Problem 6 Atomic and molecular orbitals - PianetaChimica.it
Problem 6 Atomic and molecular orbitals - PianetaChimica.it
Problem 6 Atomic and molecular orbitals - PianetaChimica.it
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33 rd International Chemistry Olympiad ∗ Preparatory <strong>Problem</strong>sc. Determine the α-disintegration rate of 210 Po <strong>and</strong> β-disintegration rate of 210 Biat that time.<strong>Problem</strong> 9 Redox reactionsa. A solution containing Sn 2+ ions is t<strong>it</strong>rated potentiometrically w<strong>it</strong>h Fe 3+ . Thest<strong>and</strong>ard reduction potentials for Sn 4+/2+ <strong>and</strong> Fe 3+/2+ are given below.Sn 4+ + 2e - = Sn 2+ E° = 0.154 VFe 3+ + e - = Fe 2+ E° = 0.771 Vi. Wr<strong>it</strong>e down the overall reaction <strong>and</strong> calculate the st<strong>and</strong>ard free energychange of the overall reaction.ii.Determine the equilibrium constant of the reaction.b. If 20 mL of 0.10 M Sn 2+ is t<strong>it</strong>rated w<strong>it</strong>h 0.20 M Fe 3+ solution, calculate thevoltage of the celli. when 5 mL of Fe 3+ solution is added.ii.at the equivalence point.iii. when 30 mL Fe 3+ of the solution is added.The saturated calomel electrode (E° S.C.E = 0.242 V) is used as the referenceelectrode in the t<strong>it</strong>ration.c. One of the important analytical methods for estimation of Cu 2+ is iodometrict<strong>it</strong>ration. In this reaction Cu 2+ is reduced to Cu + by I − <strong>and</strong> the liberated I 2 isthen t<strong>it</strong>rated w<strong>it</strong>h st<strong>and</strong>ard Na 2 S 2 O 3 solution. The redox reaction is as follows:2Cu 2+ + 4I − → 2CuI (s) + I 2 (aq)Electrode potentials of the relevant half-cells are:Cu 2+ + e - = Cu + E° = 0.153 VI 2 + 2e - = 2I − E° = 0.535 VMumbai, India, July15 15