JK-TK/Newdata huggingface dataset:
The volume of Oxygen gas at the anode is thus a half the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is thus a twice the volume of Oxygen produced at the anode. VI. Why is electrolysis of dilute sulphuric(VI) acid called “electrolysis of (acidified) water”? The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. This implies/means that water in the electrolyte is being decomposed into hydrogen and Oxygen gases. The electrolysis of dilute sulphuric acid is therefore called “electrolysis of acidified water.” VI. Explain the changes in concentration of the electrolyte during electrolysis of acidified water” The concentration of dilute sulphuric (VI) acid increases. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape. The concentration /mole of acid present in a given volume of solution thus continue increasing/rising. (ii)Electrolysis of Magnesium sulphate(VI) solution Fill the Hoffmann voltameter with dilute sulphuric(VI) acid. Connect the Hoffmann voltameter to a d.c. electric supply. Note the observations at each electrode. Answer the following questions:
29 I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) Mg SO4(aq) -> SO42-(aq) + Mg2+(aq) II. Name the ions in Magnesium sulphate(VI) solution that are attracted/move to: Cathode- Mg2+(aq) from Magnesium sulphate(VI) solution (Mg SO4) and H+(aq) from water (H2O) Anode- SO42-(aq) from Magnesium sulphate(VI) solution (Mg SO4) and OH- (aq) from water (H2O) III.
Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) Mg SO4(aq) -> SO42-(aq) + Mg2+(aq) II. Name the ions in Magnesium sulphate(VI) solution that are attracted/move to: Cathode- Mg2+(aq) from Magnesium sulphate(VI) solution (Mg SO4) and H+(aq) from water (H2O) Anode- SO42-(aq) from Magnesium sulphate(VI) solution (Mg SO4) and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) H+ ions selectively discharged instead of Mg2+ ions at the cathode) Anode 4OH- (aq) -> 2H2O(l) + O2 (g) + 4e (4OH- ions selectively discharged instead of SO42- ions at the anode) IV. Name the products of electrolysis of Magnesium sulphate(VI) solution Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound Anode-Oxygen gas (colourless gas that relights /rekindles glowing splint) V. Explain the difference in volume of products at the cathode and anode. The four(4) electrons donated/lost by OH- ions to form 1 molecule/1volume/1mole of oxygen (O2)gas at the anode are gained/acquired/accepted by the four H+(aq) ions to form 2 molecule/2volume/2mole of Hydrogen (H2)gas at the cathode. The volume of Oxygen gas at the anode is thus a half the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is thus a twice the volume of Oxygen produced at the anode. VI. Explain the changes in concentration of the electrolyte during electrolysis of Magnesium sulphate(VI) solution The concentration of dilute Magnesium sulphate(VI) solution increases. The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water.
VI. Explain the changes in concentration of the electrolyte during electrolysis of Magnesium sulphate(VI) solution The concentration of dilute Magnesium sulphate(VI) solution increases. The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products. The concentration /mole of acid present in a given volume of Magnesium sulphate(VI) solution thus continue increasing/rising. 30 The set – up below was used during the electrolysis of aqueous magnesium sulphate using inert electrodes. Name a suitable pair of electrodes for this experiment Identify the ions and cations in the solution On the diagram label the cathode Write ionic equations for the reactions that took place at the anode. Explain the change that occurred to the concentration of magnesium sulphate solution during the experience. During the electrolysis a current of 2 amperes was passed through the solution for 4 hours. Calculate the volume of the gas produced at the anode.(1 faraday 96500 coulombs and volume of a gas at room temperature is 24000cm3) One of the uses of electrolysis is electroplating What is meant by electroplating? Give tow reasons why electroplating is necessary. b) Concentration of the electrolytes 1.High concentrations of cations and/or anions at the electrodes block the ion/s that is likely to be discharged at the electrode. This is called over voltage. A concentrated solution therefore produces different products of electrolysis from a dilute one. 2. The following experiments show the influence/effect of concentration of electrolyte on the products of electrolysis. (i)Electrolysis of dilute and concentrated(brine)sodium chloride solution
31 I. Dissolve about 0.5 g of pure sodium chloride crystals in 100cm3 of water. Place the solution in an electrolytic cell. Note the observations at each electrode for 10 minutes. Transfer the set up into a fume chamber/open and continue to make observations for a further 10 minute. Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) NaCl(aq) -> Cl-(aq) + Na+(aq) II.
Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) NaCl(aq) -> Cl-(aq) + Na+(aq) II. Name the ions in sodium chloride solution that are attracted/move to: Cathode- Na+(aq) from Sodium chloride solution (NaCl) and H+(aq) from water (H2O) Anode- Cl-(aq) from sodiumchloride solution (NaCl) and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) H+ ions selectively discharged instead of Na+ ions at the cathode) Anode 4OH- (aq) -> 2H2O(l) + O2 (g) + 4e (4OH- ions selectively discharged instead of Cl- ions at the anode) IV. Name the products of electrolysis of dilute sodium chloride solution Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound Anode-Oxygen gas (colourless gas that relights /rekindles glowing splint) V. Explain the difference in volume of products at the cathode and anode. Four(4) electrons donated/lost by OH- ions to form 1 molecule/1volume/1mole of oxygen (O2)gas at the anode are gained/acquired/accepted by four H+(aq) ions to form 2 molecule/2volume/2mole of Hydrogen (H2)gas at the cathode. The volume of Oxygen gas at the anode is half the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is twice the volume of Oxygen produced at the anode. VI. Explain the changes in concentration of the electrolyte during electrolysis of sodium chloride solution The concentration of dilute sodium chloride solution increases. The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products. The concentration /moles of salt present in a given volume of sodium chloride solution continue increasing/rising. 32 II.
Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products. The concentration /moles of salt present in a given volume of sodium chloride solution continue increasing/rising. 32 II. Dissolve about 20 g of pure sodium chloride crystals in 100cm3 of water. Place the solution in an electrolytic cell. Note the observations continuously at each electrode for 30 minutes in a fume chamber/open. Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) NaCl(aq) -> Cl-(aq) + Na+(aq) II. Name the ions in sodium chloride solution that are attracted/move to: Cathode- Na+(aq) from Sodium chloride solution (NaCl) and H+(aq) from water (H2O) Anode- Cl-(aq) from sodium chloride solution (NaCl) and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 2H+(aq) + 2e -> H2(g) H+ ions selectively discharged instead of Na+ ions at the cathode) Anode 2Cl- (aq) -> Cl2(g) + 4e (Cl- ions with a higher concentration block the discharge of OH- ions at the anode) IV. Name the products of electrolysis of concentrated sodium chloride solution/brine Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound Anode-Chlorine gas(pale green gas that bleaches damp/moist/wet litmus papers) V. Explain the difference in volume of products at the cathode and anode. Two (2) electrons donated/lost by Cl- ions to form 1 molecule/1volume/1mole of Chlorine (Cl2)gas at the anode are gained/acquired/accepted by two H+(aq) ions to form 1 molecule/1volume/1mole of Hydrogen (H2)gas at the cathode. The volume of Chlorine gas at the anode is equal to the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is equal to the volume of Chlorine produced at the anode. VI.
Two (2) electrons donated/lost by Cl- ions to form 1 molecule/1volume/1mole of Chlorine (Cl2)gas at the anode are gained/acquired/accepted by two H+(aq) ions to form 1 molecule/1volume/1mole of Hydrogen (H2)gas at the cathode. The volume of Chlorine gas at the anode is equal to the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is equal to the volume of Chlorine produced at the anode. VI. Explain the changes in concentration of the electrolyte during electrolysis of concentrated sodium chloride solution/brine The concentration of concentrated sodium chloride solution/brine increases. The ratio of Cl2 (g): H2 (g) is 1:1 as they are combined in water. Water in the electrolyte is decomposed into only Hydrogen gas that escapes as products at cathode. 33 The concentration /moles of OH- (aq) and Na+ ion (as NaOH) present in a given volume of electrolyte continue increasing/rising. This makes the electrolyte strongly alkaline with high pH. As the electrolysis of brine continues the concentration of Cl- ions decrease and oxygen gas start being liberated at anode. The electrolyte pH is thus lowered and the concentration of brine starts again increasing. (ii)Electrolysis of dilute and concentrated Hydrochloric acid solution I. Prepare about 50cm3 of 0.05 M of dilute Hydrochloric acid in 100cm3 solution. Place the solution in an electrolytic cell. Note the observations at each electrode for 10 minutes. Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) HCl(aq) -> Cl-(aq) + H+(aq) II. Name the ions in dilute Hydrochloric acid solution that are attracted/move to: Cathode- H+(aq) from dilute Hydrochloric acid (HCl) and H+(aq) from water (H2O) Anode- Cl-(aq) from dilute Hydrochloric acid (HCl) and OH- (aq) from water (H2O) III.
Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) HCl(aq) -> Cl-(aq) + H+(aq) II. Name the ions in dilute Hydrochloric acid solution that are attracted/move to: Cathode- H+(aq) from dilute Hydrochloric acid (HCl) and H+(aq) from water (H2O) Anode- Cl-(aq) from dilute Hydrochloric acid (HCl) and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) H+ ions selectively discharged instead of Na+ ions at the cathode) Anode 4OH- (aq) -> H2O(l) +O2+ 4e (4OH- ions selectively discharged instead of Cl- ions at the anode) IV. Name the products of electrolysis of dilute Hydrochloric acid Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound Anode-Oxygen gas (colourless gas that relights /rekindles glowing splint) V. Explain the difference in volume of products at the cathode and anode. Four(4) electrons donated/lost by OH- ions to form 1 molecule/1volume/1mole of oxygen (O2)gas at the anode are gained/acquired/accepted by four H+(aq) ions to form 2 molecule/2volume/2mole of Hydrogen (H2)gas at the cathode. The volume of Oxygen gas at the anode is half the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is twice the volume of Oxygen produced at the anode. 34 VI. Explain the changes in concentration of the electrolyte during electrolysis of dilute Hydrochloric acid The concentration of dilute Hydrochloric acid increases. The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products.
Explain the changes in concentration of the electrolyte during electrolysis of dilute Hydrochloric acid The concentration of dilute Hydrochloric acid increases. The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products. The concentration /moles of HCl present in a given volume of dilute Hydrochloric acid continue increasing/rising. II. Prepare about 50cm3 of 2M of Hydrochloric acid in 100cm3 solution. Place the solution in an electrolytic cell. Note the observations at each electrode for 30 minutes CautionThis experiment should be done in the open/fume chamber. Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) HCl(aq) -> Cl-(aq) + H+(aq) II. Name the ions in 2M Hydrochloric acid solution that are attracted/move to: Cathode- H+(aq) from dilute Hydrochloric acid (HCl) and H+(aq) from water (H2O) Anode- Cl-(aq) from dilute Hydrochloric acid (HCl) and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) H+ ions selectively discharged instead of Na+ ions at the cathode) Anode 2Cl- (aq) -> Cl2+ 2e (OH- ions concentration is low.Cl- ions concentration is higher at the anode thus cause over voltage/block discharge of OH- ions) IV. Name the products of electrolysis of 2M Hydrochloric acid Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound Anode-Chlorine gas (Pale green gas that bleaches blue/red moist/wet/damp litmus papers) V. Explain the difference in volume of products at the cathode and anode.
Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) H+ ions selectively discharged instead of Na+ ions at the cathode) Anode 2Cl- (aq) -> Cl2+ 2e (OH- ions concentration is low.Cl- ions concentration is higher at the anode thus cause over voltage/block discharge of OH- ions) IV. Name the products of electrolysis of 2M Hydrochloric acid Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound Anode-Chlorine gas (Pale green gas that bleaches blue/red moist/wet/damp litmus papers) V. Explain the difference in volume of products at the cathode and anode. Two(2) electrons donated/lost by Cl- ions to form 1 molecule/1volume/1mole of Chlorine (Cl2)gas at the anode are gained/acquired/accepted by two H+(aq) ions to form 1 molecule/1volume/1mole of Hydrogen (H2)gas at the cathode. The volume of Chlorine gas at the anode is equal to the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is twice the volume of Chlorine produced at the anode. 35 VI. Explain the changes in concentration of the electrolyte during electrolysis of 2M Hydrochloric acid The concentration of Hydrochloric acid decreases. The ratio of H2 (g): Cl2 (g) is 1:1 as they are combined in Hydrochloric acid. Water in the electrolyte is decomposed only into Hydrogen gas that escapes as products at the cathode. There is a net accumulation of excess OH- (aq) ions in solution. This makes the electrolyte strongly alkaline with high pH. c) Nature of electrodes used in the electrolytic cell Inert electrodes (carbon-graphite and platinum) do not alter the expected products of electrolysis in an electrolytic cell. If another/different electrode is used in the electrolytic cell it alters/influences/changes the expected products of electrolysis. The examples below illustrate the influence of the nature of electrode on the products of electrolysis: (i)Electrolysis of copper(II) sulphate(VI) solution I.
c) Nature of electrodes used in the electrolytic cell Inert electrodes (carbon-graphite and platinum) do not alter the expected products of electrolysis in an electrolytic cell. If another/different electrode is used in the electrolytic cell it alters/influences/changes the expected products of electrolysis. The examples below illustrate the influence of the nature of electrode on the products of electrolysis: (i)Electrolysis of copper(II) sulphate(VI) solution I. Using carbon-graphite electrodes Weigh Carbon -graphite 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 current for about 20 minutes. Observe each electrode and any changes in electrolyte. Remove the electrodes from the electrolyte. Wash with acetone/propanone and allow them to dry. Reweigh each electrode. Sample results Mass of cathode before electrolysis 23.4 g Mass of anode before electrolysis 22.4 g Mass of cathode after electrolysis 25.4 g Mass of anode after electrolysis 22.4 g Brown solid deposit at the cathode after electrolysis - Bubbles of colourless gas that relight splint - Blue colour of electrolyte fades/become less blue - Blue colour of electrolyte fades /become less blue - Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) CuSO4(aq) -> SO42-(aq) + Cu2+(aq)
36 II. Name the ions in 1M copper(II) sulphate(VI) solution that are attracted/move to: Cathode- Cu2+ (aq) from copper(II) sulphate(VI) solution and H+(aq) from water (H2O) Anode- SO42-(aq) from copper(II) sulphate(VI) solution and OH- (aq) from water (H2O) III.
Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) CuSO4(aq) -> SO42-(aq) + Cu2+(aq)
36 II. Name the ions in 1M copper(II) sulphate(VI) solution that are attracted/move to: Cathode- Cu2+ (aq) from copper(II) sulphate(VI) solution and H+(aq) from water (H2O) Anode- SO42-(aq) from copper(II) sulphate(VI) solution and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 2Cu2+ (aq) + 4e -> 2Cu(g) Cu2+ ions are lower than H+ ions in the electrochemical series therefore selectively discharged at the cathode.) Anode 4OH- (aq) -> H2O(l) + O2+ 4e (OH- ions ions are higher than SO42- ions in the electrochemical series therefore selectively discharged at the cathode.)) IV. Name the products of electrolysis of 1M copper(II) sulphate(VI) solution Cathode-2 moles of copper metal as brown solid coat Anode-Oxygen gas (Colourless gas that relights /rekindles glowing splint) V. Explain the changes that take place at the cathode and anode. Four(4) electrons donated/lost by OH- ions to form 1 molecule/1volume/1mole of Oxygen (O2)gas at the anode are gained/acquired/accepted by two Cu2+(aq) ions to form 2 moles of brown copper solid that deposit itself at the cathode. The moles of oxygen gas at the anode is equal to the moles of copper produced at the cathode VI. Explain the changes in electrolyte during electrolysis of 1M copper (II) sulphate(VI) solution. (i)The pH of copper(II) sulphate(VI) solution lowers/decreases. The salt becomes more acidic. Water in the electrolyte is decomposed only into Oxygen gas (from the OH- ions) that escapes as products at the anode. There is a net accumulation of excess H+ (aq) ions in solution.
The salt becomes more acidic. Water in the electrolyte is decomposed only into Oxygen gas (from the OH- ions) that escapes as products at the anode. There is a net accumulation of excess H+ (aq) ions in solution. This makes the electrolyte strongly acidic with low pH. (ii) Cu2+ (aq) ions are responsible for the blue colour of the electrolyte/ copper(II) sulphate (VI) solution. As electrolysis continues, blue Cu2+ (aq) ions gain electrons to form brown Copper. The blue colour of electrolyte therefore fades/become less blue. (iii)Copper is deposited at the cathode. This increases the mass of the cathode.OH- ions that produce Oxygen gas at anode come from water. Oxygen escapes out/away without increasing the mass of anode. 37 II. Using copper electrodes 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 current for about 20 minutes. Observe each electrode and any changes in electrolyte. Remove the electrodes from the electrolyte. Wash with acetone/propanone and allow them to dry. Reweigh each electrode. Sample results Mass of cathode before electrolysis 23.4 g Mass of anode before electrolysis 22.4 g Mass of cathode after electrolysis 25.4 g Mass of anode after electrolysis 20.4 g Brown solid deposit at the cathode after electrolysis - Anode decrease insize/erodes/wear off - Blue colour of electrolyte remain blue - Blue colour of electrolyte remain blue - Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) CuSO4(aq) -> SO42-(aq) + Cu2+(aq) II.
Sample results Mass of cathode before electrolysis 23.4 g Mass of anode before electrolysis 22.4 g Mass of cathode after electrolysis 25.4 g Mass of anode after electrolysis 20.4 g Brown solid deposit at the cathode after electrolysis - Anode decrease insize/erodes/wear off - Blue colour of electrolyte remain blue - Blue colour of electrolyte remain blue - Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) CuSO4(aq) -> SO42-(aq) + Cu2+(aq) II. Name the ions in 1M copper(II) sulphate(VI) solution that are attracted/move to: Cathode- Cu2+ (aq) from copper(II) sulphate(VI) solution and H+(aq) from water (H2O) Anode- SO42-(aq) from copper(II) sulphate(VI) solution and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode Cu2+ (aq) + 2e -> Cu(s) Cu2+ ions are lower than H+ ions in the electrochemical series therefore selectively discharged at the cathode.) Anode Cu (s) -> Cu2+(aq) + 2e (Both OH- ions and SO42- ions move to the anode but none is discharged. The copper anode itself ionizes/dissolves/dissociate because less energy is used to remove an electron/ionize /dissociate copper atoms than OH- ions. 38 IV. 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.