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CIE Jun 2015 v1 Paper 4 Q5
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![(a) Atoms and ions of elements are made up from the three subatomic particles, protons, electrons and neutrons, in varying amounts. Complete the following table to show the number of each particle in 14C2. protons electrons neutrons 14C2 [2] (b) Describe the observations you would make during the reactions, if any, of the following chlorides with water. Write equations for any reactions that occur. CCl 4 observationequationGeCl 4 observationequationSnCl 4 observationequation[4] (c) Suggest a reason for any difference in the reactivities of the chlorides given in (b).[1] (d) Use data from the Data Booklet to explain why an aqueous solution of SnCl 2 reacts with Cl 2(g) but an aqueous solution of PbCl 2 does not. Write an equation for the reaction.[3]UCLES 2015 9701/41/M/J/15<br />
(e) (i) State the relationship between the Faraday constant and the Avogadro constant.[1] (ii) When a current of 1.2 A was passed through dilute sulfuric acid for 30 minutes, it was found that 130 cm3 of oxygen, measured at 25 C and 1 atm, was collected at the anode. The following reaction takes place. 2H2O(l)4H+(aq) + O2(g) + 4e Use these data and data from the Data Booklet to calculate a value for the Avogadro constant, L, by calculatingthe number of moles of oxygen produced, the number of moles of electrons needed for this, the number of coulombs passed, the number of electrons passed, the number of electrons in one mole of electrons (L).UCLES 2015 9701/41/M/J/15 L =mol1 [4] [Total: 15] [Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/June_2015_v1_QP_-_Paper_4_CIE_Chemistry_A-level_5-0.jpg)
![Page 4 Cambridge International A LevelMay/June 2015 Mark Scheme Syllabus Paper 9701 41more negative / less positive (iv) + (aq)] will decrease the [Ag Eelectrode becomes less positive or due to the common ion effect 3+ (aq)] = 0.2 mol dm [Fe + [H + ]) = 1.9 (or 1.871.89) pH = log([H ] = (c.Ka) = (0.2 x 8.9 x 10 4 ) or 1.33 x 10 2 (mol dm 3 ) (d) (a) protons electrons neutrons 14 (b) (c) (d) no reaction CCl4: GeCl4 and SnCl4: for each steamy fumes evolved or white solid produced GeCl4 + 2H2OGeO2 + 4HCl SnCl4 + 2H2OSnO2 + 4HCl Ge / Sn use dorbitals or Ge / Sn have low lying d orbitals or carbon cannot expand its octet or carbon cannot accommodate more than 4 bonded pairs = +1.69 V and Cl2 / Cl = + 1.36 V is a good reducing agent due to its smaller E value than Cl2 ora is a stronger oxidising agent than Cl2 so Pb with Cl2 reaction is not 2+ is less positive / more negative is a weaker oxidising agent than Cl2 so Sn 2+ with Cl2 reaction is 4+ 2+ 4+ 2+ 2+ 4+ 2+ / Pb / Sn = +0.15V and Pb is oxidised by Cl2 because its Eo Sn Sn or Sn or Pb feasible or Sn feasible SnCl2 + Cl2SnCl4 + Cl2Sn or Sn or SnCl2 + Cl2 + 2H2OSnO2 + 4HCl + 2Cl 4+ 2+ 4+ (e) (i) F = Le (ii) moles of O2(g) = 130 / 24 000 = 5.417 x 10 3 mol moles of electrons needed = 4 x 5.417 x 10 no. of coulombs passed = 1.2 x 30 x 60 or 2160 C no. of electrons passed = 2160 / 1.6 x 10 no. of electrons per mole = 1.35 x 10 / 2.17 x 10 19 22 or 1.35 x 10 22 3 or 2.17 x 10 2 mol 2 = 6.2 x 1023 (mol 1 1 [Total: 13] 1 1 1 1 1 1 1 1 1 1 1 1 1Cambridge International Examinations 2015 [Total: 15]<br />](https://learnah.org/wp-content/uploads/2018/08/June_2015_v1_MS_-_Paper_4_CIE_Chemistry_A-level_5.jpg)