Tiêu đề DPP 3 Solutions.pdf ✅

Class : XIIth Subject : CHEMISTRY Date : DPP No. : 3 1 (d) E ° cell= 2.303RT nF logKeq 0.295 = 0.0591 2 logKeq ∴ logKeq = 10 ∴ Keq = 1010 2 (c) k = 1 R × 1 a = 1 32 × 1.8 5.4 = 0.0104 And λ = k × V = 0.0104 × 10,000 = 104 3 (d) E° = 0.059 n logK; 4 (c) E° = 0.059 n logKc ∴ 0.295 = 0.059 2 logKc ∴ Kc = 1010 5 (a) High value for E ° red. Shows more electronegativity i.e., Zn is more electropositive than Fe. ( E ° Zn2+/Zn < E ° Fe2+ /Fe ) 6 (b) Eq. of Cu = Eq. of Ag ∴ W 63.5/2 = 1.08 108 ∴ WCu = 0.3175 g 7 (c) The cell reaction is H2(g) + I2(s)⇌2H +(aq) + 2I ―(aq) 0.7714 = 0.535 ― 0.0591 2 log [H +] 2 [I ―] 2 pH2 Topic :- Electro Chemistry Solutions
∴ pH = 3 8 (a) E ° Cr3+/Cr2+ = ― 0.41 V E °Mn3+/Mn2+ = + 1.57 V E ° Fe3+/Fe2+ = + 0.77 V E ° Co3+/Co2+ = + 1.97 V More negative value of E ° red indicates better reducing agent thus easily oxidized. Thus, oxidation of Cr2+ to Cr3+ is the easiest. 9 (b) In other cells, two liquid are not present. 10 (d) AgNO3 ∆ Ag2O ∆ Ag + O2 11 (a) Cu2+ + 2e― → Cu, E ° = 0.34 Zn2+ + 2e― → Zn, E ° = 0.76 In the cell, Cu | Cu2+ || Zn2+ | Zn anode cathode In the cell , E ° cell = E ° cathode ― E ° anode = 0.76 – (-0.34) = 1.10 V 12 (a) Here Fe acts as anode while Sn act as cathode. We know that, E ° cell = E ° cathode - E ° anode = (-0.14) – (-0.44) = - 0.14 + 0.44 = 0.30 V 14 (d) Ag + + e ― ⟶ Ag ∵ 96500 C are required to deposite Ag = 108 g ∴ 965 C are required to deposite Ag = 108 96500 × 965 = 1.08 g 15 (a) Λm = Λeq. × valency factor; For NaCl, valency factor = 1; Molecular conductivity Λm is defined as the conductance of all the ions present in a solution containing 1g molecule in it; Λeq. is defined as the conductance of all the ions present in a solution containing 1g equivalent in it. 16 (a) In electrochemical series, iron is placed below sodium, so it cannot displace sodium from
its salt solution. Hence, no reaction takes place. Fe + Na3PO4 → No reaction 17 (d) During electrolysis of NaCl(aq), H + ions are discharged at cathode and the pH of solution increases due to decrease in[H +]. 18 (d) Galvanic cell is Cu (s) | Cu2+ (aq) || Hg2+ (aq) | Hg (l) In the above cell, oxidation of copper and reduction of mercury takes place. Its cell reaction is written as Cu (s) + Hg2+ (aq) → Cu2+ (aq) + Hg (l) 19 (a) W = E. i. t 96500 = 1 × 0.4 × 30 × 60 96500 = 7.46 × 10―3 g and volume = 7.46 × 103 × 22.4 2 = 0.0836 litre 20 (c) E ° cell = 2.303RT nF logKeq E ° cell = 0.0591 n logKeq [At 298 K] 0.591 = 0.0591 1 logKeq ∴ logKeq = 10 ∴ Keq = 1 × 1010
ANSWER-KEY Q. 1 2 3 4 5 6 7 8 9 10 A. D C D C A B C A B D Q. 11 12 13 14 15 16 17 18 19 20 A. A A A D A A D D A C