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Let’s read! Pages 82 to 89. Objectives. To know how to carry out electrolysis experiments. To work out what happens to ions at each electrode. To be able to write successful half-equations. To predict the products of electrolysis. To do electrolysis calculations using the idea of a Faraday.
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Let’s read! Pages 82 to 89
Objectives • To know how to carry out electrolysis experiments. • To work out what happens to ions at each electrode. • To be able to write successful half-equations. • To predict the products of electrolysis. • To do electrolysis calculations using the idea of a Faraday
Electrolysis Lead bromide
Copy please Electrode - an electrical conductor which carries charge to or from a liquid undergoing electrolysis. Electrolysis Electrolysis is the break-down of a substance by electricity Electrolyte- a molten or aqueous solution through which an electrical current can flow. Lead bromide
Electrolysis experiments • Electrolysis only happens in: • - molten ionic liquids or • - aqueous solutions containing ions. • There must be a complete circuit. • A lamp or ammeter shows that electricity is flowing around the circuit.
Moving charges Can you stick the sheet in your book?
Draw the sentence • Electrolytes contain positive and negative ions. • During electrolysis, positive and negative electrodes are put into the electrolyte. • The positive electrode is called the anode. • The negative electrode is called the cathode. • The negative ions (called anions) are attracted to the anode. • At the anode, the negative ions lose electrons to become atoms/molecules. • The positive ions (called cations) are attracted to the cathode. • At the cathode, the positive ions gain electrons to become atoms/molecules
Draw the sentence • Electrolytes contain positive and negative ions. • During electrolysis, positive and negative electrodes are put into the electrolyte. • The positive electrode is called the anode. • The negative electrode is called the cathode. • The negative ions (called anions) are attracted to the anode. • At the anode, the negative ions lose electrons to become atoms/molecules. • The positive ions (called cations) are attracted to the cathode. • At the cathode, the positive ions gain electrons to become atoms/molecules
Cathode (-) (negative electrode) Positive ions go here (cations). As metal ions are positive, they go to the cathode. Ions gain electrons. They are reducedand become neutral atoms. Anode (+) (positive electrode) Negative ions go here (anions). As non-metal ions are negative, they go to the anode. Ions lose electrons. They are oxidised and become neutral atoms (which react together to form molecules). At the electrodes
Common ions • Li+,Na+,K+, • Mg2+,Ca2+ , Zn2+ • Cu+,Cu2+,Fe2+, Fe3+, Al3+ • NH4+ (ammonium ion) • F-, Cl-, Br-, I- • O2-, S2- • OH- (hydroxide ion), CO32-, NO3- (nitrate ion), SO42- (sulphate ion)
Questions! • Let’s try some easy questions
Half equations • Show what happens at each electrode. • Are balanced equations. • Consider the electrolysis of copper chloride: Cu2+ + 2e- Cu 2Cl- - 2e- Cl2
Let’s read again! Page 92 to 95 (Chemistry for You)
For solutions of highly reactive metals: Hydrogen gas, not the metal, is produced at the cathode. Electrolysis of solutions –Cathode
The product at the anode depends on: The negative anions present in the solution. Electrolysis of solutions – Anode
Electrolysis The Faraday
Objectives • recall that one Faraday represents one mole of electrons • calculate the amounts of the products of the electrolysis of molten salts and aqueous solutions
The Faraday • A Faraday is one mole of electrons, and is equivalent to 96 500C (Coulombs) • A current of 1A = 1C per second flowing • For example, Cu2+ + 2e Cu 1 mole of Cu2+ ions reacts with 2 Faradays of electrons, to produce 1 mole of Cu
Quantity of electricity in coulombs = current in amps x time in seconds • Q (C) = I (A) x t (s)
Example How much copper is deposited if a current of 0.2 Amps is passed for 2 hours through a copper(II) sulphate solution ?
current of 0.2 Amps is passed for 2 hours At the cathode: Cu2+(aq) + 2e- Cu(s) Q = I x t = 0.2 x (2 x 60 x 60) =1440 Coulombs 1 mole electrons = 96500 Coulombs So, 1440 / 96500 = 0.0149 moles of electrons (Faradays)
Example moles of electrons passed through circuit = 0.01492 Cu2+(aq) + 2e- Cu(s) • From equation, it takes two moles of electrons to form one mole of copper • moles copper = 0.01492 / 2 = 0.00746
Example • moles Cu = 0.00746 • mass of Cu = moles x Ar = 0.00746 x 64 = 0.4775g of Cu deposited.
‘How To” Guide • Write out relevant half equation • Work out coulombs of electrons flowing (Q = It) • Convert C into moles of electrons (Faradays) (Q/96500) • Work out moles of product using ratio from equation • Convert into mass (mass = moles x Ar)
In an electrolysis of sodium chloride solution experiment a current of 2 A was passed for 2 minutes. • (a) Calculate the volume of chlorine gas produced. • (b) What volume of hydrogen would be formed? • (c) In practice the measured volume of chlorine can be less than the theoretical value. Why?
Electrode equations: • (-) cathode 2H+ + 2e- H2 • (+) anode 2Cl- Cl2 + 2e • (a) Calculate the volume of chlorine gas produced. • Q = I x t, so Q = 2 x 2 x 60 = 240 C • 240 C = 240 / 96500 = 0.002487 mol electrons • this will produce 0.002487 / 2 = 0.001244 mol Cl2 (two electrons/molecule) • vol = mol x molar volume = 0.001244 x 24000 = 29.8 cm3 of Cl2
(b) What volume of hydrogen would be formed? • 29.8 cm3 of H2 because two electrons transferred per molecule, same as chlorine. • (c) In practice the measured volume of chlorine can be less than the theoretical value. Why? • chlorine is moderately soluble in water and also reacts with the sodium hydroxide formed.
In the electrolysis of molten sodium chloride 60 cm3 of chlorine was produced. • Calculate ... • (a) how many moles of were chlorine produced? • (b) what mass of sodium would be formed? • (c) for how long would a current of 3 A in the electrolysis circuit have to flow to produce the 60cm3 of chlorine?
(a) how many moles of chlorine produced? • 60 / 24000 = 0.0025 mol Cl2
(b) what mass of sodium would be formed? • from the electrode equations 2 mol sodium will be made for every mole of chlorine • so 0.0025 x 2 = 0.005 mol sodium will be formed. Ar(Na) = 23 • mass = mol x atomic or formula mass = 0.005 x 23 = 0.115g Na
(c) for how long would a current of 3 A in the electrolysis circuit have to flow to produce the 60cm3 of chlorine? • To produce 0.0025 mol of Cl2 you need 0.005 mol of electrons • 0.005 mol electrons = 0.005 x 96500 coulombs = 482.5 C • Q = I x t, so 482.5 = 2 x t, therefore t = 482.5 / 3 = 161 s (to nearest second)
Brine? Brine is salty water (sodium chloride solution)
Electrolysis of brine • Hydrogen is produced at the cathode • Chlorine is produced at the anode • The solution remaining is sodium hydroxide
Electrolysis of brine • Hydrogen is produced at the cathode • Chlorine is produced at the anode • The solution remaining is sodium hydroxide Cathy’s Ankles (CatHy’s AnCl)
Electrolysis of brine • Hydrogen is produced at the cathode • Chlorine is produced at the anode • The solution remaining is sodium hydroxide Copy please! Cathy’s Ankles (CatHy’s AnCl)
Electrolysis of brine • Cathode (-) 2H+(aq) + 2e- H2(g) (SODIUM IS NOT FORMED (the sodium ion is more stable than the hydrogen ion in water H2O H+ + OH-)) • Anode (+) 2Cl-(aq) – 2e- Cl2(g)
Electrolysis of brine • Cathode (-) 2H+(aq) + 2e- H2(g) (SODIUM IS NOT FORMED (the sodium ion is more stable than the hydrogen ion in water H2O H+ + OH-)) • Anode (+) 2Cl-(aq) – 2e- Cl2(g) Copy please!
Chemicals from salt Copy please!