how do you tell the difference b/t oxidation/reduction vs. acid/base rxn?

[knooch]

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First, Is it that if you see a strong/weak acid/base it's acid/base reaction?

Second, In this reaction:

How do you tell if H2H4 or N2O4 is the reducing/oxidation agent? How do you tell if the Nitrogen in "3N(2)" is from H2H4 or N2O4?

 

[leathersofa]

acid base reactions are not redox reactions. they are, if you will, "double displacement" reactions
double displacement reactions are not redox reactions

Ex: HF + HOH --> F- + H3O+
The oxidation state of F (-1), H (+1), and O(-2) is same on products and reactants side

 

[knooch]

what? doesn't make sense

 

[Romz]

I may be wrong on this one but the way I looked at the equation:

It was hard for me to first see where the charges where going. I then figured "Hmm, if I can't determine what is happening with charge distribution, let me focus on the hydrogen atoms instead.

I know that an atom is reduced when it gains electrons in the form of hydrogens. We see that the Oxygen (luckily the only one in the reaction) gains all of the hydrogens from N2H4 nd becomes reduced. N2H4 is therefore the OXIDIZING AGENT, because it becomes OXIDIZED to reduce N2O4 to water. N2 lost all it's hydrogens, and N2 in N2O4 was replaced by hydrogens, so the two lone nitrogens grouped up. Hence, the free Nitrogen in 3N2 is actually the Nitrogens from BOTH of the reactants.

This should be a lot clearer now. It took me about a minute to figure this out. No worries if it didn't click right away. Try and piece together what is going on during the Rx.

 

[FeinMS]

I may be wrong on this one but the way I looked at the equation:

It was hard for me to first see where the charges where going. I then figured "Hmm, if I can't determine what is happening with charge distribution, let me focus on the hydrogen atoms instead.

I know that an atom is reduced when it gains electrons in the form of hydrogens. We see that the Oxygen (luckily the only one in the reaction) gains all of the hydrogens from N2H4 nd becomes reduced. N2H4 is therefore the OXIDIZING AGENT, because it becomes OXIDIZED to reduce N2O4 to water. N2 lost all it's hydrogens, and N2 in N2O4 was replaced by hydrogens, so the two lone nitrogens grouped up. Hence, the free Nitrogen in 3N2 is actually the Nitrogens from BOTH of the reactants.

This should be a lot clearer now. It took me about a minute to figure this out. No worries if it didn't click right away. Try and piece together what is going on during the Rx.

N2H4 is the REDUCING agent which becomes oxidized to reduce N2O4. Smart approach to the problem though.👍 To add on to your argument, since reduction occurs via gaining of electrons (through usually gaining of H,) a compound rich in H is usually reducing agents. On the other hand, since oxidation occurs via loss of electrons (through gaining O,) a compound rich in O is usually oxidizing agent.

 

[Romz]

N2H4 is the REDUCING agent which becomes oxidized to reduce N2O4. Smart approach to the problem though.👍 To add on to your argument, since reduction occurs via gaining of electrons (through usually gaining of H,) a compound rich in H is usually reducing agents. On the other hand, since oxidation occurs via loss of electrons (through gaining O,) a compound rich in O is usually oxidizing agent.

DAMNIT I ALWAYS GET THEM ****ING CONFUSED.

Thanks for that. 👍