94420 + Interview Questions in Chemistry in Class 12

1.	Standard Gibb's energy of reaction (Delta_(r)G^(@)) at a certain temperature can be computed as Delta_(r)G^(@)=Delta_(r)H^(@)-T.Delta_(r)S^(@) and the change in the value of Delta_(r)H^(@) and Delta_(r)S^(@) for a reaction with temperature can be computed as follows Delta_(r)H_(T_(2))^(@)-Delta_(r)H_(T_(1))^(@)=Delta_(r)C_(p)^(@)(T_(2)-T_(1))Delta_(r)S_(T_(2))^(@)-Delta_(r)S_(T_(1))^(@)=Delta_(r)C_(p)^(@)ln((T_(2))/(T_(1)))Delta_(r)G^(@)=Delta_(r)H^(@)-T.Delta_(r)S^(@) "andby "Delta_(r)G^(@)=-RT lnK_(eq)Consider the following reaction : CO(g)+2H_(2)(g)subCH_(3)OH(g)Given :Delta_(f)H^(@)(CH_(3)OH),g=-201 kJ//mol Delta_(f)H^(@)(CO, g)=-114 kJ//mol S^(@)(CH_(3)OH, g)=240 J//K-mol, S^(@)(H_(2),g)=29 JK^(-1)mol^(-1)S^(@)(CO, g)=198 J//mol-K, C_(p,m)^(@)(H_(2))=28.8J//mol-KC_(p,m)^(@)(CO)=29.4 J//mol-K, C_(p,m)^(@)(CH_(3)OH)=44 J//mol-Kand ln((320)/(300))0.06,all data at 300 KDelta_(r)G^(@) at 320 K is
Answer» `-48295.2 kJ//mol` `-240.85 kJ//mol` `240.85 kJ//mol` `-81.91 kJ//mol` Solution :`Delta_(R)G_(320)^(@)=Delta_(r)H_(300)^(@)-T.Delta_(r)S_(320)^(@)` `=-87.86-(320xx(-18.58))/(1000)` `=-81.91 KJ//mol`.

Discussion

2.	Standard hydrogen electrode has a following composition :
Answer» PT ELECTRODE `[H^(+)] =1 M, p_(H_(2)) =1` bar NI electrode, `[H^(+)] =0.1 M, p_(H_(2)) = 1` bar Pt electrode, `[H^(+)] =1 M, p_(H_(2)) =10` bar Ni electrode, `[H^(+)]=1 M, p_(H_(2)) = 10` bar Answer :A

Discussion

3.	Standard hydrogen electrode is
Answer» POTENTIAL of 1 volt reference electrode indicator electrode all of these Solution :ARBITRARILY ZERO potential.

Discussion

4.	Standard heat of formation of CH_4(g), CO_2(g) and water at 25^@C are -17.9, -94.1 K and -68.3 kcal mol^-1 respectively. Calculate the heat change (in kcal) in the following reaction at 25^@C: CH_4(g)+2O_2(g)=CO_2(g)+2H_2O(g)
Answer» `-144.5` `-180.3` `-248.6` `-212.8` ANSWER :D

Discussion

5.	Standard heat of formation of CH_(4).CO_(2) and H_(2)O (l) are - 76.2, -394.8 and - 241.6 kJ mol^(-1). Amount of heat evolved by burning 1 m^(3) of CH_(4) measured at normal conditions is
Answer» `3.579 xx 10^(6) KJ` `3.579 xx 10^(4) kJ` `6.240 xx 10^(4) kJ` `6.240 xx 10^(7) kJ` ANSWER :B

Discussion

6.	Standard Gibb's energy of reaction (Delta_(r)G^(@)) at a certain temperature can be computed as Delta_(r)G^(@)=Delta_(r)H^(@)-T.Delta_(r)S^(@) and the change in the value of Delta_(r)H^(@) and Delta_(r)S^(@) for a reaction with temperature can be computed as follows Delta_(r)H_(T_(2))^(@)-Delta_(r)H_(T_(1))^(@)=Delta_(r)C_(p)^(@)(T_(2)-T_(1))Delta_(r)S_(T_(2))^(@)-Delta_(r)S_(T_(1))^(@)=Delta_(r)C_(p)^(@)ln((T_(2))/(T_(1)))Delta_(r)G^(@)=Delta_(r)H^(@)-T.Delta_(r)S^(@) "andby "Delta_(r)G^(@)=-RT lnK_(eq)Consider the following reaction : CO(g)+2H_(2)(g)subCH_(3)OH(g)Given :Delta_(f)H^(@)(CH_(3)OH),g=-201 kJ//mol Delta_(f)H^(@)(CO, g)=-114 kJ//mol S^(@)(CH_(3)OH, g)=240 J//K-mol, S^(@)(H_(2),g)=29 JK^(-1)mol^(-1)S^(@)(CO, g)=198 J//mol-K, C_(p,m)^(@)(H_(2))=28.8J//mol-KC_(p,m)^(@)(CO)=29.4 J//mol-K, C_(p,m)^(@)(CH_(3)OH)=44 J//mol-Kand ln((320)/(300))0.06,all data at 300 KDelta_(r)H^(@) at 320 K is
Answer» `-288.86 kJ//mol` `-289.1 kJ//mol` `-87.86 kJ//mol` NONE of these Solution :`Delta_(r)H^(@)" at 320 K = "Delta_(r)H_(320)^(@)+Delta_(r)C_(P)^(@)(T_(2)-T_(1))` `=-87+((-43)xx20)/(1000)=-87.86KJ//mol`.

Discussion

7.	Standard Gibb's energy of reaction (Delta_(r)G^(@)) at a certain temperature can be computed as Delta_(r)G^(@)=Delta_(r)H^(@)-T.Delta_(r)S^(@) and the change in the value of Delta_(r)H^(@) and Delta_(r)S^(@) for a reaction with temperature can be computed as follows Delta_(r)H_(T_(2))^(@)-Delta_(r)H_(T_(1))^(@)=Delta_(r)C_(p)^(@)(T_(2)-T_(1))Delta_(r)S_(T_(2))^(@)-Delta_(r)S_(T_(1))^(@)=Delta_(r)C_(p)^(@)ln((T_(2))/(T_(1)))Delta_(r)G^(@)=Delta_(r)H^(@)-T.Delta_(r)S^(@) "andby "Delta_(r)G^(@)=-RT lnK_(eq)Consider the following reaction : CO(g)+2H_(2)(g)subCH_(3)OH(g)Given :Delta_(f)H^(@)(CH_(3)OH),g=-201 kJ//mol Delta_(f)H^(@)(CO, g)=-114 kJ//mol S^(@)(CH_(3)OH, g)=240 J//K-mol, S^(@)(H_(2),g)=29 JK^(-1)mol^(-1)S^(@)(CO, g)=198 J//mol-K, C_(p,m)^(@)(H_(2))=28.8J//mol-KC_(p,m)^(@)(CO)=29.4 J//mol-K, C_(p,m)^(@)(CH_(3)OH)=44 J//mol-Kand ln((320)/(300))0.06,all data at 300 KDelta_(r)S^(@) at 300 K for the reaction is
Answer» 152.6 J/K - MOL 181.6 J/K - mol `-16 J//K - mol` None of these Solution :`Delta_(r)S^(@)" at 300 K = "S_(CH_(3)OH)^(@)-S_(CO)^(@)-2S_(H_(2))^(@)` `=240-198-(2xx29)` =42-58 `=-16J//K-mol`.

Discussion

8.	Standard Gibb's energy of reaction (Delta_(r)G^(@)) at a certain temperature can be computed as Delta_(r)G^(@)=Delta_(r)H^(@)-T.Delta_(r)S^(@) and the change in the value of Delta_(r)H^(@) and Delta_(r)S^(@) for a reaction with temperature can be computed as follows Delta_(r)H_(T_(2))^(@)-Delta_(r)H_(T_(1))^(@)=Delta_(r)C_(p)^(@)(T_(2)-T_(1))Delta_(r)S_(T_(2))^(@)-Delta_(r)S_(T_(1))^(@)=Delta_(r)C_(p)^(@)ln((T_(2))/(T_(1)))Delta_(r)G^(@)=Delta_(r)H^(@)-T.Delta_(r)S^(@) "andby "Delta_(r)G^(@)=-RT lnK_(eq)Consider the following reaction : CO(g)+2H_(2)(g)subCH_(3)OH(g)Given :Delta_(f)H^(@)(CH_(3)OH),g=-201 kJ//mol Delta_(f)H^(@)(CO, g)=-114 kJ//mol S^(@)(CH_(3)OH, g)=240 J//K-mol, S^(@)(H_(2),g)=29 JK^(-1)mol^(-1)S^(@)(CO, g)=198 J//mol-K, C_(p,m)^(@)(H_(2))=28.8J//mol-KC_(p,m)^(@)(CO)=29.4 J//mol-K, C_(p,m)^(@)(CH_(3)OH)=44 J//mol-Kand ln((320)/(300))0.06,all data at 300 KDelta_(r)S^(@) at 320 K is
Answer» 155.18 J/mol - K 150.02 J/mol - K 172 J/mol - K None of these Solution :`Delta_(R)S^(@)" at 320 K"=Delta_(r)S_(320)^(@)` `because Delta_(r)S_(320)^(@)-Delta_(r)S_(300)^(@)=Delta_(r)C_(P)^(@)LN((T_(2))/(T_(1)))` where`Delta_(r)C_(P)^(@)=44-29.4-2xx28.8` `=-43 J//K-mol` `therefore Delta_(r)S_(320)^(@)=-43ln((320)/(300))+(-16)` `=-18.74 J//K-mol`

Discussion

9.	Standard free energies of formation in kJ/mol) at 298 K are -237.2, -394.4 and -8.2 for H_(2)O(l),CO_(2)(g) and pentane (g) respectively. The value of E_(cell)^(@) for the pentane-oxygen fuel cell is
Answer» 1.968 V 2.0968V 1.0968V 0.0968V Solution :The balanced equation for pentane-oxygen cell reaction will be `C_(5)H_(12)+8O_(2)to5CO_(2)+6H_(2)O,n=32` `Delta_(r)G^(@)=[5xxDelta_(F)G^(@)(CO_(2))+6Delta_(f)G^(@)(H_(2)O)]-[Delta_(f)G^(@)(C_(5)H_(12))+8Delta_(f)G^(@)(O_(2))]` `=[5(-394.5)+6(-237.2)]-[(-8.2)+0]` `=-1972-1423.2+8.2=-3387" KJ "mol^(-1)` `Delta_(f)G^(@)=-nFE_(ceLL)^(@)` `therefore-3387000=-32xx96500xxE_(cell)^(@)` or `E_(cell)^(@)=1.0968V`

Discussion

10.	Standard free energies of formation (in kJ/mol) at 298 K are 237.2, -394.4 and -8.2 for H_(2)O(l),CO_(2)(g) and pentane (g), respectively. The value of E_(cell)^(o) for the pentane-oxygen fuel cell is
Answer» 1.0968V 0.0968 1.968V 2.0968V Solution :`C_(5)H_(12)+8O_(2)to5CO_(2)+6H_(2)O` `DELTAG^(@)=-5xx394.4-6xx237.2+8.2` `DeltaG^(@)=-nFE_(Cell)^(@)` `-3387kJ//`MOLE

Discussion

11.	Standard free energies of formation (in kJ/mol) at 298 K are -237.2, -394.4 and -8.2for H_(2)O(l), CO_(2)(g) and pentance (g) respectively. The value E_(cell)^(@) for the pentance-oxygen fuel cell is:
Answer» `1.968V` `2.0968V` `1.0968V` `0.0968V` Solution :Writing the equation for pentane-oxygen fuel CELL at respective electrodes and overall reaction, we get At Anode: `underset(("pentace"))(CH_(5)H_(12))+10H_(2)O to 5CO_(2)+32H^(+)+32e ^(-)` Al Cathode: `(8O_(2)+ 32H^(+) +32e^(-) to 16H_(2)O)/("Overall:" C_(5) H_(12) +8O _(2) to 5CO_(2) +6H _(2) O)` Calculation of`DeltaG ^(@)` for the above reaction `DeltaG^(@)=[5XX(-394.4) +6xx(-237.2)] -[-8.2]` `=-1972.0 -1423.2+8.2 =-3387.0kJ` `=-3387000` Joules. From the equation we find `n=32` Using the relation, `Delta G^(@)=-nFE_(cell)^(@)` and substituting various values, we get `-3387000=-32xx96500xxE_(cell)^(@) (F=96500C)` or `E_(cell) ^(@) =(3387000)/(32xx96500)` `=(3387000)/(3088000)or (3387)/(3088)V-1.0968V` Thus OPTION (c) is correct answer.

Discussion

12.	Standard entropy of X_(2),Y_(2) and XY_(3) are 60, 40 and 50 JK^(-1)mol^(-1), respectively. For the reaction, (1)/(2)X_(2)+(3)/(2)Y_(2)rarrXY_(3),DeltaH=-30kJ to be at equilibrium, the temperature will be
Answer» 500 K 750 K 1000 K 1250 K Solution :`(1)/(2)X_(2)+(3)/(2)Y_(2)rarrXY_(3)` `DeltaS=50-((60)/(2)+(3)/(2)xx40)=50-(30+60)=-40J//k, mol` at equilibrium `DeltaG=0` `DeltaH=TDeltaS, T=(DeltaH)/(DeltaS)=(-30xx10^(3))/(-40)=750 K`

Discussion

13.	Standard free energies fo formation (in kJ/ mol) at 298 K are -237.2 - 394.4 and -8.2 for H_(2)O(l)CO_(2)(g) and pentane (g) respecitively the value of E_(cell)^(@) for the pentane oxygen fuel cell is
Answer» 1.98 V 2.0968 V 1.0968 V 0.0968 V Solution :Given `TRIANGLE G_(H_(2)O)^(@)(I)=(-237.2 KJ)//"MOL"` `triangle_(CO_(2))^(@)(g)=-394.4 KJ//"Mol"` `triangle G_(C_(5)H_(12)(g)^(@) =- 8.2 KJ//"mol"` `C_(5)H_(12)+8O_(2) rarr 5CO_(2)+6H_(2)O` `triangle G^(@)=5xxtriangle G_(CO_(2))^(@)+6xx triangle_(H_(2)^(@)-triangle_(C_(5)(H_(12)^(@))+8xx triangle G_(O_(2))^(@)` In pentane oxygen fuel cell 32 electrons are involved `triangle G^(@)=- nFE^(@)` `-3387 xx10^(3)=- 32 xx96500xxE^(@)` `therefore16^(O)+32 E^(-) rarr 16 O^(2)` `E^(@)=(3387xx10^(3))/(32xx96500)=1.0968 V`

Discussion

14.	Standard entropy of X_(2),Y_(2)andXY_(3) are 60, 40 and 50 JK^(-1)mol^(-1), respectively. For the reaction, 1/2X_(2)+3/2Y_(2)toXY_(3),DeltaH=-30kJ to be at equilibrium, the temperature will be
Answer» 500 K 750 K 1000 K 1250 K Answer :B

Discussion

15.	Standard entropy of X_(2), Y_(2) and XY_(3) are 60, 40 and 50" J K"^(-1) mol^(-1), respectively. For the reaction, 1/2 X_(2)+3/2 Y_(2) rarr XY_(3) Delta H=-30 kJ, to be at equilibrium, the temperature will be
Answer» 500 K 750 K 1000 K 1250 K Solution :`DeltaS_("REACTION")=DeltaS_("Product")-DeltaS_("Reactant")` `=50-[3//2xx40+1//2xx60]=50-[60+30]=-4 J//K` We know `DeltaG=DELTAH-T DeltaS` at equilibrium `DeltaG=0` `0=DeltaH-T DeltaS` `DeltaH=T DeltaS` `T=(DeltaH)/(DeltaS)=((-) 30xx1000)/((-)40)=750 K`

Discussion

16.	Standard entropy of A, B and C are 30, 60 and 100 JK^-1 mol^-1 respectively. For the reaction (2A + 5Brarr 6C), If the AH of the reaction is 300 kJ then the temperature at which the reaction will become spontaneous is
Answer» 1150 K 1240 K 1000 K 1260 K Answer :1

Discussion

17.	Standard entropies of X_(2), Y_(2) and XY_(3) are 60, 40 and 50 JK^(-1) respectively (1)/(2) X_(2) + (3)/(2) Y, Delta H = -30 kJ to be at equilibrium, the temperature should be :
Answer» 1250 K 500 K 750 K 1000 K Solution :`(1)/(2) X_(2) + (3)/(2) Y_(2) hArr XY_(3)` `Delta S^(@) = Sum S_(P)^(@) - Sum S_(R)^(@)` = 50 - (30 + 60) `Delta S^(@) = - 40 JK^(-1) mol^(-1)` `Delta G = Delta H - T Delta S`, `0 = Delta H - T Delta S` `Delta H = - 30 xx 10^(3) J mol^(-1)` `T = (Delta H^(@))/(Delta S^(@))` `= (-30 xx 10^(3) J mol^(-1))/(-40 JK^(-1) mol^(-1))` = 750 K

Discussion

18.	Standard enthorpyofX_(2), Y_(2) andXY_(3) are60,40 and50 JK^(-1) mol^(-1), respectively . For the reaction , (1)/(2)X_(2)+(3)/(2)Y_(2) to XY_(3), DeltaH= - 30 kJ to be at equilibrium, the temperature ofwill be
Answer» 1250 K 500 K 750K 1000K Solution :`DeltaS = - 40 J//K` `DeltaH =TDELTAS` ` T = 750 K`

Discussion

19.	Standard enthalpy of vapourisation Delta_("vap")H^(@) for water at 100^(@)C si 40.66 kJ mol^(-1). The internal energy of vaporization of water at 100^(@)C (in kJ mol^(-1)) is(Assume water vapour to behave like an ideal gas)
Answer» `+37.56` `-43.76` `+43.76` `+40.66` SOLUTION :`DeltaH=DeltaE+Deltan(g)RT` `40.66xx1000=DeltaE+(1)xx8.314xx373` `DeltaE=37.56 kJ mol^(-1)`

Discussion

20.	Standard enthaply of vaporisation Delta_(vap)H^(theta) forwater at 100^(@)C is 40.66 kJ mol^(-1). The internal energy of vapourisation of water at 100^(@)C (in kJ mol^(-1)) is :
Answer» `+ 40.66` `+ 37.56` `- 43.76` `+ 43.76` Solution :`Delta H = Delta U + Delta n_(g) RT` `H_(2)O(l) RARR H_(2)O(g)` `Delta n_(g) = 1, Delta H = 40.66 xx 10^(3) Jmol^(-1)`, T = 373K `40660 = Delta U + 1 xx 8.314 xx 373` `40660 = Delta U + 3101.12` `Delta U = 40660 - 3101.12` `= 37.558 Jmol^(-1)` `= 37. 56 kJmol^(-1)`

Discussion

21.	Standard enthalpy of formation is zero for
Answer» DIAMOND `O_3(G)` `Brbar`(AQ) `Cl_2(g)` ANSWER :B

Discussion

22.	Standard enthalpy of combustion of cyclopropane is -2091 kJ//"mole" at 25^(@)C then calculated the enthalpy formation of cyclopropane. If DeltaH_(f)^(o)(CO_(2))= -393.5 kJ//"mole" and Delta H_(f)^(o)(H_(2)O)= -285.8 kJ//"mole".
Answer» SOLUTION :N//A

Discussion

23.	Standard enthalpy and standard entropy of vaporization of water are +40 kJ mol^(-1) and+ 120J mol^(-1) K^(-1) respectively. Vapour pressure of water 27^(@)Cis P_(H_(2)O) then the value of 5 in P_(H_(2)O will be (consider standard temperature to be 300K)
Answer» ANSWER :8

Discussion

24.	Standard enthalpy change for combustion of methane is –890 kJ "mol"^(-1) and standard entropy change for the same combustive reaction is -242.98 J.K^(-1) at 25^@C. Calculate DeltaG^@ of the reaction.
Answer» SOLUTION : 817.6 KJ `"MOLE"^(-1)`

Discussion

25.	Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are -382.64 kJ mol^(-1) and -145.6 JK^(-1) mol^(-1), respectively. Standard Gibbs energy change for the same reaction at 298 K is
Answer» `-439.3 KJ MOL^(-1)` `-523.2 kJ mol^(-1)` `-221.1 kJ mol^(-1)` `-339.3 kJ mol^(-1)` SOLUTION :`DeltaG=DeltaH-TDeltaS` `DeltaG=-382.64-(-145.6)xx10^(-3)xx298` `=-339.3 kJ mol^(-1)`

Discussion

26.	Standard EMF of the cell: Cu\|\|Cu^(2+) (1m)\|\|Ag^(+) (1m)\|Ag is 0.46 at 25^@C. Find the value of standard free energy charge for the reaction that occurs in the cell.
Answer» Solution :`DELTA G^@=-nFE^@ CELL` `n=2,F=96,500 E^@=0.46V` `=-2 times 96500 times 0.46` `=-108080 Joul e s (or)-108.08` K JOU l es

Discussion

27.	Standard emf of 0.59 V of galvanic cell in which 3 mole electron taking part in redox reaction. So for such reaction find out value of equilibrium constant ?
Answer» `10^(25)` `10^(20)` `10^(15)` `10^(30)` Solution :`LOGK=(nE_(CELL)^(@))/(0.059)=(3xx0.59)/(0.59)=30` `therefore K=10^(30)`.

Discussion

28.	Standard electrode potentials are given as under : Ti^(4+)+e^(-)toTi^(3+)E^(@)=+0.01V Fe^(3+)+e^(-)toFe^(2+)E^(@)=+0.77V Tell whether Ti^(4+) ion may be used to oxidies F^(II) to Fe^(III) .
Answer» Solution :The GIVEN reaction is : `FE^(2+)+Ti^(4+)toFe^(3+)+Ti^(3+)` The electrochemical cell based UPON above redox reaction is : `underset("Anode")Fe^(2+)\|Fe^(3+)\|\|underset("Cathode")Ti^(4+)\|Ti^(3+)` `E_("cell")^(@)=E_("cathode")^(@)-E_("anode")^(@)` `=0.01-0.77` `=-0.76V` Since `E^(@)` cell is negative, `:.Ti^(4+)` cannot be used to oxidise `Fe^(2+)` to `Fe^(3+)`.

Discussion

29.	Standard electrode potentials of Fe^(2+) +2e rarrFe andFe^(3+)+3e rarrFe are -0.440V and-0.036 V respectively. The standard electrode potential (E^@) for Fe^(3+) + e rarrFe^(2+)is :
Answer» `-0.476V` `-0.404 V` `+0.404 V` `+0.772 V` ANSWER :D

Discussion

30.	Standard electrode potentials of Fe^(2+) + 2etoFe and Fe^(3+)+ 3eto Fe are -0.440V and -0.036V respectively. The standard electrode potential (E^@) forFe^(3+) +e to Fe^(2+) is
Answer» `-0.476V` `-0.404V` `+404V` `+0.722V` ANSWER :D

Discussion

31.	E^(@) values for Fe^(3+) + 3e rarr Fe and Fe^(2+) + 2 e rarr Fe are - 0.036V and -0.44V respectively. Calculate the E^(0) and DeltaG^(0) for the cell reaction Fe + 2 Fe^(3+) rarr 3Fe^(2+) .
Answer» `-0.476V` `-0.404 V` `+0.404 V` `+0.772 V` ANSWER :D

Discussion

32.	Standared electrode potential of two half-reactions are given below : Fe^(2+) iff Fe ""E^(@) = -0.44 V , Fe^(3+) iff Fe^(2+) ""E^(@) = + 0.77V If Fe^(2+) , Fe^(3+) and Feare kept together :
Answer» `FE^(3+)` INCREASES `Fe^(3+)` decreases `Fe^(2+)//Fe^(3+)` remains unchanged `Fe^(2+)` decreases

Discussion

33.	Standard electrode potentials are : Fe^(2+)\|Fe(E^(@)=-0.44V).Fe^(3+)\|Fe^(2+)(E^(@)=0.77V)Fe^(2+).Fe^(3+)andFe blocks are kept tigether , then
Answer» `FE^(3+)` increases `Fe^(3+)` DECREASES `Fe^(2+)//Fe^(3+)` REMAINS unchanged `Fe^(2+)` decreases Answer :A

Discussion

34.	Standard electrode potentials are Fe^(2+)//Fe,""E^(@)=-0.44V Fe^(3+)//Fe^(2),""E^(@)=+0.77V Fe^(2+),Fe^(3+) and Fe blocks are kept together, then
Answer» `Fe^(3+)` increases `Fe^(3+)` decreases `Fe^(2+)//Fe^(3+)` remains unchanged `Fe^(2+)` decreases Solution :For the emf to be +ve, the following half-cell REACTIONS will occur. `FetoFe^(2+)+2e^(-),E^(@)=+0.44V` `underline(2Fe^(3+)+2e^(-)2Fe^(2+),""E^(@)=+0.77V" ")` OVERALL reaction: `Fe+2Fe^(3+)to3Fe^(2+),""E_(cell)^(@)=0.121V` Thus, `Fe^(3+)` will DECREASE.

Discussion

35.	Standard electrode potential of Sn^(4+)//Sn^(2+) couple is +0.15 V and that of Cr^(3+)//Cr is 0.85 V. When connected, the cell potential will be
Answer» 1.10 V 1.00 V 0.70 V 0.30 V Solution :`E_("CELL")^(@)=E_("CATHODE")^(@)-E_("ANODE")^(@)` `""=+0.15-(-0.85)` `""=1.00 V`

Discussion

36.	Standard electrode potential of three metals x,y and z are -1.2 V,+0.5 V and -3.0 V. The order of reducing agent of three metals are. . .
Answer» `y GT Z gt X` `YGT x gt z` `zgtxgty` `xgtygtz` SOLUTION :`zgtxgty`

Discussion

37.	Standard electrode potential of three metals X,Y and Z are -1.2V, +0.5 V and -3.0V respectively. The reducing power of these metals will be
Answer» `X gt Y gt Z` `Y gt Z gt X` `YgtXgtZ` `ZgtXgtY` Solution :Higher the reduction potential, more easily it is reduced and HENCE higher is its OXIDIZING POWER or less is its reducing power. As `E_(red)^(@)` are in the ORDER YgtXgtZ, their reducing power will be in the order ZgtXgtY.

Discussion

38.	Standard electrode potential of normal hydrogen electrode is :
Answer» 0.5 V `+1.5V` 0.0 V `+0.5V` ANSWER :C

Discussion

39.	Standardelectrode potential of ell H_(2)\|H^(+)\|\|Ag^(+)\|Ag is(Ag^(+)//Ag)^(@)=0.80 V
Answer» 0.8 V `-0.8 V` `-1.2 V` `1.2 v` Solution :`E_(cell) =E_(cathode)-E_(anode)` Given `E_(AG^(+)//Ag)^(@)=0.80 V H_(2) \|H^(+)\|\|Ag^(+)\|Ag` `THEREFORE`HYDROGEN is anode and silver is cathode `E_(cell)=E_(c )-E_(A)` =0.80 -0 =0.80 V

Discussion

40.	Standard electrode potential of An and Fe are known to be (i) -0.76V and (ii) -0.44V respectively. How does it explain that galvanization prevents rusting of iron while zinc slowly dissolves away
Answer» Since (i) is LESS than (II), zinc becomes the CATHODE and IRON the anode Since (i) is less than (ii), zinc becomes the anode and iron the cathode Since (i) is more than (ii), zinc becomes the anode and iron the cathode Since (i) is more than (ii), zinc becomes the cathode and iron the anode Answer :B

Discussion

41.	Standard electrode potential is highest for
Answer» `1/2 F_(2)(g)+E^(-) Leftrightarrow F^(-)(aq)`<BR>`1/2 Cl_(2)(g)+e^(-) Leftrightarrow Cl^(-)(aq)` `1/2 Br_(2)(g)+e^(-) Leftrightarrow Br^(-)(aq)` `1/2 I_(2)(g)+e^(-) Leftrightarrow I^(-)(aq)` Answer :A

Discussion

42.	Standard electrode potential of Ag^(+)//Ag and Cu^(+)//Cu is +0.80V and +0.34V respectively. These electrodes are joint together by salt bridge if
Answer» copper electrode WORK like anode then `E_(cell)^(o)` is +0.45V Silver electrode work like anode then `E_(cell)^(o)` is -0.34V Copper electrode work like anode then `E_(cell)^(o)` is +0.46V Silver electrode work like cathode then `E_(cell)^(o)` is-0.34V Solution :Cell reaction is `Cu_((s))+2AG^(+)toCu^(2+)+2Ag` Two half cell reaction `CutoCu^(2+)+2e^(-)"Oxidation (anode)"` `AG^(+)+e^(-)TOAG"Reduction (cathode)"` `E_(cell)=E_(o x)-E_(red)=0.80-0.34=+0.46V`

Discussion

43.	Standard electrode potential for Sn^(4+)//Sn^(2+) couple is +0.15V and that for the Cr^(3+)//Cr couple is -0.74V. These two couples in their standard state are connected to make a cell. The cell potential will be
Answer» `+1.83V` `+1.19V` `+0.89V` `+0.18V` SOLUTION :`E_(CELL)^(@)=E_("cathode (RP)")^(@)-E_("anode (RP)")^(@)` `=0.15-(-0.74)=+0.89V`

Discussion

44.	Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reaction and their standard potentials are given below MnO_(4)^(-)(aq)+8H^(+)(aq)+5e^(-)toMn^(2+)(aq)+4H_(2)O(l)""E^(@)=1.51V Cr_(2)O_(7)^(2-)(aq)+14H^(+)(aq)+6e^(-)to2Cr^(3+)(aq)+7H_(2)O(l)""E^(@)=1.38V Fe^(3+)(aq)+e^(-)toFe^(2+)(aq)""E^(@)=0.77V Cl_(2)(g)+2e^(-)to2Cl^(-)(aq)""E^(@)=1.40V Identify the only incorrect statement regarding the quantitative estimation of aqueous Fe(NO_(3))_(2)
Answer» `MnO_(4)^(-)` can be used in aqueous HCl `Cr_(2)O_(7)^(2-)` can be used in aqueous HCl `MnO_(4)^(-)` can be used in aqueous `H_(2)SO_(4)` `Cr_(2)O_(7)^(2-)` can be used in aqueous `H_(2)SO_(4)` Solution :`MnO_(4)^(-)` will oxidize `Cl^-` ION ACCORDING to the equation, `2MnO_(4)^(-)+16H^(+)+10Cl^(-)to2Mn^(2+)+8H_(2)O+5Cl_(2)uarr` The cell corresponding to this reaction is as follows: `Pt,Cl_(2)("1 ATM")\|Cl^(-)\|\|MnO_(4)^(-), Mn^(2+),H^(+)\|Pt,` `E_(cell)^(o)=1.51-1.40=0.11V` `E_(cell)^(o)` BRING +ve, `DeltaG^(o)` will be -ve and hence the above reaction is feasible. `MnO_(4)^(-)` will not only oxidize `Fe^(2+)` ion but also `Cl^(-)` ion simultaneously.

Discussion

45.	Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reactions and their standard potentials are given below: MnO_((aq))^- +8_((aq))^+ +5e to Mn_((aq))^(2+)+4H_2O_((l)), E^0=1.51V Cr_2O_(7(aq))^(2-)+14H_((aq))^+ +6e to MnO_(4(aq))^-+8H_((aq))^+ +6e to 2Cr_((aq))^(3+)+7H_2O_((l)),E^0=1.38V Fe_((aq))^(3+)+e^- toFe_((aq))^(2+),E^0=0.71V Cl_2(g)+2e^- to 2C_((aq))^-, E^0=1.40V Identify the only incorrect statement regarding the quantitive estimation of aqueous Fe(NO+3)_2
Answer» `MnO_4^-` can be USED in aqueous HCL `Cr_2O_7^(2-)` can be used in aqueous HCl `MnO_4^-` can be used in aqueous `H_2SO_4` `Cr_2O_7^(2-)` can be used in aqueous `H_2SO_4` Solution :(D) it is the case of attacked electrode that is ANODE: Cu `rarr Cu^(2+)`+2e, CATHODE : `Cu^(2+)` +2e `rarr` Cu Thus no change in CONC. Of `Cu^(2+)` ions.

Discussion

46.	Standard electrode potential (E^(0)) for OCl^(-) // Cl^(-) and Cl^(-)// (1)/(2) Cl_2are respectively 0.94 V and -1.36 V . The E^(@) value of OCl^(-) //Cl_2 will be
Answer» `-0.42 V ` `-2.20 V ` `0.52 V ` `1.04 V ` Solution :(1) `2H^(OPLUS) + 2e^(-) + OCl^(-) to Cl^(-) + H_2O , (2 X 0.94) ` (2)` Cl^(Ɵ) to (1)/(2) Cl_2 + 1e^(-) , (-1.36 xx 1)` (3) `2H^(oplus) + 1 e^(-) + OCl^(Ɵ) to (1)/(2) Cl_2 + H_2O , (1 xx E_3^(0)) (3) = (1) + (2)` ` 1 xx E_3^(0) = (2 xx 0.94 ) + (-1.36 xx 1) , E_(3)^(0)=((1.88- 1.36)/(1)) = 0.52 V `

Discussion

47.	Standard electrode potential for Sn^(4+)//Sn^(2+) couple is +0.15V and that for the Cr^(2+)//Cr couple is -0.74V. These two couples in their standard state are connected to make a cell. The cell potential will be
Answer» `+1.83V` `+1.19V` `+0.89V` `+0.18V` SOLUTION :`E_(CELL)^(@)=E_("CATHODE")^(@)-E_("ANODE")^(@)` `=+0.15-(-0.74)+0.89V`

Discussion

48.	Standard electrode potential dataare useful for understandingthe suitability of an oxidant in a redoxtitrationsomehalfcell reactionsand theirstandardpotentials are givenbelow MnO_(4)^(-)(aq)+8H^(+)(aq)+5e^(-)rarrMn^(2+)(Aq)+4H_(2)O(l)^(@)=1.51 V Cr_(2)O_(7)^(2-)(Aq)+14 H^(+)(aq)+6e^(-)rarr2cr^(3+)(aq)+7H_(2)O(l),E^(2)=1.38 V CI_(2)(g)+2e^(-)rarr2CI^(-)(aq),E^(@)=1.40 V Identify the incorrect statement regardingthe qunatitavite estimation of gaseous Fe(NO_(3))_(2)
Answer» `MnO_(4)^(-)` can beusedin aqueous HCI `Cr_(2)O_(7)^(2-)` can be used in aqueous HCI `MnO_(4)^(-)` can be usedin aqueous `H_(2)SO_(4)` `Cr_(2)O_(7)^(2-)`can be used in aqueous `H_(2)SO_(4)` Solution :The REACTION between `MnO_(4)^(-)` and HCI may be represented as follows `2MnO_(4)^(-) (AQ)+16H^(+) +10CI^(-) rarr 2Mn^(2+) (aq)+8H_(2)O(I)+5CI_(2)(g)` Thuson the BASISOF this reactionfollowingelectrochemcial cell will be represented as since `E_(cell)^(2)=e_(cathode)^(@) -E_(anode)^(2)` `E_(cell)^(@)` is positive hence `triangle G^(@)` is negativethus above celll reaction is feasible but `MnO_(4)^(-1)`ion can oxidse `FE^(2+)` to `Fe^(3+)`and `CI^(-)` to `CI_(2)` in aqueous medium also Therefore for quantivative estimiationof aqueous`Fe(NO_(3))_(2)`it is not a suitable reagent

Discussion

49.	Standard electrode potential data are useful for understanding the stability of an oxidant in a redox titration . Somehalf reactions and their standard potentials are given below : MnO_(4)^(-)(aq)+8H^(+)(g)+5e^(-)toMn^(2+)(aq)+4H_(2)O(l)E^(@)=151VCr_(2)O_(7)^(2-)(aq)+14H^(+)aq)+6e^(-)to2Cr^(3+)(aq)+7H_(2)+(l)E^(@)=138VFe^(3+)(aq)e^(-)toFe^(2+)(aq) "" E^(@)=077VCl_(2)(g)+2e^(-)to2Cl^(-)(aq) "" E^(@)=140VIdentify the only incorrect statement regarding the quantitative estimation of aqueous Fe(NO_(3))_(2).
Answer» `MnO_(4)^(-)`can be USED in aqueous HCl `Cr_(2)O_(7)^(2-)` can be used in aqueous HCl `MnO_(4)^(-)` can be used in aqueous `H_(2)SO_(4)` `Cr_(2)O_(7)^(2-)` can be used in aqueous `H_(2)SO_(4)` Answer :A

Discussion

50.	Standard electrode potenital data are useful for understanding the suitablity of an oxidant in a redox titration. Some half cell reactions and their standard potential are given below: MnO_(4(aq))^(-) + 8H_((aq.))^(+) + 5e rarr Mn_((aq.))^(2+) + 4H_(2)O_(l),E^(@) = 1.51 V Cr_(2)O_(7(aq.))^(2-) + 14H_((aq.))^(+) + 6e rarr 2Cr_((aq.))^(3+) + 7H_(2)O_(l), E^(@) = 1.38 V Fe_(aq.)^(3+) 2e^(-) rarr Fe_((aq.))^(2), E^(@) = 0.77 V Cl_(2(g)) + 2e^(-) rarr 2Cl_((aq.))^(-), E^(@) = 1.40 V Identify the only incorrect statement regarding the quantitative estimation of aqueousFe(NO_(3))_(2):
Answer» `MnO_(4)^(-)` can be used in aqueous `HCl` `Cr_(2)O_(7)^(2-)` can be used in aqueous `HCl` `MnO_(4)^(-)` can be used in aqueous `H_(2)SO_(4)` `Cr_(2)O_(7)^(2-)` can be used in aqueous `H_(2)SO_(4)` Solution :`MnO_(4)^(-)` will oxidise `Cl^(-)` ION according to equation. `Mn^(7+) + 5e rarr Mn^(2+)` `2CL^(-) rarr Cl_(2) + 2e` Thus `E_(CELL)^(@) = E_(OP_(Cl^(-)//Cl_(2)))^(@) + E_(RP_Mn^(7+)//Mn^(2))^(@)` `= -1.40 + 1.51 = 0.11 V` or reaction is feasible. `MnO_(4)^(-)` will oxidise `Fe^(2+)` to `Fe^(3+)` `Mn^(7+) + 5e rarr Mn^(2+)` `Fe^(2+) rarr Fe^(3+) + E` `E_(cell)^(@) = E_(OPFe^(2+)//Fe^(3+))^(@) + E_(RPMn^(7+)//Mn^(2+))^(@)` `= -0.77 + 1.51 = 0.74 V` or reaction is feasible. Thus `MnO_(4)^(-)` will not oxidise only `Fe^(2+)` to `Fe^(3+)` in aqueous `HCl` but it will also oxidise `Cl^(-)` to `Cl_(2)`. Suitable oxidant shoould not oxidise `Cl^(-)` to `Cl_(2)` and should oxidise only `Fe^(2+)` to `Fe^(3+)` in redox titration.

Discussion

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