Tiêu đề DPP 5 Solutions.pdf ✅

Class : XIIth Subject : CHEMISTRY Date : DPP No. : 5 1 (d) E ° cell = 0.059 n logKc 0.295 = 0.059 2 logKc ; ∵ Kc = 1010 2 (b) Fe being above Cu in electrochemical series and thus, liberates Cu from CuSO4. 3 (a) 2Al + dil .H2SO4→Al2SO4 + H2↑ 4 (a) As ‘A’ has more E ° red value than B, A will act as cathode in the galvanic cell. Hence, E ° cell = E ° cathode - E ° anode = (2.23) – (-1.43) = 2.23 + 1.43 = 3.66 V 5 (d) H2SO4 will furnish maximum H +. 6 (c) Hg is placed below H in electrochemical series. 7 (c) Eq. of H2 = Eq. of Al = 4.5 27/3 = 0.5 ∵ 1 eq. H2 = 11.2 L ∴ 0.5 eq. H2 = 5.6 L 8 (b) No. of moles of H2 = 11.2 22400 No .of equivalence of hydrogen = 1.12 × 2 22400 = 10―4 No. of Faradays required = 10―4 ∴ Current to be passed in one second = 96500 × 10―4 = 9.65 A Topic :- Electro Chemistry Solutions