[jsmith613]

http://s359.photobucket.com/albums/oo40/jsmith613/?action=view&current=Cell.jpg

In this diagram, the larger line represents the +ive side and the smaller line represents the -ive side. Therefore electrons SHOULD flow from the smaller line to the larger one. In electrolysis, this is not the case. Electrons flow from the +ive anode to the -ive cathode. Why is this? Please can someone explain.

Thanks

[Schrödinger]

In this diagram, the larger line represents the +ive side and the smaller line represents the -ive side. Therefore electrons SHOULD flow from the smaller line to the larger one.

That is the case only if you are looking at the external circuit. The direction of flow of electrons is from the -ve terminal to +ve terminal only in the conducting circuit wires. Within the cell, electrons flow from the +ve terminal to the -ve terminal.

Please correct me if I am wrong

[cliverlong]

In this diagram, the larger line represents the +ive side and the smaller line represents the -ive side. Therefore electrons SHOULD flow from the smaller line to the larger one.

That is the case only if you are looking at the external circuit. The direction of flow of electrons is from the -ve terminal to +ve terminal only in the conducting circuit wires. Within the cell, electrons flow from the +ve terminal to the -ve terminal.

Please correct me if I am wrong

In electrolysis electrons flow in the wires (external circuit), ions carry current in the electrolyte

http://www.bbc.co.uk/schools/gcsebitesize/science/add_aqa/ions/electrolysisrev1.shtml

[jsmith613]

I think I actually understand: please can someone clarify my 'epiphany'!

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Anode = +ive
Cathode = -ive

As ions are attracted towards each electrode respectively, the cathode remains 'relatively positive' to the anode. Therefore, every time one anion goes to the anode, and deposits its electrons, a cation goes to the cathode. This cation causes the negative electron to flow towards the cathode, where the cation will be discharged. The overall effect is nil as each time one electron is deposited on the anode that cathode takes that electron and discharges it. Effectively, the charge on each electrode is always the same.
When it comes to anions with a charge of 2-, the two electrons are transferred due to the presence of TWO cations so the two electrons are again 'neutralized'
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effectively: the cations will ALWAYS be attracted to the cathode as, each time an electron is deposited their the cathode 'regains' its relative negativity. But each time a cation is attracted to the cathode, the relative negativity --> relative positivity. This positivity is IMMEDIATELY canceled out by surplus electrons thus returning the charge of the cathode to -ive

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Is this correct? Please correct me if i am wrong

[cliverlong]

I think I actually understand: please can someone clarify my 'epiphany'!

Always a rewarding experience.

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<< >>
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Is this correct? Please correct me if i am wrong

You have the right ideas but they are expressed a little bit unconventionally.

Have a read of this:

http://www.docbrown.info/page01/ExIndChem/ExtraElectrochem.htm

1. Introduction  to  electrolysis - electrolytes and non-electrolytes

and you should be encouraged by your current level of understanding

Clive

[jsmith613]

Have you read of this:

http://www.docbrown.info/page01/ExIndChem/ExtraElectrochem.htm

I did read this site before publishing my question. I find that this site does not actually address my question, even though it says that it will.

To make my answer more simple:

Surplus electrons at the anode make it relatively -ive to the cathode. Electrons are pulled towards the cathode making it negative. Cations are attracted and immediately pick up the electrons, therefore, the cathode becomes -ive again.

The process happens so quickly that the change in charges are, to the human 'eye', un-noticable

[cliverlong]

In your original post you wrote

In this diagram, the larger line represents the +ive side and the smaller line represents the -ive side. Therefore electrons SHOULD flow from the smaller line to the larger one. In electrolysis, this is not the case. Electrons flow from the +ive anode to the -ive cathode. Why is this? Please can someone explain.

then after a few posts and suggestions you wrote

Have you read of this:

http://www.docbrown.info/page01/ExIndChem/ExtraElectrochem.htm

I did read this site before publishing my question. I find that this site does not actually address my question, even though it says that it will.

To make my answer more simple:

Surplus electrons at the anode make it relatively -ive to the cathode. Electrons are pulled towards the cathode making it negative. Cations are attracted and immediately pick up the electrons, therefore, the cathode becomes -ive again.

The process happens so quickly that the change in charges are, to the human 'eye', un-noticable

I refer you again back to the Doc Brown link

In section 2b there is the following text, that describes the process you are interested in, except more clearly

(-) negative cathode electrode  where reduction of the  attracted positive cations is by electron gain to form metal atoms or hydrogen [from Mn+ or H+, n = numerical positive charge]. The electrons come from the positive anode (see below).

(+) positive anode electrode where the oxidation of the atom or anion is by electron loss. Non-metallic negative anions are attracted and may be oxidised to the free element. Metal atoms of a metal electrode can also be oxidised to form positive metal ions which pass into the liquid electrolyte. The released electrons move round in the external part of the circuit to produce the negative charge on the cathode electrode.

There is no "surplus" of electrons
The cathode does not "become negative again"

The cathode is always ready to supply electrons to the ions in the electrolyte. It does this by discharging positive cations or creating negative anions

The anode is always ready to remove electrons from the ions in the electrolyte. It does this by discharging negative anions or creating positive cations.

I strongly suggest you draw a picture of the changes that are happening to ions at the surface of the electrode when you are electrolysing various ionic solutions and work out where the electrons are moving.

[jsmith613]

O.K, I think I get it now.

The cations, at the cathode, cause the electrons to flow from the anode to the cathode; the electrons are attracted by their +ive charge. The charge on the electrode is irrelevant to demonstrate the flow of electrons; it only demonstrates which ions go to which electrode.

Anode Negative ions are attracted due to the anode's positive charge. Here they DEPOSIT their electrons to become neutral

2Cl^- --> Cl₂ + 2e^-

Cathode Positive ions are attracted due to the cathode's negative charge. Here electrons are DEPOSITED from the cathode on the cations to cause them to become neutral

Ca²⁺ + 2e^- --> Ca

The electrons (from the anions) flow from the anode to the cathode, where they are be deposited onto cations to give them a neutral charge (i.e: a charge of Zero).

[jsmith613]

There is no 'relative charge'. Its simply that +ive cations attract -ive electrons.

Right?

[cliverlong]

There is no 'relative charge'. Its simply that +ive cations attract -ive electrons.

Right?

Yep, I think that is a fair way of putting it.

The reason why it happens is mainly (I believe) down to electrode potential. To turn a positive metal ion to the metal reuqires energy to be put in and the elecrode potential is a measure of that maount of energy. However , I think whenthe metal is in an aqueous solution other factors come into play - hydration maybe - but I'm at the limit of my knowledge to identify and explain exactly what those factros are.

Well done persisting with that. I recommend the chemguide website as a source of clear explanations on topics you ask questions on.

Clive

[FreeTheBee]

It is easier to remember that reduction takes place on the cathode and oxidation on the anode. The signs depend on whether you are running an electrolytic (electrolysis) or galvanic (battery) experiment. To remember this I always used the fact that it is (an o)xidation, where the first three letters form the start of anode. Sounds stupid, but it worked for me.

[cliverlong]

It is easier to remember that reduction takes place on the cathode and oxidation on the anode. The signs depend on whether you are running an electrolytic (electrolysis) or galvanic (battery) experiment. To remember this I always used the fact that it is (an o)xidation, where the first three letters form the start of anode. Sounds stupid, but it worked for me.

I like the three letter "reminder" "An o". Have a snack.

Clive