JK-TK/Newdata huggingface dataset:
Name the products of electrolysis of 1M copper(II) sulphate(VI) solution using copper electrodes. Cathode-1 moles of copper metal as brown solid coat (Cathode increase/deposits) Anode-Anode erodes/decrease in size V. Explain the changes that take place during the electrolytic process (i)Cathode -Cu2+ ions are lower than H+ ions in the electrochemical series therefore selectively discharged at the cathode. Cu2+ ions have greater tendency to accept/gain/acquire electrons to form brown copper atoms/solid that deposit itself and increase the mass/size of the cathode.The copper deposited at the cathode is pure -H+ ions accumulate around the cathode. Electrolyte thus becomes strongly acidic around the cathode. -Cu2+ ions in solution are responsible for the blue colour of electrolyte. Blue colour of electrolyte fade around the cathode. (ii)Anode Copper atom at the anode easily ionizes to release electrons. The anode therefore keeps decreasing in mass/eroding. The amount of copper that dissolve/erode is equal to the mass of copper deposited. This is called electrode ionization. Electrode ionization is where the anode erodes/decrease and the cathode deposits/increase during electrolysis. The overall concentration of the electrolyte remains constant 14.In industries electrolysis has the following uses/applications: (a)Extraction of reactive metals from their ores. Potassium, sodium ,magnesium, and aluminium are extracted from their ores using electrolytic methods. (b)Purifying copper after exraction from copper pyrites ores. Copper obtained from copper pyrites ores is not pure. After extraction, the copper is refined by electrolysing copper(II)sulphate(VI) solution using the impure copper as anode and a thin strip of pure copper as cathode.
(b)Purifying copper after exraction from copper pyrites ores. Copper obtained from copper pyrites ores is not pure. After extraction, the copper is refined by electrolysing copper(II)sulphate(VI) solution using the impure copper as anode and a thin strip of pure copper as cathode. Electrode ionization take place there: (i)At the cathode; Cu2+ (aq) + 2e -> Cu(s) (Pure copper deposits on the strip (ii)At the anode; Cu(s) ->Cu2+ (aq) + 2e (impure copper erodes/dissolves) (c)Electroplating The label EPNS(Electro Plated Nickel Silver) on some steel/metallic utensils mean they are plated/coated with silver and/or Nickel to improve their appearance(add their aesthetic value)and prevent/slow corrosion(rusting of iron). Electroplating is the process of coating a metal with another metal using an electric current. During electroplating,the cathode is made of the metal to be coated/impure. 39 Example: During the electroplating of a spoon with silver (i)the spoon/impure is placed as the cathode(negative terminal of battery) (ii)the pure silver is placed as the anode(positive terminal of battery) (iii)the pure silver erodes/ionizes/dissociates to release electrons: Ag(s) ->Ag+ (aq) + e (impure silver erodes/dissolves) (iv) silver (Ag+)ions from electrolyte gain electrons to form pure silver deposits / coat /cover the spoon/impure Ag+ (aq) + e ->Ag(s) (pure silver deposits /coat/cover on spoon) 15.The quantitative amount of products of electrolysis can be determined by applying Faradays 1st law of electrolysis. Faradays 1st law of electrolysis states that “the mass/amount of substance liberated/produced/used during electrolysis is directly proportional to the quantity of of electricity passed/used.” (a)The SI unit of quantity of electricity is the coulomb(C).
During electroplating,the cathode is made of the metal to be coated/impure. 39 Example: During the electroplating of a spoon with silver (i)the spoon/impure is placed as the cathode(negative terminal of battery) (ii)the pure silver is placed as the anode(positive terminal of battery) (iii)the pure silver erodes/ionizes/dissociates to release electrons: Ag(s) ->Ag+ (aq) + e (impure silver erodes/dissolves) (iv) silver (Ag+)ions from electrolyte gain electrons to form pure silver deposits / coat /cover the spoon/impure Ag+ (aq) + e ->Ag(s) (pure silver deposits /coat/cover on spoon) 15.The quantitative amount of products of electrolysis can be determined by applying Faradays 1st law of electrolysis. Faradays 1st law of electrolysis states that “the mass/amount of substance liberated/produced/used during electrolysis is directly proportional to the quantity of of electricity passed/used.” (a)The SI unit of quantity of electricity is the coulomb(C). The coulomb may be defined as the quantity of electricity passed/used when a current of one ampere flow for one second.i.e; 1Coulomb = 1 Ampere x 1Second The Ampere is the SI unit of current(I) The Second is the SI unit of time(t) therefore;
40 Quantity of electricity(in Coulombs) = Current(I) x time(t) Practice examples 1. A current of 2 amperes was passed through an electrolytic cell for 20 minutes. Calculate the quantity of electric charge produced. Working: Quantity of electricity(in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 2 x (20 x 60) = 2400 C 2. A current of 2 amperes was passed through an electrolytic.96500 coulombs of charge were produced. Calculate the time taken. Working: Time(t) in seconds = Quantity of electricity(in Coulombs) Current(I) in amperes Substituting = 96500 2 = 48250 seconds 3. 96500 coulombs of charge were produced after 10 minutes in an electrolytic cell . Calculate the amount of current used.
Working: Time(t) in seconds = Quantity of electricity(in Coulombs) Current(I) in amperes Substituting = 96500 2 = 48250 seconds 3. 96500 coulombs of charge were produced after 10 minutes in an electrolytic cell . Calculate the amount of current used. Working: Current(I) in amperes = Quantity of electricity(in Coulombs) Time(t) in seconds Substituting/converting time to second= 96500 10 x 60 = 160.8333 Amperes (b)The quantity of electricity required for one mole of electrons at the anode/cathode is called the Faraday constant(F). It is about 96500 Coulombs.i.e The number of Faradays used /required is equal to the number of electrons used at cathode/anode during the electrolytic process. e.g. Cu2+ require to gain 2 moles of electrons=2 Faradays =2 x 96500 coulombs of electricity at the cathode. Al3+ require to gain 3 moles of electrons=3 Faradays =3 x 96500 coulombs of electricity at the cathode Na+ require to gain 1 moles of electrons=1 Faradays =1 x 96500 coulombs of electricity at the cathode 2H+ require to gain 2 moles of electrons=2 Faradays =2 x 96500 coulombs of electricity at the cathode to form 1molecule of hydrogen gas 2O2- require to lose/donate 4 moles of electrons=4 Faradays =4 x 96500 coulombs of electricity at the anode to form 1molecule of Oxygen O2 gas. 4OH- require to lose/donate 4 moles of electrons=4 Faradays =4 x 96500 coulombs of electricity at the anode to form 1molecule of Oxygen gas and 2 molecules of water. 41 (c)The mass/amount of products at the cathode/anode is related to the molar mass of the substance and/or the volume of gases at standard/room temperature and pressure as in the below examples: Practice examples 1.Calculate the mass of copper deposited at the cathode when a steady current of 4.0 amperes is passed through copper(II)sulphate(VI) for 30 minutes in an electrolytic cell.
4OH- require to lose/donate 4 moles of electrons=4 Faradays =4 x 96500 coulombs of electricity at the anode to form 1molecule of Oxygen gas and 2 molecules of water. 41 (c)The mass/amount of products at the cathode/anode is related to the molar mass of the substance and/or the volume of gases at standard/room temperature and pressure as in the below examples: Practice examples 1.Calculate the mass of copper deposited at the cathode when a steady current of 4.0 amperes is passed through copper(II)sulphate(VI) for 30 minutes in an electrolytic cell. (Cu=63.5, 1F = 96500C) Working: Quantity of electricity(in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 4 x (30 x 60) = 7200 C Equation at the cathode: Cu2+ (aq) + 2e -> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper thus; 2 x 96500C -> 63.5 g 72000C -> 7200 x 63.5 = 2.3689 g of copper 2 x 96500 2.a)If 3.2 g of Lead were deposited when a current of 2.5 amperes was passed through an electrolytic cell of molten Lead(II)bromide for 20 minutes, determine the Faraday constant.(Pb = 207) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 2.5 x (20 x 60) = 3000 C If 3.2g of Lead -> 3000C Then 207 g of Lead -> 207 x 3000 = 194062.5 C 3.2 Equation at the cathode: Pb2+ (l) + 2e -> Pb(l) From the equation: 2 moles of electrons = 2 Faradays = 194062.5 C 1mole of electrons = 1 Faraday => 194062.5 = 97031.25 C 2 b)What is the volume of bromine vapour produced at the anode at room temperature(1mole of gas at room temperature and pressure = 24000cm3) Method 1 Equation at the anode: Br- (l) -> Br2(g) + 2e
42 From the equation: 2 moles of electrons = 2 Faradays = 194062.5 C -> 24000cm3 3000 C -> 3000 x 24000 194062.5 =371.0145cm3 Method 2 Equation at the anode: Br- (l) -> Br2(g) + 2e Mole ratio of products at Cathode: anode = 1:1 Moles of Lead at cathode = 3.2 = 0.0155moles = moles of Bromine 207 1 moles of bromine vapour -> 24000cm3 0.0155moles of Bromine -> 0.0155 x 24000 = 372 cm3 Method 3 Equation at the anode: Br- (l) -> Br2(g) + 2e Ratio of Faradays used to form products at Cathode: anode = 2:2 => 2 x 97031.25 C produce 24000cm3 of bromine vapour Then: 3000 C -> 3000 x 24000cm3 = 371.0145cm3 2 x 97031.25 3.What mass of copper remain from 2.0 at the anode if a solution of copper(II)sulphate(VI) is electrolysed using a current of 1 ampere flowing through an electrolytic cell for 20 minutes.(Cu= 63.5, 1Faraday = 96487 coulombs) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 1 x (20 x 60) = 1200 C Equation at the cathode: Cu2+ (aq) + 2e -> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C erode/dissolve a mass =molar mass of copper thus; 2 x 96500C -> 63.5 g 1200C -> 1200 x 63.5 = 0.3948g of copper deposited 2 x 96500 Mass of copper remaining = Original mass – mass dissolved/eroded => 2.0 -0.3948 = 1.6052 g of copper remain 4. Calculate the current passed if a mass of 0.234 g of copper is deposited in 4 minutes during electrolysis of a solution of copper (II)sulphate(VI).
(Cu=63.5, 1F = 96500C) Working: Quantity of electricity(in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 4 x (30 x 60) = 7200 C Equation at the cathode: Cu2+ (aq) + 2e -> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper thus; 2 x 96500C -> 63.5 g 72000C -> 7200 x 63.5 = 2.3689 g of copper 2 x 96500 2.a)If 3.2 g of Lead were deposited when a current of 2.5 amperes was passed through an electrolytic cell of molten Lead(II)bromide for 20 minutes, determine the Faraday constant.(Pb = 207) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 2.5 x (20 x 60) = 3000 C If 3.2g of Lead -> 3000C Then 207 g of Lead -> 207 x 3000 = 194062.5 C 3.2 Equation at the cathode: Pb2+ (l) + 2e -> Pb(l) From the equation: 2 moles of electrons = 2 Faradays = 194062.5 C 1mole of electrons = 1 Faraday => 194062.5 = 97031.25 C 2 b)What is the volume of bromine vapour produced at the anode at room temperature(1mole of gas at room temperature and pressure = 24000cm3) Method 1 Equation at the anode: Br- (l) -> Br2(g) + 2e
42 From the equation: 2 moles of electrons = 2 Faradays = 194062.5 C -> 24000cm3 3000 C -> 3000 x 24000 194062.5 =371.0145cm3 Method 2 Equation at the anode: Br- (l) -> Br2(g) + 2e Mole ratio of products at Cathode: anode = 1:1 Moles of Lead at cathode = 3.2 = 0.0155moles = moles of Bromine 207 1 moles of bromine vapour -> 24000cm3 0.0155moles of Bromine -> 0.0155 x 24000 = 372 cm3 Method 3 Equation at the anode: Br- (l) -> Br2(g) + 2e Ratio of Faradays used to form products at Cathode: anode = 2:2 => 2 x 97031.25 C produce 24000cm3 of bromine vapour Then: 3000 C -> 3000 x 24000cm3 = 371.0145cm3 2 x 97031.25 3.What mass of copper remain from 2.0 at the anode if a solution of copper(II)sulphate(VI) is electrolysed using a current of 1 ampere flowing through an electrolytic cell for 20 minutes.(Cu= 63.5, 1Faraday = 96487 coulombs) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 1 x (20 x 60) = 1200 C Equation at the cathode: Cu2+ (aq) + 2e -> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C erode/dissolve a mass =molar mass of copper thus; 2 x 96500C -> 63.5 g 1200C -> 1200 x 63.5 = 0.3948g of copper deposited 2 x 96500 Mass of copper remaining = Original mass – mass dissolved/eroded => 2.0 -0.3948 = 1.6052 g of copper remain 4. Calculate the current passed if a mass of 0.234 g of copper is deposited in 4 minutes during electrolysis of a solution of copper (II)sulphate(VI). (Cu= 63.5 ,1F = 96500C) Working: Equation at the cathode: Cu(s) -> Cu2+ (aq) + 2e 2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper thus;
43 63.5 g -> 2 x 96500C 0.234 g -> 0.234 x 2 x 96500 = 711.2126 C 63.5 Current(I) in amperes = Quantity of electricity(in Coulombs) Time(t) in seconds Substituting/converting time to second= 711.2126 C 4x 60 = 2.9634 Amperes 5. (a)What quantity of electricity will deposit a mass of 2.43 g of Zinc during electrolysis of a solution of Zinc (II)sulphate(VI).
Calculate the current passed if a mass of 0.234 g of copper is deposited in 4 minutes during electrolysis of a solution of copper (II)sulphate(VI). (Cu= 63.5 ,1F = 96500C) Working: Equation at the cathode: Cu(s) -> Cu2+ (aq) + 2e 2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper thus;
43 63.5 g -> 2 x 96500C 0.234 g -> 0.234 x 2 x 96500 = 711.2126 C 63.5 Current(I) in amperes = Quantity of electricity(in Coulombs) Time(t) in seconds Substituting/converting time to second= 711.2126 C 4x 60 = 2.9634 Amperes 5. (a)What quantity of electricity will deposit a mass of 2.43 g of Zinc during electrolysis of a solution of Zinc (II)sulphate(VI). (Zn= 65 ,1F = 96500C) Working: Equation at the cathode: Zn2+ (aq) + 2e -> Zn(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C erode/dissolve a mass =molar mass of Zinc thus; 65 g -> 2 x 96500 2.43 g -> 2.43 x 2 x 96500 = 7215.2308 C 65 (b)Calculate the time (in minutes) it would take during electrolysis of the solution of Zinc (II)sulphate(VI) above if a current of 4.0 Amperes is used. Time(t) in seconds = Quantity of electricity(in Coulombs) Current(I) in amperes Substituting = 7215.2308 = 1803.8077 seconds = 30.0635 minutes 4 60 6.When a current of 1.5 amperes was passed through a cell containing M3+ ions of metal M for 15 minutes, the mass at cathode increased by 0.26 g.(Faraday constant = 96500C a) Calculate the quantity of electricity used.
(a)What quantity of electricity will deposit a mass of 2.43 g of Zinc during electrolysis of a solution of Zinc (II)sulphate(VI). (Zn= 65 ,1F = 96500C) Working: Equation at the cathode: Zn2+ (aq) + 2e -> Zn(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C erode/dissolve a mass =molar mass of Zinc thus; 65 g -> 2 x 96500 2.43 g -> 2.43 x 2 x 96500 = 7215.2308 C 65 (b)Calculate the time (in minutes) it would take during electrolysis of the solution of Zinc (II)sulphate(VI) above if a current of 4.0 Amperes is used. Time(t) in seconds = Quantity of electricity(in Coulombs) Current(I) in amperes Substituting = 7215.2308 = 1803.8077 seconds = 30.0635 minutes 4 60 6.When a current of 1.5 amperes was passed through a cell containing M3+ ions of metal M for 15 minutes, the mass at cathode increased by 0.26 g.(Faraday constant = 96500C a) Calculate the quantity of electricity used. Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 1.5 x (15 x 60) = 1350 C b) Determine the relative atomic mass of metal M Equation at the cathode: M3+ (aq) + 3e -> M(s) 1350 C of electricity -> 0.26 g of metal M 3 mole of electrons = 3 Faradays = 3 x 96500 C produce a mass =molar mass of M thus; RAM of M = 0.26 g x 3 x 96500 = 55.7556(No units) 1350 7.An element “P” has a relative atomic mass 88.When a current of 0.5 amperes was passed through fused chloride of “P” for 32 minutes and 10seconds ,0.44 g
44 of “P” was deposited at the cathode.
(Zn= 65 ,1F = 96500C) Working: Equation at the cathode: Zn2+ (aq) + 2e -> Zn(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C erode/dissolve a mass =molar mass of Zinc thus; 65 g -> 2 x 96500 2.43 g -> 2.43 x 2 x 96500 = 7215.2308 C 65 (b)Calculate the time (in minutes) it would take during electrolysis of the solution of Zinc (II)sulphate(VI) above if a current of 4.0 Amperes is used. Time(t) in seconds = Quantity of electricity(in Coulombs) Current(I) in amperes Substituting = 7215.2308 = 1803.8077 seconds = 30.0635 minutes 4 60 6.When a current of 1.5 amperes was passed through a cell containing M3+ ions of metal M for 15 minutes, the mass at cathode increased by 0.26 g.(Faraday constant = 96500C a) Calculate the quantity of electricity used. Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 1.5 x (15 x 60) = 1350 C b) Determine the relative atomic mass of metal M Equation at the cathode: M3+ (aq) + 3e -> M(s) 1350 C of electricity -> 0.26 g of metal M 3 mole of electrons = 3 Faradays = 3 x 96500 C produce a mass =molar mass of M thus; RAM of M = 0.26 g x 3 x 96500 = 55.7556(No units) 1350 7.An element “P” has a relative atomic mass 88.When a current of 0.5 amperes was passed through fused chloride of “P” for 32 minutes and 10seconds ,0.44 g
44 of “P” was deposited at the cathode. Determine the charge on an ion of “P”(Faraday constant = 96500C) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 0.5 x ((32 x 60) + 10) = 965C 0.44 g of metal “P” are deposited by 965C 88g of of metal “P” are deposited by: 88 x 965= 193000 C 0.44 96500 C = 1 mole of electrons = 1 Faradays = single charge 193000 C -> 193000 = 2 moles/Faradays/charges => symbol of ion = P2+ 96500 8.
Time(t) in seconds = Quantity of electricity(in Coulombs) Current(I) in amperes Substituting = 7215.2308 = 1803.8077 seconds = 30.0635 minutes 4 60 6.When a current of 1.5 amperes was passed through a cell containing M3+ ions of metal M for 15 minutes, the mass at cathode increased by 0.26 g.(Faraday constant = 96500C a) Calculate the quantity of electricity used. Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 1.5 x (15 x 60) = 1350 C b) Determine the relative atomic mass of metal M Equation at the cathode: M3+ (aq) + 3e -> M(s) 1350 C of electricity -> 0.26 g of metal M 3 mole of electrons = 3 Faradays = 3 x 96500 C produce a mass =molar mass of M thus; RAM of M = 0.26 g x 3 x 96500 = 55.7556(No units) 1350 7.An element “P” has a relative atomic mass 88.When a current of 0.5 amperes was passed through fused chloride of “P” for 32 minutes and 10seconds ,0.44 g
44 of “P” was deposited at the cathode. Determine the charge on an ion of “P”(Faraday constant = 96500C) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 0.5 x ((32 x 60) + 10) = 965C 0.44 g of metal “P” are deposited by 965C 88g of of metal “P” are deposited by: 88 x 965= 193000 C 0.44 96500 C = 1 mole of electrons = 1 Faradays = single charge 193000 C -> 193000 = 2 moles/Faradays/charges => symbol of ion = P2+ 96500 8. During purification of copper by electrolysis 1.48 g of copper was deposited when a current was passed through aqueous copper (II)sulphate(VI) for 2 ½ hours.
Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 1.5 x (15 x 60) = 1350 C b) Determine the relative atomic mass of metal M Equation at the cathode: M3+ (aq) + 3e -> M(s) 1350 C of electricity -> 0.26 g of metal M 3 mole of electrons = 3 Faradays = 3 x 96500 C produce a mass =molar mass of M thus; RAM of M = 0.26 g x 3 x 96500 = 55.7556(No units) 1350 7.An element “P” has a relative atomic mass 88.When a current of 0.5 amperes was passed through fused chloride of “P” for 32 minutes and 10seconds ,0.44 g
44 of “P” was deposited at the cathode. Determine the charge on an ion of “P”(Faraday constant = 96500C) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 0.5 x ((32 x 60) + 10) = 965C 0.44 g of metal “P” are deposited by 965C 88g of of metal “P” are deposited by: 88 x 965= 193000 C 0.44 96500 C = 1 mole of electrons = 1 Faradays = single charge 193000 C -> 193000 = 2 moles/Faradays/charges => symbol of ion = P2+ 96500 8. During purification of copper by electrolysis 1.48 g of copper was deposited when a current was passed through aqueous copper (II)sulphate(VI) for 2 ½ hours. Calculate the amount of current that was passed.
Determine the charge on an ion of “P”(Faraday constant = 96500C) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 0.5 x ((32 x 60) + 10) = 965C 0.44 g of metal “P” are deposited by 965C 88g of of metal “P” are deposited by: 88 x 965= 193000 C 0.44 96500 C = 1 mole of electrons = 1 Faradays = single charge 193000 C -> 193000 = 2 moles/Faradays/charges => symbol of ion = P2+ 96500 8. During purification of copper by electrolysis 1.48 g of copper was deposited when a current was passed through aqueous copper (II)sulphate(VI) for 2 ½ hours. Calculate the amount of current that was passed. (Cu= 63.5 ,1F = 96500C) Working: Equation at the cathode: Cu2+ (aq) + 2e-> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper thus; 63.5 g -> 2 x 96500C 1.48 g -> 1.48 x 2 x 96500 = 4255.1181 C 63.5 Current(I) in amperes = Quantity of electricity(in Coulombs) Time(t) in seconds Substituting/converting time to second= 4255.1181C (( 2 x 60) + 30) x60 = 0.4728 Amperes 17. Practically Faraday 1st law of electrolysis can be verified as below. Verifying Faraday 1st law of electrolysis Procedure. Weigh clean copper plates electrodes. Record the masses of the electrodes in table I below. Place the electrodes in 1M copper(II) sulphate(VI) solution in a beaker. Set up an electrolytic cell. Close the switch and pass a steady current of 2 amperes by adjusting the rheostat for exactly 20 minutes.Remove the electrodes from the electrolyte. Wash with acetone/ propanone and allow them to dry.