AJC_H2_CHEM_P2_MS.pdf - Anderson Junior College H2...

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©2017AndersonJC/CHEM 1 Anderson Junior College H2 Chemistry 9729 2017 JC2 Prelim Paper 2 Suggested Solutions 1 (a) (i) 2Fe 3+ (aq) + Sn 2+ (aq) 2Fe 2+ (aq) + Sn 4+ (aq) ignore state symbols [1] (ii) E o (Fe 3+ /Fe 2+ ) = +0.77 V E o (Sn 4+ /Sn 2+ ) = +0.15 V E o cell = E o reduction E o oxidation = +0.77 – (+0.15) = +0.62 V [1] (iii) Reduction of Fe 3+ to Fe 2+ by SnC l 2 is spontaneous since the E o cell calculated in (ii) is positive. E o (Fe 2+ /Fe) = –0.44 V E o (Sn 4+ /Sn 2+ ) = +0.15 V E o cell = –0.44 – (+0.15) = –0.59 V < 0 (not spontaneous) However, the reduction of Fe 2+ to Fe by SnC l 2 is not spontaneous since the E o cell is negative. [1]: determine E o cell for reduction of Fe 2+ [1]: conclusion based on both E o cell [2] (iv) titration number 1 2 3 initial burette reading / cm 3 0.00 19.95 2.10 final burette reading / cm 3 19.95 40.05 22.15 titre / cm 3 19.95 20.10 20.05 average volume of K I used = ½(20.10 + 20.05) = 20.08 cm 3 n(K 2 Cr 2 O 7 ) required = 1000 08 . 20 x 0.100 = 0.002008 mol = 0.00201 mol [1]: correctly determine the titre for run 2 and 3 and used it to find the average titre based on hierarchy rule (i.e. within 0.05 cm 3 in this case) [1]: n(K 2 Cr 2 O 7 ) [2] (v) 6Fe 2+ (aq) + Cr 2 O 7 2– (aq) + 14H + (aq) 6Fe 3+ (aq) + 2Cr 3+ (aq) + 7H 2 O(l) ignore state symbols [1]
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