let’s read! pages 82 to 89
DESCRIPTION
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. - PowerPoint PPT PresentationTRANSCRIPT
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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Electrolysis
Lead bromide
Electrolysis
Lead bromide
Electrolysis is the break-down of a substance by electricity
Electrolyte - a molten or aqueous solution through which an electrical current can flow.
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Electrode - an electrical conductor which carries charge to or from a liquid undergoing electrolysis.
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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
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Electrolysis of zinc chloride
1. Electrolytes contain positive and negative ions.2. During electrolysis, positive and negative electrodes
are put into the electrolyte.3. The positive electrode is called the anode.4. The negative electrode is called the cathode.5. The negative ions (called anions) are attracted to the
anode.6. At the anode, the negative ions lose electrons to
become atoms/molecules.7. The positive ions (called cations) are attracted to the
cathode.8. At the cathode, the positive ions gain electrons to
become atoms/molecules
Draw the sentence
1. Electrolytes contain positive and negative ions.2. During electrolysis, positive and negative electrodes
are put into the electrolyte.3. The positive electrode is called the anode.4. The negative electrode is called the cathode.5. The negative ions (called anions) are attracted to the
anode.6. At the anode, the negative ions lose electrons to
become atoms/molecules.7. The positive ions (called cations) are attracted to the
cathode.8. At the cathode, the positive ions gain electrons to
become atoms/molecules
Draw the sentence
At the electrodes
Cathode (-)(negative electrode)
• Positive ions go here (cations).
• As metal ions are positive, they go to the cathode.
• Ions gain electrons. They are reduced and 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).
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), SO4
2- (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)
Electrolysis of solutions – Cathode
For solutions of highly reactive metals: Hydrogen gas, not the metal, is produced at the cathode.
Electrolysis of solutions – Anode
The product at the anode depends on:The negative anions present in the solution.
Electrolysis of aluminium oxide
CO2
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
1. Write out relevant half equation2. Work out coulombs of electrons flowing (Q = It)3. Convert C into moles of electrons (Faradays)
(Q/96500)4. Work out moles of product using ratio from
equation5. 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
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
Cathy’s Ankles (CatHy’s AnCl)
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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)
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Chemicals from salt
Chemicals from salt
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Let’s try some questions!