ans on electrochemical and redox reaction

Electrochemical and Redox reactionsKeys: 1 d 2 c 3 c 4 c 5 d 6 b 7 a 8 d 9 a 10 c 11 b 12 a 13 c 14 a 15 c 16 a & d 17 b 18 b 19 c 20 b 21 c 22 a 23 a 24 a 25 b 26 c 27 c 28 b 29 a 30 b 31 a 32 d 33 d 34 a 35 b 36 d 37 a 38 a 39 c 40 b 41 a 42 b 43 c 44 d 45 c 46 b 47 c 48 b 49 b 50 b Solutions: 1. The value of equivalent conductance of a strong electrolyte increases slightly with decrease in concentration. 2. 3. 4. Q = IT 6. 9. move towards cathode and the given reaction is an anodic reaction. 10. This is according to Faradays first law. 12. The strongest reducing agent is one with lowest reduction potential. 13. The liberation of cations will be in order of increasing oxidation potential, or decreasing reduction potentials. 15. Anodic reaction will be 19. The given electrode potential are oxidation potential so reduction potential are Then 23. Conductivity 25. 26. P changes its ON form and also from 27. 28. 33. In this On of H is + 1 so oxidation of N is 38. Lower the reduction potential, higher is the reducing power. 39. 41. The balanced equation is 42. In (b) H2O2 (O.N of O = -1) is getting oxidized to O2 (ON of O =0) and thus acts as reducing agent. 43. Mg2P2O7 2(+2) +2x-14 =0; x = +5. 44. In 2Ag + H2SO4Ag2SO4+2H2O+SO2 H2SO4 act as acid ( S retains its O.N = +6 in Ag2SO4) as well as oxidizing agent. 45. 2.5 N2H25 (new compd.) oxidation state of nitrogen in N2H4 = -2 oxidation no. of 2.5 moles = 2.52(-2)= -10 Loss of electrons =25 moles This means oxidation number changes from -10 to +15. It means in new compound the oxidation number of 5 moles of N = +3 49. 2 + 2x + 7(-2) =0 K Cr O X = = + 6 *****