Past HSC question (1 Viewer)

[Mr_Kap]

This equation represents a common redox reaction.

Cr2O72–(aq) + 14H+(aq) + 6Fe2+(aq) → 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l)

What is the oxidising agent in the reaction?


How would I figure this out? I know that an oxidising agent is one that reduces but how do I write out the ionic equations to find out?

 

[Drsoccerball]

This equation represents a common redox reaction.

Cr2O72–(aq) + 14H+(aq) + 6Fe2+(aq) → 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l)

What is the oxidising agent in the reaction?


How would I figure this out? I know that an oxidising agent is one that reduces but how do I write out the ionic equations to find out?

So what the question is asking basically is : "which one gains the most electrons"
Therefore the answers H+ ?

 

[sharoooooo]

to figure it out, write the oxidation numbers for each compound. then compare the oxidation numbers on each sides and find then u can find the oxidising agent.

so... for this question, on the left side, H+ has an oxidation number of +1. but if we look on the right side and find 'H', it has an oxidation number of +4 (since O has an oxidation no. = -2. So -2 x 2= -4, thus H has ox. no. of +4)
therefore, the number has increased, which means it lost e-.
Cr changed from +7 to +3. i think Cr is reduced since ox.no. has decreased. So Cr has gained e-, so it has been reduced, so Cr is the oxidising agent..




not 100% sure but yep..thats what i would do

 

[rand_althor]

The redox reaction can be broken down into a reduction and oxidation reaction. From the data sheet, you can see that there is a reaction which involves Cr₂O₇^2- which produces Cr³⁺ and another involving Fe²⁺ which produces Fe³⁺. These reactions are:
Reduction: 2Cr₂O₇^2- +14H⁺+6e^^- <-> Cr³⁺ + 7H₂O
Oxidation: Fe²⁺ <-> Fe³⁺ + e^^-

Adding these reactions will give you the redox reaction. From the reduction reaction, you can clearly see that Cr₂O₇^2- is undergoing reduction, hence Cr₂O₇^2- is the oxidising agent.

 

[Kaido]

When they ask for oxidising/reducing agent, this is always found on the left hand side of the equation.
H is always 1+, O is always 2-, (in HSC anyways)
So you can eliminate those immediately

Now oxidising agent is almost always the one undergoing reduction (a reduction in oxidation number e.g. from 3+ to 2+)
Vice versa

After this, its a matter of doing some simple addition and subtraction to figure out the oxidation states

 

[enigma_1]

No dude it's not always on the left they can trick you

 

[rand_althor]

No dude it's not always on the left they can trick you

Can you give an example of when it's on the right?