Download Electrolysis 24-1

Electrolysis
Using electricity to make a "non-spontaneous" chemical
reaction take place.
An electrolytic cell consists of two electrodes in a molten salt or
electrolyte solution. A battery or other voltage source is
attached across the electrodes and serves as an "electron
pump" drawing electrons from what would have been the
positive electrode and forcing them in at what would have been
Note: an ox : red cat ---still true
the negative
Example: chrome plating from a solution of chromium ions
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A reduction half-reaction occurs at the electrode (cathode) attached to the
negative terminal of the battery.
An oxidation half-reaction occurs at the electrode (anode) attached to the
positive terminal of the battery.
So for molten sodium chloride:
2CR-(R) 6 CR2(g) + 2e- (oxidation at the anode)
Na+(R) + e- 6 Na(s) (reduction at the cathode)
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Electrolysis in aqueous solution
More complicated because water can be electrolyzed as well
We have to learn how to decide what reduces and oxidizes
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Electrolysis in Aqueous Solution (know this well)
Electrolysis of aqueous solutions is complicated since water can be
oxidized and reduced in addition to (or instead of) the ions present in
solution:
The reactions are:
2H2O(R) + 2e- 6 H2(g) + 2OH-(aq)
[reduction at the cathode]
2H2O(R) 6 4H+(aq) + O2(g) + 4e[oxidation at the anode]
Also, many metal electrodes can be oxidized to ions during electrolysis
of aqueous solutions.
Easy shortcut method to
How to decide what the products of electrolysis
decide this
will be?
Look at all the possible half reactions:
The most easily oxidized species react at the anode
The most easily reduced species react at the cathode
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Beautiful example
Electrolysis of Aqueous NaF
***--first write down what is in solution***
We have H2O(R), Na+(aq) and F-(aq)
Data from SRP tables
2H2O(R) + 2e- 6 H2(g) + 2OH-(aq),
V
4H+(aq) + O2(g) + 4e- 6 2H2O(R),
F2(g) + 2e- 6 2F-(aq),
Na+(aq) + e- 6 Na(s),
EE -0.83
EE +1.23 V
EE +2.87 V
EE -2.71 V
Rearrange to get our possible reactants at the LHS and
arrange as possible Oxidations and Reductions
To be continued…...
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Oxidations?
Reductions?
2H2O(R) 6 4H+(aq) + O2(g) + 4e-, EE -1.23V 2H2O(R) + 2e- 6 H2(g) + 2OH-(aq),
Na+(aq) + e- 6 Na(s),
2F-(aq) 6 F2(g) + 2e-, EE -2.87 V
(Anode) Oxidations?
2H2O(R) 6 4H+(aq) + O2(g) + 4e-,
2F-(aq) 6 F2(g) + 2e-,
EE -0.83 V
EE -2.71 V
Written as an OX
EE -1.23 V
EE -2.87 V
The most easily oxidized species is the one with the most positive potential 6 this
means that in this case water will be oxidized to O2(g) (EE -1.23 V), rather than
fluoride ion be oxidized to fluorine gas (-2.87 V).
So - the "winner" at the anode is Water 6
oxidized to O2(g)
2H2O(R) 6 4H+(aq) + O2(g) + 4e-, EE -1.23 V
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(Cathode) Reductions?
2H2O(R) + 2e- 6 H2(g) + 2OH-(aq),
Na+(aq) + e- 6 Na(s),
EE -0.83 V
EE -2.71 V
The most easily reduced species is the one with the most positive potential
6 this means that in this case water ( EE -0.83 V ) will be much more easily
reduced (to H2(g)) than Na+ (EE -2.71 V) (to solid sodium, Na(s))
So - the "winner" at the cathode is Water 6
reduced to H2(g)
2H2O(R) + 2e- 6 H2(g) + 2OH-(aq), EE -0.83 V
Shortcut to remember
The reaction with the most positive E “wins”
But -- you must write possible ox as ox
and reds as reds
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So what actually happens?
Water is converted to hydrogen gas and oxygen gas and the Na+(aq)
and F-(aq) stay in solution.
2H2O(R) 6 4H+(aq) + O2(g) + 4e-,
4H2O(R) + 4e- 6 2H2(g) + 4OH-(aq),
EE -1.23 V
EE -0.83 V
Add
6H2O(R) 6 4H+(aq) + O2(g) + 2H2(g) + 4OH-(aq)
but:
4H+(aq) + 4OH-(aq) 6 4H2O(R)
What is E0 ?
-does it make sense
hint: +ve or -ve
So overall we simply have:
2H2O(R) 6 O2(g) + 2H2(g)
and the sodium fluoride undergoes no chemical change - it simply
provides ions to carry the current
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Another example
Electrolysis of Aqueous Sodium Chloride:
(The overpotential or overvoltage problem)
When we electrolyze a sodium chloride solution, we see that water is
reduced to hydrogen at the cathode exactly as we had (above) for the case
of NaF electrolysis.
4H2O(R) + 4e- 6 2H2(g) + 4OH-(aq),
versus
Na+(aq) + e- 6 Na(s),
EE -0.83 V
EE -2.71 V
At the anode, we have two possible oxidations:
2H2O(R) 6 4H+(aq) + O2(g) + 4e-,
versus
2CR-(aq) 6 CR2(g) + 2e-,
EE -1.23 V
EE -1.36 V
The argument that we used with NaF leads us to predict that we should get oxygen
gas evolved and that the chloride ion should stay in solution. What actually
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happens is that Chlorine gas is evolved! Why?
The problem is caused by Overpotential - the difference between the actual
potential to make a reaction take place and what we would predict from
tables of SRP values.
Like "an activation energy" to make the reaction occur :size of
overpotential depends on the electrode material and the reaction
Overpotential effect we find experimentally that we need to supply e.g. 1.8
V to get water to electrolyze - and this is more than enough to oxidize
chloride ion to chlorine gas:
2CR-(aq) 6 CR2(g) + 2e-,
EE -1.36 V
So the overall electrolysis reaction is:
2CR-(aq) + 2H2O(R) 6 CR2(g) +H2(g) + 2OH-(aq)
The electrolysis of brine is used to make Chlorine (and Hydrogen) gas and
Sodium Hydroxide solution.
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Example:a 1M aq. soln of ZnBr was electrolyzed using inert electrodes.
What are the products? There was a 0.5V overpotential for both oxidation
and reduction of water.
Remember to only consider reactions to things that are already in solution
Possible reductions
Zn2+ + 2e
E0 = - 0.67
Zn
versus
2H2O(R) + 2e- 6 H2(g) + 2OH-(aq), E0 = -0.83 V = - 1.33 with o.p
Possible oxidations
2Br-
E0 = - 1.065
Br2 + 2e
versus
2H2O(R) 6 4H+(aq) + O2(g) + 4e-,
E0 -1.23 V (-1.73 with o.p)
So Zinc is reduced and Br- is oxidized
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Quantitative Aspects:
Faraday's Laws of Electrolysis:
Electrolysis - Using electricity to make a
"non-spontaneous" chemical reaction take place.
"The number of moles of product formed in an Electrolytic Reaction is
proportional to the number of moles of electrons that passed through the
solution."
The Faraday Constant, F = 96485 C mol-1
relates the amount of charge which passes through a solution to the number of moles
of electrons that were involved in the redox reaction.
Each electron can reduce a cation of one +ve charge
so two will reduce two and so on: a mole of electrons will reduce a mole of C+
Charge in a mole of electrons = charge on 1 electron X avogadros number
= 1.602 x 10-19 C x 6.02 x 1023 mol-1
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= 96458 C mol
Units and Equations
1 Volt = 1 Joule per Coulomb = 1 J C-1
Coulomb: The S.I. unit of charge.
Current: Measured in Amperes, A
Current = charge / time (IN SECONDS)
units = C s-1
or
Coulombs = Amperes x time
Q=Ixt
Q = Charge in Coulombs,
I = current in Amperes
t = time in Seconds
e.g. if 1.50 A flows for 5.00 min then the charge
Z = (1.50 A)(5.00 x 60 sec) = 450 C
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From Faraday's Laws
The number of moles of electrons used in electrolysis is
equal to the total charge, Q, divided by the Faraday Constant:
Moles of Electrons = Q (C) / F(C mol-1)
= It/F
Mass produced per mole of electrons m = M/n
(M = molar mass, n = number of electrons to produce 1 formula
unit of the substance).
Combine these to get
Good one to remember
m = (ItM)/Fn
i.e. Mass deposited or evolved =
(mole of e) x (mass produced per mole of e)
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example
What mass of Cu is deposited when a current of
10.0 A is passed for 30.0 minutes through 1.00 L of a 0.100 M CuSO4 solution?
Possible Limiting Reactant Problem!!!
How much copper available to deposit?
0.100 mol = 6.35 g
i.e. no matter how much current you pass through the solution, you
cannot get more than 6.35 g Cu from it!
m = (I T M)/Fn
m = (10.0 A x 30.0 x 60 sec x 63.546 )/96485 x 2
= 5.94 g Cu
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example: work through by yourselves:
How many grams of Al(s) and Cl2(g) are produced if a cell of molten AlCl3 is
electrolyzed with 0.452 amps for 1.5 hours?
m= MIT/Fn
Al
m = (27 x 0.452 x 1.5 x 60 x 60)/ 96485 x 3
= 0.228 g
Cl2
m = (70.9 x 0.452 x 1.5 x 60 x 60)/ 96485 x 2
= 0.896 g
(0.309 L)
note well: "n" is different for the processes at the anode and cathode
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homework
1. How long would it take to produce a solution of pH
2.00 by electrolysis of 500 mL of 0.100 M AgNO3(aq) by
using a constant current of 0.240 Amperes?
soln: t = 2.0 x 103 s
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