+acid= metallic acid + hydrogen gas form

Its a displacement reaction (an acid is a substance that contains H+ ions in solution)

So the more reactive metal Zinc displaces the less reactive Iron:

Zn + FeSO4 ---> Fe + ZnSO4

 

[DatAtarLyfe]

DatAtarLyfe's equation should be Mg +2HCl ----> MgCl2 + H2 (not H2O, as acid+metal-->salt+hydrogen gas)

So Ionic equation is:

Mg + 2H+ + 2Cl- ---> Mg2+ + 2Cl- + H2

Hopefully you can write the net ionic equation now (i.e remove the spectator ions)

Whoops, wat a noob mistake.
Soz and thanks

 

[leehuan]

(Since your HSC is supposedly in 2017) Why are you being taught oxidation states in Yr 11.

 

[drsabz101]

Its a component of the metals topic in yr 11

 

[drsabz101]

Q. how to balance reaction of zinc and water

 

[Shuuya]

Q. how to balance reaction of zinc and water

Zn + 2H2O ---> Zn(OH)2 + H2

 

[drsabz101]

Thnx