Title DPP 2 Solutions.pdf ✅

Class : XIIth Subject : CHEMISTRY Date : DPP No. : 2 1 (a) In the process of electro decomposition for purification of metal, impure metal acts as anode. 2 (b) Specific conductivity (κ) = 1 R × cell constant Cell constant = κ × R = 0.0129 × 100 = 1.29 3 (b) According to Nernst equation. Ecell = E ° Cell + 0.0591 2 log [Cu 2+] [Zn 2+] Ecell = E ° Cell ― 0.0591 2 log [Zn 2+] [Cu 2+] Or y = c + ( ―m)x Thus, the slope is negative. 4 (a) In MnO― 4 the oxidation number of Mn is + 7. +7 +2 ∴ Mn + 5e ―⟶ Mn In the reaction, 5 electrons are involved hence 5 Faraday will be needed for the reduction of 1 mole of MnO― 4 . Therefore, for 0.5 mole of MnO― 4 , number of Faradays required = 2.5 F 5 (a) Anode is electrode at which oxidation occurs. 6 (b) MnO2 in Lechlanche cell. 7 (d) As Cr has maximum oxidation potential value, therefore its oxidation should be easiest 8 (d) Topic :- Electro Chemistry Solutions
More is reduction potential, more is the power to get itself reduced or greater is oxidising power. 9 (d) F = N × e 10 (c) NaCl gives Na + and Cl ― ions; At anode : Cl ― ⟶(1/2)Cl2 + e At cathode : H + ⟶(1/2)H2 +e 11 (b) Electrons flow from Zn to Cu in outside circuit and current from Cu to Zn. 12 (d) E ° cell = E ° cathode ― E ° anode Ni / Ni2+ [1.0 M] || Au3+ [1.0 M] | Au Ecell (Au3+ / Au) = 0.150 V Ecell (Ni2+ / Ni) = ― 0.25 V E ° cell = E ° cathode ― E ° anode = 0.150 – (- 0.25) = 0.15 + 0.25 = + 0.4 V 13 (b) 50 % H2SO4aqueous solution can be electrolysed by using Pt electrodes as 2H2 SO4 → 2HSO ― 4 + 2H+ 2HSO ― 4 → H2S2O8 + 2e― (at anode) 14 (d) It is fact. 15 (d) For the given cell, Ecell = E ° cell ― 0.0591 2 log [Zn2+] [Cu2+] 1. E1 = E ° cell ― 0.0591 2 log 1 0.1 = E ° cell ― 0.0591 2 2. E2 = E ° cell ― 0.0591 2 log 1 1 = E ° cell ― 0.0591 2 × 0 = E ° cell 3. E2 = E ° cell ― 0.0591 2 log 0.1 1
= E ° cell + 0.0591 2 ∴ E3 > E2 > E1 16 (a) Transport number of an ion = current carried by that ion total current carried by both the ions 17 (c) Reduction is always carried out at cathode. 18 (a) Reactions (i) Fe(s)→ Fe2+ + 2e―, E ° = + 0.44 V and ∆G° 1 = ― nE°F = ― 2 × 0.44 × F (ii) 2H+ + 2e― + 1 2 O2⟶H2O(l); E ° = +1.23 V and ∆G° 2 = ― 2 × ( + 1.23) × F Net reaction, Fe (s) + 2H+ + 1 2 O2 →Fe2+ + H2O(l) ∆G° 3 = ∆G° 1 + ∆G° 2 = ―2 × ( + 0.44) F + ( ― 2 × 1.23 × F) = ― 0.88 F × ― 2.46 F = ― 3.34 F = ― 3.34 × 96500 J = ―322.31 kJ = ― 322 kJ 19 (b) 2 faraday will deposit 2 eq. or 1 mole of Cu. 20 (a) Cl2 is placed above F2 in electrochemical series, halogen placed below replaces the other from its solution.
ANSWER-KEY Q. 1 2 3 4 5 6 7 8 9 10 A. A B B A A B D D D C Q. 11 12 13 14 15 16 17 18 19 20 A. B D B D D A C A B A