Tiêu đề DPP 4 Solutions.pdf ✅

Class : XIIth Subject : CHEMISTRY Date : DPP No. : 4 1 (c) The Gibb’s free energy change △ G and emf (E ° ) of a reversible electrochemical cell are related by the following expression. △ G = ― nFE° cell or = ― nFE 2 (b) E = E ° RP + 0.0591 n log[M+] Given, E ° RP = ― 2.36 V, [M+] = 0.1 M n = 1 (for M+ → M ) E = E ° RP + 0.0591 n log[M+] = ― 2.36 + 0.0591 1 log 0.1 = ― 2.36 + 0.0591 × ( ― 1) = ― 2.36 – 0.0591 = ― 2.419 V 3 (d) 1 faraday deposits 1 g equivalent of any substance. 64 (b) 1 a = k × R = 0.002765 × 400 = 1.106 cm―1 . 5 (d) ECell = E ° cell + 0.059 2 log [Fe 2+] [Zn 2+] 0.2905 = E ° cell + 0.059 2 log0.01 0.10 ∴ E ° cell = 0.32 No, E ° cell = 0.059 2 log10K ∴ 0.32 = 0.059 2 log10K K = 100.32/0.0295 7 (a) Ni― + 2e― →Ni (at cathode) Topic :- Electro Chemistry Solutions
Equivalent weight of Ni = mol.wt. gain electron = 58.7 2 = 29.35 i = 12A, t = 1h = 60 × 60s., Z = eq.wt. 96500 Weight of deposit Ni = Zit × efficiency 100 = 29.35 × 12 × 60 × 60 × 60 96500 × 100 = 7.883 g 68 (a) W E = i × t 96500 ∴ W E = 10―2 (Ag is monovalent) ∴ Q = i × t = 96500 × 10―2 = 965 C 69 (a) The tendency to gain electron is in the order z > y > x Thus, y + e ―→y ― x→x ― + e ― 10 (d) NaCl, KNO3, HCl are strong electrolytes but the size of H + is smallest. Smaller the size of the ions, greater is the conductance and hence greater is the conductivity (κ = C × cell constant). 11 (a) Given, i = 2.5A t = 6 min 26 s = 6 × 60 + 26 = 386s Number of coulomb passed = i × t = 2.5 × 386 = 965 C Cu2+ + 2e― ⟶Cu ∴ 2 × 96500 C charge deposits Cu = 63.5 g ∴ 965 C charge deposits Cu = 63.5 2 × 96500 × 965 = 0.3175 g 12 (c) Metal placed above in electrochemical series replaces the other from its salt solutions. 73 (c) Ecell = E ° OPZn + E ° RPCu + 0.059 2 log [Cu2+] [Zn2+]
∴ 1.1 = 0.78 + E ° RPCu + 0.059 2 1 ∴ E ° RPCu2+/Cu = 0.32 ∴ E ° RPCu2+/Cu = ―0.32V 14 (d) More the reduction potential, more is the power to get itself reduced or lesser is reducing power or greater is oxidizing power 15 (a) Quantity of current is charge, i.e., coulomb or ampere sec. 16 (d) Cobalt is anode, ie, oxidation takes place on cobalt electrode ie, cell reaction is Co + 2Ag+→Co2+ + 2Ag Ecell = E ° cell ― RT nF ln [Co2+] [Ag+] 2 Thus, less is the factor [Co2+] [Ag+] , greater is the E ° cell 17 (c) Electrolysis of water takes place as follows H2O ⇌ H+ + OH― Cathode anode At anod OH― oxidation OH + e ― 4OH ⟶ 2H2O + O2 At cathode 2H+ + 2e― Reduction H2 Given, time, t = 1930s Number of moles of hydrogen collected = 1120 × 10―3 22.4 moles = 0.05 moles ∵ 1 mole of hydrogen is deposited by = 2 moles of electrons ∵ 0.05 moles of hydrogen will be deposited by = 2 × 0.05 = 0.10 mole of electrons Charge, Q = nF = 0.1 × 96500 Charge, Q = it 0.1 × 96500 = i × 1930 i = 0.1 × 96500 1930
= 5.0 A 18 (d) 1. ∆G° = ― nFE° cell 2. E ° cell = 2.303 RT nF logKc 3. k = Ae―Ea/RT 19 (b) Cr2O 2― 7 + 6e ―⟶2Cr3+ Reduction of 1 mol of Cr2O 2― 7 to Cr3+ required 6 moles of electrons. Hence, charge required = 2 × 96500 C 20 (b) Cell constant = 1 a = length area ∴ unit is cm―1 .