I've tried and tried....

[Zebra]

i struggle so much with predicting what reaction happens where for both galvanic and electrolysis. i told myself doing questions on this area would improve my understanding of it but its not happening. I dont even think i know the concepts... I've read the textbook and I'm like oh yeah i can see that.. when i try to do a question on it im like ..GG

any help?!?!?!

and HOW COME NONE OF MY SOURCES MENTION ANYTHING ON ACIDIC FUEL CELLS...? (anyone)?

[lorelai]

The most important concepts you'll have to understand for these two topics:

In galvanic cells, spontaneous redox reactions occur. As you might recall, a redox reaction is one in which electrons are transferred from one species to another, so they produce electricity. The species that loses electrons is oxidised. The species that gains electrons is reduced.
And we can also say that that the species that loses electrons is the reductant, as it causes the reduction of the other species. And in the same way, the species that gains electrons is the oxidant, as it causes the oxidation of the other species. (This is kind of counter-intuitive, but you'll get the hang of it.)
You can tell whether a species is oxidised/reduced and hence whether it is the oxidant/reductant in the reaction by using oxidation numbers. You should know how to do this from Unit 3.
In a galvanic cell, reduction occurs at the cathode, which is the positively-charged electrode. Oxidation occurs at the anode, which is the negatively-charged electrode. (Though you'll have to see that these electrodes have no charge to begin with - their charge comes from the reaction itself.)

So:

reductant - is oxidised at the anode - loses electrons
oxidant - is reduced at the cathode - gains electrons

This, now, is where your electrochemical series comes in. It is a list of redox half-equations, arranged such that the oxidant is on the left-hand side, and the reductant is on the right-hand side. These half-equations are ordered so that strength of oxidants increases as you go up the table, and strength of reductants increases as you go down the table.
This order was established by measuring the electrical potential (potential to produce electricity when coupled with another half-cell) of each reaction against a standard hydrogen electrode (SHE). This is a half-cell (meaning that it only uses one half-equation) that is used as a reference point. We say that it produces 0.00 V of electricity. This electrical potential (EMF) comes out as the E⁰ value you can see on the right of the series.

You can predict whether a spontaneous reaction will occur by finding the oxidant and finding the reductant and seeing whether you can connect them with a line that goes from top left, to bottom right. If this can be done, then that means that the E⁰ value (EMF) of the cell is positive and thus a redox reaction is possible.

So if you want the full steps to figure out what overall equation occurs and how to figure out EMF:

1. Identify which species is being oxidised and which is being reduced.
2. Find their positions on the electrochemical series.
3. Write out the two half-equations in descending order of EMF.
4. Reverse the equation with the lowest EMF (so that the reductant goes to its oxidised form).
5. Balance out the electrons (by multiplying the equations so that they end up having the same number of electrons).
6. Put the two half-equations together (the electrons should now cancel out).
7. Calculate EMF value using this formula:

EMF = E⁰_oxidant - E⁰_reductant

For example:
A galvanic cell using zinc metal and copper(II) ions

Cu²⁺(aq) + 2e^- --> Cu(s)     E⁰ = +0.34 V
Zn²⁺(aq) + 2e^- --> Zn(s)     E⁰ = -0.76 V

The ones in bold are the species that are reacting. You can see that they are in order of descending EMF, and that they can be connected by a diagonal line with a negative gradient. It follows that Zn is the reductant and is being oxidised, and that Cu²⁺ is the oxidant and is being reduced.

Now to reverse the second equation:
Zn(s) --> Zn²⁺(aq) + 2e^-

And put the two half-equations together (the electrons are already balanced):
Cu²⁺(aq) + Zn(s) --> Cu(s) + Zn²⁺(aq)

And finally, to calculate EMF:
EMF = E⁰_oxidant - E⁰_reductant
       = (+0.34) - (-0.76)
       = 1.10 V

If you consider what this particular galvanic cell would look like:



You might notice the little e^- (electrons) with an arrow pointing to the direction in which they are moving. They are moving through what is referred to as the external circuit, which is the wires connecting the anode and cathode. In this particular diagram, the anode is to the left, and the cathode is on the right. The anode is made out of zinc, which is oxidised to zinc(II) ions in the course of the reaction. The cathode is made out of copper - the copper(II) ions are reduced to copper metal, which builds up on this cathode.

So where do the Zn²⁺ ions go, and where do the Cu²⁺ ions come from? The Cu²⁺ ions come from a Cu(NO₃)₂(aq) solution placed in the right beaker, and the Zn²⁺ ions that are produced go to a Zn(NO₃)₂ solution placed in the left beaker. (The species are deliberately separated from each other).

However, the galvanic cell cannot operate as of yet, because there is nothing connecting the two beakers of solution. In comes the salt bridge, which is either a piece of filter paper soaked with or a glass tube filled with the electrolyte. This is another solution, whose job it is to supply ions to the reaction. In this case, it is a sodium nitrate solution. Sodium nitrate dissociates in water to produce Na⁺ ions and NO₃^- ions.
As the Cu²⁺ ions are used up, there is a build-up of negative charge in the right beaker. So the Na⁺ ions are sent in from the salt bridge in order to balance this. And in the other beaker, as Zn²⁺ ions are produced, there is a build-up of positive charge. The NO₃^- ions come in from the salt bridge to correct this.
In this way, the electrolyte in the salt bridge maintains electroneutrality in the cell, meaning that there is no net build-up of charge in either of the beakers. It also completes what is known as the internal circuit, which is the part of the cell through which ions flow.

[nemolala]

you smart little cookie ^^ Nice sum very easy to follow

[Zebra]

my goodness. Thanks so much.
I would appreciate if you can post electrolytic cells as well.

thanks my chemhero

[hala_madrid]

i struggle so much with predicting what reaction happens where for both galvanic and electrolysis. i told myself doing questions on this area would improve my understanding of it but its not happening. I dont even think i know the concepts... I've read the textbook and I'm like oh yeah i can see that.. when i try to do a question on it im like ..GG

any help?!?!?!

and HOW COME NONE OF MY SOURCES MENTION ANYTHING ON ACIDIC FUEL CELLS...? (anyone)?

AMEN BROTHER. pretty sure i've spent a total of 20 hours on elecytrolysis and learnt absolutely nothing!

[lorelai]

Okay, so here goes for electrolytic cells:

What you have to realise for electrolysis is that the process occurring is the opposite of that which occurs in a galvanic cell. Whilst in galvanic cells chemical energy is transformed into electrical energy, in electrolytic cells electrical energy is transformed into chemical energy. Electrolysis still involves the transfer of electrons, but instead of producing electricity, the reactions require electricity in order to occur. Why? Because these reactions are not spontaneous, meaning they would not normally occur. They require electrons to be flowing in the opposite direction to how they normally would.
So all of this means that in an electrolytic cell, we have to have a source of electricity, such as a battery pack. This battery pack has to put into the cell a greater voltage than would be produced by the equivalent galvanic cell. Now this is where it starts to get tricky, so be warned.

The battery has a negative terminal and a positive terminal. If we want to force the reactions to go the other way to a normal redox reaction, and thus the electrons to flow in the opposite direction, the negative terminal is attached to the cathode and the positive terminal is attached to the anode. Yes, this means that the cathode becomes the negative electrode and that the anode becomes the positive electrode - the opposite to in galvanic cells.
By definition, reduction occurs at the cathode and oxidation occurs at the anode. So the same still happens in an electrolytic cell.

So:

reductant - is oxidised at the anode - the positive electrode - loses electrons
oxidant - is reduced at the cathode - the negative electrode - gains electrons

Again, this is when the electrochemical series comes in. In an electrolytic cell, all the information about this series still stands - it is simply used in a different way.

Here is some basic theory behind the operation of electrolytic cells.

Electrolytic cells can be used to electrolyse an aqueous solution. This means that the species taking place in the reaction are present in the electrolyte, which in this case is just a solution that can conduct electricity. One of these species (the reductant) will be oxidised at the anode, and another (the oxidant) will be reduced at the cathode. (Note that inert electrodes are used here, because any reactive metals used as electrodes may take place in the reaction.)
The problem is now that once we start with an aqueous solution we have more than two species that could possibly take part in the reaction: the cations (positively-charged ions), the anions (negatively-charged ions) but also water - H₂O. Now, if you look at the electrochemical series you'll see that there's more than one equation containing water - the trick is to work out which reactions are occurring and where. When there are more than two species that could react like this, the cell is said to have competing oxidants and reductants or competition at electrodes.

Here's the full steps to work out what's going to react and what's not going to react:

1. List all of the possible reactants at each electrode.
2. Find their positions on the series.
You'll find that some of them will be on the right-hand side of the series and some will be on the left. The ones on the right are the possible reductants, and the ones on the left are the possible oxidants.
3. The two species that will react will be the reductant and oxidant that are closest together. So you'll be able to draw a short line that goes from top right to bottom left between the two species reacting (this is the opposite of in galvanic cells).
The reason for this is that strength of reductants increases as you go down the table, and so the right-hand side species that is closest to the bottom will be oxidised. So, as the strength of oxidants increases as you go up the table, the left-hand side species that is closest to the top will be reduced.
4. Write out the two half-equations in descending order of EMF.
5. Reverse the half-equation with the highest EMF (again this is the opposite of in galvanic cells).
5. Balance out the electrons (by multiplying the equations so that they end up having the same number of electrons).
6. Put the two half-equations together (the electrons should now cancel out).
7. Calculate EMF value using this formula:

EMF = E⁰_oxidant - E⁰_reductant

The only difference with this step is that it now calculates how much electricity needs to be put into the cell, instead of how much electricity the cell will produce.

Here's an example of these steps being used for an aqueous solution of sodium chloride:

The possible reactants are Na²⁺, Cl^-, and H₂O.

The half-equations are:

Cl₂(g) + 2e^- ---> 2Cl^-(aq)                                      E⁰ = +1.36 V
O₂(g) + 4H⁺(aq) + 4^- ---> 2H₂O(l)       E⁰ = +1.23 V
2H₂O(l) + 2e^- ---> H₂(g) + 2OH^-(aq)   E⁰ = -0.83 V
Na⁺(aq) + e^- ---> Na(s)                                                           E⁰ = -2.71 V

See what this means? The two reactants that are closest together are H₂O and H₂O.

So these are the two half-equations that will occur:

O₂(g) + 4H⁺(aq) + 4^- ---> 2H₂O(l)      E⁰ = +1.23 V
2H₂O(l) + 2e^- ---> H₂(g) + 2OH^-(aq)   E⁰ = -0.83 V

Now to reverse the first equation:
2H₂O(l) ---> O₂(g) + 4H⁺(aq) + 4^-

And now to balance the equations. You can see that the first equation has four electrons, whilst the second has only two. This means that we need to multiply the second equation by two so that they both have four electrons:
2 x [ 2H₂O(l) + 2e^- ---> H₂(g) + 2OH^-(aq) ]
4H₂O(l) + 4e^- ---> 2H₂(g) + 4OH^-(aq)

Now put the two equations together. The acid (H⁺), the base (OH^-) and the electrons cancel out so it becomes:
2H₂O(l) ---> 2H₂(g) + O₂(g)

And finally, to calculate EMF:

EMF = E⁰_oxidant - E⁰_reductant
       = (-0.83) - (1.23)
       = -2.06 V
This means that 2.06 V needs to be supplied to the cell for its operation.

So now you're probably wondering why sodium chloride solution was used when the only reactant was going to be water - why not just use plain water? Well, there is a way to obtain sodium metal and chlorine gas by electrolysis from its salt. But water has to be eliminated from the equation.

First, though, there's another thing that has to be understood about the electrochemical series. The order of the series was established under standard conditions - concentrations of 1.0 M, temperature of 25 °C, and gases collected at pressure of 1 atm. So any significant change to these conditions may result in a different order of EMF to that in the series. When the conditions are changed, they are said to be non-standard conditions.
This is often a good thing - it allows us to obtain different products to those at standard conditions and is the basis of many applications of electrolysis.

So how to eliminate water from the above electrolysis problem? All that is needed is to use pure sodium chloride rather than an aqueous solution. However, it's not quite that simple: solid sodium chloride can't be used because it cannot conduct electricity, as it is an ionic compound and so electrons are firmly trapped in bonds involving whole charges. So we must use molten sodium chloride, which means the salt must be heated until it melts. Now a current is free to move through the structure.
This process is called electrolysis of a molten compound.

The half-equations occurring now are:
Cl₂(g) + 2e^- ---> 2Cl^-(aq)            E⁰ = +1.36 V
Na⁺(aq) + e^- ---> Na(s)                                E⁰ = -2.71 V

And the overall equation:
2Cl^-(aq) + Na⁺(aq) ---> Cl₂(g) + Na(s)

And finally the EMF:
EMF = E⁰_oxidant - E⁰_reductant
       = (-2.71) - (1.36)
       = -4.07 V

Now we have our desired products, sodium metal and chlorine gas. Notice that the electricity requirement of this modified cell is higher than that of the cell with aqueous solution.

Practical applications of electrolysis that you'll need to know are Down's cell (basically the example above with the electrolysis of molten sodium chloride), diaphragm/membrane cell (the above example with aqueous sodium chloride - though in reality chlorine gas is evolved, not oxygen), what happens when you change electrodes, and how non-standard conditions among other things can slow the rate of reaction.

[Zebra]

so kind. its a pity that i don't know how to + your respect..

[jane1234]

3. The two species that will react will be the reductant and oxidant that are closest together. So you'll be able to draw a short line that goes from top right to bottom left between the two species reacting (this is the opposite of in galvanic cells).

Um, I'm not sure that's 100% accurate.
I was always under the impression that the strongest oxidant and the strongest reductant will react, even in electrolysis. So it won't necessarily be the two reactions "closest together", it could in fact be the ones furthest apart, just as long as the strongest oxidant reacts with the strongest reductant (given appropriate concentrations).

I'm really confused now, can someone verify this?

[tony3272]

3. The two species that will react will be the reductant and oxidant that are closest together. So you'll be able to draw a short line that goes from top right to bottom left between the two species reacting (this is the opposite of in galvanic cells).

Um, I'm not sure that's 100% accurate.
I was always under the impression that the strongest oxidant and the strongest reductant will react, even in electrolysis. So it won't necessarily be the two reactions "closest together", it could in fact be the ones furthest apart, just as long as the strongest oxidant reacts with the strongest reductant (given appropriate concentrations).

I'm really confused now, can someone verify this?

It still is the strongest oxidant reacting with the strongest reductant. In the case of electrolytic cells though this happens to be the two that are closest together.



Take this electrocemical series for example. If the reactant you had were Al³⁺, H₂O and Pb_(s) then the possible half equations will be -
Oxidation
H₂O -->H₂O₂ + 2H⁺ +2e^-
H₂O --> O₂ + 4H⁺ + 4e^-
Pb_(s) --> Pb²⁺_(aq) +2e^-

Reduction:
Al³⁺ + 3e^- --> Al_s
2H₂O + 2e^- -->H₂ + 2OH^-

In order to determine which reaction will occur you simply, like always, pick the strongest reductant and oxidant.

So in this case these reactions will be:

Reduction: 2H₂O + 2e^- -->H₂ + 2OH^-
Oxidation: Pb_(s) --> Pb²⁺_(aq) +2e^-

Which are the closest together on the series.

In simple terms, for an electrolytic reaction the reduction reaction is at a lower place on the electochemical series than the oxidation reaction ( hence why it is non-spontaneous ). So in order to find the strongest oxidant it goes up ( but still being lower than the oxidation reaction) and in order to find the strongest reductant it goes down the series ( but still being lower than the reduction reaction ). Hence why the reaction that take place will be those that are closest together on the series.

[thushan]

Yup. It is just that when we talk about electrolysis, we generally talk about non-spontaneous reactions occurring. The reaction that will occur is the least "non-spontaneous." Hence, an attempt to find the strongest oxidant and strongest reductant will yield "reactants that as as close to each other as possible in series."

So basically, galvanic, or electrolytic, just find the strongest oxidant, and the strongest reductant. The only difference between the galvanic and electrolytic cell in this case is that in galvanic cells the "gradient" (if you were to draw a line between the two species that are reacting on the electrochemical series) has to be negative, and in electrolytic cells the "gradient" does not have to be negative.