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#Post#: 744--------------------------------------------------
The Voltaic Cell Zinc and Sulphuric Acid reaction in the simple
cell
DIR By: mshaugh11
Date: September 26, 2012, 5:07 pm
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I've been studying the Zinc and Sulphuric acid simple cell
(voltaic pile) reaction for the last 2 weeks, reading internet
forums and text books alike. The problem i face is that every
forum and textbook only goes so far and its awful fustrating. I
really hope someone on this forum can take their time to simply
explain the process.
Ok, so the way i see it is
Zinc atoms have 2 electrons on their outer orbital and therefore
are willing to get rid of these so that they are more stable or
less reactive since their outer shell will be complete i.e 8
electrons (simplistic approach and not looking at 2 x n2
calculations)
Sulphuric acid or should it be referred to as dilute sulphuric
acid as it would appear that in order for the H2SO4 to be
ionised, it must be diluted using H2O.
Now what actually happens with the H2O. Well according to the
text books, water or H2O molecules (covalent bonds) effectively
collide to form hydroxide OH -ive and hydronium H3O +ive ions.
Therefore water exists as a mixture of molecules, hydroxide ions
and hydronium ions.
So when the water mixture is added to the Sulphuric Acid, it
become dilute and there seems to be a reaction between the H2SO4
and the H2O, the H3O +ive and the OH -ive creating an ionidised
solution of +ive and -ive charged ions.
I need to get an understanding on what actually happens here
with the water ions and the sulphuric ions and how and why
certain atoms are more attracted to other atoms. I need to
understand why the solution eventually becomes ZnSO4 in
solution. Now why is it that certain molecules i.e acids
dissociate releasing a hydrogen ion without releasing a
hydroxide ion and what causes this?
Looking at the formula I see the H2SO4 breaks down into H+ive
and HSO4. What causes the breakdown extracting the H+ive. In the
second reaction I see the HSO4 -ive breaking down into H+ive and
SO4 -ive. Again why the extraction or breakoff of the H+.What
actually causes this.
So now we have H2SO4 dilute = 2 H +ive +SO4 --ive.
This ionidised solution then attracts the Zn electrons from the
Zn. Now the more +ions in the solution the more Zn electrons are
lost to the solution. Then these electrons attach to the H+ive
thus creating H2 gas. The solution remaining is Zn ++ which
dissolves into the solution and the SO4-- bond to form Zinc
Sulphate ZnSO4. How are these atoms bonded?
To advance this situation, my understanding is if a copper rod
is inserted into the solution (Daniell Cell) and a wire is
attached between the Zn rod and the Copper rod, a current will
flow or in atomic terms, the Zinc will become positive as its
losing electrons and its nucleus (protons) will be more
positive. How does this happen. Looking at the atomic structure
and electron orbits, i can't see how the copper becomes negative
to have a flow from the Zn -- to the Cu ++.
I'm really interested to get your help and simple explanations
on this matter
#Post#: 745--------------------------------------------------
Re: The Voltaic Cell Zinc and Sulphuric Acid reaction in the
simple cell
DIR By: Michel
Date: September 27, 2012, 3:58 am
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Water ionization is actually not complete. Infact liquid water
exists mainly as H2O molecules than H3O+ and OH- ions.
I will use a simple way of writing the ionization of water to
explain this.
2H2O <==> H30+ + OH-
If you've learnt about acids and bases, you will know that for
every one mole of water, only 10^-7 mole of H+ and OH- exist,
the majority is the H2O molecules. Thus when H2SO4 interacts
with water, the H2SO4 ionizes thus
H2SO4---> H+ + HSO4-
HSO4- ---> H+ + SO42-
in essence,
H2SO4---> 2H+ + SO42-
Here, you see that each molecule or radical ionizes to produce
its component cations and anions, an acid will ionize to produce
it's ions, in this case, H2SO4 ionizes producing H+ and
SO42-[\sup], the same way water ionizes producing H+ and OH-
(the H+ in both cases attach to water molecules forming H3O+)
H2SO4 produces SO4[sup]2- as its anion, the same way water
produces OH- as its anion. This holds for all other acids. Hope
this explains why acids cant produce OH- ions
Going into details on how this ionization takes place may not be
helpful at the moment, (you could read on acids and bases in any
general chemistry text to understand this). What is important to
know is that water primarily exists as H2O molecules, and not a
mixture of ions. The ions produced by water ionization are too
few to be considered important
--- Quote from: mshaugh11 link ---
>
>
> This ionidised solution then attracts the Zn electrons from
the Zn. Now the more +ions in the solution the more Zn electrons
are lost to the solution. Then these electrons attach to the
H+ive thus creating H2 gas. The solution remaining is Zn ++
which dissolves into the solution and the SO4-- bond to form
Zinc Sulphate ZnSO4. How are these atoms bonded?
>
> To advance this situation, my understanding is if a copper rod
is inserted into the solution (Daniell Cell) and a wire is
attached between the Zn rod and the Copper rod, a current will
flow or in atomic terms, the Zinc will become positive as its
losing electrons and its nucleus (protons) will be more
positive. How does this happen. Looking at the atomic structure
and electron orbits, i can't see how the copper becomes negative
to have a flow from the Zn -- to the Cu ++.
>
--- End Quote ---
The Zn2+ ion and the SO42- ion remain as ions moving about in
the solution, they do not literally become bonded per se, only
when the solution is evaporated, the electrostatic forces
between these ions become more pronounced, and the ZnSO4
crystals form.
In the Daniell cell, there is Zn in a solution of ZnSO4, and Cu
in a solution of CuSO4 separated by a porous partition. Some Zn
atoms in the Zn electrode ionize and go into the ZnSO4 solution
as Zn2+ the two electrons left on the electrode surface then
move through the wire to the cathode cell where they are
accepted by the Cu2+ ion in the CuSO4 solution, depositing
Copper metal on the Copper cathode.
The Zn electrode is called negative, NOT POSITIVE, because the
electrons left on it after ionization of Zn atoms give it a
negative charge. The Cu cathode is the positive electrode.
#Post#: 749--------------------------------------------------
Re: The Voltaic Cell Zinc and Sulphuric Acid reaction in the
simple cell
DIR By: mshaugh11
Date: September 27, 2012, 12:32 pm
---------------------------------------------------------
Thank you for your response. As my background is in Engineering,
im having some difficulty understanding some of your
explanations so apologies if you feel im asking questions twice.
In reply to your answers, may I ask the following questions
1. You say that there are very few free ions in water and so
water is mainly made up of molecules. Now because there are very
few ions in solution, why is it then that the H2SO4 reacts with
the H2O and the ionisation process takes place to get to your
equation of HSO4 etc.
I really think in order to fully understand this, I need to
understand how ionisation actually takes place. I accept your
comment on maybe leaving this out but I dont believe I can move
on without understanding ionisation and the actual reasons why
the water molecules and Sulphuric Acid react in the way they do.
2.Now your explanation of the Zn and the SO4 being in solution
and when the solution is evaporated, you're left with
electrostatic bonding in the form of crystals makes sense. Does
the zinc rod totally dissolve in the solution and is the
evaporation caused by the reaction between the Zn and H2SO4. Ive
seen a youtube with shows the reaction and its quite reactive or
at some point does one have to apply heat to evaporate the
solution to get the crystals.
3. Im a little confused by your explanation on the Zn rod or
electrode being negative. I had understood that if Zn lost
Electrons, that would mean that the overall charge of that atom
would be positive. There would be more protons than electrons
and therefore the overall charge on the Zn electrode would be
positive.
4. In the Daniell cell, the copper is in Copper Sulphate
solution. But coppers atomic structure is 2,8,8,8,3 so the outer
orbital has 3 electrons. In a similar way that Zn gives away 2
electrons to become more noble, why doesnt the copper give away
3 electrons and therefore also become more positive (more
protons than electrons). If this were the case, there would be
no electric current as both rods would be of the same polarity.
Again my apologies for all these questions and hopefully you
will have the patients to answer these as simply as possible but
with as much detail as is necessary. Many thanks.
#Post#: 750--------------------------------------------------
Re: The Voltaic Cell Zinc and Sulphuric Acid reaction in the
simple cell
DIR By: Michel
Date: September 27, 2012, 5:47 pm
---------------------------------------------------------
1.) I have attached a photo of the structure of H2SO4. In this
structure, you can see two OH bonds. These bonds are highly
polar because the electrons in the bond are more attracted to
the Oxygen atom than the Hydrogen atom. In effect, a hydrogen
ion can easily break off from the bond. Suppose a H+ ion breaks
off from the H2SO4 molecule, it gets hydrated by a H2O molecule
present, forming H3O+ . This can happen continuously until all
the H2SO4 dissociates(I'm desperately trying to simplify) Due to
the abundance of H2O molecules present, the attraction between
the H+ and SO42- ions is low, hence to a very large extent, they
dont combine to reproduce H2SO4. They remain as ions in
solution.
2. Whether the Zn rod will totally dissolve in the solution
depends on the amount of acid present. If the stoichiometric
amount of H2SO4 is used, the Zn rod will totally dissolve in the
dilute H2SO4.
Evaporation isn't caused by the reaction. You will have a
solution of ZnSO4 at the end of the reaction which you can
evaporate carefully to obtain the crystals.
3. The Zn rod is negative because; as the Zn atoms ionize, the
Zn2+ ion goes into solution leaving behind the 2electrons on the
Zn rod. The presence of the excess 2e- on the Zn rod makes it
negatively charge.
4.) Copper electronic configuration is 2,8,18,1 not 2,8,8,8,3.
The outermost shell has one electron, not three. Cu is a
transition metal, and we don't normally use electron shells to
explain transition metal chemistry. The d-subshell is the rule
of thumb here. Cu has two oxidation states, but the Cu2+ is the
more stable ion
#Post#: 752--------------------------------------------------
Re: The Voltaic Cell Zinc and Sulphuric Acid reaction in the
simple cell
DIR By: mshaugh11
Date: September 28, 2012, 6:48 am
---------------------------------------------------------
Michael,
For some reason, the attached file seems to be corrupt when I
down load. I've tried to open it in numerous programmes but no
good. Would you mind trying to resend as a jpg maybe.
Sorry for the inconvenience.
#Post#: 755--------------------------------------------------
Re: The Voltaic Cell Zinc and Sulphuric Acid reaction in the
simple cell
DIR By: mshaugh11
Date: September 28, 2012, 7:13 am
---------------------------------------------------------
Further to your reply and without the diagram, does
electronegativity come into play here. Is this why one atom is
more attrached to another. In the case where you say the
electrons are more attracted to the Oxygen than the Hydrogen.
Looking at the electronegativity table, H has a
electronegativity of 2.2 whereas Oxygen has 3.4. I presume that
since Sulphur is 2.6, the stronger attraction is between the S
and O and not the S and H or O and H. Am I on the correct path
or totally confused. If its not electronegativity whats causing
these attractions. :-[. If it is, then the 2.6 and the 3.4 have
a greater attraction than the 2.2 with either of the others. Now
if this is the case, why does the H break away from the O in the
(OH bond) as the electronegativity value for a O-H (2.6-2.2)
should be stronger than a H-H (2.2-2.2) bond. If Im missing the
point, maybe you could refer me to articles where I can read
more on this subject and get it clear in my head.
Relating to Q3 previous,
Am I correct in assuming in the first stage of the process, the
Zn gives 2 electrons away to the Sulphuric acid thus making the
Zn atoms in the rod, Zn++. Therefore the Zn atom has an overall
positive charge? Is this correct?
Now you say that the Zn ionises. Are you saying that there is a
second stage and in that another reaction takes place. The Zn++
starts to dissolve and something happens causing the rod to
become negative. You say that as the Zn++ ionises, it leaves 2
electrons behind. But I understood that the Zn atom had already
given off its 2 outer orbital electrons and these were taken up
by the Hydrogen to give off H gas. So why are there yet another
2 electrons left behind to return the Zn rod back to negative.
???
Q4
Ok, most of the physics books explain the orbitals in a simple
manner as having orbitals as 2, 8, 8 ,8 to simplify but ofcourse
this is not the case and only leads to confusion. I am aware of
the formula 2xn2 for calculating electrons per orbital and the
dimensional structure of the various s, p,d, f energy levels. It
just shows you, when you simplify too much, it makes things even
more confusing and incorrect.
Now being a transition metal, its valency seems to be 1, its on
the 3d level (d block) and as an oxidation agent, its happy to
lose one electron (2,8,18,1) so that Cu should be left on the
copper rod as Cu+. Why is it that it gets a double charge i.e
Cu++ then. I see that Zinc is also a transitional element and so
why is it treated differently.
#Post#: 756--------------------------------------------------
Re: The Voltaic Cell Zinc and Sulphuric Acid reaction in the
simple cell
DIR By: Michel
Date: September 28, 2012, 7:45 am
---------------------------------------------------------
I'm sorry about the photo. I was very sleepy when I made that
post, I didn't even wait to see the new page when I submitted
the post. You can do a quick google for the structure of H2SO4,
anyway, you'll see the image and the O-H bonds.
Your analogy is quite right. If you look up the structure of
H2SO4, you will see that there is no H-S bond. The S-O bond is
definitely stronger than the O-H bond. The O-H bond is HIGHLY
POLAR, due to the structure of H2SO4, which is a reason why
H2SO4 releases a proton in aqueous solutions.
#Post#: 761--------------------------------------------------
Re: The Voltaic Cell Zinc and Sulphuric Acid reaction in the
simple cell
DIR By: mshaugh11
Date: September 28, 2012, 8:46 am
---------------------------------------------------------
Oops, was replying and modifying to your previous response and I
see you replied...
Further to your reply and without the diagram, does
electronegativity come into play here. Is this why one atom is
more attrached to another. In the case where you say the
electrons are more attracted to the Oxygen than the Hydrogen.
Looking at the electronegativity table, H has a
electronegativity of 2.2 whereas Oxygen has 3.4. I presume that
since Sulphur is 2.6, the stronger attraction is between the S
and O and not the S and H or O and H. Am I on the correct path
or totally confused. If its not electronegativity whats causing
these attractions. . If it is, then the 2.6 and the 3.4 have a
greater attraction than the 2.2 with either of the others. Now
if this is the case, why does the H break away from the O in the
(OH bond) as the electronegativity value for a O-H (2.6-2.2)
should be stronger than a H-H (2.2-2.2) bond. If Im missing the
point, maybe you could refer me to articles where I can read
more on this subject and get it clear in my head.
Relating to Q3 previous,
Am I correct in assuming in the first stage of the process, the
Zn gives 2 electrons away to the Sulphuric acid thus making the
Zn atoms in the rod, Zn++. Therefore the Zn atom has an overall
positive charge? Is this correct?
Now you say that the Zn ionises. Are you saying that there is a
second stage and in that another reaction takes place. The Zn++
starts to dissolve and something happens causing the rod to
become negative. You say that as the Zn++ ionises, it leaves 2
electrons behind. But I understood that the Zn atom had already
given off its 2 outer orbital electrons and these were taken up
by the Hydrogen to give off H gas. So why are there yet another
2 electrons left behind to return the Zn rod back to negative.
Q4
Ok, most of the physics books explain the orbitals in a simple
manner as having orbitals as 2, 8, 8 ,8 to simplify but ofcourse
this is not the case and only leads to confusion. I am aware of
the formula 2xn2 for calculating electrons per orbital and the
dimensional structure of the various s, p,d, f energy levels. It
just shows you, when you simplify too much, it makes things even
more confusing and incorrect.
Now being a transition metal, its valency seems to be 1, its on
the 3d level (d block) and as an oxidation agent, its happy to
lose one electron (2,8,18,1) so that Cu should be left on the
copper rod as Cu+. Why is it that it gets a double charge i.e
Cu++ then. I see that Zinc is also a transitional element and so
why is it treated differently.
#Post#: 762--------------------------------------------------
Re: The Voltaic Cell Zinc and Sulphuric Acid reaction in the
simple cell
DIR By: Michel
Date: September 28, 2012, 11:15 am
---------------------------------------------------------
We are trying to use the electronegativity values of Oxygen and
Hydrogen to explain why the H+ easily removes from the H2SO4
molecule. The high electronegativity of the oxygen atom means
that the electrons in the bond are far nearer oxygen than
Hydrogen. Meaning it is easier for the H+ to break off, leaving
the two electrons on the Oxygen atom. In the structure of H2SO4,
the Hydrogen atoms have low electron density, meaning they can
easily break off the molecule, leaving behind their electrons,
producing H+ ions in aqueous solutions.
In a Daniell cell, we have Zn in ZnSO4 as the anode half cell,
not Zn in H2SO4. Even if you put a large Zn rod into a beaker of
dilute H2SO4, the Zn will react with the H2SO4 producing ZnSO4
and H2, you will now have Zn in ZnSO4 which you can use as your
anode half cell. Zn ionizes, producing Zn2+ ions. The Zn2+ ions
enter the ZnSO4 solution and get hydrated by water molecules
present. The 2e- are left on the Zn anode (this gives it
negative charge). The 2e- then flow through the wire to the
cathode compartment and they are accepted by the Cu2+ present,
thus depositing solid Cu on the anode.
Q4. Cu does produce Cu+, but Cu+ isn't stable, due to the high
redox potential of Cu/Cu+ system.
Zn ionizes by releasing its two valence electrons. Technically,
it isn't part of the 1st transition series, but it's included in
the list
#Post#: 775--------------------------------------------------
Re: The Voltaic Cell Zinc and Sulphuric Acid reaction in the
simple cell
DIR By: mshaugh11
Date: September 30, 2012, 7:45 am
---------------------------------------------------------
When the Zn loses its electrons, its atomic structure changes as
its outer orbital is now without 2 electrons. Is it the case
then that its no longer Zn, its an ion? It it were to become Zn
again, it would need its electrons back. Is it true therefore to
say, that its more willing to be released from its ionic
structure with its other Zn atoms and therefore is attracted
into the solution by positive ions. Is it true to say as this
happens, the rod is actually dissolving?
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