(10) WMP/Jan12/CHEM2

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1 0 Do not write outside the box 5 Iodine reacts with concentrated nitric acid to produce nitrogen dioxide (NO 2 ). 5 (a) (i) Give the oxidation state of iodine in each of the following. I 2... HIO 3... (2 marks) 5 (a) (ii) Complete the balancing of the following equation. I 2 + 0HNO 3... HIO NO H 2O 5 (b) In industry, iodine is produced from the NalO 3 that remains after sodium nitrate has been crystallised from the mineral Chile saltpetre. The final stage involves the reaction between NalO 3 and Nal in acidic solution. Half-equations for the redox processes are given below. IO 3 + 5e + 6H + 3H 2 O + l 2 2 l l 2 + e Use these half-equations to deduce an overall ionic equation for the production of iodine by this process. Identify the oxidising agent. Overall ionic equation 2 The oxidising agent... (2 marks) (0) WMP/Jan2/CHEM2

2 Do not write outside the box 5 (c) When concentrated sulfuric acid is added to potassium iodide, solid sulfur and a black solid are formed. 5 (c) (i) Identify the black solid. 5 (c) (ii) Deduce the half-equation for the formation of sulfur from concentrated sulfuric acid. 5 (d) When iodide ions react with concentrated sulfuric acid in a different redox reaction, the oxidation state of sulfur changes from +6 to 2. The reduction product of this reaction is a poisonous gas that has an unpleasant smell. Identify this gas. 5 (e) A yellow precipitate is formed when silver nitrate solution, acidified with dilute nitric acid, is added to an aqueous solution containing iodide ions. 5 (e) (i) Write the simplest ionic equation for the formation of the yellow precipitate. 5 (e) (ii) State what is observed when concentrated ammonia solution is added to this precipitate. 5 (e) (iii) State why the silver nitrate is acidified when testing for iodide ions. Question 5 continues on the next page Turn over () WMP/Jan2/CHEM2

3 2 Do not write outside the box 5 (f) Consider the following reaction in which iodide ions behave as reducing agents. Cl 2 (aq) + 2I (aq) I 2 (aq) + 2Cl (aq) 5 (f) (i) In terms of electrons, state the meaning of the term reducing agent. 5 (f) (ii) Write a half-equation for the conversion of chlorine into chloride ions. 5 (f) (iii) Suggest why iodide ions are stronger reducing agents than chloride ions. (2 marks) (Extra space)... 5 (2) WMP/Jan2/CHEM2

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5 Mark Scheme General Certificate of Education (A-level) Chemistry Unit 2: Chemistry In Action January 202 Question Marking Guidance Mark Comments 5(a)(i) M 0 M2 (+) 5 2 Accept Roman V for M2 5(a)(ii) I 2 + 0HNO 3 2HIO 3 + 0NO 2 + 4H 2 O Accept multiples 5(b) M IO 3 + 6H + + 5I 3I 2 + 3H 2 O 2 For M, ignore state symbols Credit multiples M2 NaIO 3 OR IO 3 OR iodate ions OR iodate(v) ions etc. Accept the iodine in iodate ions but NOT iodine alone Accept 2½I 2 + ½I 2 as alternative to 3I 2 Electrons must be cancelled For M2 Do not penalise an incorrect name for the correct oxidising agent that is written in addition to the formula. Accept the iodine / I in iodate ions but NOT iodine alone 5(c)(i) Iodine OR I 2 Insist on correct name or formula 5(c)(ii) H 2 SO 4 + 6H+ + 6e S + 4H 2 O SO H+ + 6e S + 4H 2 O Ignore state symbols Credit multiples Do not penalise absence of charge on the electron 5(d) hydrogen sulfide OR H 2 S OR hydrogen sulphide 0

6 Mark Scheme General Certificate of Education (A-level) Chemistry Unit 2: Chemistry In Action January 202 5(e)(i) Ag + + I AgI ONLY Ignore state symbols No multiples 5(e)(ii) The (yellow) precipitate / solid / it does not dissolve / is insoluble OR turns to a white solid OR stays the same OR no (visible/ observable) change OR no effect / no reaction ignore nothing (happens) ignore no observation 5(e)(iii) The silver nitrate is acidified to react with / remove (an)ions that would interfere with the test prevent the formation of other silver precipitates / insoluble silver compounds that would interfere with the test remove (other) ions that react with the silver nitrate react with / remove carbonate / hydroxide / sulfite (ions) Ignore reference to false positive Do not penalise an incorrect formula for an ion that is written in addition to the name. If only the formula of the ion is given, it must be correct 5(f)(i) An electron donor OR (readily) donates / loses / releases / gives (away) electron(s) Penalise electron pair donor Penalise loss of electrons alone Accept electron donator 5(f)(ii) Cl 2 + 2e 2Cl Ignore state symbols Do not penalise absence of charge on electron Credit Cl 2 2Cl 2e Credit multiples

7 Mark Scheme General Certificate of Education (A-level) Chemistry Unit 2: Chemistry In Action January 202 5(f)(iii) For M and M2, iodide ions are stronger reducing agents than chloride ions, because M Relative size of ions Iodide ions / they are larger /have more electron levels(shells) (than chloride ions) / larger atomic / ionic radius OR electron to be lost/outer shell/level (of the iodide ion) is further the nucleus OR iodide ion(s) / they have greater / more shielding OR converse for chloride ion M2 Strength of attraction for electron(s) The electron(s) lost /outer shell/level electron from (an) iodide ion(s) less strongly held by the nucleus compared with that lost from a chloride ion OR converse for a chloride ion 2 Ignore general statements about Group VII trends or about halogen molecules or atoms. Answers must be specific CE=0 for the clip if iodine ions / chlorine ions QoL CE=0 for the clip if iodide ions are bigger molecules / atoms QoL Insist on iodide ions in M and M2 or the use of it / they / them, in the correct context (or chloride ions in the converse argument) Must be comparative in both M and M2 2

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