Download The electrolytic cell contains two electrodes and either a molten salt

The electrolytic cell contains two electrodes and either a molten salt (as with
NaCl) or some other type of solution (e.g. aqueous).

The external EMF is provided by a source of direct current (sometimes
another voltaic cell, battery, or power supply)

This external current acts as an electron pump
Opposite
sign
convention
to voltaic
cells!
o
o
The electrode where oxidation occurs is still called the: __________
However, for an electrolytic cell it is labeled as: _______________ .
The electrode where reduction occurs is still called the: __________
However, for an electrolytic cell it is labeled as: _______________ .
For molten NaCl

Anode Reaction:

Cathode Reaction:

Overall Rxn:
2Cl-(aq) Cl2(g) + 2e- E0ox =
Na+(aq) + e- Na(s) E0red =
This means that an outside power source of ________
would need to be applied to force this non-spontaneous reaction to happen.
Electrolysis of Aqueous Solutions
With aqueous solutions of salts it must be considered that the electrolysis
might end up driving the reduction or oxidation of H2O(l) instead.

H2O(l) can be oxidized to produce O2(g)

H2O(l) can be reduced to produce H2(g)

If a salt is present,
the cation of the salt may be preferentially reduced compared to H2O(l).
Likewise, the anion may be preferentially oxidized compared to H2O(l)
It all depends on which one takes _____________!
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For the electrolysis of aqueous NaCl, here are the reduction potential data for the
2 possible reduction ½ reactions of H2O(l) vs Na+(aq) at the cathode:
(Note: Water value is NOT E0 because hydroxide is NOT 1M!) (E0 = -0.828V)
2H2O(l) + 2e-  H2(g) + 2OH-(aq) E red =
Na+(aq) + e- Na(s) E red =

When comparing any two reduction reactions the reaction that is favored is
the one with the more positive (or less negative) reduction potential.
In this case: ______________________________________
Here are the oxidation potential data for the possible oxidation ½ reactions of
H2O(l) vs Cl-(aq) at the anode:
(Note: Water value is NOT E0 because H+ is NOT 1M!) (E0 = 1.229V)
=
2Cl-(aq) Cl2(g) + 2e- Eox =
In this case: ________________________________
2H2O(l) 4H+(aq) + O2(g) + 4e- Eox

(It has the less negative potential.)

This analysis predicts that electrolysis of an aqueous solution of Na+(aq)
and Cl-(aq) ions would not result in the reduction of Na+ or oxidation of
Cl-, but rather, will result in
the reduction of H2O to produce hydrogen, and
the oxidation of H2O to produce oxygen.
Note: although this analysis is correct, there is another effect that can happen,
known as: ____________________________
More voltage is sometimes required, beyond the voltage predicted by the Ecell.
This may be due to kinetic rather than thermodynamic considerations.
In the case of the oxidation of H2O versus Cl- (where voltages are close),
the oxidation of H2O to produce O2(g) is thermodynamically favored,
but is kinetically quite slow. (It seems like nothing is happening)
Thus we could try to increase the voltage, to get a noticeable reaction.
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However, the oxidation of chlorine occurs at a voltage only slightly higher
than that for water. Thus, in practical terms, ___________________
______________________________________________

The redox reactions for the electrolysis of NaCl(aq) are therefore:
(reduction) 2H2O(l) + 2e-  H2(g) + 2OH-(aq)
(oxidation) 2Cl-(aq) Cl2(g) + 2e-

And the overall reaction is:
Thus, while electrolysis of molten NaCl produces chlorine gas and sodium
metal, the electrolysis of an aqueous solution of NaCl produces chlorine gas,
hydrogen gas and hydroxide ions.

What external EMF is required do drive this particular electrolysis?
(reduction; cathode) 2H2O(l) + 2e-  H2(g) + 2OH-(aq) E0red =
(oxidation; anode)
2Cl-(aq) Cl2(g) + 2e-
E0red =
E0cell = E0(cat) + E0(an) =
This value is negative, indicating that the reaction is:
as written and must be driven by an external EMF
LP#10. Determine the anode and cathode reactions for the electrolysis of aqueous
LiF. Determine that voltage that would be required to drive the reaction.
The possible oxidation ½ reactions at the anode:
2H2O(l) 4H+(aq) + O2(g) + 4e-
Eooxid =
The possible reductions ½ reactions at the cathode:
2H2O(l) + 2e-  H2(g) + 2OH-(aq) Eored =
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Write the overall reaction by summing the two preferred energy ½ reactions:
2H2O(l) 4H+(aq) + O2(g) + 4e-
Eooxid = -(0.82v)
4H2O(l) + 4e-  2H2(g) + 4OH-(aq) Eored = -0.41v
Eliminating duplícate waters on each side:
This will require _______ or more to drive the reaction.
Electroplating
Many metals are protected from corrosion by plating them with other metals.
Zinc is often plated onto the surface of an iron nail to protect it from rusting.
(i.e. __________________ nails)
This process often uses _____________________
Problem: Suppose we wanted to put a plating of nickel onto another metal.
A solution of nickel ions OR a molten sickle salt would be required.
Can we use the aqueous solution?
The possible cathode reactions for the aqueous solutions are:
Ni2+(aq) + 2e- 
2H2O(l) + 2e- 

In this case, the half-reaction with the easier to drive reduction potential is
the reduction of Nickel. Therefore, at the cathode, _________
will preferentially be reduced if we
provide an external EMF that forces
electrons onto this electrode

This means that we CAN do
electrolysis from an aqueous solution
and do not need to work with a molten
salt.
19-28
5. Quantitative Aspects of Electrolysis
Half-reactions tell us, the number of e- needed per mole of metal reduced
Na+ + e-  Na
Cu2+ + 2e-  Cu
Al3+ + 3e-  Al
1 mol e- produces __ mol Na
2 mol e- produces __ mol Cu
3 mol e- produces __ mol Al
For any half-reaction the amount of substance reduced is proportional to
the number of electrons that pass into the cell.
Remember:
Charge of 1 mole of electrons =
1 amps = 1 Coulombs
1 second
rearranged
Coulombs =
LP#11. Calculate the mass of copper produced in 1.50 hours by the electrolysis of
molten CuCl2 if the electrical current is 12.3 amps.
Step 1: Find the amount of charge transferred.
The charge in Coulombs = amps * seconds
C=
Therefore, 6.64 x 104 Coulombs (total) pass into the reduction cell (cathode)
during the reaction.
Step 2: Convert this to mol of e- transferred.
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Step 3: Convert to mol of element reduced.
Cu2+ + 2e-  Cu
(it takes 2 moles of e- to reduce 1 mole of copper ion)
Step 4: Convert to mass of element reduced.
Therefore, 21.8 grams of Copper will be produced.
19-30
COMMERCIAL VOLTAIC CELLS (aka batteries)
Lead-Storage Battery
The redox half-reactions in a lead-storage (lead acid) battery are as follows:

Cathode: PbO2(s) + HSO4-(aq) + 3H+(aq) + 2e-  PbSO4(s) + 2 H2O(l) (+1.68v)

Anode: Pb(s) + HSO4-(aq)  PbSO4(s) + H+(aq) + 2e-
(+3.56v)

Overall redox reaction:
+2.04v
PbO2(s) + Pb(s) + 2 H+(aq) + 2 HSO4-(aq)  2 PbSO4(s) + 2 H2O(l)

Sulfuric acid provides the protons and bisulfate ions:
2H2SO4(aq)  2HSO4-(aq) + 2H+(aq)

The solid electrodes (Pb and PbO2) do not produce soluble ions.
In both cases, PbSO4(s) is formed and remains attached as a solid to the
electrode(s). Thus, ions do not diffuse from one half-cell to the other.
Therefore, the two electrodes can be placed in the same container of acid.

Water is produced and sulfuric acid is consumed during the reaction.

The EMF per "cell" under standard conditions is:
_____ 2.04V
Question: If a lead storage cell only produces about 2 volts, how does our car
battery produce 12 volts?
6 cells can be combined end-to-end (i.e. in series) to produce about 12 V.
19-31

This is a reversible reaction. If electrical current is applied in the opposite
direction (this is the job of a generator or alternator in your car) the
electrodes are regenerated.
2PbSO4(s) + 2H2O(l)  Pb(s) + PbO2(s) + 4H+(aq) + 2SO42-(aq)
Dry Cell
These are your basic (not alkaline) type battery.
It was invented in 1866 by Leclaché.
It consists of
Anode: a zinc can as the anode
Cathode: a graphite rod in the center
surrounded by a paste of manganese dioxide,
ammonium & zinc chlorides
A simple version of the half-reactions is as follows:
Cathode:

2NH4+(aq) + 2MnO2(s) + 2e-  Mn2O3(s) + 2NH3(aq) + H2O(l)
Anode:

Zn(s)  Zn2+(aq) + 2e
The MnO2 solid precipitates out on the surface of the inert graphite

This is not reversible, so the battery cannot be recharged (the cathode
reaction products diffuse throughout the cell)
Alkaline dry cell

In an "alkaline" type battery the ammonium chloride is replace by potassium
hydroxide (KOH).

This provides more useable voltage and greater capacity than the typical dry
cell

Cathode:

Anode:
H2O(l) + 2MnO2(s) + 2e- . ___________________
Mn2O3(s) + 2OH-(aq) + H2O(l) ____
Zn(s) + 2OH-(aq)  Zn(OH)2(aq) + 2e- ___________
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Nickel Cadmium (Nicad cell)
This type of cell uses a cadmium anode and a cathode of NiO(OH)(s) in an electrolyte of KOH.

Cathode: 2NiO(OH)(s) + 2H2O(l) + 2e-  2Ni(OH)2(s) + 2OH-(aq) _

Anode:

The solid products do not diffuse throughout the cell.

Thus, the redox reaction is reversible (i.e. like the lead acid cell, the nickel
cadmium cell is reversible)

No gases are produced, so the cell can be sealed and is rechargeable!
Cd(s) + 2OH-(aq)  Cd(OH)2(s) + 2e- _________________
These have become less common in recent years because of Cd toxicity, disposal problems, and the
tendency to develop “memory” issues requiring complete discharge before recharging.
Nickel Metal Hydride (NiMH) battery
This uses the same cathode reaction as the NiCad battery, but a different anode reaction.
In the anode, hydrogen atoms held in a metal alloy are oxidized.

Cathode: 2NiO(OH)(s) + 2H2O(l) + 2e-  2Ni(OH)2(s) + 2OH-(aq) _

Anode:

These were also an improvement because they carried twice the energy for
the same weight and are still rechargeable.
MH(s) + OH-(aq)  M(s) + 2H2O(l) + e- ______________

Lithium Ion Battery
The newest and most expensive type of
rechargeable battery. It has the highest energy
density. It works differently than the other
batteries we have studied so far and the details
of its operation are beyond the scope of this
course. A general overview of how it works it
described on page 916-917 of Tro (4th Ed).
19-33
Fuel Cells


A fuel cell is essentially a battery with a continuous supply of reactants
A common reaction being utilized in fuel cells is the reduction of oxygen
by hydrogen
.

o
Cathode: O2(g) + 2H2O(l) + 4e-  4OH-(aq) ___
o
Anode:
o
Overall reaction:
2H2(g) + 4OH-(aq)  4H2O(l) + 4e- ____
2H2(g) + O2(g)  2H2O(l) ____
This is currently a very expensive way to generate energy, but is extremely
efficient and compact.
It's broadest application to date has been to provide electricity (and drinking
water) for spacecraft.
19-34