Jp Xii Physical&inorganic Chemistry (16) - Prev Chaps + Inorg. Chem.pdf

PHY./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2016 NO. 25 & 26

Course : VIJETA (JP)

This DPP is to be discussed in the week (13.07.2015 to 18.07.2015)

DPP No. # 25 (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]

Hº of which of the following reactions is zero?

fuEu esa ls fdl vfHkfØ;k ds fy, Hº dk eku 'kwU; gS \ Sol. 2.

(A) H2(g)  2H+(g) + 2e (B) 2H(g)  2H+(g) + 2e +  (C*) H2(g) + aq  2H (aq) + 2e (D) 2H(g) + aq  2H+(aq) + 2e + Standard enthalpy of formation of H (aq) is taken to be zero. Which of the following orders is correct with respect to the property indicated against each ? (A) NH3 > SbH3 ; Boiling point. (B) S8 > P4 ; Percentage of p-character in hybrid orbitals forming X – X bonds (i.e. S – S or P – P bonds). (C*) CH3C  CCH3 < CH3CH2C  CH ; Dipole moment. (D) C22– < N2 ; Number of  bonds. fuEu Øeksa] ds vkxs fn;s x;s xq.kksa ds lanHkZ esa dkSulk Øe lgh gS ? (A) NH3 > SbH3 ; DoFkukad

fcUnqA (B) S8 > P4 ; X – X ca/kksa ds (vFkkZr~ S – S vFkok P – P ca/k) fuekZ.k djus okys ladfjr d{kdksa esa p y{k.kksa dk izfr'krA (C*) CH3C  CCH3 < CH3CH2C  CH ; f}/kqzo vk?kw.kZA (D) C22– < N2 ;  ca/kksa dh la[;kA Sol.

3.

(C) In But-2-yne vectors cancels each other ; so  = 0 while But-1-yne has some dipole moment. (C) C;qV-2-vkbu esa lfn'k ,d nwljs dks fujLr dj nsrs gS ; vr%  = 0 tcfd C;qV-1-vkbu dk dqN f}/kqzo vk?kw.kZ 

gksrk gSA



The bond moment C – F and C – Cl are 1.39 D and 1.47 D respectively, because (A) F is more electronegative than Cl (B) Cl is more electronegative than F. (C) The electron affinity of Cl is greater than F. (D*) The bond distance of C–F bond is lesser than the bond distance of C–Cl bond. 

C–F

rFkk



C – Cl

ca/kks ds ca/k vk?kw.kZ Øe'k% 1.39 D rFkk 1.47 D gS] D;ksafd -

(A) F , Cl dh

rqyuk vf/kd fo|qr_.kkRed gSA (B) Cl, F dh rqyuk vf/kd fo|qr_.kkRed gSA (C) Cl dh bySDVªkWu xzg.k ,UFkSYih F dh rqyuk esa vf/kd gSA (D*) C–F ca/k dh ca/k yEckbZ] C–Cl ca/k yEckbZ dh rqyuk esa de gksrh gSA 4.

Which process occurs in the electrolysis of an aqueous tin(II) chloride solution at a tin anode?

dkSulh fØ;k,¡ gksxh] tc fVu (II) DyksjkbM ds tyh; foy;u dk fVu ,uksM+ ij fo|qr vi?kVu djk;k tkrk gS \ (A*) Sn = Sn2+ + 2e– (C) 2H2O = O2 + 4H+ + 4e–

(B) 2Cl- = Cl2 + 2e– (D) none of these mijksDr

esa ls dksbZ ugha

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PAGE NO.- 1

Sol.

Solution contains Sn2+, Cl–, H O. 2 Anode is tin anode.

Eo

Sn2 / Sn

E oO / H O = 1.23V 2 2

= –0.14V

Eo

Cl2 / Cl–

= 1.36V..

o From the SRP values, it is clear that the reaction for which SRP is least will take place at anode ESn2 / Sn is least. So, Sn  Sn2+ + 2e– Ans.

5.

In the electrolysis of an aqueous nickel(II) sulphate solution, the process 2H2O = O2 + 4H+ + 4e- occurs at the anode. The material of construction of the anode is (A) nickel (B*) gold (C) copper (D) none of these fudy (II) lYQsV ds tyh; foy;u esa fo|qr vi?kVu djkus ij] ,suksM+ ij 2H2O = O2 + 4H+ + 4e– fØ;k gksrh gS ,suksM+

dk fuekZ.k dkSu ls inkFkZ ls gksrk gSA (A) fudy (B*) xksYM Sol.

(C) dkWij

(D)

mijksDr esa ls dksbZ ugha

The solution contains Ni2+, SO42–, H2O

E o 2 Ni

/ Ni

Eo

= 0.23V

S2O28 – / SO24–

= 2.05V

Eo 3 E o 2 E oO / H O = 1.23V = 1.4V = 0.34V 2 2 Au / Au Cu / Cu If Nickel electrode is used, then Nickel will get oxidised as its SRP is least. So Nickel is not anode. If gold electrode is used, then water will get oxidised as SRP of gold is more than SRP of water. So gold is the anode. If copper electrode is used, then copper will get oxidised as SRP of Cu is less than that of water. 6.

In the electrolysis of an aqueous potassium sulphate solution, the pH of the solution in the space near an electrode increased. Which pole of the current source is the electrode connected to? (A) The positive pole (B) Could be either pole (C*) The negative pole (D) Cannot be determined ,d tyh; iksVsf'k;e lYQsV foy;u ds fo|qr vi?kVu esa bysDVªkWM ds ikl ds LFkku esa foy;u dk pH c<+k;k tkrk gSA

fo|qr L=kksr ds dkSu ls /kqzo ls bysDVªkWM dks tksM+k x;k gS \ (A) /kukRed /kzqo (B) fdlh Hkh /kzqo ls (C*) _.kkRed /kqzo +

(D) Kkr

ugha fd;k tk ldrk gSA

2–

Sol.

K2SO4(aq.) contains K , SO4 , H2O At anode ; water gets oxidized 2H2O O2 + 4n+ + 4e– as H+ are produced, region around anode has lesser pH. At cathode, water get reduced. 2H2O + 2e– H2 + 2OH– as OH– are produced, region around cathode has higher pH. Cathode is negatively charged.

7.

In the electrolysis of an aqueous solution of a salt, the pH in the space near one of the electrodes increased. A solution of which salt is being electrolyzed? (A) none of the following (B) CuCl2 (C) Cu(NO3)2 (D*) KCl ,d yo.k ds ,d tyh; foy;u ds fo|qr vi?kVu esa ,d bysDVªkWM ds ikl ds LFkku ds foy;u dk pH c<+ tkrk gS] fdl

yo.k ds foy;u dk fo|qr vi?kVu djk;k tk,xk \ (A) fuEu esa ls dksbZ ugha (B) CuCl2

(C) Cu(NO3)2

Sol.

In CuCl2 solution, At anode either 2Cl– Cl2 + 2e–

8.

Number of electrons lost during electrolysis of 0.355 g of Cl¯ is 0.355 g Cl¯ ds fy, fo|qr vi?kVu ds nkSjku fdrus bysDVªkWu fudysxsa \ (A) 0.01

(B*) 0.01 N0

(C) 0.02 N0

(D*) KCl

0.01 (D) 2 N 0

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PAGE NO.- 2

9.

In the electrolysis of a copper(II) chloride solution, the mass of the cathode increased by 3.2 g. W hat occurred at the copper anode ? (A*) 0.05 mol of Cu2+ passed into the solution (B) 0.56 L of O2 was liberated 2+ (C) 0.1 mol of Cu passed into the solution (D) 0.112 L of Cl2 was liberated dkWij (II) DyksjkbM foy;u ds fo|qr vi?kVu ij dSFkksM+ dk nzO;eku 3.2 g c<+rk gSA dkWij ds ,suksM ij D;k ik;k tkrk gS \ (A*) Cu2+ ds 0.05 eksy foy;u esa tkrs gS (B) O2 ds 0.56 L eqDr gksrs gS (C) Cu2+ ds 0.1 eksy foy;u esa tkrs gS (D) Cl2 ds 0.112 L eqDr gksrs gS

10.

The passage of a constant current through a solution of dilute H 2SO4 with ‘Pt’ electrodes liberated 336 cm 3 of a mixture of H2 and O2 at S.T.P. The quantity of electricity that was passed is ,d ruq H2SO4 foy;u esa ls ‘Pt’ bysDVªkWM ds lkFk fu;r /kkjk izokfgr dh tkrh gS rks STP ij H2 rFkk O2 dk 336 cm 3

feJ.k eqDr gksrk gS rks izokfgr gksus okyh /kkjk dh ek=kk gksxh \ (A) 96500 C Sol.

(C*) 1930 C

(D)

1 faraday QSjkMs 100

1 2 e H2O   H2 + 2 O2 2 × 96500 C  22.4 × 1000 cm 3 H2 2  96500  224 C 22.4  1000 = 1930 C.

11.

(B) 965 C

 224 cm3 H2

A very thin copper plate is electro-plated with gold using gold chloride in HCl. The current was passed for 20 min. and the increase in the weight of the plate was found to be 2g. [Au = 197]. The current passed was ,d cgqr iryh dkWij dh IysV dks fo|qr ysiu ds fy, xksYM ds lkFk HCl esa xksYM DyksjkbM dks ysdj fd;k tkrk gSA èkkjk dks 20 feuV rd izokfgr fd;k tkrk gS rFkk IysV dk Hkkj c<+rk gSA ;g ik;k x;k gS fd ;g 2g gSA rc izokfgr dh x;h

/kkjk fuEu gS& Sol.

(A) 0.816 amp Gold chloride AuCI3 m= =

12.

(B) 1.632 amp

(C*) 2.448 amp

(D) 3.264 amp

M ×I×t 96500  n

2  96500  3 = 2.448A. 197  20  60

What is the formula of a compound of niobium and nitrogen that crystallizes in a hexagonal closest packed array of nitrogen atoms with niobium atoms in half of the tetrahedral holes ?

uk;ksfc;e o ukbVªkstu ds ,d ;kSfxd dk lw=k D;k gS tks prq”Qydh; fNnzkas ds vk/ks esa uk;ksfc;e ijek.kqvksa ds lkFk ukbVªkstu ijek.kqvksa ds ,d “kV~Hkqth; fufcM ladqy O;wg esa fØLVyhd`r gksrs gksa \ Sol.

(A) Nb4N (B) Nb2N Effective number of N = 6 Effective number of Nb = 6 × 2 ×

13.

Sol.

(C*) NbN

(D) NbN2

1 =6 2

 Formula = Nb6N6 or = NbN F-centres are (A*) the electrons trapped in anionic vacancies (B) the electrons trapped in cation vacancies (C) non-equivalent sites of stoichiometric compound (D) all of the above F-dsUnz gS % (A*) _.kk;fud fjfDr;ksa esa mifLFkr bySDVªkWu (B) /kuk;fud fjfDr;ksa esa mifLFkr bySDVªkWu (C) jllehdj.kferh; ;kSfxd ds fy, vrqY;rk LFky (D) mijksDr lHkh (A) 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.- 3

14.

The total no. of triple points in the phase diagram given below is/are :

uhps fn;s x;s izkoLFkk fp=k esa f=kd fcUnqvksa dh dqy la[;k gS%

(A) 1 (A) 1 15.

(B) 2 (B) 2

(C*) 3 (C*) 3

(D) No triple point (D) dksbZ f=kd fcUnq ugh

According to Henry’s law, the solubility of a gas in a given volume of liquid increases with increase in :

gsujh fu;e ds vuqlkj] fn;s x;s nzo ds vk;ru es xSl dh foys;rk fuEu ds c<us ds lkFk c
16.

(A) Temperature

(B*) Pressure

(C) Both (A) and (B)

(D) None of these

(A) rki

(B*) nkc

(C) (A) o (B)

(D) buesa

nksuks

ls dksbZ ugh

Identify the INCORRECT statement (A) Orbitals of a subshell have same value of ( n + ) but the orbitals having same (n + ) value may not belong to same subshell. (B*) The orbital having value of  = 2 will have 2 orientations in space. (C) 3p and 3d orbitals have directional character. (D) Orbits in Bohr’s model are two dimensional in nature. (E) All are correct.

vlR; dFku dks igpkfu;sA (A) fdlh

midks'k esa mifLFkr d{kdksa ds fy,] ( n + ) ds eku leku gksrs gSa] ijUrq d{kd ftlds fy, ( n + ) ds eku leku

gS] os leku midks'k ls lEcfU/kr gksa] ;g vko';d ugha gSA (B*) og

d{kd ftlds fy;s  dk eku 2 gS] og f=kfoe (space) esa nks izdkj ds vfHkfoU;kl (orientations) n'kkZrk gSA

(C) 3p ,oa 3d d{kd

Sol.

fn'kkRed izd`fr (directional character) ds gksrs gSaA

(D) cksgj

ds ijek.kq izfr:i esa dks'kksa (Orbits) dh izd`fr f}fofe; (two dimensional) gksrh gSA

(E) lHkh

lgh gSA

n denotes shell and  denotes subshell

 (n + ) value will be same for all orbitals of a subshell. Orbital having value of  = 2 will have 5 orientations in space (d xy , dyz , dzx , d x 2  y 2 , d z 2 ) Only s-orbital is non-directional

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PAGE NO.- 4

Sol.

n

dks'k dks n'kkZrk gS rFkk  midks'k dks n'kkZrk gSA

 ,d midks'k ds lHkh d{kdksa ds fy, (n + ) ds eku leku gksxsaA

d{kd ftlds fy, dk eku 2 gS] mlds fy, f=kfoe esa dqy 5 izdkj ds vfHkfoU;kl gksxsaA (d xy , dyz , dzx , d x 2  y 2 , d z 2 )

dsoy s-d{kd vfn'kkRed gksrs gaSA 17.

Which of the following is paramagnetic ?

fuEu esa ls dkSulk vuqpqEcdh; gS \ Sol. Sol.

18.

(A*) O2– (B) CN– (C) NO+ – O2 superoxide has one unpaired electron by MOT. 1s2, *1s2, 2s2, *2s2, 2pz2, 2px2 = 2py2, *2px2 = *2py1 MOT ds vuqlkj O2– ,d v;qfXer bysDVªkWu j[krk gSA 1s2, *1s2, 2s2, *2s2, 2pz2, 2px2 = 2py2, *2px2 = *2py1

(D) CO

Which complex is likely to show optical activity:

dkSuls ladqy }kjk izdkf'kd leko;ork n'kkZuk laHkkfor gSA (A) Trans-[Co(NH3)4Cl2]+ (C*) Cis-[Co(NH3)2(en)2]3+ (A) foi{k -[Co(NH3)4Cl2]+ (C*) lei{k-[Co(NH3)2(en)2]3+ 19.

(B) [Cr(H2O)6]3+ (D) Trans-[Co(NH3)2(en)2]3+ (B) [Cr(H2O)6]3+ (D) foi{k -[Co(NH3)2(en)2]3+

Which of the following statements is correct with respect to the metal carbonyls of Ist transition series? (A) As M – C  bonding increases, the C – O bond length increases. (B) As positive charge on the central metal atom increases, the C – O bond length increases. (C) As electron density on the central metal atom increases, the C – O bond length increases. (D*) (A) and (C) both

izFke laØe.k Js.kh ds /kkrq dkcksZfuyksa ds lUnHkZ esa fuEu esa ls dkSulk dFku lgh gS\ (A) tSls&tSls M – C  ca/ku c<+rk gS] C – O ca/k yEckbZ c<+rh gSA (B) tSls&tSls dsfUnz; /kkrq ijek.kq ij /kukos'k c<+rk gS] C – O ca/k yEckbZ c<+rh gSA (C) tSls&tSls dsfUnz; /kkrq ijek.kq ij bysDVªkWu ?kuRo c<+rk gS] C – O ca/k yEckbZ c<+rh gSA (D*) (A) rFkk (C) nksuksa Sol.

Order of C – O bond strength : [Mn(CO)6 ]  [Cr(CO)6 ]  [V(CO)6 ]  [Ti(CO)6 ]2 and [Ni(CO) 4 ]  [Co(CO)4 ]  [Fe(CO)4 ]2 (A) True Statement (B) As + ve charge on the central metal atom increases, the less readily the metal can donate electron density into the  * orbitals of CO ligand to weaken the C - O bond. (C) In the carbonylate anions, the metal has a greater electron density to be dispersed, with the result that M - C  bonding is enhanced and the C - O bond is diminished in strength. C – O ca/k lkeF;Z dk Øe % [Mn(CO)6 ]   [Cr(CO)6 ]  [V(CO)6 ]  [Ti(CO)6 ]2  and[Ni(CO)4 ]  [Co(CO)4 ]  [Fe(CO)4 ]2  (A) lR; dFku (B) tSls&tSls dsfUnz;

/kkrq ijek.kq ij /kukos'k c<+rk gS] oSls&oSls /kkrq }kjk CO fyxs.M ds  * d{kdksa esa eqDr (donate) bysDVªkWuksa dk ?kuRo de gksrk tkrk gSA vr% C - O ca/k nqcZy gks tkrk gSA (C) dkcksZfuysV _.kk;uksa esa /kkrq ij vf/kd bysDVªkWu ?kuRo forfjr (dispersed) gksrk gSA ifj.kkeLo:i M - C  ca/ku c<+ tkrk gS rFkk C - O ca/k lkeF;Z ?kV tkrh gSA

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PAGE NO.- 5

20.

A current is passed through 2 voltameters connected in series. The first voltameter contains XSO 4 (aq.) and second has Y2SO4 the relative atomic masses of X and Y are in the ratio of 2 : 1. The ratio of the mass of X liberated to the mass of Y liberated is Js.kh esa la;ksftr 2 oksYVehsVj esa ls /kkjk izokfgr dh tkrh gSA izFke oksYVehVj] XSO4 (tyh;) ;qDr gSa rFkk f}rh; Y2SO4 ;qDr gSaA X rFkk Y dk vkisf{kd ijek.kq nzO;eku 2 : 1 vuqikr esa gSaA eqDr gq, X rFkk eqDr gq, Y ds nzO;ekuksa dk vuqikr

fuEu gSA Sol.

(A*) 1 : 1 XSO4 (aq.) Y2SO4 (aq.) Mol.mass X 2 Mol. massY = 1

(B) 1 : 2



(C) 2 : 1

eq. wt. of X =

(D) None of the above mijksDr

mol. mass X 2

eq. wt. of Y =



esa ls dksbZ ugha

mol. mass Y 2

mass of Y eq.wt. of Y mol.mass Y 2 mass of X = eq.wt. of X = mol.mass X × 1 = 1:1.

DPP No. # 26 (JEE-ADVANCED) Total Marks : 84

Max. Time : 49 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.8 Multiple choice objective ('–1' negative marking) Q.9 to Q.12 Comprehension ('–1' negative marking) Q.13 to Q.16 Integer type Questions ('–1' negative marking) Q.17 to Q.18 ChemINFO : 5 Questions ('–1' negative marking) Q.19 to Q.23

1.

(3 (4 (3 (4 (4

marks, 2 min.) marks, 2 min.) marks, 2 min.) marks 3 min.) marks, 2 min.)

[24, [16, [12, [12, [20,

16] 08] 06] 09] 10]

A very thin copper plate is electro-plated with gold using gold chloride in HCl. The current was passed for 20 min. and the increase in the weight of the plate was found to be 2g. [Au = 197]. The current passed was ,d cgqr iryh dkWij dh IysV dks fo|qr ysiu ds fy, xksYM ds lkFk HCl esa xksYM DyksjkbM dks ysdj fd;k tkrk gSA èkkjk dks 20 feuV rd izokfgr fd;k tkrk gS rFkk IysV dk Hkkj c<+rk gSA ;g ik;k x;k gS fd ;g 2g gSA rc izokfgr dh x;h

/kkjk fuEu gS& Sol.

(A) 0.816 amp Gold chloride AuCI3 m= =

(B) 1.632 amp

(C*) 2.448 amp

(D) 3.264 amp

M ×I×t 96500  n

2  96500  3 = 2.448A. 197  20  60

2.

In the electrolysis of an aqueous SnCl2 solution, 4.48 L of chlorine at STP were liberated at the anode. The mass of tin deposited at the cathode was (M of Sn = 118.5) tyh; SnCl2 foy;u ds fo|qr vi?kVu esa ,suksM ij STP ij 4.48 L Dyksjhu fu"dkflr gksrh gS rks dSFkksM ij fu{ksfir gksus okys fVu dk nzO;eku gksxk \ (M of Sn = 118.5) (A) 119 g (B) 79.3 g (C) 47.4 g (D*) 23.7 g

3.

What must be the concentration of Ag+ in an aqueous solution containing Cu2+ = 1.0 M so that both the 0 0 metals can be deposited on the cathode simultaneously. Given that E Cu Cu2  = – 0.34 V and E Ag 

0.812 V, T = 298 K (A) nearly 10–19 M

,d tyh; foy;u tks fd

(B) 10–12 M Cu2+ = 1.0 M j[krk

(C*) 10–8 M

4.

(B) 10–12 M

=

(D) nearly 10–16 M

gSa esa Ag+ vk;u dh lkUnzrk D;k gksuh pkfg, dh nksuksa /kkrq dSFkksM+ ij ;qxir

:i ls ¼lkFk&lkFk½ fu{ksfir gks] fn;k x;k gS E 0Cu Cu2  = – 0.34 V rFkk E 0Ag (A) yxHkx 10–19 M

Ag

Ag

(C*) 10–8 M

= 0.812 V, T = 298K (D) yxHkx 10–16 M

Electrolytic reduction of 6.15 g of nitrobenzene using a current effeciency of 40% will require which of the following quantity of electricity. [C = 12, H = 1, N = 14, O = 16] 40% n{krk dh /kkjk iz;qDr dj ukbVªkscsUthu ds 6.15 g dk oS|qr vi?kVuh; vip;u ds fy, vko';d fo|qr /kkjk dh ek=kk fuEu esa ls dkSulh gksxhA [C = 12, H = 1, N = 14, O = 16] (A*) 0.75 F (B) 0.15 F (C) 0.75 C (D) 0.125 C 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.- 6

Sol. 6.15 = 0.05 mole of mitro benzene 123

V.F = 6 electricity of charge required if efficiency is 100% = 0.05 × 6 = 0.3 F But efficiency is 40% charge required =

0. 3 = 0.75. 0. 4

5.

Electrolysis of a solution of HSO 4 ions produces S 2O8 . Assuming 75% current efficiency, what current should be employed to achieve a production rate of 1 mole of S 2O 8 per hour ? HSO4 vk;uksa ds foy;u ds fo|q r vi?kVu ij S 2O8 vk;u mRiUu gks rs gSa A 75% fo|qr n{krk ekurs gq ,] 1 eks y S 2O8 izfr ?k.Vk dh mRiknu nj dks iz kIr djus a ds fy, fdruh /kkjk iz okfgr djuh gks xh& (A*)  71.5 amp (B) 35.7 amp (C) 142.96 amp (D) 285.93 amp

6.

Aqueous solution of Na2SO4 containing a small amount of HPh (Phenolphthalein) is electrolysed using Ptelectrodes. The colour of the solution after some time will : (A*) remain colourless (B) change from pink to colourless (C) change from colourless to pink (D) remain pink

lw{e ek=kk esa HPh ¼fQukWY¶Fksfyu½ ;qDr Na2SO4 ds tyh; foy;u esa Pt-bysDVªkWM dk iz;ksx djds fo|qr vi?kVu fd;k tkrk gSA dqN le; i'pkr foy;u dk jax gksxk % (A*) jaxghu jgsxk (B) xqykch ls jaxghu gks tk,xk (C) jaxghu ls xqykch gks tk,xk (D) xqykch jgsxk Sol.

Remains colourless since solution is almost neutral & Hph shows pink colour only in basic medium. Cathode :

2H2O + 2e  H2 + 2OH–.

Anode :

2H2O  O2 + 4H+ + 4e–.

both OH– equivalent = H+ equivalent.

jaxghu cuk jgsxk pwafd foy;u yxHkx mnklhu gS rFkk Hph dsoy {kkjh; ek/;e esa xqykch jax iznf'kZr djrk gSA dSFkksM : 2H2O + 2e  H2 + 2OH–. ; ,suksM : 2H2O  O2 + 4H+ + 4e–. nksuksa OH– ds rqY;kad = H+ ds rqY;kad 7.

What will happen to the state of a substance represented by this phase diagram when its temperature and pressure are changed from 60ºC and 0.2 atm to 40ºC and 1.0 atm ? fuEu izkoLFkk fp=k }kjk n'kkZ; x;s inkFkZ dh voLFkk D;k gksxh] tc bldk rki rFkk nkc 60º rFkk 0.2 atm ls 40ºC rFkk 1.0 atm rd ifjofrZr fd;k tkrk gSA

(A) liquid to solid (A) nzo ls Bksl

(B*) gas to solid (B*) xSl ls Bksl

(C) solid to liquid (C) Bksl ls nzo

(D) solid to gas (D) Bksl ls xSl

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PAGE NO.- 7

8.

Sol.

The emf of the cell Ag | Ag+ (1 M) || I– (1 M) | AgI | Ag is E The solubility product of AgI can be expressed as : lSy Ag | Ag+ (1 M) || I– (1 M) | AgI | Ag dk emf, E gS] AgI dk

foys;rk xq.kkad fuEu izdkj ls O;Dr fd;k tkrk gS%

nF  E   E  (C) ln K = nF (A) Ks = 2.303 RT log E (B) ln K = nF  s  T   E + – For the cell Ag | Ag || I , AgI ; Ag LHS electrode Ag  Ag+ + e– RHS electrode AgI + e–  Ag + I– Cell reaction, AgI Ag+ + I– The equilibrium constant (K) = The solubility product Ks G = – nFE Gº = – 2.303 RT log K

nFE (D*) log Ks = 2.303 RT

nFE log K = 2.303 RT 9*.

Just sufficient Al2(SO4)3 (s) is added to a 0.4 M solution of BaCl2 such that all the BaSO 4 is precipitated completely (volume of solution is maintained constant then) : (A*) Freezing point of the original solution is raised (B) Boiling point of the original solution is raised (C*) Vapour pressure of the original solution is raised (D) No change occurs in the colligative properties of solutions 0.4 M BaCl2 foy;u esa Al2(SO4)3 dh f
gS ¼foy;u dk vk;ru fLFkj j[krs gq,½ rks (A*) okLrfod foy;u dk fgekad fcUnq c
Which of the following are correct statements ? (A*) vander Waals constant ‘a’ is a measure of attractive force (B*) van der Waals constant ‘b’ is also called co-volume or excluded volume (C*) ‘b’ is expressed in L mol–1 (D*) ‘a’ is expressed in atm L2 mol–2

fuEu esa ls lgh dFku dkSulk gS \ (A*) okUMjokWYl~ fu;rkad ‘a’ vkd"kZ.k cy dk ekiu djrk gSaA (B*) okUMjokWYl~ fu;rkad ‘b’ dks lg&vk;ru vFkok vlfEefyr vk;ru ¼fudkyk x;k vk;ru½ Hkh dgrs gSaA (C*) ‘b’ dks L mol–1 esa O;Dr djrs gSa& (D*) ‘a’ dks atm L2 mol–2 esa O;Dr djrs gaSA Sol.

Factual Question.

11._*

Which of the following statements is/are correct : (A*) Among NaF & NaBF4, NaF is relatively less soluble in water. (B*) Larger negative value of lattice enthalpy is a favourable condition for formation of ionic bond. (C*) In general, ionic compounds do not conduct electricity in solid state. (D) The water solubility of alkalline earth metal carbonates increases on moving down the group. fuEu esa ls dkSulk@dkSuls dFku lgh gS : (A*) NaF ,oa NaBF4 esa ls] NaF ty esa vkisf{kd :i ls de ?kqyu'khy gksrk gSA (B*) tkyd ,UFkSYih dk cM+k _.kkRed eku vk;fud ca/k fuekZ.k ds fy, vuqdwy ifjfLFkfr gSA (C*) lkekU;r;k] vk;fud ;kSfxd Bksl voLFkk esa oS|qr /kkjk dk pkyu ugha djrs gSaA (D) {kkjh; e`nk /kkrq ds dkcksZusVksa dh tyh; foys;rk] oxZ esa uhps dh vksj tkus ij c<+rh gSA (A) Larger the size difference between ions, greater will be the solubility. (B) Larger negative value of lattice enthalpy is a favourable condition for formation of ionic bond as larger release of energy causes greater stability of lattice. (C) In general, ionic compounds conduct electricity in molten state and aqueous medium.

Sol.

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PAGE NO.- 8

g y-

(A) vk;uksa (B) tkyd

12.*

Which of the following do not exist ?

ds e/; vkdkj esa ftruk T;knk vUrj gksxk] foys;rk mruh gh T;knk gksxhA ,UFkSYih dk cM+k _.kkRed eku vk;fud ca/k fuekZ.k ds fy, vuqdqy ifjfLFkfr gksrh gS] D;ksafd cgqr vf/kd ÅtkZ eqDr gksus ds dkj.k tkyd (lattice) vis{kkd`r vf/kd LFkk;h gksrh gSA (C) lkekU;r;k] vk;fud ;kSfxd xfyr voLFkk ,oa tyh; ek/;e esa oS|qr /kkjk dk pkyu djrs gSaA fuEu esa ls fdldk vfLrRo ugha gS \

Sol.

Sol.

(A*) SH6 (B*) HFO4 (C*) FeI3 (D) HClO3 (A) With hydrogen sulphur does not undergo sp3d2 hybridisation because of larger difference in energies between s, p and d-orbitals. Sulphur show +6 oxidation state with highly electronegative elements like O and F. (B) As fluorine is smaller and more electronegative than oxygen. (C) I– being stronger reducing agent reduces Fe3+ to Fe2+.

lYQj] gkbMªkstu ds lkFk sp3d2 ladj.k ugha cukrk D;ksa fd buds e/; s, p rFkk d-d{kdksa esa ÅtkZ dk vUrj vf/kd gksrk gSA (A) lYQj vf/kd fo|qr _.kkRed rRo tSls O ,oa F ds lkFk +6 vkWDlhdj.k voLFkk fn[kkrk gSA (B) ¶yksjhu] vkWDlhtu dh rqyuk esa vf/kd fo|qr_.kkRed gksrk gS rFkk ¶yksjhu dk vkdkj NksVk gksrk gSA (C) I– izcy vipk;d gksrk gS tks fd Fe3+ dk vip;u Fe2+ esa dj nsrk gSA

COMPREHENSION # (Q.13 to Q.16) If an electrolytic solution consists of more than two ions and electrolysis is done, it is observed that all the ions are not discharge at the electrodes simultaneously but certain ions are liberate at the electrodes in perferential discharge theory. It states that if more than one type of ions are attracted towards a particular electrode, then the one discharged is the ion which required least energy. The potential at which the ion is discharge or deposited on the apropriate electrode is termed as discharge or depositon potential (E). If current of 1 ampere is passed through an aqueous (1 litre) electrolytic solution having [Zn2+] = [Cu2+] = [Mn2+] 0 0 0 = 0.03 M. The standard electrode potential, E Cu2  / Cu = 0.34 V, E Zn / Zn2  = 0.762 V and EMn / Mn2  = 1.8 V are

observed and 1 Faraday = 96500 coulomb.

,d oS?kqr vi?kV~; foy;u esa] nks ls vf/kd vk;u mifLFkr gSa rFkk bldk fo|qr vi?kVu djk;k tkrk gS] rc ;g ik;k x;k fd] lHkh vk;u ,d lkFk bysDVªkWM ij fujkosf'kr ugha gksrs gS] ijUrq dqN vk;u bysDVªksM ij] çkFkfedrk fujkos'ku fl)kar ds vuqlkj eqDr gksrs gaSA blds vuqlkj] ;fn nks ;k nks ls vf/kd vk;u ,d gh bysDVªksM dh vksj vkdf"kZr gksrs gS] rc og vk;u fujkosf'kr gksrk gS] ftlds fy, de ÅtkZ dh vko';drk gksrh gSaA foHko ftl ij] ,d vk;u mi;qDr ¼lgh½ bysDVªksM ij] fujkosf'kr ;k fu{ksfir gksrk gS] bl foHko dks fujkos'ku ;k fu{ksi.k foHko (E) dgrs gSA ;fn ,d oS?kqr vi?kV~; foy;u ds (1 yhVj ) foy;u ftlesa [Zn2+] = [Cu2+] = [Mn2+] = 0.03 M mifLFkr gSa esa ls 1 ,Eih;j dh /kkjk çokghr dh tkrh gsA ekud bysDVªkWM foHko Øe'k% = 96500 dwyke gSaA 13.

0 E0Cu 2  / Cu = 0.34 V, E0Zn / Zn2  = 0.762 V rFkk EMn = 1.8 V / Mn 2 

Which of the following metal is almost first deposited at cathode ? (A*) Cu (B) Zn (C) Mn

çsf{kr fd;s x;s rFkk 1 QSjkMs

(D) All of these.

dSFkksM ij fuEu esa ls dkSulh /kkrq loZçFke fu{ksfir gksrh gS \ Sol.

g y14.

Sol.

(A*) Cu (B) Zn The reduction potential of Cu2+ > Zn2+ > Mn2+. vip;u foHko dk Øe Cu2+ > Zn2+ > Mn2+ gSA

(D) mDr

lHkh /kkrq;saA

What would be the approximate concentration of metal ions Cu2+, Zn2+ and Mn2+ after 96.5 min respectively? 96.5 feuV ds i'pkr~ Cu2+, Zn2+ rFkk Mn2+ vk;uksa dh lkanzrk;sa Øe'k% D;k gksxh ¼yxHkx½ \ (A) 0.01 M , 0.03 M, 0.001 M (B) 0.01 M, 0.03 M, 0.03 M (C*) 0 M, 0.03 M, 0.03 M (D) 0.003 M, 0.03 M, 0 M. Since, no. of Faraday used = No of equivalent deposited = (n-factor) × mole deposited  mole deposited =

g y-

(C) Mn

no. of Faraday used n – factor

=

i t 1 1 96.5  60 × = = 30 × 10 –3 = 0.03. 96500 n 96500  2

pwafd] ç;qDr QSjkMs dh la[;k = fu{ksfir /kkrq ds rqY;kadks dh la[;k = (n-dkjd) × fu{ksfir eksy

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PAGE NO.- 9

 15.

Sol.

fu{ksfir eksy =

Sol.



1 10  60 1 × = 0.0031. 96500 2

fu{ksfir eksy =

1 10  60 1 × = 0.0031. 96500 2

 50    i. t 0.5 i. t  100  96500 = 96500

volume at NTP =



mole of O2=

D;k gS \

0.5  1 96 .5  60 1 0 .03 × = 96500 4 4

0 .03 × 22.4 = 0.168 L = 168 mL. 4

fu"dkflr O2 ds rqY;kdksa dh la[;k = ç;qDr QSjkMs dh la[;k

=  17.

(D*) Zn ds 0.003 eksy

If current efficiency is 50%, what volume of O 2 gas evolved at NTP after 96.5 minutes ? ;fn fo|qr /kkjk dh n{krk 50% gS] rc NTP ij] 96.5 feuV ds i'pkr~ fu"dkflr O2 xSl dk vk;ru (A) 332 mL (B*) 168 mL (C) 84 mL (D) 42 mL No.of equivalent of O2 evolved = no.of Faraday used

=

g y-

(D*) 0.003 mole of Zn

;fn Zn ds fu{ksfir gksus ds fy, le; t gSA  t = 100.5 – 96.5 = 4 feuVA 

16.

i t 1 1 96 .5  60 × = = 30 × 10–3 = 0.03. 96500 n 96500  2

After 106.5 minute the metal deposited is approximately ? (A) 0.012 mole of Cu (B) 0.03 mole of Zn (C) 0.03 mole of Mn 106.5 feuV ds i'pkr~] fu{ksfir /kkrq ¼yxHkx½ gS \ (A) Cu ds 0.012 eksy (B) Zn ds 0.03 eksy (C) Mn ds 0.03 eksy If time for the deposition of Zn be t  t = 106.5 – 96.5 = 10 minute.  mole deposited =

g y-

=

n–

 50    i. t 0.5 i. t  100  96500 = 96500

NTP ij

vk;ru =



O2 ds

eksy =

0.5  1 96 .5  60 1 0.03 × = 96500 4 4

0.03 × 22.4 = 0.168 L = 168 mL. 4

Consider an fcc lattice sectioned along the shaded plane shown in figure. One such plane is shown in figure. Give the number of atoms in plane DBF which are in direct contact with an atom placed at point C of metallic fcc lattice. vko`fÙk esa n'kkZ;s vuql kj Nk;kafdr ryksa dh vksj fcc tkyd Hkkx dk voyksdu dhft;sA vkÑfr esa bl izdkj dk ry n'kkZ;k x;k gSA ry DBF esa ijek.kqvksa dh la[;k nhft;s tks /kkfRod fcc tkyd ds fcUnq ij ,d ijek.kq ds lkFk uhps

lEidZ esa j[ks x;s gSaA

Ans. Sol.

3 Face centres of planes ABCD, BCFG, DCFE and C form a tetrahedral void. Hence C rests directly upon three atoms. ry ABCD, BCFG, DCFE o C ,d prq"Qydh; fjfDr;k¡ cukrs gSaA vr% C lh/ks rhuksa ijek.kqvksa ij j[kk tkrk gSA

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PAGE NO.- 10

Sol.

Calculate total number of possible isomers (structrural and stereoisomers) of the compound [Ir(en)2(NH3)(NO2)]2+ ;kSfxd [Ir(en)2(NH3)(NO2)]2+ ds fy, lHkh laHko leko;oh;ksa ¼lajpukRed rFkk f=kfoe leko;oh½ dh la[;k Kkr djksA 6

Optically active 3. Cis-NH3 / NO2 (d + ) 4. Cis-NH3/ONO (d + )

g y-

Optically inactive 1. trans-NH3 / NO2 2. trans-NH3/ONO 6

çdk'k vlfØ; 1. foi{k(trans)-NH3 / NO2 2. foi{k (trans)-NH3/ONO

çdk'k lafØ; 3. lei{k (Cis)-NH3 / NO2 (d + ) 4. lei{k (Cis)-NH3/ONO (d + )

18.

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.- 11

ChemINFO-1.18

ELECTROCHEMISTRY

Daily Self-Study Dosage for mastering Chemistry

Corrosion

Corrosion is one of the most significant problems faced by advanced industrial societies. It is estimated that about 30-40% of iron and steel produced annually is used just to replace the rusted iron materials. Corrosion toches all appliances used inside and outside home, on the road,on the sea, in plants and in aerospace vehicles. corrosion is a spontaneous slow chemical interaction of metal or alloy with its environment, resulting in the formation of one of its compounds such as oxide, hydrated oxide, carbonate, sulphide, sulphate, etc. Corrosion can be defined as the deterioration of materials by chemical interaction with their environmnet. Corrosion of metals occurs either by direct chemical attack or by electrochemical attack on the metal by the corrosive environment. The most familiar example of corrosion is rusting of iron exposed to the atmospheric conditions. 2Fe(s) +

3 O (g) + xH2O(I)  Fe2O3 · xH2O(s) 2 2 (Chemical composition of rust)

The rusting of iron is a complex chemical reaction that involves both oxygen and moisture (see Fig. 3.20). Iron does not rust in pure water that is oxygen free, and it does not rust in pure oxygen in the absence of moisture. This suggests that the corrosion process is electrochemical in nature. At one place on the surface, iron becomes oxidized in the presence of water and enters solution as Fe 2+ and this is the anodic location of the metal surface. Reaction taking place at the anode is

Fig. Corrosion of iron Fe(s)  Fe2+ (aq) + 2e – The electrons that are released when the iron is oxidized travel through the metal to some other place where the iron is exposed to oxygen. This is the cathodic region of the metal surface where reduction takes place, and oxygen is reduced to give hydroxide ion. The reaction taking place at the cathode is

1 O (aq) + H2O(I) + 2e–  2OH– (aq) 2 2 The iron (II) ions that are formed at the anodic regions gradually diffuse through the water and eventually contact the hydroxide ions. This causes a precipitate of Fe(OH) 2 to form, which is very easily oxidized by O 2 to give Fe(OH)3. This hydroxide readily loses water. In fact, complete dehydration gives the oxide, 2Fe(OH)3  Fe2O3 + 3H2O When partial dehydration of the Fe(OH)3 occurs, rust is formed. It has a composition that lies between that of the hydroxide and that of the oxide, Fe2O3, and is usually referred to as a hydrated oxide. Its formula is generally represented as Fe2O3 · xH2O. Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 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.- 12

ChemINFO-1.18

ELECTROCHEMISTRY la{kkj.k

Daily Self-Study Dosage for mastering Chemistry

vk/kqfud vkS|ksfxd leqnk;ksa esa ,d eq[; leL;k la{kkj.k (Corrosion) gSA ;g crkrk gS fd okf"kZd mRiUu yxHkx 30-40% vk;ju rFkk LVhy tax yxs vk;ju inkFkksZ dks gVkus ds fy, iz;qDr gksrs gSaA la{kkj.k ?kj ds vUnj rFkk ckgj lHkh midj.kksa eas] lM+dksa ij] leqnzksa ij] ikniksa esa rFkk ok;oh; ifg;ksa esa iz;qDr gksrk gSA la{kkj.k] /kkrq vFkok blds ok;qe.My ds lkFk feJ /kkrq dh ,d Lor% eUn jklk;fud vU;ksU; fØ;k gS] blds ifj.kkeLo:i vkWDlkbM] ty;ksftr vkWDlkbM] dkcksZusV] lYQkbM] lYQsV bR;kfn ds leku blds ,d ;kSfxd dk fuekZ.k gksrk gSA la{kkj.k dks buds ok;qe.My ds lkFk jklk;fud vU;ksU; fØ;k }kjk inkFkksZ ds voufr (deterioration) ds :i esa ifjHkkf"kr dj ldrs gSaA inkFkksaZ dk la{kkj.k ok;qe.My }kjk /kkrq ij fo|qr jklk;fud okrkoj.k }kjk vFkok izR;{k jklk;fud vkØe.k }kjk izkIr gksrk gSA la{kkj.k dk lokZf/kd ikfjokfjd mnkgj.k ok;qe.Myh; ifjfLFkfr;ksa eas [kqyk NksM+us ij vk;ju dk tax yxuk gSA 2Fe(s) +

3 O (g) + xH2O(I)  2 2

Fe2O3 · xH2O(s) (tax

dk jklk;fud laxBu) vk;ju dk tax yxuk ,d tfVy jklk;fud vfHkfØ;k gS ftlesa vkWDlhtu rFkk ueh (fp=k ns[ksa) nksuksa gksrs gSaA vk;ju ij 'kq) ty esa tax ugha yxrk gS D;ksafd ;g vkWDlhtu eqDr gksrk gS rFkk bl ij ueh dh vuqifLFkfr eas 'kq) vkWDlhtu esa tax ugha yxrk gSA ;g crkrk gS fd laa{kkj.k izØe ,d oS|qr jklk;fud izd`fr dk gksrk gSA /kkrq dh ,d lrg ij vk;ju ty dh mifLFkfr eas vkWDlhd`r gksrk gS rFkk Fe2+ ds :i esa foy;u esa izos'k djrk gS rFkk ;g /kkrq lrg dh ,uksfMd fLFkfr gSA ,uksM ij gksus okyh vfHkfØ;k gS ; Fe(s)  Fe2+ (aq) + 2e–

fp=k % ykSgs dk la{kkj.k bysDVªkWu tks fu"dkflr gksrs gS] tc vk;ju vkWDlhd`r gksrk gS] rc ;g bysDVªkWu dqN ,sls vU; LFkku ij /kkrq ds }kjk xfr'khy gksrs gSa] tgk¡ vk;ju dks vkWDlhtu esa [kqyk NksMrs gSaA ;g /kkrq lrg dk dSFkksfMd {ks=k gS] tgk¡ vip;u gksrk gS ,oa vkWDlhtu vipf;r gksdj gkbMªkWDlkbM vk;u nsrh gSA dSFkksM ij gksus okyh vfHkfØ;k gS % 1 O (aq) + H2O(I) + 2e–  2OH– (aq) 2 2

vk;ju (II) vk;u rc curs gSa tc ;g ,uksfMd {ks=k ij ty ds }kjk /khjs&/khjs folfjr gksrs gS rFkk vUr esa gkbMªkWD lkbM vk;uksa ds lEidZ eas vkrs gSaA blds dkj.k Fe(OH)2 dk vo{ksi curk gS] tks vf/kd vklkuh ls O2 }kjk vkWDlhd`r gksdj Fe(OH)3 nsrk gSA ;g gkbMªkWDlkbM rsth ls ty fu"dkflr djrk gSA okLro esa iw.kZ futZyhdj.k vkWDlkbM nsrk gSA 2Fe(OH)3  Fe2O3 + 3H2O

tc Fe(OH)3 dk vkaf'kd futZyhdj.k izkIr gksrk gS rks tax curk gSA ;g ,slk laxBu curk gS ftleas gkbMªkWDlkbM rFkk vkWDlkbM] Fe2O3 ds e/; ;g izkIr gksrk gS rFkk ;g izk;% ty;ksftr vkWDlkbM ds :i eas gksrk gS] blds lw=k dks lkekU;r% Fe2O3 · xH2O ds :i eas iznf'kZr djrs gSaA Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 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.- 13

19.

Rusting of iron is (A) A decomposition process (C*) An electrochemical

(B) A photochemical process (D) A reduction process

20.

In the rusting of iron, which of the following cell reactions occurs at the cathode ? (A) Fe2+ | Fe (B*) O2 | H 2O (C) Fe 3+ | F2+ (D) Fe | Fe3+

21.

Which of the following statements regarding rusting of iron is/are correct ? (A*) It takes palce is moist air (B*) It is stopped in CO 2 atmosphere (C*) It produces Fe (III) oxide (D) It is an electrochemical process.

22.

In which of the following will the corrosion of iron be most rapid ? (A) In pure water (B) In pure oxygen (C) In air and moisture

23.

19.

20.

(D*) In air and saline water

The rusting of iron takes place as follows : 2H  + 2e– + 1/2 O2  H2O(l) ;

E– = +1.23 V

Fe2+ + 2e–  Fe(s) ; Calculate G– for the net process : (A*) –322 kJ mol–1 (B) –161 kJ mol–1

E– = –0.44 V

ykSgs ij tax yxuk ,d gS%& (A) fo;kstu (decompostion) izØe (C*) fo|qr jklk;fud izØe

(C) –152 kJ mol–1

(B) izdk'kjklk;fud (D) vip;u

(D) –76 kJ mol–1

izØe

izØe

ykSgs ij yxus es fuEUk es ls dkSulh lsy vfHkfØ;k dSFkksM ij gksrh gS \ (A) Fe2+ | Fe

(B*) O2 | H 2O

(C) Fe 3+ | F2+

(D) Fe | Fe3+

21.

fuEu es ls dkSulk dFku ykSgs ij tax yxus ds lUnHkZ es lgh gS@gSa \ (A*) ;g ue ok;q es gksrk gSA (B*) ;g CO2 ds okrkoj.k es cUn tkrk gSA (C*) ;g Fe (III) vkWDlkbM cukrk gSA (D) ;g ,d fo?kqrjlk;fud izØe gSA

22.

fuEUk es ls fdlesa ykSgs dk l{kkj.k vf/kd rsth ls gksxk \ (A) 'kq) ty esa (B) 'kq) vkWDlhtu esa (C) ok;q o ueh esa

23.

(D*) ok;q

o yo.kh; ty esa

ykSgs ij tax yxuk fuEu izdkj ls gksrk gS%& 2H  + 2e– + 1/2 O2  H2O(l) ;

E– = +1.23 V

Fe2+ + 2e–  Fe(s) ; lEiw.kZ izØe ds fy, G– dh x.kuk dhft, (A*) –322 kJ mol–1 (B) –161 kJ mol–1

E– = –0.44 V (C) –152 kJ mol–1

(D) –76 kJ mol–1

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