Electrolysis Using electricity to make a "non-spontaneous" chemical reaction take place.

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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 the negative Note: an ox : red cat ---still true Example: chrome plating from a solution of chromium ions 24-1

2 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 CR 2 (g) + 2e - (oxidation at the anode) Na + (R) + e - 6 Na(s) (reduction at the cathode) 24-2

3 Electrolysis in aqueous solution More complicated because water can be electrolyzed as well We have to learn how to decide what reduces and oxidizes 24-3

4 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: 2H 2 O(R) + 2e - 6 H 2 (g) + 2OH - (aq) [reduction at the cathode] 2H 2 O(R) 6 4H + (aq) + O 2 (g) + 4e - [oxidation at the anode] Also, many metal electrodes can be oxidized to ions during electrolysis of aqueous solutions. How to decide what the products of electrolysis 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 Easy shortcut method to decide this 24-4

5 Beautiful example Electrolysis of Aqueous NaF ***--first write down what is in solution*** We have H 2 O(R), Na + (aq) and F - (aq) Data from SRP tables 2H 2 O(R) + 2e - 6 H 2 (g) + 2OH - (aq), EE V 4H + (aq) + O 2 (g) + 4e - 6 2H 2 O(R), EE V F 2 (g) + 2e - 6 2F - (aq), EE V Na + (aq) + e - 6 Na(s), EE V Rearrange to get our possible reactants at the LHS and arrange as possible Oxidations and Reductions To be continued

6 Oxidations? 2H 2 O(R) 6 4H + (aq) + O 2 (g) + 4e -, EE -1.23V 2F - (aq) 6 F 2 (g) + 2e -, EE V Reductions? 2H 2 O(R) + 2e - 6 H 2 (g) + 2OH - (aq), Na + (aq) + e - 6 Na(s), EE V EE V (Anode) Oxidations? 2H 2 O(R) 6 4H + (aq) + O 2 (g) + 4e -, 2F - (aq) 6 F 2 (g) + 2e -, EE V EE V Written as an OX The most easily oxidized species is the one with the most positive potential6this means that in this case water will be oxidized to O 2 (g) (EE V), rather than fluoride ion be oxidized to fluorine gas (-2.87 V). So - the "winner" at the anode is Water 6 oxidized to O 2 (g) 2H 2 O(R) 6 4H + (aq) + O 2 (g) + 4e -, EE V 24-6

7 (Cathode) Reductions? 2H 2 O(R) + 2e - 6 H 2 (g) + 2OH - (aq), Na + (aq) + e - 6 Na(s), EE V EE V The most easily reduced species is the one with the most positive potential 6 this means that in this case water ( EE V ) will be much more easily reduced (to H 2 (g)) than Na + (EE V) (to solid sodium, Na(s)) So - the "winner" at the cathode is Water 6 reduced to H 2 (g) 2H 2 O(R) + 2e - 6 H 2 (g) + 2OH - (aq), EE V Shortcut to remember The reaction with the most positive E wins But -- you must write possible ox as ox and reds as reds 24-7

8 So what actually happens? Water is converted to hydrogen gas and oxygen gas and the Na + (aq) and F - (aq) stay in solution. 2H 2 O(R) 6 4H + (aq) + O 2 (g) + 4e -, 4H 2 O(R) + 4e - 6 2H 2 (g) + 4OH - (aq), EE V EE V Add 6H 2 O(R) 6 4H + (aq) + O 2 (g) + 2H 2 (g) + 4OH - (aq) but: 4H + (aq) + 4OH - (aq) 6 4H 2 O(R) What is E 0? -does it make sense hint: +ve or -ve So overall we simply have: 2H 2 O(R) 6 O 2 (g) + 2H 2 (g) and the sodium fluoride undergoes no chemical change - it simply provides ions to carry the current 24-8

9 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. 4H 2 O(R) + 4e - 6 2H 2 (g) + 4OH - (aq), versus Na + (aq) + e - 6 Na(s), EE V EE V At the anode, we have two possible oxidations: 2H 2 O(R) 6 4H + (aq) + O 2 (g) + 4e -, versus 2CR - (aq) 6 CR 2 (g) + 2e -, EE V EE 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 happens is that Chlorine gas is evolved! Why? 24-9

10 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 CR 2 (g) + 2e -, EE V So the overall electrolysis reaction is: 2CR - (aq) + 2H 2 O(R) 6 CR 2 (g) +H 2 (g) + 2OH - (aq) The electrolysis of brine is used to make Chlorine (and Hydrogen) gas and Sodium Hydroxide solution

11 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 Zn e Zn E 0 = versus 2H 2 O(R) + 2e - 6 H 2 (g) + 2OH - (aq), E 0 = V = with o.p Possible oxidations 2Br - Br 2 + 2e E 0 = versus 2H 2 O(R) 6 4H + (aq) + O 2 (g) + 4e -, E V (-1.73 with o.p) So Zinc is reduced and Br - is oxidized 24-11

12 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 = 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 = x C x 6.02 x mol -1 = C mol

13 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 = I x t 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 24-13

14 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) 24-15

15 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 M CuSO 4 solution? Possible Limiting Reactant Problem!!! How much copper available to deposit? 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 )/96485 x 2 = 5.94 g Cu 24-16

16 example: work through by yourselves: How many grams of Al(s) and Cl 2 (g) are produced if a cell of molten AlCl 3 is electrolyzed with amps for 1.5 hours? m= MIT/Fn Al Cl 2 m = (27 x x 1.5 x 60 x 60)/ x 3 m = (70.9 x x 1.5 x 60 x 60)/ x 2 = g = g (0.309 L) note well: "n" is different for the processes at the anode and cathode 24-16

17 homework 1. How long would it take to produce a solution of ph 2.00 by electrolysis of 500 ml of M AgNO 3 (aq) by using a constant current of Amperes? soln: t = 2.0 x 10 3 s 24-17