Title DPP 7 Solutions.pdf ✅

Class : XIIth Subject : CHEMISTRY Date : DPP No. : 7 2 (b) In Galvanic cell (Daniel cell) the electrical energy is produced from chemical reactions. At anode Zn → Zn2+ + 2e― (oxidation) At cathode Cu2+ + 2e― ⟶Cu (reduction) Cell reaction Zn + Cu2+ ⟶ Zn2+ + Cu Or Zn(s) + CuSO4(aq) → Cu(s) + ZnSO4(aq) 3 (b) Λ∞ AcOH = Λ∞ AcONa + Λ∞ HCl ― Λ∞ NaCl = 91.0 + 426.2 – 126.5 = 390.7 4 (b) The metal with more E ° OP is oxidised. 5 (d) Ecell = E ° cell ― 0.05915 n logQ For standard hydrogen electrode, E ° cell = 0.00V ∴ Ecell = ― 0.05915 n logQ Given, pH = 1.0 ∴ [H +] = 1 × 10―1 Ecell = ― 0.05915 n log 1 [H +] [∵ The reaction occurring is 2H+ + 2e― → H2 ] = + 0.05915 1 log(H +) = 0.05915 log(10―1 ) = - 0.05915 V = - 59.15 mV 6 (c) Λ ° eq = κ × 1000 normality Topic :- Electro Chemistry Solutions
= 0.005 × 1000 0.01 = 500 ohm―1 cm2 equiv―1 7 (a) E ° OP of Mg > E ° OP of Al. 8 (a) For the given cell, reaction is Zn + Fe2+→Zn2+ + Fe E = E° ― 0.0591 n log C1 C2 or, E° = E + 0.0591 n log C1 C2 = 0.2905 + 0.0591 2 log 10―2 10―3 = 0.32 V E° = 0.0591 2 logKc ∴ logKc = 0.32 × 2 0.0591 = 0.32 0.0295 Kc = 10 0.32 0.295 9 (d) When Alead storage battery is discharged, the following cell reactions take place. At anode Pb + H2SO4 ⟶ PbSO4 + 2H+ + 2e― At cathode PbO2 + 4H+ + SO2― 4 + 2e― ⟶ PbSO4 + 2H2O 10 (d) 2H+ + 2e― → H2 According to Nernst equation, E = E ° + 0.0591 n log 1 [H +] 2 E = 0 ― 0.0591 2 log[H +] 2 = - 0.0591 pH 11 (a) E ° Fe 2+/Fe = ―0.441 V E ° Fe 3+/Fe = ― 0.771 V E ° cell = E ° OPFe/Fe2+ + E ° RPFe3+/Fe2+ (See redox change) = + 0.441 + 0.771 = 1.212 V
12 (b) E ° OPZn > E ° OPCu or E ° RPZn < E ° RPCu 13 (b) H2SO4 is strong electrolyte. 14 (c) Λv = Λ 0 100 ∴ α = Λv Λ 0 = Λ 0 100Λ 0 = 0.01 15 (b) 1 2 H2| H + || Ag+ | Ag E ° cell = E ° cathode ― E ° anode = E ° Ag+/Ag ― E ° H+/ 1 2 H2 = (0.80) ― (0.0) = 0.80 V 16 (a) Ions move towards opposite electrodes due to coulombic forces of attraction. 17 (c) More is E ° RP, more is the tendency to get reduced. E ° RP for Ag is maximum. 18 (d) E ° OP for Li/Li + is maximum in these. 19 (b) 250mL of 1 M AgNO3 contain = 250 1000 = 0.25 mole AgNO3 ∵ Electricity required to liberate 1 g equivalent of metal = 96500 C ∴ Electricity required to liberate 0.25 g equivalent of metal = 96500 × 0.25 1 = 24125 C 20 (b) 1 faraday = 1 eq. of Cu = 1/2 mole Cu = N/2 atoms of Cu.
ANSWER-KEY Q. 1 2 3 4 5 6 7 8 9 10 A. C B B B D C A A D D Q. 11 12 13 14 15 16 17 18 19 20 A. A B B C B A C D B B