Question

Alright, now I have a very straightforward question regarding whether certain metal atoms (out of a list) will not be able to spontaneously react with HBr, and to determine this, I'm supposed to use a Standard Reduction table---

but the table doesn't have HBr listed in it. @aaronq , could you help me with this? Posting of the table and exact prompt momentarily.

Answer 1

Question prompt and answer: http://i.imgur.com/aoSE3BR.png

Reduction table: http://i.imgur.com/shANPX1.png

Answer 2

Again, I just don't have an idea of how to compare these while HBr isn't listed in the Reduction Table. What do I do?

Answer 3

Do I assume that Hydrogen, having a Reduction Potential of zero, doesn't affect the value for HBr, and thus,

Answer 4

It would be any strong acid, the reduction looks like:

\(2H^++2e^-\rightarrow H_2\)

Answer 5

\[E^\circ \big(\rm HBr\big) = E^{\circ}\big(Br\big)\]

Answer 6

Aw, darn, lol, nevermind. I liked my pet theory.

Answer 7

So what is the general statement or idea/theorem/postulate behind this, no strong acids can bond with certain metals, or?

Answer 8

it's about the reduction potentials. If the reduction is negative it will react with \(H^+\) reducing it to \(H_2\), if it's positive then no reaction.

positive and negative is relative to \(H^+\), look at the table

Answer 9

This comes down to calculating the standard cell potential

\(E_{cell}=E_{red}-E_{anode}\)

Answer 10

So anything involving H, are you saying, basically? Because what I'm confused about is how to infer a value from HBr. Thank you nonetheless so far.

Answer 11

anything involving an acid, like HCl, \(H_2SO_4\), etc, will make \(H_3O^+\) in solution (water) .. which is simplified to \(H^+\).

So what im saying is for you compare the \(E^o\) of the metals. If they're below \(H^+\)'s \(E^o\) , then a reaction will take place.

Answer 12

lol i hope that makes sense. let me know if you want me to break it down further

Answer 13

It makes sense on a mechanical, "Monkey see, monkey do" level for me, but I'm under severe time constraints and trying to make sure I cover all of my bases before my test in an hour and a half.

I don't really understand it well and would appreciate further explanation, but I also haven't legitimately taken time to try and understand your responses, or at least I think that's the case, I could also just really not get it.

Answer 14

well, ultimately, you know what is happening right?

redox reactions = transfers of electrons

When two atoms compete for an electrons, the one with a greater "want" or "need" (i.e. higher effective nuclear charge) will get it.

That's essentially what these reduction potentials mean.

If you notice, near the bottom of the table, you have elements from group 1 and 2, like Li and Mg. These can't really "hold" on to their electrons because of the shielding electrons closer to the nucleus provide, in other words, these valence electrons are masked by core electrons.

In contrast, on the top of the table you'll see very electronegative atoms, like F.

This all makes sense when you compare the values of the reduction potential table to the positions of the elements on the periodic table.

Answer 15

Yeah, that makes sense entirely, alright.

Answer 16

k cool