DPP- 55 to 56 Physical Chemistry With Answer

DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 55 to 56 Class : XIII Course : DPP No.1 Total Marks : 39 Max. Time : 39 min. Single choice Objective (no negative marking) Q.1 to Q.9 (3 Marks, 3 Min.) [27, 27] Multiple choice objective (no negative marking) Q.10 (4 marks 4 min.) [4, 4] Subjective Questions (no negative marking) Q.11 to Q.12 (4 marks 4 min.) [8, 8] ANSWER KEY DPP No.-55 1. 2 2. 1 3. 1 4. 3 5. 4 6. 2 7. (a) 1 ; (b) 2 8. 4 9. 2 10. 1,2,4 11. – 0.0785 V. 12. E° = 0.782 V 1. Which of the following concentration cells will produce maximum Ecell at 298 K. [Take pH = 1.0 atm in each case.] 298KijfuEuesalsdkSulklkUnzrklsyvf/kdreEcell mRiUudjsxkA [izR;sdifjfLFkfresa pH2 =1.0atmç;qDrdhft,] (1) Pt | H (g) | H+ (0.01 M) || H+ (0.1 M) | H (g) | Pt (2*) Pt | H (g) | NH Cl (0.01 M) || HCl (0.1 M) | H (g) | Pt 2 2 2 4 2 (3) Pt | H (g) | H+ (0.1 M) || H+ (0.2 M) | H (g) | Pt (4) Pt | H (g) | H+ (pH = 0.0) || H+ (pH = 0.0) | H (g) | Pt. 2 2 2 2 [H ]anode Sol. Ecell = –0.0591log [H ]cathode For (1) Ecell = 0.0591 V (2) Ecell = 0.273 V [H+]Anode = = = 1 (3) Ecell = – 0.0591 log 2 2. Given that (at T = 298 K) 1 = 0.0591 log 2 = 0.0178 V ; (4) Ecell = – 0.0591 log 1 = 0. Cu(s) | Cu2+ (1.0 M) || Ag+ (1.0 M) | Ag(s) Eº Zn(s) | Zn2+ (1.0 M) || Cu2+ (1.0 M) | Cu(s) Eº = 0.46 V = 1.10 V Then Ecell for Zn | Zn2+ (0.1 M) || Ag+ (1.0 M) | Ag at 298 K will be (1*) 1.59 V (2) 1.53 V (3) 2.53 V (4) cannot be calculated due to insufficient data (T= 298K ij)fn;k x;k gS Cu(s) | Cu2+ (1.0 M) || Ag+ (1.0 M) | Ag(s) Eº = 0.46 V lSy Zn(s) | Zn2+ (1.0 M) || Cu2+ (1.0 M) | Cu(s) Eº lSy = 1.10 V rc 298 K ij Zn | Zn2+ (0.1 M) ||Ag+ (1.0 M) |Ag ds fy, E gksxk (1*) 1.59 V (2) 1.53 V (3)2.53V(4)vk¡dMsvi;kZIrgksusdsdkj.kx.kukughdhtkldrhgSA 3. What is the value of the reaction quotient, Q, for the cell - Ni(s) | Ni(NO3)2 (0.190M) | | KCl(0.40M)| Cl2(g, 0.10 atm) |Pt(s) lsy Ni(s) |Ni(NO3)2 (0.190M) | |KCl(0.40M)| Cl2(g,0.10atm) |Pt(s)ds fy, vfHkfØ;k xq.kkad Q dk eku D;k gksxk\ (1*) 3 x 10–1 (2) 1.3 x 10–1 (3) 8.0 x 10–2 (4) 3.0 x 10–2 Sol. Ni (s) + Cl2  Ni2+ (aq) + 2Cl–. [Ni2 ] [Cl– ]2 Q = Cl2 = (0.19)(0.4)2 0.1 = 0.304. 4. For the cell Pt | H2(g) | solution X || KCl (saturated) | Hg2 Cl2 | Hg | Pt the observed EMF at 25°C was 600 mV. When solution X was replaced by a standard phosphate buffer with pH = 7.00, the EMF was 718 mV. Find the pH of solution X. lsyPt | H2(g) | foy;u X || KCl (lra`Ir) | Hg2 Cl2 | Hg | Pt ds fy, 25°Cijizsf{krfo-ok-cy600mVgSaApH=7.00dslkFktc,dekudQkWLQsVcQj}kjkfoy;uXdksfoLFk firfd;k tkrkgksa]rksfo-ok-cy718mVgksrkgSaAfoy;uXdkpHKkrdhft,A (1) 3 (2) 4 (3*) 5 (4) 6 Sol. H  2H+ + 2e– Hg Cl + 2e–  2Hg + 2Cl– –––––––––––––––––––––––––––––––– H + Hg Cl  2H+ + 2Cl– + 2Hg E = E° + 0.059 initial 2 log 1 [H ]2[Cl ]2 = 0.6 E = E° + 0.059 final 2 log 1 (107 )2[Cl ]2 = 0.718 subtracting we get 0.118 = 0.059 log [H ] 10 7 [H ]  10 7 = 100 [H+] = 10–5  pH = 5. 5. The dissociation constant for [Ag(NH ) ]+ into Ag+ and NH is 10–13 at 298 K. If E0  = 0.8 V, then E0 for the half cell [Ag(NH ) ]+ + e–  Ag + 2NH will be 298K ij [Ag(NH ) ]+ dk Ag+ rFkk NH ds fy, fo;kstu fu;rkad 10–13gSA ;fn E0  =0.8Vgks] rks v)Z lsy 3 2 3 Ag Ag [Ag(NH ) ]+ + e–  Ag+ 2NH3 ds fy, Eº gksxkA (1) 0.33 V (2) – 0.33 V (3) – 0.033 V (4*) 0.033 V Sol. [Ag(NH3)2]+ + e–  Ag+ + 2 NH3 Kd = 10–13 Gº = RT ln K Ag+ + e–  Ag 1 Gº d = – 1 × F × 0.8 [Ag(NH ) ]+ + e–  Ag + 2NH Gº = Gº + Gº 3 2 3 3 1 2 – 1 × F × Eº = – RT ln kd – 1 × F × 0.8 Eº = – RT F ln kd + 0.8 ⇒ 8.314  298 96500 × 2.303 × (–13) + 0.8 Eº = 0.0313 V. 6. You are given the following cell at 298 K with Eºcell = 1.10 V Zn(s) | Zn2+ (C ) || Cu++ (C ) | Cu(s) 1 2 where C1 and C2 are the concentration in mol/lit then which of the following figures correctly correlates Ecell as a function of concentrations.. C1 x-axis : log C2 and y-axis : Ecell Eºcell =1.10 Vds lkFk 298Kij fuEu lsy fn;k tkrk gSA Zn(s) | Zn2+ (C ) || Cu++ (C ) | Cu(s) tgk¡C1rFkkC2lkUnzrkeksy@yhVjgSrclkUnzrkdsQyuds:iesaEcell dslkFklghrjglsdkSulacaf/krgSA C1 x-v{k: log C2 rFkk y-v{k: Ecell (1) (2*) (3) (4) Sol. Zn(s)  Zn2+ + 2e– Cu2+ + 2e–  Cu(s) Zn(s) + Cu2+  Zn2+ + Cu(s) E = Eº – 0.0591 log C1 cell cell 2 C2 0.0591 Since slope is –ve i.e. – 2 0.0591 P (2*) Zn | Zn2+ (C ) || Zn2+ (C ) | Zn ; C > C 2 2 1 2 P1 1M P2 (3) Pt (Cl ) | Cl– (C ) || Cl– (C ) | Pt (Cl ) ; C > C 1 2 2 1 (4*) Pt (H ) | HCl (C ) || HCl (C ) | Pt (H ) ; C > C 2 1 2 2 2 1 2 1 2 2 2 1 1 atm 1 atm 1 atm 1 atm Sol. For spontaneity, Ecell > 0 Eºcell = 0 for concentration cell. (1) Anode : H2  Zn+ + 2e– Cathode : Zn+ + 2e–  H2 H2 / anode  E = – 0.0591 log H2 / cathode H2 / cathode = – 0.0591 log P2 cell 2 = +ve H2 / anode 2 P1 (2) Anode : Zn (s)  Zn2+ + 2e– Cathode : Zn2+ + 2e–  Zn (s) Zn2+ / cathode  Zn2+ / anode 0.0591 Ecell = – 2 = +ve. gy% Lorrk ds fy, Ecell > 0 Zn2 / anode log Zn2 / cathode 0.0591 = – 2 log C1 2 lkUnzrk lsy ds fy, Eºcell =0 (1) ,uksM: H2  Zn+ + 2e– dSFksM: Zn+ + 2e–  H H2 /,uksM  H2 /dSFkksM E = – 0.0591 log H2 / cathode = – 0.0591 log P2 cell 2 = +ve H2 / anode 2 P1 (2) ,uksM: Zn (s)  Zn2+ + 2e– dSFksM: Zn2+ + 2e–  Zn (s) Zn2+ /dSFkksM  Zn2+ /,uksM 0.0591 Ecell = – 2 = +ve. Zn2 / anode log Zn2 / cathode 0.0591 = – 2 log C1 2 11. Calculate E° and E for the cell Sn | Sn2+ (1M) || Pb2+ (10–3 M) | Pb , E° (Sn2+ / Sn) = – 0.14V. Eº (Pb2+ / Pb) – 0.13V. What do you infer from cell EMF? Sn| Sn2+ (1M)||Pb2+ (10–3 M) |Pb, lsy ds fy, E°rFkk E dh x.kuk dhft, lsy dsEMF ls D;k fu"d"kZ fudyrk gSaA [ ;fnE° (Sn2+ / Sn) = – 0.14V] Eº (Pb2+ / Pb) – 0.13V]. Ans. E°cell = + 0.01 V, Ecell = – 0.0785 V, correct representation is PbPb2+ (10–3 M)Sn2+ (1M)Sn. Sol. Sn | Sn2+ (1 M) | | Pb2+ (10–3 M) | Pb. E0 2 = – 0.14 V. andrFk E0 2  = – 0.13 V. ; Eº = – 0.13 + 0.14 = 0.01 V. (Sn / Sn) 0.0591 (Pb 1 / Pb) cell Ecell = 0.01 – 2 log 10–3 = – 0.0785 V. 12. At 25°C the value of K for the equilibrium Fe3+ +Ag Fe2+ + Ag+ is 0.5 mol/litre. The standard electrode potential for Ag+ + e Ag is 0.8 V. What is the standard potential for Fe3+ + e Fe2+ ? lkE; ds fy,Fe3+ +Ag Fe2+ +Ag+, 25°C ij K dk eku 0.5 mol/litre gSA Ag+ + e Ag ds fy, ekud bysDVªkWMfoHko0.8VgSrksFe3+ +e Fe2+ dsfy,ekudfoHkoD;kgksxk\ Ans. E° = 0.782 V Sol. O = Eºcell – 0.06 1 log 0.5 Eºcell = 0.06 log 0.5 = – 0.018 EºC – EºA = – 0.018 EºC = – 0.018 + 0.8 = 0.782 V. DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 56 Class : XIII Course : DPP No.2 Total Marks : 33 Max. Time : 33 min. Single choice Objective (no negative marking) Q.1 to Q.6 (3 Marks, 3 Min.) [18, 18] BooSt YoUr PreViouS ConCept Single choice Objective (no negative marking) Q.7 to Q.11 (3 Marks, 3 Min.) [15, 15] ANSWER KEY DPP No.-56 1. 4 2. 1 3. 3 4. 3 5. 3 6. 1 7. 1 8. 3 9. 3 10. 2 11. 2 1. pH of the solution in the anode compartment of the following cell at 25ºC is x when Ecell – Eºcell = 0.0591 V, Pt(H ) (1 atm) | pH = x | | Ni2+ (1 M) | Ni x is : 25ºC ij fuEu lsy ds fy, tc E –Eº =0.0591VgS] rc lsy ds ,uksMhd Hkkx esa foy;u dh pH,x gSA lsy lsy Pt(H ) (1 atm) | pH = x | | Ni2+ (1 M) | Ni rcxgS% (1) 4 (2) 3 (3) 2 (4*) 1 Sol. E = Eº –0.0591 log [H]2 2 cell cell 2 [Ni ][H2 ] cell º cell –0.0591 T log [H+] = 0.0591 × pH  pH = 1 H  2H+ + 2e– Ni2+ + 2e–  Ni(s) H + Ni2+  2H+ + Ni(s). 2. In the electrolysis of a copper(II) chloride solution, the mass of the cathode increased by 3.2 g. What occurred at the copper anode? (1*) 0.05 mol of Cu2+ passed into the solution (2) 0.56 L of O was liberated (3) 0.1 mol of Cu2+ passed into the solution (4) 0.112 L of Cl was liberated dkWij(I)DyksjkbMfoy;udsfo|qrvi?kVuijdSFk sM+dknzO;eku3.2gc<+rkgSAdkWijds,suksMijD;kik;ktkrkgS\ (1*)Cu2+ ds0.05eksyfoy;uesatkrsgSa (2)O ds0.56LeqDrgksrsgSa (3)Cu2+ ds0.1eksyfoy;uesatkrsgSa (4)Cl ds0.112LeqDrgksrsgSa 3. The passage of a constant current through a solution of dilute H2SO4 with ‘Pt’ electrodes liberated 336 cm3 of a mixture of H2 and O2 at S.T.P. The quantity of electricity that was passed is ,druqH2SO4foy;uesals‘Pt’bysDVªkWMdslkFkfu;r/kkjkizokfgrdhtkrhgSrksSTPijH2rFkkO2dk336cm3 feJ.keqDrgksrkgSrksizokfgrgksusokyh/kjkdhek=k gksxh\ 1 (1) 96500 C (2) 965 C (3*) 1930 C (4) 100 faradayQSjkMs Sol. [3] H2O 2e  H + 1 O 2 2 2 × 96500 C  22.4 × 1000 cm3 H2 2  96500  224 22.4 1000 C   224 cm3 H2 = 1930 C. 4. A very thin copper plate is electro-plated with gold using gold chloride (AuCl3) in HCl. The current was passed for 20 min. and the increase in the weight of the plate was found to be 2g. The current passed was – [Au = 197, Cl = 35.5]. - ,d cgqr iryh dkWij dh IysV dk xksYM ds lkFk fo|qr ysiu HCl esa xksYM DyksjkbM (AuCl3) dks ysdj fd;k tkrk gS rFkk èkkjk dks 20 feuV rd izokfgr fd;k tkrk gS rks ;g ik;k x;k fd IysV dk 2g Hkkj c<+ tkrk gS rks izokfgr dh x;h èkkjk gksxh gS – [Au = 197, Cl = 35.5] (1) 0.816 amp (2) 1.632 amp (3*) 2.448 amp (4) 3.264 amp Sol. Gold chloride AuCI3 M m = 96500  n × I × t 2  96500  3  = 197  20  60 = 2.448A. 5. What must be the concentration of Ag+ in an aqueous solution containing Cu2+ = 1.0 M so that both the metals can be deposited on the cathode simultaneously. Given that E0 2  = – 0.34 V and E0  Ag = 0.812 V, T = 298 K (1) nearly 10–19 M (2) 10–12 M (3*) 10–8 M (4) nearly 10–16 M ,dtyh;foy;utksfdCu2+=1.0Mj[krkgSarksAg+vk;udhlkUnzrkD;kgksuhpkfg,dhnksuksa/k rqdSFk sM+ij;qxir :i ls¼lkFk&lkFk½ fu{ksfir gks] fn;k x;kgS E0 2  =–0.34VrFkk E0  Ag =0.812V,T=298K (1) yxHkx 10–19 M (2) 10–12 M (3*) 10–8 M (4) yxHkx 10–16 M 6. Electrolytic reduction of 6.15 g of nitrobenzene into aniline using a current effeciency of 40% will require which of the following quantity of electricity. [C = 12, H = 1, N = 14, O = 16] (1*) 0.75 F (2) 0.15 F (3) 0.75 C (4) 0.125 C 40%n{krkdh/k jkiz;qDrdjukbVªkscsUthuds6.15gds,uhyhuesaoS|qrvi?kVuh;vip;udsfy,vko';dfo|qr /kkjkdh ek=kkfuEu esals dkSulhgksxh\ [C=12,H=1,N=14,O=16] (1*) 0.75 F (2) 0.15 F (3) 0.75 C (4) 0.125 C Sol. 6.15 123 = 0.05 mole of nitro benzene V.F = 6 electricity of charge required if efficiency is 100% = 0.05 × 6 = 0.3 F But efficiency is 40% 0.3 charge required = 0.4 = 0.75. gy V.F = 6 6.15 ukbVªkscsUthudseksy= 123 =0.05 ;fn n{krk 100%gSrks fo|qr/kkjkdkvko';d vkos'k=0.05×6=0.3F ijUrqn{krk40%gS 0.3 vr% vk'o;d vkos'k = 0.4 = 0.75. BooSt YoUr PreViouS ConCept 7. Magnesium cation has plarisation power close to that of : eSXuhf'k;e/kuk;udh/kqzo.k{kerkyxHkxfdldscjkcjgksrhgSA (1*) Li+ (2) Na+ (3) K+ (4) Cs+ 8. Which of the following compound is form when the products of electrolysis of brine solution at anode and cathode are made to contact : tcczkbufoy;udsfo|qrvi?kV~ulsdSFk sMrFk ,uksMijizkIrmRiknksadse/;la;kstugksrkgS]rcfuEuesalsdkSulk ;kSfxdizkIrgksrkgSA (1) NaCl (2) H2 (3*) NaOCl (4) Cl2 9. When dry ammonia gas is passed over heated sodium (in absence of air) the product formed is : (1) sodium hydride (2) sodium nitride (3*) sodamide (4) sodium cyanamide tcxeZlksfM;e¼ok;qdhvuqifLFkresa½dsÅijls'kq"dveksfu;kxSldksizokfgrfd;ktkrkgSrksmRiknizkIrgksrkgSA (1)lksfM;egkbMªkbM (2)lksfM;eukbVªkbM (3*)lksMkekbM (4)lksfM;elk;usekbM 10. At high temperature, nitrogen combines with CaC2 to give : (1) calcium cyanide (2*) calcium cyanamide (3) calcium carbonate (4) calcium nitride mPprkiekuijukbVªkstuCaC2dslkFkla;ksftrgksdjnsrhgSA (1)dSfY'k;elk;ukbM (2*)dSfY'k;elk;usekbM (3)dSfY'k;edkcksZusV (4)dSfY'k;eukbVªkbM 11. Which of the following compounds does not have similarity in their structural aspect ? fuEuesalsdkSulk;kSfxdbudhlajpukdslkFklekurkughaj[krkgS\ (1) FeSO4. 7H2O (2*) Na2CO3 . 7H2O (3) MgSO4. 7H2O (4) ZnSO4 . 7H2O