Here we’ll work through an example of a type 2 electrolytic cell - Electrolysis of an aqueous...

Here we’ll work through an example of a type 2 electrolytic cell - Electrolysis of an aqueous solution using unreactive or inert electrodes.

Electrolytic Cells

Type 2-Example 1

We’re given that an aqueous solution of copper(II) chloride (CuCl2(aq)) is electrolyzed using inert carbon electrodes.

An aqueous solution of copper(II) chloride (CuCl2(aq)) is electrolyzed using inert carbon electrodes. a) Write the half-reaction taking place at the

cathode (with its E°).b) What is the product at the cathode?c) Write the half-reaction taking place at the

anode (with its E°).d) What is the product at the anode?e) Write the overall redox equation (with its E°)f) What is the minimum voltage required?

We can visualize this solution as mobile copper 2+ ions, chloride ions, and a large number of water molecules, all moving randomly.

Power Supply

+ –

e–

e–

e–

e–

e–

e–

e–

e–

e–

e–

e–

e–

e–

e–

e–

e–

1 M CuCl2(aq)

–+

Carbon Carbon

Anode Cathode

Cl–

Cu2

+

Cl–

Cu2

+ H

HO

H

HO

H

HO

H

HO

H

HO

H

HO H

HO

H

HO

H

HO

H

HO

H

HO

H

HO Cl–H

HO

Cl–

Question (a) asks us to write the half-reaction taking place at the cathode and include its E naught value.




In this solution, we have Cu 2+ ions

An aqueous solution of copper(II) chloride (CuCl2(aq)) is electrolyzed using inert carbon electrodes. a) Write the half-reaction taking place at the cathode (with its E°).

Cu2+ H2O Cl–

Cl minus ions

Cu2+ H2O Cl–


And aqueous means, we have water, which we’ll write in the middle.

Cu2+ H2O Cl–


We write a C minus here for the cathode

Cu2+ H2O Cl–

C–


And an A+ here for the anode.

Cu2+ H2O Cl–

C– A+


At the cathode (click) either copper (II) ions will be reduced, or (click) water will be reduced.

C–?

Cu2+ H2O Cl–

A+

?


Remember, if we have an aqueous solution, (click) for the cathode, we use the overpotential arrow on the left side of the table

For the Cathode

In Aqueous Solution

Any cations ABOVE this overpotential arrow WILL be reduced from an aqueous solution, even though water is present.

Cations ABOVE this arrow

WILL be reduced from aqueous

solution

For the Cathode

In Aqueous Solution

And any cations BELOW this overpotential arrow WILL be NOT reduced from an aqueous solution. If these are present, Water will be reduced instead.

Cations ABOVE this arrow

WILL be reduced from aqueous

solution

Cations BELOW this arrowWill be NOT reduced from

aqueous solution

For the Cathode

In Aqueous Solution

Here’s the black arrow on the left we use for the reduction overpotential of neutral water.


We see that the half-reaction for the reduction of Cu2+ ions to Cu solid is far above the black arrow on the left.


So we’ll place a checkmark by the copper (II) ion indicating that it will be reduced at the cathode, rather than water.

C–?

Cu2+ H2O Cl–

A+

?


The half-reaction for the reduction of Cu2+ is Cu2+ + 2e– Cu(s).


Reduction of Cu2+

And because it is not reversed, it’s E naught value is the same as it is on this table, +0.34 volts


Reduction of Cu2+

We’ll make a note of the cathode half-reaction up here.


Cathode: Cu2+ + 2e– Cu(s) E° = 0.34 V

Question (b) asks us for the product at the cathode.





The half-reaction at the cathode is Cu2+ + 2e– Cu(s) ,

An aqueous solution of copper(II) chloride (CuCl2(aq)) is electrolyzed using inert carbon electrodes. b) What is the product at the cathode?


Cathode: Cu2+ + 2e– Cu(s)

So the product at the cathode is solid copper.

An aqueous solution of copper(II) chloride (CuCl2(aq)) is electrolyzed using inert carbon electrodes. b) What is the product at the cathode?


Cathode: Cu2+ + 2e– Cu(s)

Product at the Cathode is Solid

Copper

Question (c) asks us for the half-reaction taking place at the anode, along with its E naught value.





Let’s go back to this diagram, and this time we’ll concentrate on the anode. Two possible half-reactions at the anode are (click) the oxidation of water and (click) the oxidation of chloride ions.

Cu2+ H2O Cl–

C– A+

An aqueous solution of copper(II) chloride (CuCl2(aq)) is electrolyzed using inert carbon electrodes. c) Write the half-reaction taking place at the anode (with its E°).

? ?

When we’re considering the anode, we go to the right side of the reduction table. (click) In aqueous solutions for the anode half-reaction, we use this overpotential arrow on the right.

For the Anode

In Aqueous Solution

Remember, how it works is any anions BELOW this overpotential arrow on the right WILL be oxidized from aqueous solution at the anode of an electrolytic cell.

For the Anode

In Aqueous Solution

Anions BELOW this arrow

WILL be oxidized from aqueous

solution

And any anions ABOVE this overpotential arrow on the right will NOT be oxidized from aqueous solution. If they are present, water will be oxidized instead.

For the Anode

In Aqueous Solution

Anions BELOW this arrow

WILL be oxidized from aqueous

solution

Anions ABOVE this arrowWill be NOT oxidized from aqueous

solution

We see that that chloride ion is right below the overpotential arrow.

So that means we can put a checkmark by the chloride in this diagram, indicating that the chloride ion will be oxidized at the anode rather than water.

Cu2+ H2O Cl–

C– A+


? ?

So we use this half-reaction and since oxidation is occurring, (click) we reverse it to write the half-reaction at the anode,

Anode: 2Cl– Cl2(g)+ 2e– E° = –1.36 V


It is 2 Cl minus

Anode: 2Cl– Cl2(g)+ 2e– E° = –1.36 V


Gives Cl2 gas

Anode: 2Cl– Cl2(g)+ 2e– E° = –1.36 V


Plus 2 electrons.

Anode: 2Cl– Cl2(g)+ 2e– E° = –1.36 V


Because the half-reaction taking place is an oxidation (click) the E naught value on the table +1.36 V, (click)must have its sign switched, so the E naught value for the half-reaction at the anode, the oxidation of chloride ions, is (click) –1.36 V

Anode: 2Cl– Cl2(g)+ 2e– E° = –1.36 V


So now we’ve answered question c, the half reaction at the anode is 2Cl– Cl2(g)

+ 2e–

and its E naught value is –1.36 Volts.

Anode: 2Cl– Cl2(g)+ 2e– E° = –1.36 V



Question (d) asks us to state the product at the anode.




Cathode: Cu2+ + 2e– Cu(s) E° = 0.34 VAnode: 2Cl– Cl2(g)+ 2e– E° = –1.36 V

To answer this question, we’ll write the anode half-reaction,


An aqueous solution of copper(II) chloride (CuCl2(aq)) is electrolyzed using inert carbon electrodes. d) What is the product at the anode?

Anode: 2Cl– Cl2(g) + 2e–

Anode: 2Cl– Cl2(g)+ 2e– E° = –1.36 V

And we see the product at the anode is chlorine gas.


An aqueous solution of copper(II) chloride (CuCl2(aq)) is electrolyzed using inert carbon electrodes. d) What is the product at the anode?

Anode: 2Cl– Cl2(g) + 2e–

Product at the Anode is Chlorine

Gas

Anode: 2Cl– Cl2(g)+ 2e– E° = –1.36 V

In Question (e) we’re asked to write the overall redox equation along with its E naught value.





We do this by writing the half-reactions at the cathode and anode, with their E naught values.


An aqueous solution of copper(II) chloride (CuCl2(aq)) is electrolyzed using inert carbon electrodes. e) Write the overall redox equation (with its E°)

Anode: 2Cl– Cl2(g) + 2e– E° = –1.36 V

Cathode: Cu2+ + 2e– Cu(s) .E° = 0.34 V

Anode: 2Cl– Cl2(g)+ 2e– E° = –1.36 V

To get the overall redox equation, we add these two half-reactions up.



Anode: 2Cl– Cl2(g) + 2e– E° = –1.36 V


: Cu2+ + 2Cl– Cu(s) + Cl2(g) E° = –

1.02 V

Overall Redox

Equation

Anode: 2Cl– Cl2(g)+ 2e– E° = –1.36 V

We see that the electrons gained by the copper ion are equal to those lost by the chloride ions


Anode: 2Cl– Cl2(g) + 2e– E° = –1.36 V


: Cu2+ + 2Cl– Cu(s) + Cl2(g) E° = –

1.02 V

Overall Redox

Equation

So we can cancel out the electrons.


Anode: 2Cl– Cl2(g) + 2e– E° = –1.36 V


: Cu2+ + 2Cl– Cu(s) + Cl2(g) E° = –

1.02 V

Overall Redox

Equation

On the left side, we have Cu2+ and 2Clminus,


Anode: 2Cl– Cl2(g) + 2e– E° = –1.36 V


: Cu2+ + 2Cl– Cu(s) + Cl2(g) E° = –

1.02 V

Overall Redox

Equation

And on the right side, we have Cu (solid) and Cl2 (gas). You can see that in this electrolysis, we have decomposed the compound CuCl2 into its elements in elemental form


Anode: 2Cl– Cl2(g) + 2e– E° = –1.36 V


: Cu2+ + 2Cl– Cu(s) + Cl2(g) E° = –

1.02 V

Overall Redox

Equation

To get the E naught value for the overall redox reaction, we add 0.34 Volts and negative 1.36 volts, to give us (click) negative 1.02 volts.


Anode: 2Cl– Cl2(g) + 2e– E° = –1.36 V


: Cu2+ + 2Cl– Cu(s) + Cl2(g) E° = –

1.02 V

Overall Redox

Equation

This value, the –1.02 Volts, is called the E naught value for the overall redox reaction and it can also be called the overall cell potential.


Anode: 2Cl– Cl2(g) + 2e– E° = –1.36 V


: Cu2+ + 2Cl– Cu(s) + Cl2(g) E° = –

1.02 V

Overall Redox

Equation

This is called the E° for the redox reaction, or the overall cell potential

The negative value confirms that the reaction is non-spontaneous, which is the case for all electrolytic cells. An external voltage must be applied to keep these operating.


Anode: 2Cl– Cl2(g) + 2e– E° = –1.36 V


: Cu2+ + 2Cl– Cu(s) + Cl2(g) E° = –

1.02 V

Overall Redox

Equation

This is called the E° for the redox reaction, or the overall cell potential

Question (f) asks us to state the minimum voltage required to make this reaction occur.





The minimum voltage required for an electrolytic cell is the positive voltage needed to overcome the negative tendency for the overall redox reaction to occur, so in this case it is (click) + 1.02 volts

An aqueous solution of copper(II) chloride (CuCl2(aq)) is electrolyzed using inert carbon electrodes. f) What is the minimum voltage required?

Anode: 2Cl– Cl2(g) + 2e– E° = –1.36 V


: Cu2+ + 2Cl– Cu(s) + Cl2(g) E° = –

1.02 V

Overall Redox

Equation

Minimum Voltage Required = 1.02 V

It should be noted that even though 1.02 volts is the MINIMUM voltage needed, in actual practice,(click)the voltage used for an effective reaction is considerably higher than this.

An aqueous solution of copper(II) chloride (CuCl2(aq)) is electrolyzed using inert carbon electrodes. f) What is the minimum voltage required?

Anode: 2Cl– Cl2(g) + 2e– E° = –1.36 V


: Cu2+ + 2Cl– Cu(s) + Cl2(g) E° = –

1.02 V

Overall Redox

Equation

Minimum Voltage Required = 1.02 V For an effective reaction, the voltage used is much

higher

Here we’ll work through an example of a type 2 electrolytic cell - Electrolysis of an aqueous...

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