18.11 Electrolysis and Electrolytic Cells We ve seen that galvanic cells can give us useful electricity from spontaneous chemical reactions. The rever...
18.11 Electrolysis and Electrolytic Cells We ve seen that galvanic cells can give us useful electricity from spontaneous chemical reactions. The reverse of every spontaneous chemical reaction is non-spontaneous. However, if we apply electric current to a chemical system, it is possible to force non-spontaneous chemical reactions occur in a process called electrolysis, in what we call electrolytic cells. The potential we apply to the system must be greater than that for the spontaneous reaction, and must be applied in the opposite direction.
The reaction below is not spontaneous since Ecell is negative Cu(s) + Fe2+ (aq) Cu2+(aq) + Fe(s) Ecell = -0.79V The reaction can be forced to proceed in the direction indicated by applying an external voltage larger than +0.748V. The cell reaction of this type is called electrolysis
Electrolysis of Molten sodium Chloride Figure 18.15 shows us an electrolytic cell from which we get sodium from molten sodium chloride (NaCl (l) ). In many ways, it looks like a galvanic cell, except that we do not connect two half-cells with a salt bridge. Instead we place both electrodes in the same container. The other difference is that we place a battery in the circuit to force electrons in the direction opposite to that in which they would normally flow. In this case the negative terminal of the battery forces electrons into an electrode which attracts the Na+ ions and reduces them to metallic sodium. At the other electrode, Cl- ions are attracted to a positive electrode, where they are oxidized into Cl2 gas.
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Anode (+) Cathode (-)
2 Cl- (l) Cl2 (g) + 2 e+ 2 Na (l) + 2 e- 2 Na (s)
Overall 2 Cl- (l) + 2 Na+ (l)
Cl2 (g) + 2 Na (s)
-1.36 V -2.71 V Ecell = - 4.07 V Eappl = > + 4.07 V
Since the anode is the electrode where the oxidation takes place, while the cathode is where reduction takes place, The charge on these electrodes is reversed from a galvanic cell! The anode is positive, while the cathode is negative. To make the electrolysis reaction go we need to apply an external voltage of > +4.07 V
Electrolysis of aqueous NaCl When aqueous NaCl is electrolyzed there is now, in addition to the Na+ and Cl- ions, water present that also may be oxidized or reduced. The possible cathode reactions are: Na+ + e Na Eored = -2.71V 2H2O + 2e H2(g) + 2OH-(aq) Eored = -0.83V It is much easier to reduce water than Na+ ions so the only product at the cathode is H2 gas. The possible anode reactions are: 2Cl- (aq) Cl2(g) + 2e Eoox = -1.36V H2O O2(g) + 4H+(aq) + 4e Eoox = -1.23V At first glance it appears that it should be easier to oxidize water than Cl- ion but the E values required are not standard values. But the deciding factor is the overvoltage, the extra voltage that must be applied to get it to occur at the rate at which it would occur in an ideal system. This can be as high as 1V for the oxidation of water. Thus by careful control of conditions we can ensure that Cl2 is produced. Cell reaction:
2NaCl(aq) + 2H2O
2Na+(aq) + 2OH-(aq) + H2(g) + Cl2(g)
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Electrolysis of water to produce H2(g) and O2(g) .....this is usually done with a non reacting electrolyte added to the water (pure water has too high an electrical resistance to permit a reasonable current flow). Usually something like Na2SO4 is added (SO42- ion is harder to oxidize than water and Na+ is harder to reduce than water) cell reaction: 2H2O (l) 2 H2(g) + O2(g) anode: 2H2O (l) O2(g) + 4H+(aq) + 4e cathode 4H2O (l) + 4e 2H2(g) + 4OH-(aq) + also: 4H (aq) + 4OH (aq) 4H2O(l) ---------------------------------------------------2H2O (l) 2 H2(g) + O2(g)
This reaction is not spontaneous since Eocell is negative. An applied voltage Eapplied > +2.057V is required to drive the reaction in the direction indicated.
Problem 18.18 Predict the half-cell reactions that occur when aqueous solutions of the following salts are electrolysed in a cell with inert electrodes. What is the overall cell reaction in each case? a) LiCl Since we are performing the electrolysis in water, there are two possible half reactions at each electrode. At the cathode (reduction reactions), they are Li+ (aq) + e- Li (s) 2 H2O (l) + 2 eH2 (g) + 2 OH- (aq)
E = -3.04 V E = -0.83 V
At the anode, the potential oxidation reactions are 2 Cl- (aq) 2 H2O (l)
Cl2 (g) + 2 eO2 (g) + 4 H+ (aq) + 4 e-
E = -1.36 V E° = -1.23 V
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Combining these half-reactions, we can identify four possible reactions 2 Li+ (aq) + 2 Cl- (aq) Li (s) + Cl2 (g) 4 Li+ (aq) + 2 H2O (l) 4 Li (s) + O2 (g) + 4 H+ (aq) 2 H2O (l) + 2 Cl- (aq) H2 (g) + 2 OH- (aq) + Cl2 (g) 6 H2O (l) 2 H2 (g) + 4 OH- (aq) + O2 (g) + 4 H+ (aq)
E E E E
= -4.40 V = -4.27 V = -2.19 V = -2.06 V
Since the last two reactions are the same as we saw for the electrolysis of aqueous sodium chloride, (overvoltage occurring for oxid. of water) we know the reaction will be 2 H2O (l) + 2 Cl- (aq)
H2 (g) + 2 OH- (aq) + Cl2 (g)
E = -2.19
b) CuSO4 Ans: done in class
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