Jp Xii Physical&inorganic Chemistry (13) - Prev Chaps + Inorg. Chem + Electrochemistry.pdf

PHY./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2016 NO. 20 & 21

Course : VIJETA (JP)

This DPP is to be discussed in the week (22-06-2015 to 27-06-2015)

DPP No. # 20 (JEE-MAIN) Total Marks : 60

Max. Time : 40 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.20

1.

(3 marks, 2 min.)

[60,40]

The virial equation for 1 mole of a real gas is written as 

A

B

C



PV = RT 1  V  2  3  .....to higher power of n V V   W here A,B and C are known as virial cofficients. If Vander waal's equation is written in virial form , then what will be the value of A. 1 eksy

okLrfod xSl ds fy, fofj;y lehdj.k fuEu gSA 

A

B

C

PV = RT 1  V  2  3  ..... n V V 

  

;gk¡ A,B rFkk C fofj;y fLFkjkad gSA ok.Mj okWy lehdj.k dks ;fn fofj;y lehdj.k ds :i esa fy[kk tk;s rks A dk eku gksxkA (A) a – 2.

b RT

(B) RT –

a b

(C*) b –

a RT

(D)

RT  b a

W hat is the correct increasing order of liquifiability of the gases shown as in above graph ?

mijksDr oØ esa fn[kk;s vuqlkj xSlksa ds nzohdj.k dh lqxerk dk c
Sol.

(A*) H 2 < N 2 < CH 4 < CO 2 (B) CO2 < CH 4 < N 2 < H 2 (C) H 2 < CH 4 < N 2 < CO2 (D) CH 4 < H 2 < N 2 < CO2 Order of Vander waals constant CO 2 > CH 4 > N 2 > H 2  ease of liqufication CO 2 > CH 4 > N 2 > H 2 okW.Mj okWYl fLFkjkadksa dk Øe CO2 > CH4 > N2 > H2 

3.

nzohdj.k dh lqxerk CO2 > CH4 > N2 > H2

CaCN 2 has : (A*) 2  bonds , 2  bonds (C) 1  bond , 2  bonds CaCN 2 es a gksr s gS % (A*) 2  ca /k , 2  ca/k (C) 1  ca/k , 2  ca /k

(B) 3  bonds , 1  bond (D) 3  bonds , 2  bonds (B) 3  ca/k , 1  ca /k (D) 3  ca/k , 2  ca/k

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4.

Which of the following statements is correct ? (A) N2F3+ is planar at each nitrogen atom. (C) The shape of N(SiMe3)3, is trigonal planar. fuEu esa ls dkSulk dFku lgh gS ?

ukbVªkstu ijek.kq ij N2F3+ leryh; gSA (C) N(SiMe3)3 dh vkd`fr f=kdks.kh; leryh; gSA (A) izR;sd

(B)

(B) In N3H, the H – N – N bond angle is exactly of 120°. (D*) (A) and (C) both. (B) N3H esa] H – N – N cU/k

dks.k ,dne 120° ik;k tkrk gSA

(D*) (A) rFkk (C) nksuksaA

Sol.

(A)

(C)

5.

The species that do not contain peroxide ions are : (A*) PbO2 (B) H2O2 (C) SrO2

(D) BaO2

Lih'kht tks ijkWDlkbM vk;u ugha j[krh gS % Ans. Sol.

g y-

(A*) PbO2 (B) H2O2 (C) SrO2 (D) BaO2 (A) Only H2O2 (hydrogen peroxide), BaO 2 (Barium peroxide) and SrO2 (Strontium peroxide) contain peroxide ions. So (A) is the correct choice. dsoy H2O2 ¼gkbMªkstu ijkWDlkbM½] BaO2 ¼csfj;e vkWDlkbM½ o SrO2 ¼LVªkWfU'k;e ijkWDlkbM½ ijkWDlkbM vk;u j[krs

gSA vr% (A) lgh fodYi gSA 6.

A solution of sodium metal in liquid ammonia is strongly reducing due to the presence of (A) sodium atoms (B) sodium hydride (C) sodium amide (D*) solvated electrons

fuEu esa ls fdldh mifLFkfr ds dkj.k] lksfM;e /kkrq dk nzo veksfu;k esa foy;u] çcy vipk;d gksrk gS % (A) lksfM;e ijek.kq (B) lksfM;e gkbMªkbM (C) lksfM;e ,ekbM (D*) foyk;fdr bysDVªku W Ans. Sol.

g y-

(D) (D) The free ammoniated electrons make the solution of Na in liquid NH3 is a very powerful reducing agent. The ammonical solution of an alkali metal is rather favoured as a reducing agent than its aqueous solution because in aqueous the alkali metal being highly electropositive quite stable, provided no catalyst (transition metal) is present. (D) eqDr veksfu;ke; bysDVªkWu] nzo NH3 esa Na foy;u dks çcy vipk;d cuk nsrs gSA {kkj /kkrq dk veksfu;ke;

foy;u blds tyh; foy;u dh vis{kk vipk;d dh rjg O;ogkj djrk gSA D;ksafd tyh; foy;u esa {kkj /kkrq mPp fo?kqr/kukRed gksrh gS blds dkj.k ;g vf/kd LFkk;h gksrh gSA bl vfHkfØ;k esa dksbZ Hkh mRçsjd ¼laØe.k /kkrq½ ç;qDr ugha gksrs gSaA 7.

Molecular formula of Glauber`s salt is : (A) MgSO4.7H2O (B) CuSO4.5H2O

(C) FeSO4.7H2O

(D*) Na2SO4.10H2O

(C) FeSO4.7H2O

(D*) Na2SO4.10H2O

Xykscj yo.k dk v.kqlw=k gS % Ans. Sol.

g y8.

(A) MgSO4.7H2O (B) CuSO4.5H2O (D) Glauber`s salt is Na2SO4. 10H2O. Xykscj yo.k Na2SO4. 10H2O gSA

The oxidation of SO2 by O2 to SO3 is an exothermic reaction. the yield of SO 3 will be maximum if (A) temperature is increased and pressure is kept constant. (B*) temperature is reduced and pressure is increased. (C) both temperature and pressure are increased. Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

PAGE NO.- 2

Sol.

(D) both temperature and pressure are reduced. O2 }kjk SO2 dk SO3 esa vkWDlhdj.k ,d Å"ek{ksih vfHkfØ;k gS] SO3 dh yfC/k (yield) vf/kdre gksxh (A) rki esa o`f) rFkk nkc fu;r j[kk tk,A (B*) rki esa deh rFkk nkc esa o`f) dh tk,A (C) rki rFkk nkc nksuksa esa o`f) dh tk,A (D) rki rFkk nkc nksuksa esa deh dh tk,A Le-chatelier principle

;fn µ

2SO2(g) + O2(g) 2 SO3 (g) rg = – 1 Pressure increases  Reaction goes in forward direction. H = – Ve Temperature decrases  Reaction goes in forward direction 9.

In Galvanic cell : (A*) Chemical reaction produces electrical energy (B) Electrical energy produces chemical reaction (C) Reduction occurs at anode (D) Oxidation occurs at cathode

xSYosuhd lsy esa & (A*) jklk;fud vfHkfØ;k fo|qr ÅtkZ mRiUu djrh gS (B) fo|qr ÅtkZ jklk;fud vfHkfØ;k mRiUu djrh gS (C) ,suksM+ ij vip;u gksrk gS (D) dSFkksM+ ij vkWDlhdj.k gksrk gS Sol.

Galvanic cells are electro chemical cells which convert chemical energy into electrical energy.

10.

The standard oxidation potentials, E°, for the half-reaction are as v)Z vfHkfØ;k ds fy, ekud vkWDlhdj.k foHko E° fuEu gSA Zn = Zn2+ + 2e– ; E° = + 0.76 V Fe = Fe2+ + 2e– ; E° = + 0.41 V the E°cell is Fe+2 + Zn Zn2+ + Fe is (A) –0.35 V (B*) + 0.35 V (C) +1.17 V Eºcell = [EºA – EºC]OP = 0.76 – 0.41 = 0.35 V.

Sol. 11.

rks E°lsy gksxkA (D) – 1.17 V

From the following E° values of half cells v)Z lsy ds fy, E° dk eku fuEu gS& (i) A + e–  A¯ ; E° = – 0.24 V (ii) B¯ + e–  B2– ; E° = + 1.25 V – 3– (iii) C¯ + 2e  C ; E° = –1.25 V (iv) D + 2e–  D2– ; E° = + 0.68 V What combination of two half cells would result in a cell with the largest potential

dkSuls nks v)Z lsyks dk la;kstu] vf/kdre foHko okys lsy dk fuekZ.k djsxk \

Sol.

12.

(A*) (ii) and (iii) (B) (ii) and (iv) (C) (i) and (iii) (A*) (ii) rFkk (iii) (B) (ii) rFkk (iv) (C) (i) rFkk (iii) Eºcell = [EºC – EºA]RP = 1.25 + 1.25 = 2.50 V. Eºcell = [EºC – EºA]RP = 1.25 – (–1.25) = 2.50 V (Maximum value)

(D) (i) and (iv) (D) (i) rFkk (iv)

The Ni/Ni2+ and F–/F2 electrode potentials are listed as +0.25 V and –2.87 V respectively (with respect to the standard hydrogen electrode). The cell potential when these are coupled under standard conditions is (A) 2.62 V and dependent on the reference electrode chosen. (B*) 3.12 V and independent of the reference electrode chosen. (C) 3.12 V and dependent on the reference electrode chosen. (D) 2.62 V and independent of the reference electrode chosen. Ni/Ni2+ rFkk F–/F2 ds fy, bysDVªkWM foHko Øe'k% +0.25 V rFkk –2.87 V fn;k x;k gS (ekud gkbMªkstu bysDVªkWM ds lUnHkZ esa) tc ;g ekud ifjfLFkfr;ksa esa la;ksftr fd, tkrs gS rks lsy foHko gksxk \ (A) 2.62 V rFkk pqus x, ekud bysDVªkWM ij fuHkZj djrk gSA (B*) 3.12 V rFkk pqus x, ekud bysDVªkWM ls Lora=k gksrk

gSA Sol. 13.

(C) 3.12 V rFkk pqus x, ekud bysDVªkWM ij fuHkZj djrk gSA (D) 2.62 V rFkk pqus x, ekud bysDVªkWM Electrode potential values depend on reference electrode chosen but not cell potential. E° for some half cell reactions are given below dqN v)Z lsy vfHkfØ;kvksa ds fy, E° uhps fn;s x;s

ls Lora=k gksrk gSA

gSaA

Sn+4 + 2e–  Sn2+ ; E° = 0.151 V 2Hg2+ + 2e  Hg2+2 ; E° = 0.92 V Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

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PbO2 + 4H+ + 2e–  Pb2+ + 2H2O ; E° = 1.45 V based on the given data which statement is correct.

fn;s x;s vkdMksa ds vk/kkj ij dkSulk dFku lgh gSA (A) Sn4+ is a stronger oxidising agent than Pb+4 (B*) Sn2+ is a stronger reducing agent than Hg22+ (C) Hg2+ is a stronger oxidising agent than Pb4+ (D) Pb+2 is a stronger reducing agent than Sn2+ (A) Sn4+ , Pb+4 fd rqyuk esa izcy vkWDlhdkjd vfHkdeZd gSA (B*) Sn2+ , Hg22+ fd rqyuk esa izcy vipk;d vfHkdeZd gSA (C) Hg2+ , Pb4+ fd rqyuk esa izcy vkWDlhdkjd vfHkdeZd gSA (D) Pb+2 , Sn2+ fd rqyuk esa izcy vipk;d vfHkdeZd gSA Sol.

14.

Sn2+  Sn+4 + 2e–

E° = –0.15 V

Hg22+  2Hg2+ + 2e– Sn2+ ] Hg22+ Sn2+ ] Hg22+ fd rqyuk esa izcy

E° = –0.92 V

vipk;d vfHkdeZd gSA

For the cell prepared from electrode A and B, electrode A :

Cr2O 72 Cr 3 

0 , Ered = + 1.33 V and electrode B:

Fe3

, E 0red = 0.77 V, which of the following statement is not correct? Fe2 (A) The electrons will flow from B to A (in the outer circuit) when connections are made. (B) The standard emf of the cell will be 0.56 V. (C) A will be positive electrode. (D*) None of the above.

bysDVªkWM A o B ls cus ,d lSy ds fy,] bysDVªkWM+ A :

Sol.

15.

Cr2O 27  Cr 3 

, Eºred = + 1.33 V

o bysDVªkWM+ B:

Fe3  Fe 2

Eºred = 0.77 V gSA fuEu esa ls dkSulk okD; lgh ugh gS & (A) tc la;ksftr fd;k tkrk gSA ] rks bysDVªkWu B ls A dh vksj ¼ckà; ifjiFk esa½ izokfgr gkssaxs A (B) lsy dk ekud emf 0.56 V gksaxk A (C) A /kukRed bysDVªkWM gksxkA (D*) buesa ls dksbZ ughaA Eº(Cr2O72–  Cr3+) = 1.33V (A) Eº (Fe3+  Fe2+) = 0.77V (B) When these two electrodes are connected, then the electrode with higher reduction potential act as cathode. So (A) is cathode (B) is anode. e– flow from anode to cathode in external circuit. Eºcell = 1.33 – 0.77 = 0.56V cathode (A) is a +ve electrode in electro chemicall cell. The standard reduction potentials at 25°C for the following half reactions are given against each 25ºC ij fuEu v)Z vfHkfØ;kvksa ds fy, ekud vip;u foHko izR;sd ds lkeus fn;k x;k gSA Zn2+(aq) + 2e¯ Zn(s), –0.762 V Cr3+(aq) + 3e¯ Cr(s), –0.740 V 2H+ + 2e¯ H2(g), 0.00 V Fe3+ + e¯ Fe2+, 0.77 V Which is the strongest reducing agent -

dkSulk izcyre vipk;d gSA Ans. 16.

,

(A*) Zn (B) Cr (C) H2(g) Higher oxidation potential, higher reducing tendency. Hydrogen gas will not reduce (A) heated cupric oxide (C) heated stannic oxide

(D) Fe3+(aq)

(B) heated feric oxide (D*) heated aluminium oxide

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    [ E Sn 4 /Sn 2  0.15V ; E Cu 2 /Cu  0.167 V ; E Fe3 / Fe2  0.771V ; E Al 3 / Al  1.67 V ]

gkbMªkstu xSl fdls vipf;r ugha djsxh& (A) xeZ D;wfizd vkWDlkbM dks (C) xeZ LVsfud vkWDlkbM dks  [ E Sn 4 /Sn 2

Sol.

EHº  / H

2

 0.15V ;

E Cu 2 /Cu

 0.167 V ;

(B) xeZ Qsfjd vkWDlkbM dks (D*) xeZ ,Y;qfefu;e vkWDlkbM E Fe3 / Fe2

 0.771V ;

E Al 3 / Al

dks

 1.67 V ]

= 0 V..

Hydrogen gas will reduce those metals which have reduction potential greater than H 2 gas. 17.

Four colourless salt solutions are placed in separate test tubes and a strip of copper is dipped in each. Which solution finally turns blue ? (use data from electrochemical series)

pkj jaxghu yo.k foy;uks dks i`Fkd µ i`Fkd ij[kuyh esa j[kk x;k gS rFkk izR;sd esa dkWij dh iryh iV~Vhdk Mqcks;h tkrh gS dkSulk foy;u vUrr% uhyk gks tkrk gS (fo|qr jklk;fud Js.kh ls vk¡dM+ks dks iz;qDr dhft, ) Sol.

18.

(A) Pb(NO3)2 (B*) AgNO3 (C) Zn(NO3)2 (D) Cd(NO3)2 Cu will reduce to those metal which have higher reduction potential than Cu. So, Ag+ has higher potential. Red hot carbon will remove oxygen from the oxide XO and YO but not from ZO. Y will remove oxygen from XO. Use this evidence to deduce the order of activity of the three metals X, Y, and Z putting the most active first yky rIr dkcZu vkWDlkbM XO rFkk YO ls vkWDlhtu dks i`Fkd djrk gS ysfdu ZO ls ugha] Y, XO ls vkWDlhtu dks i`Fkd djrk gS bl rF; dks iz;qDr dj rhu /kkrqvksa X, Y, rFkk Z dh fØ;k'khyrk ds Øe dks fu/kkZfjr dhft, rFkk vf/kd fØ;k'khy

dks igys fyf[k, \ Sol.

19.

(A) XYZ (B*) ZYX (C) YXZ (D) ZXY As carbon can reduce XO and YO but not ZO, SRP of X and Y are more than carbon. Y can reduce XO,which means SRP of X is more than Y. Hence the order of SRP is SRP of X > SRP of Y > SRP of C > SRP of Z. More active metal has lesser SRP. So the order of activity of metals is Z>Y>X Ans. W hich statement about standard reduction potentials is correct  (A*) EH / H2 = Zero at all temperature  (B) ED  / D2 = zero at 298 K

(C) A redox reaction is feasible if sum of SRP of oxidant and that of reductant is a positive quantity (D) K 2Cr2O7 (acid) is stronger oxidising agent than KMnO 4 (acid)  [Given : EMnO–4 / Mn2  = 1.51 V

;

E Cr O 2 – / Cr 3  = 1.33 V] 2 7

ekud vip;u foHko ds fy, dkSulk dFku lgh gS&  (A*) E H / H = fdlh Hkh rki ij 'kwU; 

2

(B)

ED  / D 2

= 'kwU; (298 K

rki ij)

(C) ,d jsMkDl vfHkfØ;k gksrh gS ;fn vkWDlhdkjd (D) K2Cr 2O 7 ( vEy) ( KMnO 4 ( vEy) dh rqyuk esa

ds SRP rFkk vipk;d ds SRP dk ;ksx ,d /kukRed ek=kk gS izcyre vkWDlhdkjd vfHkdeZd gSA

  [Given : EMnO –4 / Mn2  = 1.51 V; E Cr2O 27 – / Cr 3  = 1.33 V]

Sol.

(A) It is a convention that SRP of standard hydrogen electrode is zero at all tempertures. (B) is incorrect. (C) the difference should be +ve quantity.

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20.

(D) KMnO4 is stronger oxidising agent (SRP values). The temperature defining the standard electrode potential is (A) 298 K (B) 273 K (C) 373 K (D*) any temperature can be selected but it must remain constant and species must be in their standard states.

ekud bysDVªkWM foHko dks ifjHkkf"kr djus ds fy, rki gSA (A) 298 K (D*) fdlh Hkh

(B) 273 K

(C) 373 K

rkieku dk p;u fd;k tk ldrk gS ysfdu ;g fu;r jguk pkfg, rFkk Lih'kht viuh ekud voLFkk esa gksuh pkfg,A Sol.

Any temperature can be selected.

DPP No. # 21 (JEE-ADVANCED) Total Marks : 76

Max. Time : 46 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.8 Multiple choice objective ('–1' negative marking) Q.9 to Q.14 Subjective Questions ('–1' negative marking) Q.15 to Q.18 ChemINFO : 3 Questions ('–1' negative marking) Q.19 to Q.21

1.

(3 (4 (4 (4

marks, marks, marks, marks,

2 2 3 2

min.) min.) min.) min.)

[24,16] [24,12] [16,12] [12, 06]

Calculate H° for this reaction.

nh xbZ vfHkfØ;k ds fy, H° dk eku Kkr dhft,] C2H2 (g) + 2H2 (g)  C2 H6 (g) Hcombustion, kJ mol–1 C2H2 (g) H2(g) C2H6 (g)

,  H C 2 H 2 (g )

– 1300 – 286 – 1560

k J m o l –1

H 2 (g ) C 2 H 6 (g )

Sol.

(A*) – 312 kJ (B) – 26 kJ (C) + 26 kJ H°rxn = (H°C)R – (H°C)P = [– 1300 – 2 × 286] – 1560 = – 312.5 kJ

2.

S1 : (a)NH3 > (a)H2 O [(a) is Vanderwaal's constant]

– 1300 – 286 – 1560

(D) + 312 kJ

S2 : Pressure of the real gas is more than the ideal gas for same temperature and volume of the container. S3 : Compresssibilty factor for H2 (g) is never less than unity at any temperature. S4 : The value of critical temperature (T C) is greater for H2O than O2. S1 : (a)NH3 > (a)H2 O [(a) okUMWjoky

fLFkjkad gS] S2 : ik=k es leku rki rFkk vk;ru ds fy, okLrfod xSl dk nkc] vkn'kZ xSl dh rqyuk esa vf/kd gksrk gSA S3 : fdlh rki ij H2 (g) ds fy, laihM~;rk&xq.kkad] dHkh Hkh bdkbZ ls de ugha gksrk gSA S4 : O2 dh rqyuk esa H2O ds fy, ØkfUrd rki (TC) dk eku vf/kd gksrk gSA (A) T F F T

(B*) F F F T

(C) F T T F

(D) T T T F

3.

In a half-cell contaning [Tl3+] = 0.1 M and [Tl+] = 0.01M, the cell potential is –1.2496 V for the reaction Tl+  Tl3+ + 2e– . The standard reduction potential of the T+3 /T+1 couple at 25°C is ,d v)Z&lsy ftlesa [Tl3+] = 0.1 M rFkk [Tl+] = 0.01M mifLFkr gS] vfHkfØ;k Tl+  Tl3+ + 2e– ds fy, lsy foHko –1.2496 V gSA 25°C ij T+3 /T+1 ;qXe ds fy, ekud vip;u foHko gksxk \ (A) 144 V (B) 0.61 V (C) 2.44 V (D*) 1.22 V

Sol.

T3+ + 2e–  T+

ECell = + 1.2496 V

ECell = 1.2496 V = EºCell –

0.01 0.0591 log 0. 1 2

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PAGE NO.- 6

4.

EºCell = 1.22 V. The standard electrode potential for the reactions, Ag+(aq) + e¯ Ag(s) Sn2+(aq) + 2e¯ Sn(s) at 25°C are 0.80 volt and –0.14 volt, respectively. The emf of the cell Sn|Sn 2+ (1 M) ||Ag+ (1M)|Ag is(A) 0.66 volt (B) 0.80 volt (C) 1.08 volt (D*) 0.94 volt Ag+(aq) + e¯ Ag(s) Sn2+(aq) + 2e¯ Sn(s) vfHkfØ;kvksa ds fy, ekud bysDVªkWM foHko 25°C ij Øe'k% 0.80 oksYV rFkk –0.14 oksYV gSA Sn|Sn2+ (1 M) ||Ag+ (1M)|Ag lsy ds fy, E.M.F. gksxkA (A) 0.66 oksYV (B) 0.80 oksYV (C) 1.08 oksYV (D*) 0.94 oksYV [ Zn2  ] 0.0591 – log . 2 [ Ag ]2

= 0.8 – (– 0.14) –

1 0.0591 log 2 = 0.94 V.. 2 1

Sol.

Ecell = Eºcell

5.

Given that (at T = 298 K) Cu(s) | Cu2+ (1.0 M) || Ag+ (1.0 M) | Ag(s) Eºcell = 0.46 V Zn(s) | Zn2+ (1.0 M) || Cu2+ (1.0 M) | Cu(s) Eºcell = 1.10 V Then Ecell for Zn | Zn2+ (0.1 M) || Ag+ (1.0 M) | Ag at 298 K will be (A*) 1.59 V (B) 1.53 V (C) 2.53 V (D) cannot be calculated due to insufficient data (T = 298 K ij) fn;k x;k gS Cu(s) | Cu2+ (1.0 M) || Ag+ (1.0 M) | Ag(s) EºlSy = 0.46 V 2+ 2+ Zn(s) | Zn (1.0 M) || Cu (1.0 M) | Cu(s) EºlSy = 1.10 V rc 298 K ij Zn | Zn2+ (0.1 M) || Ag+ (1.0 M) | Ag ds fy, ElSy gksxk (A*) 1.59 V (B) 1.53 V (C) 2.53 V (D) vk¡dMs vi;kZIr gksus

Sol.

Eo

Ag / Ag

– Eo

Cu2 / Cu

o (+) E

Cu2 / Cu

– Eo

ds dkj.k x.kuk ugha dh tk ldrh gSA

 0.46 V

Zn2 / Zn

 1.1V

___________________________

Eo

Ag / Ag

– Eo

Zn2 / Zn

 1.56 V

Zn/Zn2+ (0.1M) || Ag+ (1.0M) | Ag Eºcell = 1.56 Ecell = Eºcell –

0.059 log Q. n

Cell reaction : Zn Zn2+ + 2e– + (Ag+ + e– Ag) × 2 ___________________________ Zn + 2Ag+ Zn2+ + 2Ag Q= 6.

Sol.

[ Zn2 ]

(0.1) = = 0.1 (1) 2 [ Ag  ] 2

 Ecell = Eºcell –

0.059 0.059 log 0.1 = 1.56 – × – 1 = 1.59V 2 2

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, 0.10 atm) |Pt(s) ds (A*) 3 x 10–1 (B) 1.3 x 10–1 (C) 8.0 x 10–2 2+ – Ni (s) + Cl2  Ni (aq) + 2Cl . Q=

Ans.

fy, vfHkfØ;k xq.kkad Q dk eku D;k gksxk\ (D) 3.0 x 10–2

[Ni2 ] [Cl– ]2 (0.19 ) (0.4 )2 = = 0.304. PCl2 0. 1

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PAGE NO.- 7

7.

The dissociation constant for [Ag(NH3)2]+ into Ag+ and NH3 is 10–13 at 298 K. If E 0Ag

= 0.8 V, then E0 for

[Ag(NH3)2]+ + e–  Ag + 2NH3 will be

the half cell

298K ij [Ag(NH3)2]+

dk Ag+ rFkk NH3 ds fy, fo;kstu fu;rkad 10–13 gSA ;fn E 0Ag 

[Ag(NH3)2]+ + e–  Ag + 2NH3 ds Sol.

Ag

Ag

= 0.8 V gks]

rks v)Z lsy

fy, Eº gksxkA

(A) 0.33 V (B) – 0.33 V [Ag(NH3)2]+ + e–  Ag+ + 2 NH3

(C) – 0.033 V Kd = 10–13

(D*) 0.033 V

G1º = RT ln Kd

Ag+ + e–  Ag

G3º = – 1 × F × 0.8

[Ag(NH3)2]+ + e–  Ag + 2NH3

G3º = G1º + G2º

– 1 × F × Eº = – RT ln k d – 1 × F × 0.8 Eº = –

RT ln k d + 0.8 F



8.314  298 × 2.303 × (–13) + 0.8 96500

Eº = 0.0313 V. 8.

You are given the following cell at 298 K with Eºcell = 1.10 V Zn(s) | Zn2+ (C1) || Cu++ (C2) | Cu(s) 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 C and y-axis : Ecell 2 Eºcell = 1.10 V ds lkFk 298 K ij fuEu lSy fn;k tkrk gSA Zn(s) | Zn2+ (C1) || Cu++ (C2) | Cu(s) tgk¡ C1 rFkk C2 eksy@yhVj esa lkUnzrk gS rc lkUnzrk ds Qyu

C1 x-v{k : log C 2

rFkk y-v{k : Ecell

(A)

Sol.

ds :i esa Ecell ds lkFk lgh rjg ls dkSu lacaf/kr gSA

(B*)

(C)

(D)

Zn(s)  Zn2+ + 2e– Cu2+ + 2e–  Cu(s) Zn(s) + Cu2+  Zn2+ + Cu(s) Ecell = Eºcell –

C1 0.0591 log C 2 2

Since slope is –ve i.e. –

0.0591 2

pw¡fd
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PAGE NO.- 8

9.*

a (dm 6 bar mol–2) b (dm3 mol–1) Gas A 6.5 0.055 Gas B 2 0.01 then which statements is/are INCORRECT : (Where symbols used here have their usual meanings) (A*) VC (A) < VC (B) (B) PC (A) < PC (B) (C) TC (A) < TC (B) (D*) Boyle temperature of A > Boyle temperature of B. a (dm 6 bar mol–2)

b (dm3 mol–1)

xSl A 6.5 0.055 xSl B 2 0.01 rc fuEu esa ls dkSulk@dkSuls dFku ¼lEcU/k½ lgh ugha gS@gSa % (tgk¡ ij mi;ksx esa fy;s x;s fpUg vius lkekU; vFkZ j[krs gSa)

Sol.

10.*

(A*) VC (A) < VC (B) (C) TC (A) < TC (B) VC = 3b  (VC)A < (VC)B

a

PC =

27b 2

 (PC)A  (PC)B

TC =

8a 27bR

 (TC)A  (TC)B

Tb =

a bR

 (Tb)A > (Tb)B

(B) PC (A) < PC (B) (D*) A dk ckW;y rki > B dk

(Tb is Boyle temperature) (Tb ckW;y

ckW;y rki

rkieku gS)

In which of the following molecules central atom involve expansion of octet ?

fuEu esa ls fdu v.kqvksa esa dsUnzh; ijek.kq esa v"V~d dk izlkj gksrk gS \ Sol. Sol.

(A*) PCl5 (B*) SO4– – (C) NH3 (D*) ClO3– (C) In NO3– molecule, nitrogen can not expand its octet due to non availability of vacant d-orbital. In (A), (B), (D) central atom expand octet by exciting electron in outermost vacant d-orbital. (C) NO3– v.kq esa] ukbVªkstu esa v"Vd dk izlkj ugh gksrk gS D;ksafd fjDr d- d{kd miyC/k ugha gksrk gSA (A), (B), (D) esa

dsUnzh; ijek.kq esa v"Vd dk izlkj bysDVªkWu dks ckáre fjDr d-d{kd mRrsftr djds fd;k tkrk gSA 11.*

Which statement(s) is/are correct ? (A*) Polarising power refers to cation. (B*) Polarisability refers to anion. (C*) Small cation is more efficient to polarise anion. (D*) Molecules in which cation having pseudo inert gas configuration are more covalent.

dkSulk dFku lR; gS \ (A*) /kzqo.k lkeF;Z] /kuk;u ls lEcfU/kr gSA (B*) /kzqoh;rk] _.kk;u ls lEcfU/kr gSA (C*) NksVk /kuk;u] _.kk;u dks vf/kd izHkkoh /kqzoh; cukrk gSA (D*) v.kq ftuesa /kuk;u Nn~e vfØ; xSl foU;kl j[krk gS] vf/kd lgla;kstd gksrk gSA Sol.

Factual according to Fajan's Rule.

Qt+ku fu;e ds vuqlkjA 12.*

When zeolite, which is hydrated sodium aluminium silicate, is treated with hard water the sodium ions are exchanged with (A) H+ ions (B*) Ca2+ ions (C) SO42 – ions (D*) Mg2+ ions

tc ft;ksykbV] tks fd lksfM;e ,Y;qfefu;e lhfydsV gS dks dBksj ty ds lkFk mipkfjr djrs gS rc lksfM;e vk;uks dk fofue; dkSuls vk;uks ls gksrk gSA (A) H+ ions (vk;u) (B*) Ca2+ ions (vk;u) (C) SO42 – ions (vk;u) (D*) Mg2+ ions (vk;u) Ans. Sol.

(B, D) Na2 AI2 Si2 O8 . xH2O + Ca2+  CaAI2 Si2 O8 . xH2O + 2Na+ Na2 AI2 Si2 O8 . xH2O + Mg2+  MgAI2 Si2 O8 . xH2O + 2Na+

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PAGE NO.- 9

13.*

Highly pure dilute solution of sodium in liquid ammonia : (A*) shows blue colour. (B*) exhibits electrical conductivity. (C) produces sodium amide. (D) produces hydrogen gas. Na dk

nzo veksfu;k esa vR;Ur 'kq) ruq foy;u % (A*) uhyk jax n'kkZrk gSA (C) lksfM;e ,ekbM cukrk gSA

(B*) oS|qr

pkydrk iznf'kZr djrk gSA (D) gkbMªkstu xSl mRiUu djrk gSA

Sol.

M + (x + y) NH3  [M (NH3)x]+ + [e(NH3)y]– The electrical conductivity is due to the solvated ions and blue colour is due to solvated electrons.

gy

M + (x + y) NH3  [M (NH3)x]+ + [e(NH3)y]–

oS/kqr pkydrk foyk;dhd`r vk;uks ds dkj.k izkIr gksrh gS rFkk uhyk jax foyk;dhd`r bysDVªkWu ds dkj.k izkIr gksrk gSA 14.*

The standard reduction potentials of some half cell reactions are given below : PbO2 + 4H+ + 2e– Pb2+ + 2H2O E0 = 1.455 V – + – 2+ MnO4 + 8H + 5e Mn + 4H2O E0 = 1.51 V 4+ – 3+ Ce + e Ce E0 = 1.61 V + – H2O2 + 2H + 2e 2H2O E0 = 1.71 V Pick out the correct statement : (A*) Ce+4 will oxidise Pb2+ to PbO (B*) MnO- will oxidise Pb2+ to PbO 4

2

(C*) H2O2 will oxidise Mn+2 to MnO-4

2

(D) PbO2 will oxidise Mn+2 to MnO-4

dqN v)Z lsy vfHkfØ;kvksa ds ekud vip;u foHko uhps fn, gSa & PbO2 + 4H+ + 2e– MnO4– + 8H+ + 5e– Ce4+ + e– Ce3+ + H2O2 + 2H + 2e–

Pb2+ + 2H2O Mn2+ + 4H2O 2H2O

E0 = 1.455 V E0 = 1.51 V E0 = 1.61 V E0 = 1.71 V

lgh dFku dks pqfu;s & (A*) Ce+4 , Pb2+ dks PbO2 esa

vkWDlhd`r djsxk

(C*) H2O2 , Mn+2 dks MnO-4 esa

vkWDlhd`r djsxk

(B*) MnO-4 , Pb2+ dks PbO2 esa (D) PbO2 , Mn+2 dks MnO-4

vkWDlhd`r djsxk

esa vkWDlhd`r djsxk

Sol.

From given SRP values.

15.

Depression of freezing point of 0.01 molal aqueous CH 3COOH solution is 0.02046°. 1 molal aqueous urea solution freezes at – 1.86°C. Assuming molality to be equal to molarity, calculate pH of CH 3COOH solution. 0.01 eksyy tyh; CH3COOH foy;u ds fgekad fcUnq dk voueu 0.02046° gSA ,d eksyy tyh; ;wfj;k foy;u – 1.86°C ij terk gSA ekuk fd eksyyrk eksyjrk ds cjkcj gS rc CH3COOH foy;u dh pH Kkr fdft;sA 3 For urea Tf = k f × m

Ans. Sol.

Tf 1.86 = = 1.86 m 1 Now for CH3COOH Tf = i kf m or

kf =

so

i=

Now so Now

so

0.02046 = 1.1 1.86  0.01 i=1+  = 1.1 – 1 = 0.1 CH3COOH CH3COO– C 0 C – C C [H+] = C = 0.01 × 0.1 = 0.001 pH = 3.

+

H+ 0 C

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PAGE NO.- 10

Sol.

;wfj;k ds fy, Tf = k

;k

kf =

Tf 1.86 = = 1.86 m 1

vr% CH3COOH ds fy, Tf = i kf m 0.02046 = 1.1 1.86  0.01

vr%

i=

rc

i=1+

vr%

 = 1.1 – 1 = 0.1

rc

CH3COOH

CH3COO–

C

0

0

C

C

C – C

+

H+

+

[H ] = C = 0.01 × 0.1 = 0.001

vr% 16.

pH = 3.

For aqueous solution of how many of the following compounds / mixtures, does the pH remains constant even upon dilution : (1) NH4Cl

(2) Na2CO3

(3) A 1 : 2 molar ratio mixture of Na2S and HCl.

(4) A 5 : 2 molar ratio mixture of NaOH and H3PO4.

(5) A 5 : 4 molar ratio mixture of CH3COONa and HCl.

(6) NaH2PO4

(7) A 2 : 1 molar ratio mixture of HCl and NaHCO3.

(8) CH3COONH4

(9) A 4 : 3 molar ratio mixture of NH4OH and HCl.

fuEufyf[kr esa ls fdrus ;kSfxdksa @ feJ.kksa ds tyh; foy;u dh pH, foy;u dks ruq djus ij Hkh fu;r jgrh gS % (1) NH4Cl

(2) Na2CO3

(3) Na2S rFkk HCl dk 1 : 2 eksyj

vuqikr dk ,d feJ.k

(4) NaOH rFkk H3PO4 dk 5:2 eksyj

vuqikr dk ,d

feJ.k (5) CH3COONa rFkk HCl dk 5:4 eksyj (7) HCl rFkk NaHCO3 dk 2 : 1 eksyj (9) NH4OH rFkk HCl dk 4 : 3 eksyj Ans.

5

Sol.

(1) pH =

vuqikr dk ,d feJ.k (6) NaH2PO4

vuqikr dk ,d feJ.k

(8) CH3COONH4

vuqikr dk ,d feJ.k

1 (pKw – pKb – logC) 2

;

(2) pH =

1 (pKw + pKa + logC) 2 2

(3) Na2S + 2HCl  2NaCl + H2S



pHWA =

1 (pKa – logC) 1 2

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PAGE NO.- 11

NaOH + H3PO4  NaH2PO4 + H2O

(4) i

5

2

f

3

0

2

2

NaOH + NaH2PO4  Na2HPO4 + H2O i

3

2

f

1

0

2

2

NaOH + Na2HPO4  Na3PO4 + H2O



i

1

2

f

0

1

Buffer solution ¼cQj

foy;u½

1

pH = pKa + log10



3

[PO 34– ] [HPO 24– ]

CH3COONa + HCl  CH3COOH + NaCl

(5) i

5

4

f

1

0



(6)

1

pH =

Acidic buffer ¼vEyh;

4

4



cQj½

pH = pKa + log10

[CH3 COO – ] [CH3 COOH]

pK a2  pK a1 2 HCl + NaHCO3  NaCl + H2CO3

(7) i

2

1

f

1

0

1

1

For a mixture of WA + SA, major H+ ions will come from HCl. WA + SA ds  (8)

pH =

,d feJ.k ds fy, H+ vk;u eq[;r% HCl ls vkrs gSA

pH = – log10[H+]HCl

1 (pKw + pKa – pKb) 2 NH4OH + HCl  NH4Cl + H2O

(9) i

4

3

f

1

0



3

3

[NH4 ] Basic buffer ¼{kkjh; cQj½  pOH = pKb + log10 [NH4 OH]

Clearly, pH of solutions 4, 5, 6, 8 and 9 is not affected by dilution.

Li"V gS fd foy;u 4, 5, 6, 8 rFkk 9 dh pH, foy;u dks ruq djus ij çHkkfor ugha gksrh gSA

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PAGE NO.- 12

17.

Determine the heat of hydrogenation of ethylene from the following data .

fuEufyf[kr vk¡dM+ksa ds vk/kkj ij ,Fkhyhu ds gkbMªkstuhdj.k dh Å"ek Kkr dhft, % Bond

Bond energy

Bond

Bond energy

ca/k

ca/k ÅtkZ

ca/k

ca/k ÅtkZ

HH

104 Kcal / mol

CC

80 Kcal / mol

CH

99 Kcal / mol

C=C

145 Kcal / mol

Ans.

 29 Kcal / mol

Sol.

C2H4(g) + H2  C2H6 H = (H)sup – (H)req H = [145 + 104] – [80 + 2 × 99] H =  29 Kcal / mol

18.

In how many following species, the central atoms have two lone pairs of electrons ?

fuEu esa ls fdruh Lih'kht esa dsUnzh; ijek.kq nks ,dkdh bysDVªkWu ;qXe j[krk gS \ XeF4

XeF5–

F2SeO2

ICl4–

SCl2

OSF4

XeF3+

XeOF4

ClOF3

Ans.

5

Sol.

XeF4

XeF5–

F2SeO2

XeF3+

XeOF4

ClOF3

ICl4–

SCl2

OSF4

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PAGE NO.- 13

ChemINFO-1.15

ELECTROCHEMISTRY Galvanic Cell & Salt Bridge

Daily Self-Study Dosage for mastering Chemistry

A device used to convert chemical energy produced in a redox reaction into electrical energy is called an electrochemical cell or simply chemical cell. These are also called galvanic cells or voltaic cells, after the name of Luigi Galvani (1780) and Alessandro Volta (1800). If the electrodes in a galvanic cell are not connected internally, initially the cell generates electrical energy. The electrodes get polarized with passage of time and the cell stops working. For continuous of electrical energy, the electrodes need to be internally connected. A salt bridge is usually used to connect the electrodes of a galvanic cell. It is easy to construct and consists of a gell drawn into a clean U-tube. The gel is formed by adding agar-agar into boiling water containing a inert electrolyte like KCl. Generally, KCl is preferred because the ionic conductances of K + and Cl– are almost the same. The gel holds the electrolyte in the tube as shown in below Fig.

Fig : Salt bridge and electrolyte As the reactions proceed in the galvanic cell, there will be accumulation of particular ions around the electrodes. This results in a slowdown of the electrode reactions. From the electrolyte of the salt bridge, say KCl, the K+ and Cl– ions migrate towards the end of the salt bridge K+ ions would migrate towards the reducation electrode (cathode) while Cl– ions would migrate towards the oxidation electrode (anode). At the oxidation electrode (anode), ions would polarize (attract) some of the oppositely charged ions towards themselves. Simultaneously, the K+ ions would move towards the reduction electrode (cathode) and polarize the negatively charged ions surrounding the metal. Thus, the cell works as usual. Hence, a salt bridge is helpful in the following ways. 1. It brings about internal contact between the electrodes. 2. It minimizes the liquid junction potential. 3. It minimizes polarization. Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 19.

In which of the following cells salt bridge is not needed ? (A*) Pb | PbSO4(s) | H2SO4 | PbO2 | Pb (B*) Cd | CdO(s) | KOH(aq) | NiO2(s) | Ni (C*) Fe(s) | FeO(s) | KOH (aq.), Ni2O3(s) | Ni (D) Zn | ZnSO 4 | CuSO4 | Cu

20.

The reaction 1/2 H2(g) + AgCl(s)  H(aq) + Cl–(aq) + Ag(s) occurs in the galvanic cell (A) Ag | AgCl(s) | KCl (sol) | AgNO3 | Ag (C*) Pt | H2(g) | HCl (sol) | AgCl(s) | Ag

21.

(B) Pt | H2(g) | HCl(sol) | AgNO3(sol) |Ag (D*) Pt | H2(g) | KCl(sol) | AgCl(s) |Ag

A saturated solution of KNO3 is used to make salt bridge because (A) The velocity of K is greater than that of NO 3–. (B) The velocity of NO3– is greater than that of K (C*) The velocities of K and NO3– are nearly the same (D) KNO 3 is highly soluble in water.

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PAGE NO.- 14

ChemINFO-1.15

fo|qr jlk;u Galvanic Cell & Salt Bridge

Daily Self-Study Dosage for mastering Chemistry

og lSy ftls jsMkWDl vfHkfØ;k ls mRiUu jklk;fud ÅtkZ dks oS|qr ÅtkZ esa :ikUrfjr djus ds fy, iz;qDr fd;k tkrk gS ,d oS|qr jklk;fud lSy ;k lkekU; jklk;fud lSy dgykrk gSA bUgsa Y;wxh XkSYosuh (1780) rFkk ,yslsUMªks oksYVk (1800) ds uke ds i'pkRk~ xSYosfud lSy ;k oksYVsbZd lSy Hkh dgk tkrk gSA ;fn xsYosfud lSy esa bysDVªkWM vkUrfjd :i ls tqM+s gq, ugh gS rks izkjEHk esa lSy fo|qr ÅtkZ mRiUu djrk gSA bysDVªkWM le; xqtjus ds lkFk&lkFk èkqzohd`r gksrk gS rFkk lSy dk;Z djuk can dj nsrk gSA lrr~ fo|qr ÅtkZ ds fy, bys DVªkWMksa dk vkUrfjd :i ls tqM+k gksuk vko';d gksrk gSA ,d yo.k lsrq izk;% xsYosfud lSy ds bysDVªkWMksa dks tksMus ds fy, iz;qDr gksrk gSA ;g vklkuh ls cuk;k tk ldrk gS rFkk ,d LoPN U ufydk esa tSy dk cuk gksrk gSA tSy dks KCl ds leku vfØ; fo|qr vi?kV~; okys ,d mcyrs gq, ty esa vxkj&vxkj esa feykdj cuk;k tkrk gSA lkekU;r% KCl dks izkFkfedrk nh tkrh gS] D;ksafd K+ rFkk Cl– dh vk;fud pkydrk;sa izk;% leku gksrh gSA tSy fuEu fp=k esa n'kkZ;s vuqlkj ufydk esa bysDVªksMksa dks jksdrk gSA

Fig : yo.k lsrq rFkk oS|qr vi?kV~; tSls gh xsYosfud lSy esa vfHkfØ;k,sa gksrh gS] ;gk¡ bysDVªksMksa ds pkjksa vkSj fo'ks"k vk;uksa dk laxzg.k gksxkA blds ifj.kkeLo:i bysDVªkWM vfHkfØ;k;sa èkheh gks tkrh gSA yo.k lsrq KCl ds oS|qr vi?kV~; ls K+ rFkk Cl– vk;u yo.k lsrq ds fljs dh vksj xeu djrs gSA K+ vk;u] vip;u bysDVªkWM (dSFkksM) dh vksj tcfd Cl– vk;u] vkWDlhdj.k bysDVªkWM (,uksM) dh vksj xeu djrs gSA vkWDlhdj.k bysDVªkWM (,uksM) ij vk;u buds Lo;a dh vksj dqN foifjr vkosf'kr vk;uksa dks èkzqoh;d`r ¼vkdf"kZr½ djsaxAsa blds lkFk gh K+ vk;u] vip;u bysDVªkWM ¼dSFkksM½ dh vksj xfr djsxas rFkk /kkrq ds pkjks vksj mifLFkr _.kkosf'kr vk;uksa dks /kzqoh;d`r djsxsaA bl izdkj lSy lkekU; :i esa dk;Z djrk gSA blfy, ,d yo.k lsrq fuEu izdkjksa esa lgk;d gksrk gS& 1.

;g bysDVªkWMksa ds eè; vkUrfjd lEidZ ykrk gSA 2. ;g æo lfUèk foHko dks U;wure djrk gSA 3. ;g èkzqoh;dj.k dks U;wure djrk gSA Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 19.

fuEu es ls dkSuls lsy es yo.k lsrq dh vko';drk ugha gksrh gSA (A*) Pb | PbSO 4(s) | H2SO4 | PbO2 | Pb (B*) Cd | CdO(s) | KOH(aq) | NiO 2(s) | Ni (C*) Fe(s) | FeO(s) | KOH (aq.), Ni2O3(s) | Ni (D) Zn | ZnSO4 | CuSO4 | Cu

20.

21.

KNO3 dk lUriz

foy;u yo.k lsrq cukus es iz;qDRk gksrk

gS D;ksfd (A) K dh xfr'khyrk NO3– dh rqyuk es vf/kd gksrh gSA (B) NO3– dh xfr'khyrk Kdh rqyuk esa vf/kd gksrh gSA (C*) K rFkk NO3– dh xfr'khyrk yxHkx leku gksrh gSA (D) KNO3 ty esa vf/kd foys;'khy gksrk gSA

vfHkfØ;k 1/2 H2(g) + AgCl(s)  H(aq) + Cl– (aq) + Ag(s) xSyosfud lsy es gksrh gSA (A) Ag | AgCl(s) | KCl (sol) | AgNO 3 | Ag (B) Pt | H2(g) | HCl(sol) | AgNO 3(sol) |Ag (C*) Pt | H2(g) | HCl (sol) | AgCl(s) | Ag (D*) Pt | H2(g) | KCl(sol) | AgCl(s) |Ag Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

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