JK-TK/Newdata huggingface dataset:
If the overall redox reaction is not possible/ not feasible/ forced, it has a negative (-) Eᶿ Calculation examples on Eᶿ Calculate the Eᶿ value of a cell made of: a)Zn and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram) Zn2+ (aq) + 2e ->Zn(s) Eᶿ = -0.77V(lower Eᶿ/ Left Hand Side diagram) Zn(s) ->Zn2+ (aq) + 2e Eᶿ = +0.77(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿoxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.34 – (- 0.77) = +1.10V Overall redox equation: Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) Eᶿ = +1.10V Overall conventional cell representation: Zn(s) / Zn2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) Eᶿ = +1.10V Overall conventional cell diagram: Voltmeter(1.10V) 1M Cu2+ (aq) 1M Zn2+ (aq) Zn2+
18 Zinc and copper reaction has a positive(+) overall Eᶿ therefore is possible/feasible and thus Zinc can displace/reduce Copper solution. b)Mg and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram) Mg2+ (aq) + 2e ->Mg(s) Eᶿ = -2.37V(lower Eᶿ/ Left Hand Side diagram) Mg(s) ->Mg2+ (aq) + 2e Eᶿ = +2.37(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.34 – (- 2.37) = +2.71V Overall redox equation: Cu2+ (aq) + Mg(s) -> Mg2+ (aq) + Cu(s) Eᶿ = +2.71V Overall conventional cell representation: Mg(s) / Mg2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) Eᶿ = +2.71V c)Ag and Pb From the table above: 2Ag+ (aq) + 2e -> 2Ag(s) Eᶿ = +0.80V(higher Eᶿ /Right Hand Side diagram) Pb2+ (aq) + 2e ->Pb(s) Eᶿ = -0.13V(lower Eᶿ/ Left Hand Side diagram) Pb(s) ->Pb2+ (aq) + 2e Eᶿ = +0.13(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.80 – (- 0.13) = +0.93V Overall redox equation: 2Ag+ (aq) + Pb(s) -> Pb2+ (aq) + 2Ag(s) Eᶿ = +0.93V Overall conventional cell representation: Pb(s) / Pb2+ (aq) 1M, // 1M,2Ag+ (aq) / Ag(s) Eᶿ = +0.93V d)Chlorine and Bromine From the table above: 2e + Cl2(g) ->2Cl- (aq) Eᶿ = +1.36V(higher Eᶿ /Right Hand Side diagram) 2e + Br2(aq) ->2Br- (aq) Eᶿ = +0.13V(lower Eᶿ/ Left Hand Side diagram) 2Br- (aq) -> Br2(aq) + 2e Eᶿ = -0.13(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = - 0.13 – (- 1.36) = +1.23V Overall redox equation:
19 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2(aq) Eᶿ = +1.23V Overall conventional cell representation: Cl2(g) / 2Cl- (aq) 1M, // 1M, 2Br- (aq) / Br2(aq) Eᶿ = +1.23V Chlorine displaces bromine from bromine water. When chlorine gas is thus bubbled in bromine water, the pale green colour fades as displacement takes place and a brown solution containing dissolved bromine liquid is formed. This reaction is feasible /possible because the overall redox reaction has a positive Eᶿ value.
b)Mg and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram) Mg2+ (aq) + 2e ->Mg(s) Eᶿ = -2.37V(lower Eᶿ/ Left Hand Side diagram) Mg(s) ->Mg2+ (aq) + 2e Eᶿ = +2.37(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.34 – (- 2.37) = +2.71V Overall redox equation: Cu2+ (aq) + Mg(s) -> Mg2+ (aq) + Cu(s) Eᶿ = +2.71V Overall conventional cell representation: Mg(s) / Mg2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) Eᶿ = +2.71V c)Ag and Pb From the table above: 2Ag+ (aq) + 2e -> 2Ag(s) Eᶿ = +0.80V(higher Eᶿ /Right Hand Side diagram) Pb2+ (aq) + 2e ->Pb(s) Eᶿ = -0.13V(lower Eᶿ/ Left Hand Side diagram) Pb(s) ->Pb2+ (aq) + 2e Eᶿ = +0.13(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.80 – (- 0.13) = +0.93V Overall redox equation: 2Ag+ (aq) + Pb(s) -> Pb2+ (aq) + 2Ag(s) Eᶿ = +0.93V Overall conventional cell representation: Pb(s) / Pb2+ (aq) 1M, // 1M,2Ag+ (aq) / Ag(s) Eᶿ = +0.93V d)Chlorine and Bromine From the table above: 2e + Cl2(g) ->2Cl- (aq) Eᶿ = +1.36V(higher Eᶿ /Right Hand Side diagram) 2e + Br2(aq) ->2Br- (aq) Eᶿ = +0.13V(lower Eᶿ/ Left Hand Side diagram) 2Br- (aq) -> Br2(aq) + 2e Eᶿ = -0.13(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = - 0.13 – (- 1.36) = +1.23V Overall redox equation:
19 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2(aq) Eᶿ = +1.23V Overall conventional cell representation: Cl2(g) / 2Cl- (aq) 1M, // 1M, 2Br- (aq) / Br2(aq) Eᶿ = +1.23V Chlorine displaces bromine from bromine water. When chlorine gas is thus bubbled in bromine water, the pale green colour fades as displacement takes place and a brown solution containing dissolved bromine liquid is formed. This reaction is feasible /possible because the overall redox reaction has a positive Eᶿ value. e)Strongest oxidizing agent and the strongest reducing agent.
When chlorine gas is thus bubbled in bromine water, the pale green colour fades as displacement takes place and a brown solution containing dissolved bromine liquid is formed. This reaction is feasible /possible because the overall redox reaction has a positive Eᶿ value. e)Strongest oxidizing agent and the strongest reducing agent. From the table above: 2e + F2(g) ->2F- (aq) Eᶿ = +2.87V(highest Eᶿ /strongest oxidizing agent) 2e + 2K+ (aq) ->2K (aq) Eᶿ = -2.92V(lowest Eᶿ/ strongest reducing agent) 2K (aq) -> 2K+ (aq) + 2e Eᶿ = +2.92V (reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = +2.87 – (-2.92) = +5.79V Overall redox equation: F2(g) + 2K(s) -> 2F- (aq) + 2K+ (aq) Eᶿ = +5.79V Overall conventional cell representation: 2K(s) / 2K+ (aq),1M, // 1M, 2F- (aq) / F2(g) Eᶿ = +5.79V The redox reactions in an electrochemical/voltaic is commercially applied to make the: (a)Dry /primary/Laclanche cell. (b)Wet /secondary /accumulators. (a)Dry/primary/Laclanche cell Examine a used dry cell. Note the positive and the negative terminal of the cell. Carefully using a knife cut a cross section from one terminal to the other. The dry cell consist of a Zinc can containing a graphite rod at the centre surrounded by a paste of; -Ammonium chloride -Zinc chloride -powdered manganese (IV) oxide mixed with Carbon.
Note the positive and the negative terminal of the cell. Carefully using a knife cut a cross section from one terminal to the other. The dry cell consist of a Zinc can containing a graphite rod at the centre surrounded by a paste of; -Ammonium chloride -Zinc chloride -powdered manganese (IV) oxide mixed with Carbon. Zinc acts/serve as the negative terminal where it ionizes/dissociates:
20 Zn(s) -> Zn2+(aq) + 2e Ammonium ions in ammonium chloride serve as the positive terminal where it is converted to ammonia gas and hydrogen gas. 2NH4+(aq) + 2e -> 2NH3(g) + H2(g) Ammonia forms a complex salt / compound /(Zn(NH3) 4)2+ (aq) / tetramminezinc(II) complex with the Zinc chloride in the paste. Manganese (IV) oxide oxidizes the hydrogen produced at the electrodes to water preventing any bubbles from coating the carbon terminal which would reduce the efficiency of the cell. Ammonium chloride is used as paste because the solid does not conduct electricity because the ions are fused/not mobile. Since the reactants are used up, the dry /primary /Laclanche cell cannot provide continous supply of electricity.The process of restoring the reactants is called recharging. b)Wet/Secondary/Accumulators 1.
Ammonium chloride is used as paste because the solid does not conduct electricity because the ions are fused/not mobile. Since the reactants are used up, the dry /primary /Laclanche cell cannot provide continous supply of electricity.The process of restoring the reactants is called recharging. b)Wet/Secondary/Accumulators 1. Wet/Secondary/Accumulators are rechargeable unlike dry /primary /Laclanche cells.Wet/Secondary/Accumulators are made up of: (i)Lead plate that forms the negative terminal (ii)Lead(IV) oxide that forms the positive terminal 2.The two electrodes are dipped in concentrated sulphuric(VI) acid of a relative density 1.2/1.3 3.At the negative terminal,lead ionizes /dissolves; Pb(s) -> Pb2+ + 2e 4.At the positive terminal,
21 (i) Lead(IV) oxide reacts with the hydrogen ions in sulphuric(VI)acid to form Pb2+ (aq) ions; PbO2(s) + 4H+(aq) + 2e -> Pb2+ (aq) + H2O(l) (ii) Pb2+ (aq) ions formed instantly react with sulphate (VI) ions/ SO42- (aq) from sulphuric (VI)acid to form insoluble Lead(II) sulphate (VI). Pb2+ (aq) + SO42- (aq) -> PbSO4(s) 5.The overall cell reaction is called discharging PbO2(s) +Pb(s) + 4H+(aq) + 2SO42- (aq)-> 2PbSO4(s) + 2H2O(l) Eᶿ = +2.0V 6.The insoluble Lead(II) sulphate (VI) formed should not be left for long since fine Lead(II) sulphate (VI) will change to a course non-reversible and inactive form making the cell less efficient. As the battery discharges ,lead and lead(IV)oxide are depleted/finished/reduced and the concentration of sulphuric(VI)acid decreases. 7.
Pb2+ (aq) + SO42- (aq) -> PbSO4(s) 5.The overall cell reaction is called discharging PbO2(s) +Pb(s) + 4H+(aq) + 2SO42- (aq)-> 2PbSO4(s) + 2H2O(l) Eᶿ = +2.0V 6.The insoluble Lead(II) sulphate (VI) formed should not be left for long since fine Lead(II) sulphate (VI) will change to a course non-reversible and inactive form making the cell less efficient. As the battery discharges ,lead and lead(IV)oxide are depleted/finished/reduced and the concentration of sulphuric(VI)acid decreases. 7. During recharging, the electrode reaction is reversed as below: 2PbSO4(s) + 2H2O(l) ->PbO2(s) +Pb(s) + 4H+(aq) + 2SO42- (aq) 8. A car battery has six Lead-acid cells making a total of 12 volts. (iii)ELECTROLYSIS (ELECTROLYTIC CELL) 1.Electrolysis is defined simply as the decomposition of a compound by an electric current/electricity. A compound that is decomposed by an electric current is called an electrolyte. Some electrolytes are weak while others are strong. 2.Strong electrolytes are those that are fully ionized/dissociated into (many) ions. Common strong electrolytes include: (i)all mineral acids (ii)all strong alkalis/sodium hydroxide/potassium hydroxide. (iii)all soluble salts
22 3.Weak electrolytes are those that are partially/partly ionized/dissociated into (few) ions. Common weak electrolytes include: (i)all organic acids (ii)all bases except sodium hydroxide/potassium hydroxide. (iii)Water 4. A compound that is not decomposed by an electric current is called nonelectrolyte. Non-electrolytes are those compounds /substances that exist as molecules and thus cannot ionize/dissociate into(any) ions . Common non-electrolytes include: (i) most organic solvents (e.g.
A compound that is not decomposed by an electric current is called nonelectrolyte. Non-electrolytes are those compounds /substances that exist as molecules and thus cannot ionize/dissociate into(any) ions . Common non-electrolytes include: (i) most organic solvents (e.g. petrol/paraffin/benzene/methylbenzene/ethanol) (ii)all hydrocarbons(alkanes /alkenes/alkynes) (iii)Chemicals of life(e.g. proteins, carbohydrates, lipids, starch, sugar) 5. An electrolytes in solid state have fused /joined ions and therefore do not conduct electricity but the ions (cations and anions) are free and mobile in molten and aqueous (solution, dissolved in water) state. 6.During electrolysis, the free ions are attracted to the electrodes. An electrode is a rod through which current enter and leave the electrolyte during electrolysis. An electrode that does not influence/alter the products of electrolysis is called an inert electrode. Common inert electrodes include: (i)Platinum (ii)Carbon graphite Platinum is not usually used in a school laboratory because it is very expensive. Carbon graphite is easily/readily and cheaply available (from used dry cells). 7.The positive electrode is called Anode.The anode is the electrode through which current enter the electrolyte/electrons leave the electrolyte 8.The negative electrode is called Cathode. The cathode is the electrode through which current leave the electrolyte / electrons enter the electrolyte 9. During the electrolysis, free anions are attracted to the anode where they lose /donate electrons to form neutral atoms/molecules. i.e. M(l) -> M+(l) + e (for cations from molten electrolytes) M(s) -> M+(aq) + e (for cations from electrolytes in aqueous state / solution / dissolved in water) The neutral atoms /molecules form the products of electrolysis at the anode. This is called discharge at anode
23 10. During electrolysis, free cations are attracted to the cathode where they gain /accept/acquire electrons to form neutral atoms/molecules.
M(l) -> M+(l) + e (for cations from molten electrolytes) M(s) -> M+(aq) + e (for cations from electrolytes in aqueous state / solution / dissolved in water) The neutral atoms /molecules form the products of electrolysis at the anode. This is called discharge at anode
23 10. During electrolysis, free cations are attracted to the cathode where they gain /accept/acquire electrons to form neutral atoms/molecules. X+ (aq) + 2e -> X(s) (for cations from electrolytes in aqueous state / solution / dissolved in water) 2X+ (l) + 2e -> X (l) (for cations from molten electrolytes) The neutral atoms /molecules form the products of electrolysis at the cathode. This is called discharge at cathode. 11. The below set up shows an electrolytic cell. BatteryAnode(+)Cathode(-)ElectrolyteSimple set up of electrolytic cellGaseous product at anodeGaseous product at cathode
24 12. For a compound /salt containing only two ion/binary salt the products of electrolysis in an electrolytic cell can be determined as in the below examples: a)To determine the products of electrolysis of molten Lead(II)chloride (i)Decomposition of electrolyte into free ions; PbCl2 (l) -> Pb 2+(l) + 2Cl-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); Pb 2+(l) + 2e -> Pb (l) (Cation / Pb 2+ gains / accepts / acquires electrons to form free atom)
25 (iii)At the anode/positive electrode(+); 2Cl-(l) -> Cl2 (g) + 2e (Anion / Cl- donate/lose electrons to form free atom then a gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid lead metal. II.At the anode pale green chlorine gas.
BatteryAnode(+)Cathode(-)ElectrolyteSimple set up of electrolytic cellGaseous product at anodeGaseous product at cathode
24 12. For a compound /salt containing only two ion/binary salt the products of electrolysis in an electrolytic cell can be determined as in the below examples: a)To determine the products of electrolysis of molten Lead(II)chloride (i)Decomposition of electrolyte into free ions; PbCl2 (l) -> Pb 2+(l) + 2Cl-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); Pb 2+(l) + 2e -> Pb (l) (Cation / Pb 2+ gains / accepts / acquires electrons to form free atom)
25 (iii)At the anode/positive electrode(+); 2Cl-(l) -> Cl2 (g) + 2e (Anion / Cl- donate/lose electrons to form free atom then a gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid lead metal. II.At the anode pale green chlorine gas. b)To determine the products of electrolysis of molten Zinc bromide (i)Decomposition of electrolyte into free ions; ZnBr2 (l) -> Zn 2+(l) + 2Br-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); Zn 2+(l) + 2e -> Zn(l) (Cation / Zn2+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 2Br-(l) -> Br2 (g) + 2e (Anion / Br- donate/lose electrons to form free atom then a liquid molecule which change to gas on heating) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid Zinc metal. II.At the anode red bromine liquid / red/brown bromine gas.
II.At the anode pale green chlorine gas. b)To determine the products of electrolysis of molten Zinc bromide (i)Decomposition of electrolyte into free ions; ZnBr2 (l) -> Zn 2+(l) + 2Br-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); Zn 2+(l) + 2e -> Zn(l) (Cation / Zn2+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 2Br-(l) -> Br2 (g) + 2e (Anion / Br- donate/lose electrons to form free atom then a liquid molecule which change to gas on heating) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid Zinc metal. II.At the anode red bromine liquid / red/brown bromine gas. c)To determine the products of electrolysis of molten sodium chloride (i)Decomposition of electrolyte into free ions; NaCl (l) -> Na +(l) + Cl-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); 2Na+(l) + 2e -> Na (l) (Cation / Na+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 2Cl-(l) -> Cl2 (g) + 2e (Anion / Cl- donate/lose electrons to form free atom then a gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid sodium metal. II.At the anode pale green chlorine gas.
II.At the anode red bromine liquid / red/brown bromine gas. c)To determine the products of electrolysis of molten sodium chloride (i)Decomposition of electrolyte into free ions; NaCl (l) -> Na +(l) + Cl-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); 2Na+(l) + 2e -> Na (l) (Cation / Na+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 2Cl-(l) -> Cl2 (g) + 2e (Anion / Cl- donate/lose electrons to form free atom then a gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid sodium metal. II.At the anode pale green chlorine gas. 26 d)To determine the products of electrolysis of molten Aluminium (III)oxide (i)Decomposition of electrolyte into free ions; Al2O3 (l) -> 2Al 3+(l) + 3O2-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); 4Al 3+ (l) + 12e -> 4Al (l) (Cation / Al 3+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 6O2-(l) -> 3O2 (g) + 12e (Anion /6O2- donate/lose 12 electrons to form free atom then three gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid aluminium metal. II.At the anode colourless gas that relights/rekindles glowing splint. 13. For a compound /salt mixture containing many ions in an electrolytic cell, the discharge of ions in the cell depend on the following factors: a) Position of cations and anions in the electrochemical series 1. Most electropositive cations require more energy to reduce (gain electrons) and thus not readily discharged. The higher elements /metals in the electrochemical series the less easily/readily it is discharged at the cathode in the electrolytic cell.
For a compound /salt mixture containing many ions in an electrolytic cell, the discharge of ions in the cell depend on the following factors: a) Position of cations and anions in the electrochemical series 1. Most electropositive cations require more energy to reduce (gain electrons) and thus not readily discharged. The higher elements /metals in the electrochemical series the less easily/readily it is discharged at the cathode in the electrolytic cell. Table I showing the relative ease of discharge of cations in an electrolytic cell K+(aq) + e -> K(s) (least readily/easily discharged) Na+(aq) + e -> Na(s) Ca2+(aq) + 2e -> Ca(s) Mg2+(aq) + 2e -> Mg(s) Al3+(aq) + 3e -> Al(s) Zn2+(aq) + 2e -> Zn(s) Fe2+(aq) + 2e -> Fe(s) Pb2+(aq) + 2e -> Pb(s) 2H+(aq) + 2e -> H2(g) (hydrogen is usually “metallic”) Cu2+(aq) + 2e -> Cu(s) Hg2+(aq) + 2e -> Hg(s) Ag+(aq) + e -> Ag(s) (most readily/easily discharged) 2.The OH- ion is the most readily/easily discharged anion . All the other anionic radicals(SO42- ,SO32- ,CO32- ,HSO4- ,HCO3- ,NO3- ,PO43-)are not/never discharged. The ease of discharge of halogen ions increase down the group. Table II showing the relative ease of discharge of anions in an electrolytic cell
27 4OH- (aq) -> 2H2O(l) + O2 (g) + 4e (most readily/easily discharged) 2 I-(aq) -> I2(aq) + 2e 2 Br-(aq) -> Br2(aq) + 2e 2 Cl-(aq) -> Cl2(aq) + 2e 2 F-(aq) -> F2(aq) + 2e SO42- ,SO32- ,CO32- ,HSO4- ,HCO3- ,NO3- ,PO43- not/never/rarely discharged.
All the other anionic radicals(SO42- ,SO32- ,CO32- ,HSO4- ,HCO3- ,NO3- ,PO43-)are not/never discharged. The ease of discharge of halogen ions increase down the group. Table II showing the relative ease of discharge of anions in an electrolytic cell
27 4OH- (aq) -> 2H2O(l) + O2 (g) + 4e (most readily/easily discharged) 2 I-(aq) -> I2(aq) + 2e 2 Br-(aq) -> Br2(aq) + 2e 2 Cl-(aq) -> Cl2(aq) + 2e 2 F-(aq) -> F2(aq) + 2e SO42- ,SO32- ,CO32- ,HSO4- ,HCO3- ,NO3- ,PO43- not/never/rarely discharged. 3.(a)When two or more cations are attracted to the cathode, the ion lower in the electrochemical series is discharged instead of that which is higher as per the table I above. This is called selective/preferential discharge at cathode. (b)When two or more anions are attracted to the anode, the ion higher in the electrochemical series is discharged instead of that which is lower as per the table I above. This is called selective/preferential discharge at anode. 4.The following experiments shows the influence /effect of selective/preferential discharge on the products of electrolysis: (i)Electrolysis of acidified water/dilute sulphuric(VI) acid 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. Electrolytic cell set up during electrolysis of acidified water/dilute sulphuric(VI) acid BatteryAnode(+)Cathode(-)ElectrolyteSimple set up of electrolytic cellGaseous product at anodeGaseous product at cathode Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq)
28 H2 SO4(aq) -> SO42-(aq) + 2H+(aq) II.
Electrolytic cell set up during electrolysis of acidified water/dilute sulphuric(VI) acid BatteryAnode(+)Cathode(-)ElectrolyteSimple set up of electrolytic cellGaseous product at anodeGaseous product at cathode Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq)
28 H2 SO4(aq) -> SO42-(aq) + 2H+(aq) II. Name the ions in acidified water that are attracted/move to: Cathode- H+(aq) from either sulphuric(VI) acid (H2 SO4) or water (H2O) Anode- SO42-(aq) from sulphuric (VI) acid (H2 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) 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 acidified water. 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. Why is electrolysis of dilute sulphuric(VI) acid called “electrolysis of (acidified) water”?