Electrolysis 3

Electrolysis An electrolysis is the inverse of an electrochemical cell. A non-spontaneous reaction is caused by the passage of an electric current through a solution. By passing a DC current through the an electrolyte, the reaction can be made to proceed in the reverse or non-spontaneous direction 4

Electrolysis The reactions at the anode and cathode depend on the relative reduction potentials of the solute and the solvent. The substance produced at the cathode depends on the cation that has the higher (more positive) reduction potential The substance produced at the anode depends on the cation that has the lower (more negative) reduction potential 5

Electrolysis of Molten NaCl If sodium chloride is heated to its melting point, then the resulting liquid contains mobile ions. This is a way of producing sodium metal. 6

Electrolysis of concentrated sodium chloride solution The electrolysis of brine solution results in the reduction of water to hydrogen gas rather than sodium ion to sodium metal 7

Electrolysis of Water The electrolysis of water requires a small amount of sulfuric acid to be added. Hydrogen and oxygen are produced in a 2 to 1 ratio. 8

Electrolysis of copper sulfate with a copper electrodes To electroplate a metal, the object to be plated is made the cathode and the metal to be plated is the anode. The electrolyte is a solution containing the cation to be plated. 9

Electrolysis of Water Water itself can be readily electrolysed Since Kw of water is very low, H+ is added to help carry current Anode (oxidation, moves to left): ½ O2(g) + 2 H+(aq) + 2e- ⇌ H2O(l) Eo = +1.23V Cathode (reduction, moves to right): 2 H2O(l) + 2e- ⇌ H2(g) + 2 OH-(aq) Eo = -0.83 V Overall*: 2 H2O(l)  2 H2(g) + O2(g) *Note: the reverse of this is the reaction that takes place in a hydrogen fuel cell and has an Eocell of 2.06 V.

Electrolysis of Aqueous Solutions Electrolysis of molten salts is simple: Cathode  Metal Anode  Non Metal Electrolysis of aqueous solutions is less straightforward due to the fact that the water can compete with the salts to undergo electrolysis. At the Anode: If an anion has an Eo more positive than +1.23 V, water will be oxidised instead of the anion Bubbles of O2 gas will be formed and H+ ions will enter solution This happens for anions including (but not only): fluoride and sulphate Note: Although chlorine has a potential of +1.36 V, it WILL be discharged in aqueous solution., contrary to what you might expect. The reasons for this are complex and are to do with changes in Cl- concentrations at the anode once current starts to flow. At the Cathode: If a cation has an Eo more negative than -0.83 V, water will be reduced instead of the cation Bubbles of H2 gas will be formed and OH- ions will enter solution This happens for cations including (but not only): lithium, potassium, sodium and magnesium

Example 1: What are the products of the electrolysis of aqueous copper sulphate? Consider the standard electrode potentials: At the cathode: 2 H2O(l) + e- ⇌ H2(g) + 2 OH-(aq) -0.83 V Cu2+(aq) + 2e- ⇌ Cu(s) +0.34 V Copper less negative than water so Cu2+ is reduced, cathode gains coating of metallic copper At the anode: ½ O2(g) + 2 H+(aq) + 2e- ⇌ H2O(l) +1.23 V S2O82–(aq) + 2e– ⇌ 2 SO42– +2.01 V Sulphate is more positive than water so water is oxidised, bubbles of O2 gas produced and solution becomes acidic Overall reaction: Cu2+(aq) + SO42-(aq) + 2H2O(l)  Cu(s) + SO42-(aq) + 2H+(aq) + O2(g)

Example 2: Electrolysis of Sodium Chloride Consider the standard electrode potentials: At the cathode: 2 H2O(l) + e- ⇌ H2(g) + 2 OH-(aq) -0.83 V Na+(aq) + e- ⇌ Na(s) -2.71 V Sodium more negative than water so water is reduced, cathode produces bubbles of H2 gas and solution becomes more alkaline At the anode: ½ O2(g) + 2 H+(aq) + 2e- ⇌ H2O(l) +1.23 V ½ Cl2(g) + e- ⇌ Cl- +1.36 V Chlorine is more positive than water, so you expect it water to be oxidised, but chloride is due to reasons stated earlier. Overall reaction: 2Na+(aq) + 2Cl-(aq) + 2H2O(l)  2Na+(aq) + 2OH-(aq) + H2(g) + Cl2(g)

Predict the products at the anode and cathode for electrolysis of the following aqueous solutions. For each one, write the overall equation for the reaction. KCl NiSO4 PbI2 ZnCl2 LiOH

ELECTROLYTE At the CATHODE At the ANODE Molten Sodium Bromide   Concentrated solution Potassium Iodide Molten Copper(II) Iodide Concentrated solution of Iron Sulfate Molten Zinc Bromide Concentrated solution of Magnesium Nitrate Concentrated solution of Silver chloride Dilute solution of Silver Chloride Concentrated solution of Silver Sulfide

ELECTROLYTE At the CATHODE At the ANODE Molten Sodium Bromide  Sodium Bromine Concentrated solution Potassium Iodide  Hydrogen  Iodine Molten Copper(II) Iodide  Copper Iodine  Concentrated solution of Iron Sulfate  Iron  Oxygen Molten Zinc Bromide  Zinc Bromine  Concentrated solution of Magnesium Nitrate Concentrated solution of Silver chloride  Silver  Chlorine Dilute solution of Silver Chloride Concentrated solution of Silver Sulfide

Summary Questions

Answers

Aqueous Electrolysis

Factors that affect electrolysis products Charge on the ion Less mass produced when an ion must be reduced from 2+ compared to 1+ because it takes 2 times as many electrons Current Higher current = more mass produced b/c more electrons transferred in given time Time Isn’t this self explanatory?

Current depends on Voltage and Resistance Electricity Terms Ohm’s Law: V = I R Current depends on Voltage and Resistance Charge (Q) Units: Coulomb 1 electron carries 1.6 x10-19 Coulombs Voltage (V) Units: Volts or Joules/Coulomb This measure the energy transferred per unit of charge that goes through the wire. Example: 1 electron goes through the wire when the voltage is 1 volt: 1.6 x 10-19 Joules of energy was released (fyi: this is an electron volt (eV)!) Current (I) Units: Charge / time : 1 Coulomb/sec = 1 Ampere (amp) The faster electrons move through the wire, the higher the current Resistance (R) Limits how fast electrons can move through the wires do to the chemical make-up of the material. Some materials are good conductors (low resistance) and some are poor conductors (high resistance) Resistance in a cell can be affected by temperature, concentration, length of wires, etc.

Cell Potential Cell potential is measured in volts (V). J 1 V = 1 C C = Coulomb, unit of charge, 1 electron = 1.6x10-19 C

Summary of equations V = J/C C = current x time (Faraday’s First Law) 1 mole = 96,500 C (Faradays constant,F ) The charge of one electron = 1.6 x 10-19 C Use mole ratios for electrons (just as you do for quantitative chemistry) (Faradays Second Law) G = −nFE Reaction is spontaneous: ΔG < 0

Example Step 1 Charge = current*time 2*(15*60) =1800C Step 2 1 mole = 96,500 C 1800/96500 =0.01865 moles of electrons Step 3 Mole ratio Cu2+ +2e  Cu 0.01865/2 =0.00933mol Cu -0.00933*63.55 =0.593g of Cu Answer part (b) – 1.19g of Cu

Quanititive Electrolysis

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