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Balancing Cr with H+
dogman:
--- Quote from: Orcio_87 on April 30, 2022, 03:31:48 PM ---
--- Quote ---And that's to say, with which reagents am I working here? I mean, is my equation truly just:
"Cr + H+"? That can't be right, can it? I'm sure there's more going on, that's not explicitly written.
--- End quote ---
Did you heard about the oxidation-reduction reactions ?
Here H+ is an oxidant and the Cr is the reductant. You don't need anything else.
--- End quote ---
Hmm...So you claim I truly need absolutely nothing but Cr and H+ to end up with Cr^+3 and hydrogen gas? Am I getting you right?
I posted this question in another forum, and received this reply, which showcases the use of water? (while I did encourage posters to somehow find a way to add water to solving the problem, I am asserting that I wanted clarification over what is necessary in the process, since it was heavily implied that water is somehow related by follow-up questions...)
And in an attempt to not be as vague and confusing as the lecturer who came up with this ambiguous question, I will also update the post to feature the entirety of the question, just so some context may be added, perhaps...
EDIT:
Well, seems like I can't edit my OP, so:
The metal chromium (Cr) reacts with an acid to produce Cr+3 and hydrogen gas.
* Write the balanced reaction.
* with the reaction of the metal with an excess of acid, 94.7 ml of hydrogen gas is created.The gas is picked up over water with pressure of 754 mmHg and temperature of 20C.
Calculate the mass of the metal that reacted. Show your calculations. (water vapor in 20C is 24 mmHg)
* Explain how the volume of the gas will change, if Cr{+2} would be produced (and not Cr{+3}).
* Calculate the number of present water molecules in the volume of the gas-phase (sorry, kind of confused how to translate this one ><).
* Calculate the ratio between the avg speed of the hydrogen gas molecules and between the avg speed of the water-vapor molecules that are present in the same temperature.[/list]
Borek:
Putting aside some details which can make the thing quite confusing, reactions between metals and acids all follow the same scheme (this is just a skeletal reaction, not intended to be balanced):
Me + H+ + counteranion :rarrow: Men+ + H2 + counteranion
where H+ + counteranion is a lousy way of writing "dissociated acid".
Note that in the counteranion doesn't change, so we can safely ignore it and write so called "net ionic equation":
Me + H+ :rarrow: Men+ + H2
(again, not balanced and not intended to be balanced, just a scheme).
This can easily get tricky when you have to deal with the acid dissociation (not every acid is fully dissociated), when the metal is less reactive (so needs something stronger than H+ to be oxidized), plus technically there is no "naked" H+ in water solutions, we typically assume they are present as H3O+ (H+ and a water molecule) - which is not entirely true either.
The part about water in your question has nothing to do with the reaction itself, gas collected over water is always saturated with water vapor. You are given partial pressure of the water vapor so you should use it to in your calculations.
Orcio_87:
--- Quote ---Hmm...So you claim I truly need absolutely nothing but Cr and H+ to end up with Cr^+3 and hydrogen gas? Am I getting you right?
--- End quote ---
Right. Look at the table:
https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)
Cr ----> Cr3+ + 3 e-
H+ + e- ----> 1/2 H2
All you have to do is multiply second equation and add them together to make electrons will cancel out.
dogman:
Hmm...Okay.
From what I understand, is that I am missing a buncha study subjects. I'll have to address this to my lecturer. :s
Thanks everyone! =)
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