Tiêu đề DPP 6 Solutions.pdf ✅

Class : XIIth Subject : CHEMISTRY Date : DPP No. : 6 1 (b) Electrode potential of cell must be positive for spontaneous reaction. Zn2+ ⟶Zn ; E ° = ― 0.76 V Cu2+ ⟶Cu ; E ° = ― 0.34 V Redox reaction is Zn → Zn2+ + 2e― (oxidation) Cu2+ + 2e― →Cu (reduction) Zn + Cu2+ → Zn2+ + Cu Ecell = E ° cathode ― E ° anode = - 0.34 – (- 0.76) = + 0.42 V Ecell is positive , so above reaction is feasible. 2 (c) Among given elements, D has the minimum reduction potential (- 2.37 V) hence, it can displace all other from their salts. 3 (a) Cr/Cr3+ (0.1 M) || Fe2+ (0.01 M) | Fe Oxidation half-cell; Cr → Cr3+ + 3e― × 2 Reduction half-cell; Fe2+ + 2e― → Fe × 3 Net cell reaction; 2Cr + 2Fe2+ → 2Cr3+ + 3Fe (n = 6) E ° cell = E ° oxidation ― E ° reduction = 0.72 – 0.42 = 0.30 V E ° cell = E ° cell ― 0.0591 n log [Cr3+] 2 [Fe2+] 3 = 0.30 ― 0.0591 6 log (0.1)2 (0.01)3 = 0.30 ― 0.0591 6 log 10―2 10―6 = 0.30 – 0.0591 6 log 104 Ecell = 0.2606 V Topic :- Electro Chemistry Solutions
4 (c) A thin film of Cr2O3 is formed on Cr Surface. 5 (b) The unit of electrochemical equivalent (Z) is g/C. w = Z.i.t ∴ Z = w i.t g/C 6 (d) The elements which are below H2 in electrochemical series, cannot displaceH2 . ∵Out of Li+, Sr2+, Al3+ and Ag+, Ag+ is below H2 in electrochemical series, so Ag+ cannot displaceH2. 7 (b) As the reduction potential of Zn is less than that of Ag, hence Zn will act as anode when Acell is made using them. Hence , the correct reaction will be Zn(s) → Zn2+ (aq) + 2e― (oxidation) 2Ag+ (aq) + 2e― →2Ag (s) Zn (s) + 2Ag+ (aq)→ Zn2+ (aq) + 2Ag (s) 8 (a) W ∝ i × t and W = Z × i × t. 9 (b) Cell constant = k C = 0.0212 × 55 = 1.166 cm―1 10 (c) Reducing power, ie, the tendency to lose electrons increases as the reduction potential decreases 11 (a) 1. Reducing character ∝ 1 reduction potentials 2. Oxidizing power of halogen decreases from F2 to I2 because their reduction potentials decreases from fluorine to iodine. 3. The reducing power of hydrogen halides increases from hydrogen chloride to hydrogen iodide since, the stability of the H – X bond decreases in the same order. Hence, all statements are correct. 12 (d) If E° = 0, then ∆G° = ― nE° F = 0. 13 (d) E ° cell = E ° cathode ― E ° anode ∴ 2.46 = ( + 0.80) ― E ° Al3+ /Al Or E ° Al3+ /Al = 0.80 – 2.46 = ― 1.66 V
14 (d) E° for reaction in (d) = E ° OPBr + E ° RP1 = ―1.09 + ( ―0.54) = ― 1.63 V Since, E° is negative and thus, reaction is non-spontaneous. 16 (b) ∆G° = ―nFE° ∆G° = ―2.303 RT logKc ∴ nFE° = 2.303 RT logKc logKc = nFE° 2.303 RT = 2 × 96500 × 0.295 2.303 × 8.314 × 298 logKc = 9.97 ∴ Kc = 1 × 1010 17 (c) The molar conductivity of potassium hexacyanoferrate (II) i.e., K4[Fe(CN)6 ] is highest because it gives maximum number of ions on ionization. K4[Fe(CN)6 ]⟶ 4K+ + [Fe(CN)6] 4― 19 (a) The metals having higher negative value of standard reduction potential are placed above hydrogen in electrochemical series. The metals places above hydrogen has a great tendency to donate electrons or oxidising power. The metals having great oxidizing power are strongest reducing agent. Zn has higher negative value of standard reduction potential. Therefore, it is the strongest reducing agent. 20 (d) w = 60 g i = 5A Equivalent weight of Ca = atomic weight valency = 40 2 = 20 According to first law of Faraday electrolysis w = Zit = equivalent weight 96500 × i × t ∴ 60 = 20 96500 × 5 × t t = 96500 × 60 20 × 5 s = 96500 × 60 20 × 5 × 60 × 60 h = 16.08 h
ANSWER-KEY Q. 1 2 3 4 5 6 7 8 9 10 A. B C A C B D B A B C Q. 11 12 13 14 15 16 17 18 19 20 A. A D D D C B C A A D