TJC H2 CHEM P2 MARK SCHEME Prelim

Mark Scheme (2011 H2 Structured Prelims) 2011 Prelims JC2 H2 Chemistry Paper 2 [Turn Over 1 1 (a) HO reaction Mg(s) + Cu(s) + 2HCl (aq) CuC l2(aq) + H2(g) + Mg(s) MgCl2(aq) + H2(g) + Cu(s)  By Hess’ law, HO reaction = HO 1 - HO 2 = (x - y) kJ mol-1  Correct energy cycle with balanced equations and state symbols [2] (b) Example: Assuming 100 cm 3 of acid is used and a temperature raise of 10 oC is to be achieved. macidCT = m Mg/Mr(Mg) x 460 000 100 x 4.2 x 10 = m Mg x 460 000/24.3 m Mg = (24.3)[100 x 4.2 x 10)/460000] = 0.22 g No. of moles of acid required to completely react off all Mg(s) = 2(0.22/24.3) = 0 . 0 1 8 1 m o l No. of moles of HCl in 100 cm3 = 5 x 100/1000 = 0.5 mol Hence, there is sufficient acid in 100 cm3 to react with all Mg(s). Marking points [Justification]:  Appropriate change in temperature (any value in the range 5 – 15 oC)  Appropriate mass and volume of acid used to achieve the change in temperature.  Appropriate mass and volume of acid in terms of appropriate quantity i.e. not too small or large amount (at least 0.1 g and 20 cm3) Procedure:  Use a 100 cm 3 measuring cylinder to measure out 100 cm 3 of HC l(aq) HO 2 HO 1

Mark Scheme (2011 H2 Structured Prelims) 2011 Prelims JC2 H2 Chemistry Paper 2 [Turn Over 2 acid into a styrofoam cup. Record the initial temperature of the solution.  Using an electronic balance, weigh out accurately about 0.2 g of powdered Mg(s) into a dry empty weighing bottle.  Use a -10 to 100 oC thermometer to measure the temperature of the 100 cm3 of HCl(aq) acid in the styrofoam cup.  Pour the powered Mg(s) into the Styrofoam cup containing the HCl(aq), stir and measure the highest temperature reached. Record the final temperature reached.  Reweigh the weighing bottle to determine the exact amount of Mg(s) that was poured into the styrofoam cup. Marking points [Procedures]:  Correct sequence of steps that will lead to useful data.  Selection of apparatus is appropriate and types of apparatus used are clearly stated. (Note: this mark is only awarded if the suggested procedure is appropriate.) [5] (c) Marking points [Tables]:  Correct headers and units for both tables to record data about mass and temperature. Mass of weighing bottle + Mg(s) / g Mass of weighing bottle + residual Mg(s) after transferring / g Mass of Mg(s) / g Initial temperature of HCl(aq) / oC Final temperature of HCl(aq) / oC Temperature change / oC [1] (d) Mass of Mg(s) = a g

Mark Scheme (2011 H2 Structured Prelims) 2011 Prelims JC2 H2 Chemistry Paper 2 [Turn Over 3 T = b oC Volume of acid = c cm 3 Heat change = mC T = c x 4.2 x b = 4.2(cb) J No. of moles of Mg(s) used = a/24.3 mol Therefore H O 1 = - 4.2(cb) ÷ a/24.3 = - 102.06 (cb/a) J mol -1 Marking points [Data Collection]:  Shows correct way to calculate heat change.  Shows correct way to calculate HO 1. [2] (e) The acid used is corrosive as it is rather concentrated 5 mol dm -3, one should avoid any contact with the acid by wearing gloves. Or The reaction is vigorous and might splash out of the styroform cup, one should wear gloves and goggles. [1] (f) G = H - TS A reaction can be considered feasible if G is negative. S of the reaction should be positive as the reaction result in the formation of gas which is consid ered more disorderly. H for this reaction is positive. It is possible for G to be negative at high temperature where │TS │ │H│. [1] [Total: 12] 2 (a) (i) ●8H+ + 8I- + H2SO4 → 4I2 + H2S + 4H2O (ii) ●Violet vapour of iodine, trace white fumes of H I, and a pungent hydrogen sulfide (smells like rotten eggs) will be evolved. (iii) ●It is to cool the beaker surface so that the iodine vapour can be deposited as a solid.

Mark Scheme (2011 H2 Structured Prelims) 2011 Prelims JC2 H2 Chemistry Paper 2 [Turn Over 4 (iv) ●When I2 is in contact with I-, the reddish-brown I3 - ion would be formed. (v) ●The yield would likely be lower as some of the iodine would react with the sodium hydroxide. ●3I2 + 6NaOH → 5NaI + NaIO3 + 3H2O (I2 + 2NaOH → NaI + NaIO + H2O is also acceptable) [6] (b) (i) ●H2O2 oxidises the I -, to aqueous I2, so a brown solution would be obtained. ●H2O2 + 2I- + 2H+ → I2 + 2H2O (ii) - 2 θ Cl /ClE =+1.36V 22 2 θ HO / HOE =+1.77V ●1 mark for both values quoted ●Since 22 2 θ HO / HOE is more positive than - 2 θ Cl /ClE , H2O2 can oxidise chloride to chlorine while it itself is reduced to H 2O. Hence, the oxidation of iodide may not be complete. (iii) ●Both cyclohexane and iodine have simple molecular structures, are non-polar and have van der Waals’ (vdW) forces of attraction between their molecules. ●When iodine dissolves in cyclohexane, the energy released from the vdW forces formed between iodine and cyclohexane is enough to overcome the vdW forces between iodine molecules and the vdW forces between cyclohexane. ●On shaking with cyclohexane, I2 will dissolve in the colourless cyclohexane to form a violet organi c layer. Brown colour of aqueous layer fades. (iv) ●Cyclohexane is flammable, and can cause a fire to break out / iodine may sublime and escape. (v) ●Dilute aqueous NH 3 should be added. AgC l is soluble in NH 3(aq) but AgI is not. If there was significant amount of silver chloride in the precipitate obtained, much of th e precipitate dissolved upon adding aqueous NH3. (vi) ●Sunlight likely catalysed the decomposition of silver chloride / redox reaction between the Ag+ and Cl- ions to form silver metal and chlorine

Mark Scheme (2011 H2 Structured Prelims) 2011 Prelims JC2 H2 Chemistry Paper 2 [Turn Over 5 gas. ●Silver metal atoms are likely to be formed in nanoclusters in the range of 70 nm, producing the purplish colouration. [11] (c) ●2CuI2(s) → 2CuI(s) + I2(aq) [1] [Total: 18] 3 (a) (i) Anode: Ag(s)  Ag+(aq) + e- Cathode: Ag+(aq) + e-  Ag(s) (ii) Explanation: ●As current passes through this electrolytic cell, the silver electrode will get oxidized to form Ag + ions which enter the solution. Ag+ ions in the electrolyte will get reduced to silver metal and plates onto the spoon at the cathode. Observation: Overtime as current passes through the electrolytic cell, the ●silver electrode become thinner and the object is plated with a layer of silver coating. [3] (b) ●For every Ag + ion reduced and plated onto the object, one Ag + ion will be released into the electrolyte as the silver electrode is oxidized to release Ag + ion. [1] (c) (i) ●Effervescence of oxygen gas at the positive electrode. (ii) 2H 2O  O2 + 4H+ + 4e- Ag+ + e-  Ag (x4) ●Overall: 2H2O + 4Ag+  O2 + 4H+ + 4Ag (iii) No. of mol of Ag = 10/107 = 0.0935mol ●No. of mol of e- transferred = 0.0935mol Total Quantity of charge = 0.0935 x 96500 = 9018.7C Q = It 9018.7 = 5(t) ●t= 1803.7s = 0.5hrs ● ●

Mark Scheme (2011 H2 Structured Prelims) 2011 Prelims JC2 H2 Chemistry Paper 2 [Turn Over 6 (iv) C(s) + O 2(g)  CO2(g) The graphite electrode will react with O2 gas/ be ●oxidized by O2 gas to form CO2 and thus needs to be replaced after some time. [6] [Total: 10] 4 (a) ● C is ● D is [2] (b) (i) Rate = k [C]x [NH3]y = 61 1 2 2amount of C H CH NH t Hence, xy 3 1[C] [NH ] t or  xy 13 1