LABORATORY REPORT FOR EXPERIMENT 1 HYDROCARBON AND HYDROXY COMPOUNDS A. REACTIONS OF HYDROCARBONS OBJECTIVE(S): To study the properties of HC and Hydroxy compounds DATA/RESULTS Observation Test Cyclohexane, C6H12 1 Addition of water 2 Ignition 3 Bromine (CH2Cl2) Formation of 2 layers Burned with less sooty flame Cyclohexene, C6H10 Formation of 2 layers Burned with more sooty flame Brown color of bromine remains Brown color of bromine discharged (b) In sunlight Brown color of bromine discharged Brown color of bromine discharged 4 KMnO4(aq) H+(aq), cold The purple color of Mn(VII) ions remains The purple color of Mn(VII) ions turn to colorless Mn(II) ions 5 KMnO4(aq) The purple color of Mn(VII) ions remains The purple color of Mn(VII) ions turn to colorless Mn(II) ions (a) In the dark H+(aq), hot [2] any two completed testing DISCUSSION Addition of water : Hydrocarbons insoluble in H2O , HC is a non polar substance and less dense than water [1] (any one point) 1 Ignition : C6H12 + 9 O2 → 6 CO2 + 6 H2O C6H10 + 17/2 O2 → 6 CO2 + 5 H2O more sooty flame Bromination: C6H12 + Br2 in CH2Cl2 uv C6H10 + Br2 in CH2Cl2 → C6H11Br + HBr C6H10Br2 Oxidation: C6H10 + MnO4- cold C6H10(OH)2 + Mn2+ @ MnO2 (brown ppt) purple C6H10 + MnO4- colorless heat HOOCCH2CH2CH2CH2COOH + Purple Mn2+ colorless [1] on observation OR equation 2 B. REACTIONS OF ALCOHOLS OBJECTIVE(S): DATA/RESULTS: Observation Test 1-propanol 2-methyl 2-butanol -2-propanol 1 PCl5(s) White fumes produced White fumes produced White fumes produced 3 K2Cr2O7(aq) The orange color changes to green/blue color The orange color changes to green/blue color The orange color remains 4 Lucas test No white ppt formed at room temperature White ppt formed after 5 minutes White ppt formed immediately (10-30 mins) (<5 mins) No formation of yellow ppt wt antiseptic smell yellow ppt wt antiseptic smell No formation of yellow ppt wt antiseptic smell 5 Iodoform test [2] any two completed testings DISCUSSION: PCl5 : CH3CH2CH2OH + PCl5 → CH3CH2CH2Cl + POCl3 + HCl (White fume) CH3CH2CH(OH)CH3 + PCl5 → CH3CH2CH(Cl)CH3 + POCl3 + HCl (White fume) CH3C(OH)(CH3)2 + PCl5 → CH3C(Cl)(CH3)2 + POCl3 + HCl (White fume) Oxidation, K2Cr2O7 (aq) : CH3CH2CH2OH + Cr2O72- + 8H+ → CH3CH2COOH + 2Cr3+ + 7H2O orange green CH3CH2CH(OH)CH3 + Cr2O72- + 8H+ → CH3CH2COCH3 + 2Cr3+ + 7H2O orange green Lucas test: 3 CH3CH2CH(OH)CH3 + ZnCl2 + HCl → CH3CH2CH(Cl)CH3 + H2O + [Zn(OH)Cl2] cloudy CH3C(OH)(CH3)2 + ZnCl2 + HCl → CH3C(Cl)(CH3)2 + H2O + [Zn(OH)Cl2] cloudy rate of formation of cloudiness: 30 > 20 >10 alcohols Iodoform test: CH3CH2CH(OH)CH3 + I2 + NaOH → CH3CH2COO-Na+ + CHI3 + NaI + H2O [1] yellow ppt with antiseptic smell CONCLUSION [1] (any 1) 1. Cyclohexene is an unsaturated compound and is more reactive than cyclohexane. 2. Cyclohexene undergoes halogenation with or without uv light and can be oxidized. Whereas cyclohexane only undergoes halogenation with the presence of uv light and cannot be oxidized. 3. Confirmatory test for alcohol: PCl 5 4. Only 1o and 2o alcohol can be oxidized. 5. Iodoform test is used to test methyl alcohol. REFLECTION ‘See you not that Allâh sends down water (rain) from the sky, and We produce therewith fruits of varying colours, and among the mountains are streaks white and red, of varying colours and (others) very black’ (35:27) This verse shows the variety of Allah creation in this world. Each and every creation has its own significance and play roles in the world. REFERENCES 1. Tan Yin Toon, Chemistry for Matriculation Semester 2, Oxford Fajar Sdn.Bhd. (pg. 114-210) 2. Holy Quran; Text and Translation by Abdullah Yusuf Ali (pg 426) TOTAL MARKS 8 4 LABORATORY REPORT FOR EXPERIMENT 2 CARBONYL COMPOUND AND PHENOL OBJECTIVE(S):To study the properties of carbonyl and phenol A. REACTIONS OF CARBONYL COMPOUNDS DATA/RESULTS: Test Observation Ethanal Propanone Benzaldehyde 1 Brady’s reagent Yellow orange precipitate is formed Yellow orange precipitate is formed Yellow orange precipitate is formed 2 K2Cr2O7(aq) Orange colour of K2Cr2O7(aq) turns green Orange colour of K2Cr2O7(aq) remains Orange colour of K2Cr2O7(aq) turns green 3 Tollen’s reagent Silver mirror is formed on the wall of the test tube/grayish black precipitate The solution remains colourless Silver mirror is formed on the wall of the test tube/grayish black precipitate 4 Fehling’s reagent Blue colour Blue colour remains Blue colour remains Yellow precipitate with anticeptic smell is formed Colourless solution changes to brick red 5 Iodoform test Yellow precipitate with anticeptic smell is formed [any one correct testing and observations √1M] 5 DATA/RESULTS: Observation Test Unknown 1 Brady’s reagent Yellow orange precipitate is formed 2 K2Cr2O7(aq) Orange colour remains 3 Tollen’s reagent The solution remains colourless 4 Fehling’s reagent Blue colour remains 5 Iodoform test Yellow precipitate with antiseptic smell is formed Class/Family of unknown: Ketone @ aldehyde [any one correct testing and observations √1M] B. REACTIONS OF PHENOL DATA/RESULTS Test Observations Phenol 1 Addition of water 2 Acidity Emulsion (cloudiness) is formed and colourless homogeneous solution formed in hot water bath. (a) Litmus paper (a)Blue litmus paper turns red (b) (i) NaOH (b)(i) homogeneous solution formed (ii) HCl (ii) emulsion formed 3 Iron(III) chloride Blue violet/purple solution formed 4 Bromine water White precipitate formed/ brown colour of bromine discharged [any one correct testing and observations √1M] 6 DISCUSSIONS: Reaction of carbonyl compound 1. Brady’s Reagent [√1M for equation OR observation] a. Ethanal O2 N O CH3 CH + O2 N NH2 NH NO2 CH3 CH NNH N2 O + H2 O ethanal-2,4-dinitrophenylhydrazone (yellow precipitate ) b. Propanone O2 N O CH3 CCH3 + O2 N NO2 NH2 NH CH3 C(CH3 ) NNH N2 O + H O 2 propanone-2,4-dinitrophenylhydrazone (yellow precipitate) c. Benzaldehyde O2 N C6H5COH + O2 N NO2 NH2 NH C6H5CH NNH N2 O + H O 2 benzaldehyde-2,4-dinitrophenylhydrazone (yellow precipitate) 2. K2Cr2O7 [√1M for equation OR observation] a. Ethanal O CH CH 3 O K C r O /H +/w ar m 2 2 7 CH COH 3 ethanoic acid + Cr3+ (green/blue) b. Benzaldehyde O O CH COH + K2 Cr2 O7 /H /warm + Cr3+ (green/blue) benzoic acid 7 3. Tollen’s reagent [√1M for equation OR observation] a. Ethanal O O CH3 CH + + 2 [Ag(NH3 )2 ] + - - 3 OH CH3 CO 2Ag(s)+ 4NH3 + + H2 O Silver mirror b. Benzaldehyde O O CH CO - + + 2 [Ag(NH3 )2 ] + - 3 OH + 2Ag(s)+ 4NH3 + H2 O Silver mirror 4. Fehling’s reagent [√1M for equation OR observation] ~ positive for aliphatic aldehyde a. Ethanal O O CH3 CH 5. + 2Cu2+ + - 5OH + Cu2 O (s) Brick red precipitate - CH3 CO + 3 H2 O Iodoform test [√1M for equation OR observation] a. Ethanal O CH3 CH O + 3I 2 + 4 NaOH CH3 CONa + 3 NaI + H2 O + CHI3 Yellow precipitate with antiseptic smell 8 b. Propanone O O CH3 CCH3 + 3I2 + 4 NaOH CH3 CONa + 3 NaI + H2 O + CHI3 Yellow precipitate with antiseptic smell Reaction of Phenol 1. Addition of water: Phenol partially soluble in water. √1M 2. Acidity (a) phenol is an acidic compound ⇌ C6H5OH + H2O C6H5O- + H3O+ phenoxide ion (b) (i) C6H5OH + NaOH C6H5O- + Na+ + H2O phenoxide ion(soluble) (ii) C6H5O- + H+ C6H5OH cloudy as phenol precipitate out 3. Iron (III) chloride OH OH FeCl3 + FeCl3 → Blue violet soln 9 4. Bromine in water [√1M for equation OR observation] OH OH Br Br Br2 (aq) Br White ppt (2,4,6-tribromophenol) CONCLUSION: 1. Carbonyl compounds undergo nucleophilic addition reaction 2. Aldehyde can be oxidized/reducing agent to carboxylic acid whereas ketones cannot be oxidized 3. Ethanal and propanone can be distinguish from benzaldehyde by using Iodoform test due to the carbonyl group being bonded to the methyl group 4. Phenol is an acidic compound which partially soluble in water. REFLECTION: ‘He knows that which goes into the earth and that which comes forth from it, and that which descend from the heaven and that which ascends to it. And He is the Most Merciful, the Oft¬Forgiving ‘ (34:2) Allah knows everything that happens around us, we as the caliph should observe the change in the reaction to know actually happening. REFERENCES: 1. Tan Yin Toon, Chemistry for Matriculation Semester 2, Oxford Fajar Sdn.Bhd. (pg. 219-250) 2. Holy Quran; Text and Translation by Abdullah Yusuf Ali (pg 416) TOTAL MARKS 10 10 LABORATORY REPORT FOR EXPERIMENT 3 HESS’S LAW - ADDITIVITY OF HEAT OF REACTION OBJECTIVE : To calculate the heat of reaction, H based on the Hess’s law RESULTS : Experimental data Reaction I Reaction II Reaction III 2.0000 2.0000 - 1 Mass of solid NaOH (g) 2 Volume NaOH (mL) - - 50.0 3 Volume water (mL) 100.0 - - 4 Volume HCl (mL) - 100.0 50.0 5 Total mass of solution (g) 102.0 102.0 150.0 6 Final temperature (C) 25.6 31.7 26.0 7 Initial temperature (C) 20.9 20.6 20.7 8 Change in temperature, T (C) 4.7 11.1 5.3 9 Heat, qreaction (kJ) 2.0 4.73 3.3 0.050000 0.050000 0.0500 -40 -94.6 -66 10 No. of moles of NaOH/mol 11 H (kJ/mol) [√1M for H (kJ/mol) with –ve sign and correct arrangement (Rxn II>Rxn1 &RxnIII) ] Answers to the questions 1. (a) NaOH(s) + H+(aq) + Cl-(aq) H2O(l) + Na+(aq) + Cl-(aq) Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) 11 √1M √1M (b) Heat of reaction I + heat of reaction III = [-40 + (-72)] kJ/mol = -106 kJ/mol Heat of reaction II = -94.6 kJ/mol 2. % Error = Theoretical H – Experimental H x 100% Theoretical H = (94.6 - 106 )kJ/mol 100% 94.6 kJ/mol = 12 % √1M DISCUSSIONS: Theory √1M Heat, q is a transfer of energy between two objects at different temperature Enthalpy change, H is the energy changes that occur in a chemical reaction at constant pressure. First law of thermodynamic: - The law of conservation of energy - Energy can be changed from one form into another - It can neither be created nor destroyed Hess’s law states that when reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. Hsoln is the enthalpy change when 1 mol of solute dissolves in a solvent. Observations √1M As the reaction occurs, the temperature of the solution increases/exothermic rxn. Sources of error √1M Did not stir the solution simultaneously. Did not add the NaOH(s) into the solution immediately. Did not cover the styroform cup properly, thus some heat escape to the surrounding. 12 Precautionary steps √1M NaOH(s) is very hygroscopic (easily absorbed moisture from the air). Thus weigh it and proceed to the next step without delay. Constantly stir the solution. The two solutions must be at the same temperature. Cover the styroform cup properly to avoid the escape of heat to the surrounding. CONCLUSION: The experimental value for the heat of reaction, H for the reaction between NaOH(s) + HCl(aq) H2O(l) + NaCl(aq) is -92.8 kJ/mol Based on the Hess’s law, the heat of reaction, H for the reaction between NaOH(s) + HCl(aq) H2O(l) + NaCl(aq) is -85.1 kJ/mol REFLECTION: ‘The Day that we roll up the heavens like a scroll rolled up for books(completed)even as We produced the first creation, so shall We produced a new one: a promise We have undertaken; truly shall We fulfill it.’ (21:104) This is about the Day of Judgment, where Allah promised how he created at the beginning, He created the same for the last. BIBLIOGRAPHY: 1. Brown, Lemay and Bursten; Chemistry: The Central Science, 10 th edition, Pearson Prentice Hall, page 176-178 2. Holy Quran; Text and Translation by Abdullah Yusuf Ali, page 393. TOTAL MARKS 8 13 LABORATORY REPORT FOR EXPERIMENT 4 DISTURBING THE POSITION OF EQUILIBRIUM – LE CHATELIER’S PRINCIPLE OBJECTIVES : √1M - To analyze the effect of concentration (common ion) and temperature on the equilibrium position. - To explain the change of equilibrium position based on the Le Chatelier’s principle RESULTS : PART A: Effect of Temperature Change on a Physical System Test conditions Observations Cooling to 0.0C White crystals formed √1M Warming to room temperature White crystals dissolved PART B: Common Ion Effect on a Chemical System Test tube Conditions Observations 1 Equilibrium without stress Orange 2 Increase H+ added from H2SO4 Darker orange 3 Decrease H+ from added NaOH Yellow √1M 4 After addition of H2SO4 to solution that was added NaOH previously 14 Darker orange PART C: Common Ion Effect on a Chemical System Test tube Ions added Observations 1 Control Orange 2 Fe3+ Darker orange √1M 3 SCN- Blood red 4 K+ and Cl- 5 OH- Light yellow OR Light orange 6 Ag+ Pale yellow /cloudy Pale yellow OR Pale orange DISCUSSIONS: Theory Equilibrium occurs when the concentrations of reactants and products remain constant. Equilibrium is established when the rate of forward reaction equals the rate of backward reaction. Factors that affect equilibrium: - Concentration - Temperature - Pressure - Catalyst Le Chatelier’s principle states that if a system at equilibrium is subjected to a change (stress), the equilibrium tends to shift so as to minimize the effect of change (relieve the stress). Observations and explanations PART A √1M The equilibrium as written is endothermic in the forward direction. When the temperature is decreased, white crystal formed. This shows that the equilibrium has shifted to the left (in the exothermic direction), therefore releasing heat. When the temperature is increased, white crystal dissolves as the equilibrium shifts to the right. PART B√1M Adding more acid will increase the concentration of H +(aq) ions. The equilibrium position will shift to the right where some of the extra H + ions added react with the CrO42- ions to form orange Cr2O72- ions and so minimize the increased in the concentration of H+(aq). 15 Adding base (containing OH- ions) will reduce the H+(aq) concentration by neutralization: H+(aq) + OH-(aq) H2O(l) The equilibrium position will shift to the left to minimize the decrease in H+(aq) ion concentration, forming yellow CrO42- ions. PART C√2M Adding more FeCl3 will increase the concentration of Fe3+(aq) ions. As a response, the equilibrium position will shift to the right where some of the extra Fe3+ ions added react with the SCN- ions to form blood red [Fe(SCN)]2+ ions and so minimize the increased in the concentration of Fe 3+(aq). Adding more KSCN will increase the concentration of SCN -(aq) ions. As a response, the equilibrium position will shift to the right where some of the extra SCN- ions added react with the Fe3+ ions to form blood red [Fe(SCN)]2+ ions and so minimize the increased in the concentration of SCN-(aq). Adding KCl (containing Cl - ions) will reduce the Fe3+(aq) concentration by a reaction: Cl-(aq) + Fe3+(aq) FeCl3(s) The equilibrium position will shift to the left to increase the Fe3+(aq) ion concentration again, forming pale yellow solution. Adding NaOH (containing OH- ions) will reduce the Fe3+(aq) concentration by a reaction: Fe3+(aq) + OH-(aq) Fe(OH)3(s) The equilibrium position will shift to the left to minimize the decrease in Fe3+(aq) ion concentration, forming yellow solution. Adding AgNO3 (containing Ag+ ions) will reduce the SCN-(aq) concentration by a reaction: Ag+(aq) + SCN-(aq) AgSCN(s) The equilibrium position will shift to the left to minimize the decrease in SCN(aq) ion concentration, forming yellow solution. Sources of error Do not wait until the brown precipitate of Fe(OH) 3 and white solid of AgSCN to settle down , thus difficult to identify the color the resulting solution. Use the same dropper for different solution may cause contamination. Precautionary steps Wait until the brown precipitate of Fe(OH) 3 and white solid of AgSCN to settle down to see the color of the solution clearly. 16 CONCLUSION : An increased in temperature shifts equilibrium in the endothermic direction; a @decrease in temperature shifts equilibrium in the exothermic direction. Equilibrium will shift to the right when the concentration of a reactant is increased or the concentration of the product is decreased. @ Equilibrium will shift to the left when the concentration of a product is increased or the concentration of the reactant is decreased. REFLECTION: ‘Glory to Allah who created in pairs all things that the earth produces, as well as their own (human) and things of which they have no knowledge’ (Yaasin:36) This experiment is one proved that everything created is in pairs. Like this experiment that observed the equilibrium system, forward and reverse reaction, there would be shift to the left or right in order to achieve a new equilibrium. BIBLIOGRAPHY: 1. Chemistry, Principle and Reactions, Masterton and Hurley, 5th edition, page 335. 2. Brown, Le May and Bursten; Chemistry: The Central Science, 10 th edition, Pearson Prentice Hall, page 649-655. TOTAL MARKS 8 17 LABORATORY REPORT FOR EXPERIMENT 5 ACID-BASE TITRATION OBJECTIVES: To determine the molarity of acetic acid used To obtain the shape the titration curve using interface To determine the pH of solution at equivalence point through the graph RESULTS: Molarity of NaOH used/M 0.100 M Volume of NaOH used/mL (At equivalence point) 25.60 mL Volume of CH3CO2H used/mL 25.0 mL Table 1: Before equivalence point Volume of 0.00 2.00 4.00 6.00 NaOH added/mL pH of 3.9 4.4 4.7 4.9 solution Volume of added NaOH pH of solution 10.00 12.00 14.00 16.00 5.1 5.3 5.4 5.5 5.7 18.00 mL 20.00 mL 22.00 mL 23.00 mL 6.0 6.2 6.3 6.4 Volume of 2 drops 2 drops 2 drops NaOH (23.50 (23.60 (23.70 added mL) mL) mL) pH of 6.9 7.0 7.1 solution √1M pH value at equivalence point 2 drops (23.10 mL) 6.5 2 drops (23.20 mL) 6.6 2 drops (23.30 mL) 6.7 2 drops (23.40 mL) 6.8 At end point (permanent pale pink) Volume = ___25.60_ mL 9.1 Table 2: After equivalence point Volume of 2.00 mL 4.00 mL 6.00 mL 8.00 mL NaOH added pH of 11.9 12.4 12.6 12.7 solution Volume of NaOH added pH of solution 8.00 20.00 mL 13.1 18 10.00 mL 12.00 mL 14.00 mL 16.00 mL 18.00 mL 12.8 12.9 12.9 13.0 13.0 CALCULATIONS: 1. Calculate the number of moles of NaOH from this experiment. Moles NaOH = (0.100 mol/L)(25.60 mL)(1 L/1000 mL) = 2.56 × 10-3 mol √1M 2. Calculate the molarity of acetic acid used in this experiment. Moles CH3COOH = (2.56 × 10-3 mol NaOH)(1 mol CH3COOH/ 1 mol NaOH) = 2.56 × 10-3 mol Molarity CH3COOH = (2.56 × 10-3 mol)(25.0 mL) (1 L/1000 mL) = 0.102 M √1M 3. Based on your graph, what is the pH of solution at equivalence point? Label it on your graph. √1M graph with correct shape The pH of solution at equivalence point is 8.1 4. Is the pH of the solution acidic, basic or neutral? Explain your answer briefly. Basic √1M The acetate ion that is dissociated from the sodium acetate salt can undergo hydrolysis in water to produce OH ion. √1M DISCUSSIONS: Theory Titration curve is a plot of pH against volume of added base or added acid. It shows the pH changes that occur when an aqueous acid reacts with an aqueous base. Phenophtalein is used as an indicator because its color changes around pH 810. Observation √1M As NaOH is added to the acid, a gradual increase in the pH is observed until the solution gets close to the equivalence point. Near the equivalence point, a rapid change in pH occurs. Beyond the equivalent point, where more bases have been added than acid, more gradual changes in pH are observed. The end point is reached when the solution turns to permanent pale pink, which is about 25.0 mL NaOH added. The changes of pH around the equivalent point occur at pH 7-11. Sources of error Do not rinse the pH sensor with distilled water before and after measuring each value of pH of solution. Precautionary step Rinse the pH sensor with distilled water before and after measuring each value of pH of solution. Constantly stir the solution. 19 CONCLUSION: √1M The molarity of CH3COOH from the experiment is 0.102 M REFLECTION ‘And He has cast in the earth anchorages (mountains standing firm) so that it should not reel with you, and rivers and roads that possibly you would be guided’ (16:15) This verse mentions many signs of creation that give evidence of the divinity of Allah as reflected in the greatness of His creation. The Surah also mentions the many blessings of Allah on His worshippers in His perfect knowledge, the greatness of His wisdom and His precise planning. REFERENCES/BIBLIOGRAPHY: 1. Silberberg, Chemistry: the Molecular Nature of Matter and Change, 5 th Edition, page 841-850. 2. Holy Quran; Text and Translation by Abdullah Yusuf Ali, page 626. TOTAL MARKS 8 20 LABORATORY REPORT FOR EXPERIMENT 6 REDUCTION POTENTIALS IN MICRO–VOLTAIC CELL OBJECTIVES √1M To measure the reduction potential of micro-voltaic cells To rank metals according to measured reduction potentials RESULTS Data Table 1: Copper as the Reference Metal No. Combination Potential (V) Metal for Red (+) Tip Metal for Black (-) Tip 1 Copper/Zinc 0.784 Copper Zinc 2 Copper/Lead 0.201 Copper Lead 3 Copper/Silver 0.796 Silver Copper 4 Copper/Iron 0.507 Copper Iron √1M All values +ve Data Table 2: Rank the Metals Metal Lowest (-) Reduction Potential, Eo (V) Zinc -0.784 V Iron -0.507 V Lead -0.201 V Copper 0.000 V Silver +0.796 V Highest (+) Reduction Potential, Eo (V) √1M For arrangement of metals according to measured potentials *For reference: Theoretical electrode reduction potentials Eo Metal Zinc Iron Lead Copper Silver Eo (V) - 0.762 - 0.409 - 0.127 + 0.339 + 0.799 21 Data Table 3: Predictions and Results Half-Cell Combination Predicted Potentials (V) Measured Potential (V) Percent Error (%) Zinc/Lead -0.201 – (-0.784) = 0.583 0.576 1.20 Zinc/Silver 0.796 – (-0.784) = 1.580 1.523 3.61 Zinc/Iron -0.507 – (-0.784) = 0.277 0.292 5.40 Lead/Silver 0.796 – (-0.201) = 0.997 0.938 5.92 Lead/Iron -0.201 – (-0.507) = 0.306 0.315 2.94 Iron/Silver 0.796 – (-0.507) = 1.303 1.280 1.77 √1M for any correct calculation in predicted potential columns QUESTIONS In a galvanic cell, what happen at the anode and cathode? Anode: Oxidation √1M Cathode: Reduction √1M What is the direction of electron flow? Explain. Electrons flow out from the negative anode (electrons are released in the oxidation process) through the external wire to the positive cathode (where the electrons are accepted in the reduction process). √1M DISCUSSIONS Theory A voltaic cell uses a spontaneous oxidation-reduction reaction to produce electrical energy. Placing a piece of metal into a solution containing a cation of the metal produces half-cells. In this micro-version of a voltaic cell, the half-cell is a small piece of metal placed into three drops of corresponding cation solution on a piece of filter paper. A porous barrier or a salt bridge normally separates the two half-reactions. Here, the salt bridge is made from several drops of aqueous sodium nitrate (NaNO3) placed on the filter paper linking the two half-cells. 22 Observations When the tip of the red(+) end of the voltage sensor is touched to one metal and the tip of the black(-) end is touched to the other metal the positive voltage reading is observed. By the way, the ends of the voltage sensor is reversed if the voltage drops to negative value. Sources of error The piece of metal is not completely sand with sand paper. The drops of each solution are not enough. The addition of sodium nitrate solution is not enough. Precautions √2M Top side of the metals should be kept dry. Carefully sand each piece of metal on both sides. Dampen the filter paper with more NaNO3 solution from time to time during the experiment. The voltage reading must be positive as the redox reaction that occurs in the voltaic cell is spontaneous. CONCLUSION The reduction potential sequence for the metals is: Zinc, Iron, Lead, Copper, Silver. REFLECTION √1M ‘For all there will be degrees (or ranks) according to what they did. And your Lord is not unaware of what they do’ (6:132) Allah has arranged and ranked everything in the world at their own place, shows that Allah is very precise in everything (Al-Khabir-The Most Meticulous) BIBLIOGRAPHY 1. Silberberg, Chemistry: the Molecular Nature of Matter and Change, 5 th Edition, page 933-943. 2. Holy Quran; Text and Translation by Abdullah Yusuf Ali, page 130. TOTAL MARKS 10 END 23