Title DPP 8 Solutions.pdf ✅

Class : XIIth Subject : CHEMISTRY Date : DPP No. : 8 1 (a) The metals, present below hydrogen in the electrochemical series, cannot liberate hydrogen from the dilute acids. Among the given metal only Ag is present below hydrogen in electrochemical series, so it does not evolve hydrogen withdil HCl. Ag + dil. HCl →No reaction 2 (b) Any cell (like fuel cell), works when potential difference is developed. 3 (a) tc = uc ua + uc , ta = ua ua + uc Where, ua and uc are speed of ion and tc and ta are transport number of cation and anions respectively of an electrolyte. Thus, tc + ta = 1 4 (b) We know that 1 Faraday charge liberates one gram- equivalent of a metal, hence 0.5 F charge will liberate = 0.5 × 23 = 11.50 g of sodium (E = 23) 5 (b) Current (i) = 1.5 A Time (t) 10 min = 10 × 60 = 600 s Quantity of electricity passed Q = i × t = (1.5 A) × (600 s) = 900 C Copper is deposited as Cu 2+ + 2e ―⟶Cu(s) 2 moles of electrons or 2 × 96500 C of current deposit copper = 63.56 g 900 C of current will deposit copper = 63.56 2 × 96500 × 900 Topic :- Electro Chemistry Solutions
= 0.296 g 6 (b) Ionic mobility depends upon the charge to size ratio of ion. The ionic size in case of hydrated cation is K +(aq) < Na+(aq) < Li+(aq) 7 (b) Eq. of A = Eq. of B = Eq. of C or 2.1 7/n1 = 2.7 27/n2 = 7.2 48/n3 0.3 n1 = 0.1n2 = 0.15 n3 ∴ n1 = n2 3 = n3 2 If n1 = 1 then n2 = 3, n3 = 2 8 (d) The electrode, which shows colour change during redox process is called indicator electrode. 9 (c) Molar conductivity or molar conductance (Λm) = κ × V Λm = κ × 1000 Cm Where, Cm is molar concentration (mol L ―1 ) ∴ Molar conductance (Λm) ∝ (1 C ) 10 (b) Rusting of iron is catalyzed by moist air. 11 (b) 2H2O + 2e― → H2 + 2OH― For 0.01 mole H2, 0.02 mole of electrons are consumed charge required = 0.02 × 96500 C = i × t Time required = 0.02 × 96500 10 × 10―3 = 19.3 × 104 s 12 (b) Metal having higher E ° OP replaces the other from its solution. 13 (a) Eq. of Ag = Eq. of H2; W 108 = 5600 × 2 22400 × 1 ∴ WAg = 54 g 14 (b) Ag+ + e ― →Ag 9650 C = 0.1 F = 0.1 equivalent Ag
= 0.1 mol Ag = 10.8 g Ag 15 (d) More or +ve is E ° Op for an electrode more is its reducing power and vice ― versa. 17 (b) BeCl2 is predominantly more covalent among halides of alkaline earth metals. 18 (b) In CuSO4,change is Cu 2+ +2e ⟶Cu; In CuCN, change is Cu + + e ⟶Cu; Thus, W ∝ ECu, which is more in CuCN. 19 (c) E ° nE ° Mn2+ + 2e― →Mn -1.18 V -2.36 Mn3+ + e ― → Mn2+ 1.51 V 1.51 V Mn3+ + 3e― → Mn ― 0.28 ― 0.85 20 (a) 2H ― ⟶H2 +2e; Hydrogen in CaH2 is ⎯ ve.
ANSWER-KEY Q. 1 2 3 4 5 6 7 8 9 10 A. A B A B B B B D C B Q. 11 12 13 14 15 16 17 18 19 20 A. B B A B D C B B C A