JK-TK/Newdata huggingface dataset:
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. The copper rod is therefore relatively more positive with respect to Zinc rod.
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. The copper rod is therefore relatively more positive with respect to Zinc rod. When the two half cells are connected , electrons therefore flow from the negative Zinc rod through the external wire to be gained by copper ions.
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. The copper rod is therefore relatively more positive with respect to Zinc rod. When the two half cells are connected , electrons therefore flow from the negative Zinc rod through the external wire to be gained by copper ions. This means a net accumulation/increase of Zn2+ positive ions on the negative half cell and a net decrease in Cu2+ positive ions on the positive half cell. The purpose of the salt bridge therefore is: (i)complete the circuit (ii)maintain balance of charges /ions on both half cells.
When the two half cells are connected , electrons therefore flow from the negative Zinc rod through the external wire to be gained by copper ions. This means a net accumulation/increase of Zn2+ positive ions on the negative half cell and a net decrease in Cu2+ positive ions on the positive half cell. The purpose of the salt bridge therefore is: (i)complete the circuit (ii)maintain balance of charges /ions on both half cells. For the negative half cell the NO3- /Cl- from salt bridge decrease/neutralise the increased positive(Zn2+) ion. For the positive half cell the Na+ / K+ from salt bridge increase the decreased positive(Cu2+) ion. The voltmeter should theoretically register/read a 1.10Volts as a measure of the electromotive force (e.m.f) of the cell .Practically the voltage reading is lowered because the connecting wires have some resistance to be overcomed. A combination of two half cells that can generate an electric current from a redox reaction is called a voltaic/electrochemical cell. By convention a voltaic/electrochemical cell is represented; M(s) / M2+(aq) // N2+ (aq) / N(s) (metal rod of M)(solution ofM)(solution ofN)(metal rod ofN) Note; a)(i)Metal M must be the one higher in the reactivity series. (ii)It forms the negative terminal of the cell. (iii)It must diagrammatically be drawn first on the left hand side when illustrating the voltaic/electrochemical cell. b)(i)Metal N must be the one lower in the reactivity series. (ii)It forms the positive terminal of the cell. (iii)It must diagrammatically be drawn second/after/ right hand side when illustrating the voltaic/electrochemical cell. Illustration of the voltaic/electrochemical cell. (i)Zn/Cu cell
13 1. Zinc rod ionizes /dissolves to form Zn2+ ions at the negative terminal Zn(s) -> Zn2+(aq) + 2e 2. Copper ions in solution gain the donated electrons to form copper atoms/metal Cu2+(aq) + 2e -> Cu(s) 3.Overall redox equation Cu2+(aq) + Zn(s) -> Zn2+(aq) + Cu(s) 4.cell representation.
(i)Zn/Cu cell
13 1. Zinc rod ionizes /dissolves to form Zn2+ ions at the negative terminal Zn(s) -> Zn2+(aq) + 2e 2. Copper ions in solution gain the donated electrons to form copper atoms/metal Cu2+(aq) + 2e -> Cu(s) 3.Overall redox equation Cu2+(aq) + Zn(s) -> Zn2+(aq) + Cu(s) 4.cell representation. Zn(s) / 1M, Zn2+(aq) // 1M,Cu2+(aq) / Cu(s) E0 = +1.10 V 5.cell diagram (ii)Mg/Cu cell 1. Magnesium rod ionizes /dissolves to form Mg2+ ions at the negative terminal Mg(s) -> Mg2+(aq) + 2e 2. Copper ions in solution gain the donated electrons to form copper atoms/metal Cu2+(aq) + 2e -> Cu(s) 3.Overall redox equation Cu2+(aq) + Mg(s) -> Mg2+(aq) + Cu(s) 4.cell representation. Mg(s) / 1M, Mg2+(aq) // 1M,Cu2+(aq) / Cu(s) E0 = +2.04 V 5.cell diagram. Voltmeter
14 (iii)Fe/Cu cell 1. Magnesium rod ionizes /dissolves to form Mg2+ ions at the negative terminal Fe(s) -> Fe2+(aq) + 2e 2. Copper ions in solution gain the donated electrons to form copper atoms/metal Cu2+(aq) + 2e -> Cu(s) 3.Overall redox equation Cu2+(aq) + Fe(s) -> Fe2+(aq) + Cu(s) 4.cell representation. Fe(s) / 1M, Fe2+(aq) // 1M,Cu2+(aq) / Cu(s) E0 = +0.78 V 5.cell diagram. Fe Fe++ V
15 (iv)Ag/Cu cell 1. Copper rod ionizes /dissolves to form Cu2+ ions at the negative terminal Cu(s) -> Cu2+(aq) + 2e 2.
Fe(s) / 1M, Fe2+(aq) // 1M,Cu2+(aq) / Cu(s) E0 = +0.78 V 5.cell diagram. Fe Fe++ V
15 (iv)Ag/Cu cell 1. Copper rod ionizes /dissolves to form Cu2+ ions at the negative terminal Cu(s) -> Cu2+(aq) + 2e 2. Silver ions in solution gain the donated electrons to form silver atoms/metal 2Ag+(aq) + 2e -> 2Ag(s) 3.Overall redox equation 2Ag+(aq) + Cu(s) -> Cu2+(aq) + 2Ag(s) 4.cell representation. Cu(s) / 1M, Cu2+(aq) // 1M,2Ag+(aq) / 2Ag(s) E0 = +0.46 V 5.cell diagram. Standard electrode potential (Eᶿ) The standard electrode potential (Eᶿ) is obtained if the hydrogen half cell is used as reference. The standard electrode potential (Eᶿ) consist of inert platinum electrode immersed/dipped in 1M solution of (sulphuric(VI) acid) H+ ions. Hydrogen gas is bubbled on the platinum electrodes at: (i)a temperature of 25oC (ii)atmospheric pressure of 101300Pa/101300Nm-2/1atm/760mmHg/76cmHg (iii)a concentration of 1M(1moledm-3) of sulphuric(VI) acid/ H+ ions and 1M(1moledm-3) of the other half cell. Hydrogen is adsorbed onto the surface of the platinum. An equilibrium/balance exist between the adsorbed layer of molecular hydrogen and H+ ions in solution to form a half cell.
Hydrogen gas is bubbled on the platinum electrodes at: (i)a temperature of 25oC (ii)atmospheric pressure of 101300Pa/101300Nm-2/1atm/760mmHg/76cmHg (iii)a concentration of 1M(1moledm-3) of sulphuric(VI) acid/ H+ ions and 1M(1moledm-3) of the other half cell. Hydrogen is adsorbed onto the surface of the platinum. An equilibrium/balance exist between the adsorbed layer of molecular hydrogen and H+ ions in solution to form a half cell. ½ H2 (g) ==== H+ (aq) + e The half cell representation is: Pt,½ H2 (g) / H+ (aq), 1M Cu Ag V Cu++ Ag+ Cu(s) + 2Ag+ (aq) Cu2+(aq) + 2Ag(s)
16 The standard electrode potential (Eᶿ) is thus defined as the potential difference for a cell comprising of a particular element in contact with1M solution of its own ions and the standard hydrogen electrode. If the other electrode has a higher/greater tendency to lose electrons than the hydrogen electrode, the electrode is therefore negative with respect to hydrogen electrode and its electrode potential has negative (Eᶿ) values. If the other electrode has a lower/lesser tendency to lose electrons than the hydrogen electrode, the electrode is therefore positive with respect to hydrogen electrode and its electrode potential has positive (Eᶿ) values.
½ H2 (g) ==== H+ (aq) + e The half cell representation is: Pt,½ H2 (g) / H+ (aq), 1M Cu Ag V Cu++ Ag+ Cu(s) + 2Ag+ (aq) Cu2+(aq) + 2Ag(s)
16 The standard electrode potential (Eᶿ) is thus defined as the potential difference for a cell comprising of a particular element in contact with1M solution of its own ions and the standard hydrogen electrode. If the other electrode has a higher/greater tendency to lose electrons than the hydrogen electrode, the electrode is therefore negative with respect to hydrogen electrode and its electrode potential has negative (Eᶿ) values. If the other electrode has a lower/lesser tendency to lose electrons than the hydrogen electrode, the electrode is therefore positive with respect to hydrogen electrode and its electrode potential has positive (Eᶿ) values. Table showing the standard electrode potential (Eᶿ) of some reactions Reaction (Eᶿ) values in volts F2 (g)+ 2e -> 2F- (aq) +2.87 H2 O2 (aq)+ H+ (aq) +2e -> H2 O (l) +1.77 Mn O4- (aq)+ 4H+ (aq) +3e -> MnO2 (s) +H2 O (l) +1.70 2HClO (aq)+ 2H+ (aq) +2e -> Cl2 (aq) +2H2 O (l) +1.59 Mn O4- (aq)+ 4H+ (aq) +5e -> Mn2+ (aq) +H2 O (l) +1.51 Cl2 (g)+ 2e -> 2Cl- (aq) +1.36 Mn O2 (s)+ 4H+ (aq) +2e -> Mn2+ (aq) +2H2 O (l) +1.23 Br2 (aq)+ 2e -> 2Br- (aq) +1.09 NO3- (aq)+ 2H+ (aq) + e -> NO2 (g) + H2 O (l) +0.80 Ag+ (aq) + e -> Ag(s) +0.80 Fe3+ (aq) + e -> Fe2+ (aq) +0.77 2H+ (aq)+ O2 (g) -> H2 O2 (aq) +0.68 I2 (aq)+ 2e -> 2I- (aq) +0.54 Cu2+ (aq) + 2e -> Cu(s) +0.34 2H+ (aq) + 2e -> H2(g) +0.00 Pb2+ (aq) + 2e -> Pb(s) -0.13 Fe2+ (aq) + 2e -> Fe(s) -0.44 Zn2+ (aq) + 2e -> Zn(s) -0.77 Al3+ (aq) + 3e -> Al(s) -1.66 Mg2+ (aq) + 2e -> Mg(s) -2.37 Na+ (aq) + e -> Na(s) -2.71 K+ (aq) + e -> K(s) -2.92 Note: (i)Eᶿ values generally show the possibility/feasibility of a reduction process/oxidizing strength.
(ii)The element/species in the half cell with the highest negative Eᶿ value easily gain / acquire electrons. It is thus the strongest oxidizing agent and its reduction process is highly possible/feasible. The element/species in the half cell with the lowest positive Eᶿ value easily donate / lose electrons. 17 It is thus the strongest reducing agent and its reduction process is the least possible/feasible. (iii)The overall redox reaction is possible/feasible is it has a positive (+) Eᶿ. 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.
17 It is thus the strongest reducing agent and its reduction process is the least possible/feasible. (iii)The overall redox reaction is possible/feasible is it has a positive (+) Eᶿ. 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.
(iii)The overall redox reaction is possible/feasible is it has a positive (+) Eᶿ. 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.