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Topic: Yielding metal from aqueous solution  (Read 3795 times)

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Offline shalikadm

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Yielding metal from aqueous solution
« on: March 29, 2013, 09:08:53 AM »
We've learned , "Na,Zn are above Hydrogen in the electrochemical series so they can't be yielded from an aqueous solution of the metal cations.
eg:Na from NaCl(aq)"

But I've read that Mn,Co,Ni can be obtained from their aqueous solutions.Yet they are above Hydrogen in the electrochemical series.

So what the actual rule applies there ?
Is there something more than being above Hydrogen ?
« Last Edit: March 30, 2013, 08:20:19 AM by Arkcon »

Offline Dan

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Re: Yielding metal from aqueous solution
« Reply #1 on: March 30, 2013, 02:40:41 PM »
You can't isolate sodium metal from aqueous solution because it reacts with water:

Na(s) + H2O :rarrow: NaOH(aq) + H2(g)

In order to understand this, you should look at the half equation for the reduction of water:

H2O + 2e- :rarrow: H2 + 2-OH    E = -0.83 V
Na+ + e- :rarrow: Na     E = -2.7 V

So, it should follow that any metal with E > -0.83 V should be thermodynamically stable in water (see data table) - Ni and Co fall into this category.

Only if the solution is acidic can you compare with the standard hydrogen electrode (0 V). You can use the tables to explain why Zn does not react with water but dissolves in acidic solutions, for example.

Some metals that are thermodynamically unstable in water might not seem to react with it - Mn and Fe are examples. There is a reason for this. Hint: Sugar is thermodynamically unstable in oxygen, so why doesn't it spontaneously combust when exposed to air?


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Offline shalikadm

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Re: Yielding metal from aqueous solution
« Reply #2 on: March 30, 2013, 10:08:56 PM »
2 H2O + 2 e− ::equil:: H2(g) + 2 OH−    E= −0.8277 V
Ni2+ + 2 e−  ::equil:: Ni(s)    E= −0.25 V
2 H+ + 2 e−  ::equil:: H2(g)     E=  0.0000 V
So if we want to do some Nickel plating,we have to use a neutral solution,correct ?

2 H2O + 2 e− ::equil:: H2(g) + 2 OH−    E= −0.8277 V
Zn2+ + 2 e−  ::equil::    E= −0.7618 V
2 H+ + 2 e−  ::equil:: H2(g)     E=  0.0000 V
In the same way can't we gain Zn from a neutral Zn2+ solution?

Some metals that are thermodynamically unstable in water might not seem to react with it - Mn and Fe are examples.
Mn2+ + 2 e−  ::equil:: Mn(s)    E= −1.185 V
2 H2O + 2 e−  ::equil:: H2(g) + 2 OH−    E= −0.8277 V
Fe2+ + 2 e−  ::equil:: Fe(s)    E= −0.44 V
I see Mn above H2O,so it seems thermodynamically unstable in water.it's ok.But I see Fe below H2O.why you say Fe is thermodynamically unstable?


There is a reason for this. Hint: Sugar is thermodynamically unstable in oxygen, so why doesn't it spontaneously combust when exposed to air?
Is it something related to Gibbs Free Energy?
Is it because the ΔG for the reaction between Sugar and O2 is positive ?

Offline Corribus

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Re: Yielding metal from aqueous solution
« Reply #3 on: March 31, 2013, 01:20:34 AM »
Mn2+ + 2 e−  ::equil:: Mn(s)    E= −1.185 V
2 H2O + 2 e−  ::equil:: H2(g) + 2 OH−    E= −0.8277 V
Fe2+ + 2 e−  ::equil:: Fe(s)    E= −0.44 V
I see Mn above H2O,so it seems thermodynamically unstable in water.it's ok.But I see Fe below H2O.why you say Fe is thermodynamically unstable?
Because water is not what is actually doing the oxidation of iron.  There's another ingredient involved.

Is it something related to Gibbs Free Energy?
Is it because the ΔG for the reaction between Sugar and O2 is positive ?
The Gibbs energy change tells you what the relative concentrations of products and reactants will be at equilibrium.  But it does not tell you how long equilibrium will take to reach.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Dan

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Re: Yielding metal from aqueous solution
« Reply #4 on: March 31, 2013, 06:46:47 AM »
But I see Fe below H2O.why you say Fe is thermodynamically unstable?

Fe(OH)2 + 2e- :rarrow: Fe + 2-OH     E = -0.89
2H2O + 2e- :rarrow: H2 + 2-OH     E = -0.83
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Offline shalikadm

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Re: Yielding metal from aqueous solution
« Reply #5 on: March 31, 2013, 08:18:11 AM »
Fe(OH)2 + 2e- :rarrow: Fe + 2-OH     E = -0.89
2H2O + 2e- :rarrow: H2 + 2-OH     E = -0.83
Fe(OH)2(s) + 2 e− i ::equil:: Fe(s) + 2 OH−     −0.89
Fe3O4(s) + 8 H+ + 8 e−  ::equil:: 3 Fe(s) + 4 H2O     +0.085
FeO42− + 3 e− + 8 H+  ::equil:: Fe3+ + 4 H2O     +2.20
Fe2+ + 2 e−  ::equil:: Fe(s)     −0.44
Fe3+ + 3 e−  ::equil:: Fe(s)     −0.04
Several reactions containing Fe,I don't know how to select exactly which one goes with the explanation.
How you chose ,Fe(OH)2(s) + 2 e− i ::equil:: Fe(s) + 2 OH−     −0.89
for the explanation?

Because water is not what is actually doing the oxidation of iron.  There's another ingredient involved.
What's that ?..Is  OH- involved...I think we are talking about a neutral solution of Fe2+

The Gibbs energy change tells you what the relative concentrations of products and reactants will be at equilibrium.  But it does not tell you how long equilibrium will take to reach.
Is it the Activation energy...Sugar and O2 doesn't get the sufficient energy to make product ?

ps :I think that I'm surfing net a lot right now....My exams are near...So I think I have to give up..I think this is somewhat above A Levels..We've not taught about these things...In my text books,I saw that Ni,Co,Mn can be gained from an aqueous solutions but not Zn,Na....I decided to put here because It's better to have proper understanding about why it's happening like this rather than memorizing the thing.But I know that forum rules won't let you give me a direct answer... :-\

Offline Corribus

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Re: Yielding metal from aqueous solution
« Reply #6 on: March 31, 2013, 10:09:34 AM »
What's that ?..Is  OH- involved...I think we are talking about a neutral solution of Fe2+
Here's a hint: they call it oxidation for a reason. ;)

Is it the Activation energy...Sugar and O2 doesn't get the sufficient energy to make product ?
Now you're on to something.  Thermodynamic stability and kinetic stability are two different things...
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

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