JK-TK/Newdata huggingface dataset:
Zn(s) -> Zn2+(aq) + 2e (oxidation/donation of electrons) Cu2+(aq) + 2e -> Cu(s) (reduction/gain of electrons) Cu2+(aq) + Zn(s) -> Zn2+(aq) + Cu(s) (redox/both donation and gain of electrons) 4. Fe(s) -> Fe2+(aq) + 2e (oxidation/donation of electrons) 2Ag+(aq) + 2e -> 2Ag(s) (reduction/gain of electrons) 2Ag+(aq) + Fe(s) -> Fe2+(aq) + 2Ag(s) (redox/both donation and gain of electrons) 5. Zn(s) -> Zn2+(aq) + 2e (oxidation/donation of electrons) Cl2(g) + 2e -> 2Cl-(aq) (reduction/gain of electrons) Cl2(g) + Zn(s) -> Zn2+(aq) + 2Cl-(aq) (redox/both donation and gain of electrons) 6. 2Mg(s) -> 2Mg2+(aq) + 4e (oxidation/donation of electrons) O2(g) + 4e -> 2O2-(aq) (reduction/gain of electrons) O2(g) + 2Mg(s) -> 2Mg2+(aq) + 2O2-(aq) (redox/both donation and gain of electrons)
4 Note (i)The number of electrons donated/lost MUST be equal to the number of electrons gained/acquired. (i)During displacement reaction, the colour of ions /salts fades but does not if displacement does not take place. e.g a)Green colour of Fe2+(aq) fades if Fe2+(aq) ions are displaced from their solution. Green colour of Fe2+(aq) appear if Fe/iron displaces another salt/ions from their solution. b)Blue colour of Cu2+(aq) fades if Cu2+(aq) ions are displaced from their solution and brown copper deposits appear. Blue colour of Cu2+(aq) appear if Cu/copper displaces another salt/ions from their solution. c)Brown colour of Fe3+(aq) fades if Fe3+(aq) ions are displaced from their solution.
b)Blue colour of Cu2+(aq) fades if Cu2+(aq) ions are displaced from their solution and brown copper deposits appear. Blue colour of Cu2+(aq) appear if Cu/copper displaces another salt/ions from their solution. c)Brown colour of Fe3+(aq) fades if Fe3+(aq) ions are displaced from their solution. Brown colour of Fe3+(aq) appear if Fe/iron displaces another salt/ions from their solution to form Fe3+(aq). (iii)Displacement reactions also produce energy/heat. The closer/nearer the metals in the reactivity/electrochemical series the less energy/heat of displacement. (iv)The higher the metal in the reactivity series therefore the easier to loose/donate electrons and thus the stronger the reducing agent. 4. (a)In terms of oxidation number: i) Oxidation is increase in oxidation numbers. ii) Reduction is decrease in oxidation numbers. iii) Redox is simultaneous increase in oxidation numbers of one species/substance and a decrease in oxidation numbers of another species/substance. iv) Reducing agent is the species that undergoes oxidation, therefore increases its oxidation number. v) Oxidizing agent is the species that undergoes reduction, therefore increases its oxidation number. (b)The idea/concept of oxidation numbers uses/applies the following simple guideline rules: Guidelines /rules applied in assigning oxidation number 1.Oxidation number of combined Oxygen is always -2 except in peroxides (Na2O2/H2O2) where its Oxidation number is -1
5 2.Oxidation number of combined Hydrogen is always +1except in Hydrides (NaH/KH) where its Oxidation number is -1 3.All atoms and molecules of elements have oxidation number 0 (zero) Atom Oxidation number Molecule Oxidation number Na 0 Cl2 0 O 0 O2 0 H 0 H2 0 Al 0 N2 0 Ne 0 O3 0 K 0 P3 0 Cu 0 S8 0 4.All combined metals and non-metals have oxidation numbers equal to their valency /oxidation state e.g.
iv) Reducing agent is the species that undergoes oxidation, therefore increases its oxidation number. v) Oxidizing agent is the species that undergoes reduction, therefore increases its oxidation number. (b)The idea/concept of oxidation numbers uses/applies the following simple guideline rules: Guidelines /rules applied in assigning oxidation number 1.Oxidation number of combined Oxygen is always -2 except in peroxides (Na2O2/H2O2) where its Oxidation number is -1
5 2.Oxidation number of combined Hydrogen is always +1except in Hydrides (NaH/KH) where its Oxidation number is -1 3.All atoms and molecules of elements have oxidation number 0 (zero) Atom Oxidation number Molecule Oxidation number Na 0 Cl2 0 O 0 O2 0 H 0 H2 0 Al 0 N2 0 Ne 0 O3 0 K 0 P3 0 Cu 0 S8 0 4.All combined metals and non-metals have oxidation numbers equal to their valency /oxidation state e.g. Metal/non-metal ion Valency Oxidation state Oxidation number Fe2+ 2 -2 -2 Fe3+ 3 -3 -3 Cu2+ 2 -2 -2 Cu+ 1 +1 +1 Cl- 1 -1 -1 O2- 2 -2 -2 Na+ 1 +1 +1 Al3+ 3 +3 +3 P3- 3 -3 -3 Pb2+ 2 +2 +2 5.Sum of oxidation numbers of atoms of elements making a compound is equal zero(0) e.g.
v) Oxidizing agent is the species that undergoes reduction, therefore increases its oxidation number. (b)The idea/concept of oxidation numbers uses/applies the following simple guideline rules: Guidelines /rules applied in assigning oxidation number 1.Oxidation number of combined Oxygen is always -2 except in peroxides (Na2O2/H2O2) where its Oxidation number is -1
5 2.Oxidation number of combined Hydrogen is always +1except in Hydrides (NaH/KH) where its Oxidation number is -1 3.All atoms and molecules of elements have oxidation number 0 (zero) Atom Oxidation number Molecule Oxidation number Na 0 Cl2 0 O 0 O2 0 H 0 H2 0 Al 0 N2 0 Ne 0 O3 0 K 0 P3 0 Cu 0 S8 0 4.All combined metals and non-metals have oxidation numbers equal to their valency /oxidation state e.g. Metal/non-metal ion Valency Oxidation state Oxidation number Fe2+ 2 -2 -2 Fe3+ 3 -3 -3 Cu2+ 2 -2 -2 Cu+ 1 +1 +1 Cl- 1 -1 -1 O2- 2 -2 -2 Na+ 1 +1 +1 Al3+ 3 +3 +3 P3- 3 -3 -3 Pb2+ 2 +2 +2 5.Sum of oxidation numbers of atoms of elements making a compound is equal zero(0) e.g. Using this rule ,an unknown oxidation number of an atom in a compound can be determined as below: a) CuSO4 has- -one atom of Cu with oxidation number +2( refer to Rule 4) -one atom of S with oxidation number +6 ( refer to Rule 4) -six atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in CuSO4 = (+2 + +6 + (-2 x 6)) = 0 b) H2SO4 has- -two atom of H each with oxidation number +1( refer to Rule 2) -one atom of S with oxidation number +6 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in H2SO4 = (+2 + +6 + (-2 x 4)) = 0
6 c) KMnO4 has- -one atom of K with oxidation number +1( refer to Rule 4) -one atom of Mn with oxidation number +7 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in KMnO4 = (+1 + +7 + (-2 x 4)) = 0 Determine the oxidation number of: I.Nitrogen in; -NO => x + -2 = 0 thus x = 0 – (-2) = + 2 The chemical name of this compound is thus Nitrogen(II)oxide -NO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Nitrogen(IV)oxide -N2O => 2x + -2 = 0 thus 2x = 0 – (-2) = +2/2= +1 The chemical name of this compound is thus Nitrogen(I)oxide II.
(b)The idea/concept of oxidation numbers uses/applies the following simple guideline rules: Guidelines /rules applied in assigning oxidation number 1.Oxidation number of combined Oxygen is always -2 except in peroxides (Na2O2/H2O2) where its Oxidation number is -1
5 2.Oxidation number of combined Hydrogen is always +1except in Hydrides (NaH/KH) where its Oxidation number is -1 3.All atoms and molecules of elements have oxidation number 0 (zero) Atom Oxidation number Molecule Oxidation number Na 0 Cl2 0 O 0 O2 0 H 0 H2 0 Al 0 N2 0 Ne 0 O3 0 K 0 P3 0 Cu 0 S8 0 4.All combined metals and non-metals have oxidation numbers equal to their valency /oxidation state e.g. Metal/non-metal ion Valency Oxidation state Oxidation number Fe2+ 2 -2 -2 Fe3+ 3 -3 -3 Cu2+ 2 -2 -2 Cu+ 1 +1 +1 Cl- 1 -1 -1 O2- 2 -2 -2 Na+ 1 +1 +1 Al3+ 3 +3 +3 P3- 3 -3 -3 Pb2+ 2 +2 +2 5.Sum of oxidation numbers of atoms of elements making a compound is equal zero(0) e.g. Using this rule ,an unknown oxidation number of an atom in a compound can be determined as below: a) CuSO4 has- -one atom of Cu with oxidation number +2( refer to Rule 4) -one atom of S with oxidation number +6 ( refer to Rule 4) -six atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in CuSO4 = (+2 + +6 + (-2 x 6)) = 0 b) H2SO4 has- -two atom of H each with oxidation number +1( refer to Rule 2) -one atom of S with oxidation number +6 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in H2SO4 = (+2 + +6 + (-2 x 4)) = 0
6 c) KMnO4 has- -one atom of K with oxidation number +1( refer to Rule 4) -one atom of Mn with oxidation number +7 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in KMnO4 = (+1 + +7 + (-2 x 4)) = 0 Determine the oxidation number of: I.Nitrogen in; -NO => x + -2 = 0 thus x = 0 – (-2) = + 2 The chemical name of this compound is thus Nitrogen(II)oxide -NO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Nitrogen(IV)oxide -N2O => 2x + -2 = 0 thus 2x = 0 – (-2) = +2/2= +1 The chemical name of this compound is thus Nitrogen(I)oxide II. Sulphur in; -SO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Sulphur(IV)oxide -SO3 => x + (-2 x3)= 0 thus x = 0 – (-6) = + 6 The chemical name of this compound is thus Sulphur(VI)oxide -H2SO4 = ((+1 x 2) + x + (-2 x 4)) thus x= 0-( +2 +-8) =+6 The chemical name of this compound is thus Sulphuric(VI)acid -H2SO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Sulphuric(IV)acid III.
Metal/non-metal ion Valency Oxidation state Oxidation number Fe2+ 2 -2 -2 Fe3+ 3 -3 -3 Cu2+ 2 -2 -2 Cu+ 1 +1 +1 Cl- 1 -1 -1 O2- 2 -2 -2 Na+ 1 +1 +1 Al3+ 3 +3 +3 P3- 3 -3 -3 Pb2+ 2 +2 +2 5.Sum of oxidation numbers of atoms of elements making a compound is equal zero(0) e.g. Using this rule ,an unknown oxidation number of an atom in a compound can be determined as below: a) CuSO4 has- -one atom of Cu with oxidation number +2( refer to Rule 4) -one atom of S with oxidation number +6 ( refer to Rule 4) -six atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in CuSO4 = (+2 + +6 + (-2 x 6)) = 0 b) H2SO4 has- -two atom of H each with oxidation number +1( refer to Rule 2) -one atom of S with oxidation number +6 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in H2SO4 = (+2 + +6 + (-2 x 4)) = 0
6 c) KMnO4 has- -one atom of K with oxidation number +1( refer to Rule 4) -one atom of Mn with oxidation number +7 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in KMnO4 = (+1 + +7 + (-2 x 4)) = 0 Determine the oxidation number of: I.Nitrogen in; -NO => x + -2 = 0 thus x = 0 – (-2) = + 2 The chemical name of this compound is thus Nitrogen(II)oxide -NO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Nitrogen(IV)oxide -N2O => 2x + -2 = 0 thus 2x = 0 – (-2) = +2/2= +1 The chemical name of this compound is thus Nitrogen(I)oxide II. Sulphur in; -SO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Sulphur(IV)oxide -SO3 => x + (-2 x3)= 0 thus x = 0 – (-6) = + 6 The chemical name of this compound is thus Sulphur(VI)oxide -H2SO4 = ((+1 x 2) + x + (-2 x 4)) thus x= 0-( +2 +-8) =+6 The chemical name of this compound is thus Sulphuric(VI)acid -H2SO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Sulphuric(IV)acid III. Carbon in; -CO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus carbon(IV)oxide -CO => x + -2 = 0 thus x = 0 – -2 = + 2 The chemical name of this compound is thus carbon(II)oxide -H2CO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Carbonic(IV)acid IV.Manganese in; -MnO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Manganese(IV)oxide -KMnO4 = ((+1 + x + (-2 x 4)) thus x= 0-( +1 +-8) =+7 The chemical name of this compound is thus Potassium manganate(VII) V.Chromium in; - Cr2O3 => 2x + (-2 x 3)= 0 thus 2x = 0 – (-6) = +6 / 2= +3 The chemical name of this compound is thus Chromium(III)oxide -K2Cr2O7 => (+1 x 2) + 2x + (-2 x7)= 0 thus 2x = 0 – +2 +-14 = +12 / 2= +6 The chemical name of this compound is thus Potassium dichromate(VI) -K2CrO4 => (+1 x 2) + x + (-2 x4)= 0 thus 2x = 0 – +2 +-8 = +12 / 2= +6
7 The chemical name of this compound is thus Potassium chromate(VI) 6.The sum of the oxidation numbers of atoms of elements making a charged radical/complex ion is equal to its charge.
Sulphur in; -SO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Sulphur(IV)oxide -SO3 => x + (-2 x3)= 0 thus x = 0 – (-6) = + 6 The chemical name of this compound is thus Sulphur(VI)oxide -H2SO4 = ((+1 x 2) + x + (-2 x 4)) thus x= 0-( +2 +-8) =+6 The chemical name of this compound is thus Sulphuric(VI)acid -H2SO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Sulphuric(IV)acid III. Carbon in; -CO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus carbon(IV)oxide -CO => x + -2 = 0 thus x = 0 – -2 = + 2 The chemical name of this compound is thus carbon(II)oxide -H2CO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Carbonic(IV)acid IV.Manganese in; -MnO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Manganese(IV)oxide -KMnO4 = ((+1 + x + (-2 x 4)) thus x= 0-( +1 +-8) =+7 The chemical name of this compound is thus Potassium manganate(VII) V.Chromium in; - Cr2O3 => 2x + (-2 x 3)= 0 thus 2x = 0 – (-6) = +6 / 2= +3 The chemical name of this compound is thus Chromium(III)oxide -K2Cr2O7 => (+1 x 2) + 2x + (-2 x7)= 0 thus 2x = 0 – +2 +-14 = +12 / 2= +6 The chemical name of this compound is thus Potassium dichromate(VI) -K2CrO4 => (+1 x 2) + x + (-2 x4)= 0 thus 2x = 0 – +2 +-8 = +12 / 2= +6
7 The chemical name of this compound is thus Potassium chromate(VI) 6.The sum of the oxidation numbers of atoms of elements making a charged radical/complex ion is equal to its charge. Using this rule ,the oxidation number of unknown atom of an element in a charged radical/complex ion can be determined as in the examples below; a) SO42- has- -one atom of S with oxidation number +6( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in SO42- = ( +6 + (-2 x 4)) = -2 The chemical name of this radical is thus sulphate(VI) ion b) NO3- has- -one atom of N with oxidation number +4( refer to Rule 4) -three atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in NO3- = ( +4 + (-2 x 3)) = -1 The chemical name of this radical is thus nitrate(IV) ion. Determine the oxidation number of: I.Nitrogen in; -NO2- => x + (-2 x2)= -1 thus x = -1 – (-4) = + 3 The chemical name of this compound/ion/radical is thus Nitrate(III)ion II.
Carbon in; -CO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus carbon(IV)oxide -CO => x + -2 = 0 thus x = 0 – -2 = + 2 The chemical name of this compound is thus carbon(II)oxide -H2CO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Carbonic(IV)acid IV.Manganese in; -MnO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Manganese(IV)oxide -KMnO4 = ((+1 + x + (-2 x 4)) thus x= 0-( +1 +-8) =+7 The chemical name of this compound is thus Potassium manganate(VII) V.Chromium in; - Cr2O3 => 2x + (-2 x 3)= 0 thus 2x = 0 – (-6) = +6 / 2= +3 The chemical name of this compound is thus Chromium(III)oxide -K2Cr2O7 => (+1 x 2) + 2x + (-2 x7)= 0 thus 2x = 0 – +2 +-14 = +12 / 2= +6 The chemical name of this compound is thus Potassium dichromate(VI) -K2CrO4 => (+1 x 2) + x + (-2 x4)= 0 thus 2x = 0 – +2 +-8 = +12 / 2= +6
7 The chemical name of this compound is thus Potassium chromate(VI) 6.The sum of the oxidation numbers of atoms of elements making a charged radical/complex ion is equal to its charge. Using this rule ,the oxidation number of unknown atom of an element in a charged radical/complex ion can be determined as in the examples below; a) SO42- has- -one atom of S with oxidation number +6( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in SO42- = ( +6 + (-2 x 4)) = -2 The chemical name of this radical is thus sulphate(VI) ion b) NO3- has- -one atom of N with oxidation number +4( refer to Rule 4) -three atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in NO3- = ( +4 + (-2 x 3)) = -1 The chemical name of this radical is thus nitrate(IV) ion. Determine the oxidation number of: I.Nitrogen in; -NO2- => x + (-2 x2)= -1 thus x = -1 – (-4) = + 3 The chemical name of this compound/ion/radical is thus Nitrate(III)ion II. Sulphur in; -SO32- => x + (-2 x3)= -2 thus x = -2 – (-6) = + 4 The chemical name of this compound/ion/radical is thus Sulphate(IV)ion III.
When a metal rod/plate is put in a solution of its own salt, some of the metal ionizes and dissolve into the solution i.e. M(s) -> M+(aq) + e ( monovalent metal) M(s) -> M2+(aq) + 2e ( divalent metal) M(s) -> M3+(aq) + 3e ( Trivalent metal) The ions move into the solution leaving electrons on the surface of the metal rod/plate. 2.The metal rod becomes therefore negatively charged while its own solution positively charged. As the positive charges of the solution increase, some of them recombine with the electrons to form back the metal atoms M+(aq) + e -> M(s) ( monovalent metal) M2+(aq) + 2e -> M(s) (divalent metal) M3+(aq) + 3e -> M(s) (Trivalent metal) 3. When a metal rod/plate is put in a solution of its own salt, it constitutes/forms a half-cell. The tendency of metals to ionize differ from one metal to the other. The difference can be measured by connecting two half cells to form an electrochemical/voltaic cell as in the below procedure: To set up an electrochemical /voltaic cell To compare the relative tendency of metals to ionize Place 50cm3 of 1M Zinc(II) sulphate(VI) in 100cm3 beaker. Put a clean zinc rod/plate into the solution. Place 50cm3 of 1M Copper(II) sulphate(VI) in another 100cm3 beaker. Put a clean copper rod/plate of equal area (length x width) with Zinc into the solution. Connect/join the two metals(to a voltmeter) using connecting wires. Dip a folded filter paper into a solution of Potassium nitrate(V) or sodium(I) chloride(I) until it soaks.
Put a clean copper rod/plate of equal area (length x width) with Zinc into the solution. Connect/join the two metals(to a voltmeter) using connecting wires. Dip a folded filter paper into a solution of Potassium nitrate(V) or sodium(I) chloride(I) until it soaks. Use the folded soaked filter paper to connect/join the two solutions in the two beakers.The whole set up should be as below Repeat the above procedure by replacing: (i)Zinc half cell with Magnesium rod/plate/ribbon dipped in 50cm3 of IM magnesium (II) sulphate(VI) solution V
11 (ii)Zinc half cell with Silver rod/plate/coin dipped in 50cm3 of IM silver(I) nitrate(V) solution (iii)Copper half cell with Iron rod/plate/spoon dipped in 50cm3 of IM Iron (II) sulphate(VI) solution Record the observations in the table below Changes on the 1st metal rod (A) Changes on the 2nd metal rod (B) Changes on the 1st solution (A(aq)) Changes on the 2nd solution (B(aq)) Voltage/voltmeter reading(Volts) Using Zn/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Zinc(II)sulphate (VI)colour remain colourless Blue Copper (II)sulphate (VI)colour fades.
Connect/join the two metals(to a voltmeter) using connecting wires. Dip a folded filter paper into a solution of Potassium nitrate(V) or sodium(I) chloride(I) until it soaks. Use the folded soaked filter paper to connect/join the two solutions in the two beakers.The whole set up should be as below Repeat the above procedure by replacing: (i)Zinc half cell with Magnesium rod/plate/ribbon dipped in 50cm3 of IM magnesium (II) sulphate(VI) solution V
11 (ii)Zinc half cell with Silver rod/plate/coin dipped in 50cm3 of IM silver(I) nitrate(V) solution (iii)Copper half cell with Iron rod/plate/spoon dipped in 50cm3 of IM Iron (II) sulphate(VI) solution Record the observations in the table below Changes on the 1st metal rod (A) Changes on the 2nd metal rod (B) Changes on the 1st solution (A(aq)) Changes on the 2nd solution (B(aq)) Voltage/voltmeter reading(Volts) Using Zn/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Zinc(II)sulphate (VI)colour remain colourless Blue Copper (II)sulphate (VI)colour fades. Brown solid/residue/ deposit 0.8 (Theoretical value=1.10V) Using Mg/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Magnesium(II) sulphate(VI) colour remain colourless Blue Copper (II)sulphate (VI)colour fades Brown solid/residue/ deposit 1.5 (Theoretical value=2.04V) Using Ag/Cu half cell -The rod increase in size /mass /deposited -silver coin/ rod /plate increase in size /mass/ deposited Blue Copper (II)sulphate (VI)colour remains Silver(I)nitrate (V)colour remain colourless 0.20 (Theoretical value=0.46V) Using Fe/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Iron(II)sulphate (VI)colour becomes more green Blue Copper (II)sulphate (VI)colour fades.Brown solid/residue/ deposit 0.60 (Theoretical value=0.78V) From the above observations ,it can be deduced that: (i)in the Zn/Cu half-cell the; -Zinc rod/plate ionizes /dissolves faster than the copper rod/plate to form Zn2+
12 Ionic equation Zn(s) -> Zn2+(aq) + 2e -blue copper ions in the Copper (II)sulphate solution gains the donated electrons to form brown copper metal/atoms Ionic equation Cu2+(aq) + 2e -> Cu(s) This reaction shows /imply the Zinc rod has a higher tendency to ionize than copper.The Zinc rod has a higher net accumulation of electrons and is more negative compared to the copper rod which has lower accumulation of electrons.