94420 + Interview Questions in Chemistry in Class 12

1.	Calculate the reduction potential of the following electrode at 298 K : Pt Cl_(2) (2.5 atm) HCl (0.01 M), E^(Theta) Cl_(2) \| 2 Cl^(-) = 1.36 V .
Answer» Solution :The electrode REACTION is : `Cl_(2) (g) + 2E^(-) to 2 Cl^(-) (aq)` `E (Cl_(2)\|2Cl^(-)) = E^(THETA) (Cl\|2Cl^(-)) - ((0.059))/(2) "log" ([Cl^(-)]^(2))/(p (Cl_(2)))` `p(Cl_(2)) = 2.5` atm , `[Cl^(-)] = 0.1 M` (same as the CONCENTRATION of HCl) `E^(Theta) (Cl_(2) \| 2Cl^(-)) = 1.36` V `THEREFORE E(Cl_(2) \|Cl^(-)) = 1.36 - ((0.059))/(2) "log" ((0.1)^(2))/((2.5))` `= 1.36 + 0.0295 xx 4.398 ` `= 1.36 + 0.13 = 1.49` V

Discussion

2.	Calculate the reduction potential of a half-cell containing of platinum electrode immersed in 2.)Mfe^(2+) and 0.2MFe^(3+), Given:- E_(Fe^(3+)//Fe^(2+))^(o)=0.771V Fe^(3+) +e^(-)toFe^(2+)
Answer» 0.653V 0.889V 0.683V 2.771V Solution :`E_(CELL)=E_(cell)^(0)-(RT)/(NF)xx2.303"log"([Fe^(2+)])/([Fe^(3+)])` `E_(cell)=0.771-(8.314xx2.303xx298)/(96500)"log"(2)/(0.02)` `=0.771-(0.0591xx2)=0.771-0.1182` `E_(cell)=0.653V`.

Discussion

3.	Calculate the reduction potential of a half-cell consisting of platinum electrode immersed in 2.0 M Fe^(2+) and 0.02 M Fe^(3+). Given E_(Fe^(3+)//Fe^(2+))^(@)=0.771V
Answer» 0.653 V 0.889 V 0.683 V 2.771 V Solution :`FE^(3+)+E^(-)toFe^(2+)` `E_(RED)=E_(Fe^(2+)//Fe^(2+))^(@)-(0.0591)/(2)"log"([Fe^(2+)])/([Fe^(3+)])` `=0.771-0.0591"log"(2.0)/(0.02)` `=0.771-0.591xxlog100` `=0.771-0.0591xx2=0.653V`.

Discussion

4.	Calculate the reduction potential for the following half cells at 25^@C Pt\|Fe^(2+) (0.1M)-Fe^(3+) (0.01M), E_(Fe^(3+),Fe^(2+))^@=+0.77V
Answer» Solution :`FE^(3+)+e LEFTRIGHTARROW Fe^(2+)` (reduction) `E_(Fe^(3+),Fe^(2+))=E_(Fe^(3+),Fe^(2+))-0.0591/n LOG""([Fe^(2+)])/([Fe^(3+)])` `=0.77- 0.0591/1 log"" 0.1/0.01=0.7109V`

Discussion

5.	Calculate the reduction potential for the following half cells at 25^@C Mg\|Mg^(2+) (1 times 10^-4 M), E_(Mg,Mg^(2+))^@=+2.36V
Answer» SOLUTION :`Mg^(2+)+2e LEFTRIGHTARROW Mg`(REDUCTION) `E_(Mg^(2+),Mg=E_(Mg^(2+),Mg))^@+0.0591/2log(1 times 10^-4)`.......[see example] `=-2.36 - (4 times 0.0591)/2=-2.4782 V`

Discussion

6.	Calculate the reduction potential for the following half cells at 25^@C Cl_2\|Cl^(-)(2 times 10^-5 M), E_(Cl_2,Cl^-)^@=+1.36V
Answer» SOLUTION :`1/2 Cl_2+e leftrightarrow Cl^-` (reduction) `E_(Cl_2,Cl^-)=E_(Cl_2,Cl^-)^@-0.0591/n LOG [Cl^-]` [ See example 1] `=1.36 -0.0591/1 log (2 times 10^-5)=1.6377 V`

Discussion

7.	Calculate the ratio of Z in bcc of fcc?
Answer» SOLUTION :`Z(BC C) =2` `Z(FC C) =4` `THEREFORE 2/4=0.5`

Discussion

8.	Calculate the ratio of the rate of diffusion of gas A to B if their pressure are in the ratio 2:1 and their molecules if travel with same rms speed then ratio of their kinetic energies is 1:9
Answer» ANSWER :6

Discussion

9.	An alpha-particle and a proton are accelerated from rest through the same potential difference V. Find the ratio of de-Broglie wavelength associated with them.
Answer» `(1)/(2)` `(1)/(sqrt2)` `(1)/(2SQRT2)` `2sqrt2` Answer :C

Discussion

10.	Calculate the ratio of the oxidized to the reduced form at half-cell potential of 0.1 volt for the half cell Fe^(3+) , Fe^(2+) \| Pt , E_(Fe^(3+), Fe^(2+))^@ = 0.7591
Answer» SOLUTION :`1.41 XX 10^11`

Discussion

11.	Calculate the ratio of separation between successive (1 0 0), (11 0) and (111) lattice planes in a cubic cell.
Answer» Solution :We have, `d=a/sqrt(h^(2) + K^(2) + l^(2))` THUS, `d_(100) = a/sqrt(1^(2) + 0^(2) + 0^(2)) =a` `d_(110) = a/sqrt(1^(2) + 1^(2) + 0^(2)) = a/sqrt(2)` and `d_(111) = a/sqrt(1^(2) + 1^(2) + 1^(2)) = a/sqrt(3)` Hence, `d_(100): d_(110): d_(111) =a:a/sqrt(2) : a/sqrt(3) =1: 1/sqrt(2) : 1/sqrt(3)`

Discussion

12.	Calculatethe ratio of pHof a solutioncontaining 1 moleof CH_(3)COONa+ 1mole of HCl perlitreandof othersolution containing 1 moleof HClper litreand of other solutioncontaining 1 moleCH_(3)XCOoNa+ 1moleof aceticacid per litre .
Answer» Solution :CaseI : pH when1 mole`CH_(3)COONA`and 1 moleHCl are present . `{:(,CH_(3)COONa,+,HCL,to,CH_(3)COOH,+,NaCl),(" Before reaction",1,,1,,0,,0),("After reaction",0,,0,,1,,1):}` ` :. [ CH_(3)COOH] = 1 M ` ` :. [H^(+)]= C .alpha= C sqrt(((K_(s))/(C))) = sqrt(K_(a).C) = sqrt((K_(s))) "" :.C = 1 ` ` :. pH_(1) = -1/2 logK_(a)` CaseII . pH when1 mole`CH_(3)COONa`and 1 molemole of `CH_(3)COOH` , a buffersolution ` :.pH_(2) = - log K_(a)+log. (["salt"])/(["acid"])"":.["Salt"]=1M` ` pH_(2) = -logK_(a) "":. ["Acid"] = 1 M` ` :. (pH_(1))/(pH_(2))=1/2`

Discussion

13.	Calculate the ratio of concentration of HCOO^(-) & OCN^(-) ions in a solution containing 0.1MHCOOH (K_(a)=1.8xx10^(-4)) and 0.1MHOCN(K_(a_(1))=4xx10^(-4)) .if simplest ratio is a:b reprot your answer as (a+b).
Answer» Solution :`[H^(+)]` is the same for equilibria of both acids. Now ,let `[HCOO^(-)]=aM` and `[OCN^(-)]=b M` `:.([H^(+)]XXA)/(0.1)=1.8xx10^(-4)&([H^(+)]xxb)/(0.1)=4XX10^(-4)rArr (a)/(b)=(9)/(20)` So `(a+b)=29`

Discussion

14.	Calculate the ratio of ([Ag^(+)])/([Ag(NH_(3))_(2)]^(+)) in 0.2M solution of NH_(3). If the stability constant for the complex [Ag(NH_(3))_(2)]^(+) is 1.7xx10^(7)
Answer» Solution :The STABILITY constant for the comples `[Ag(NH_(3))_(2)]^(+)` is, `beta=1.7xx10^(7)`, overall dissociation constant is the reciprocal of overall stability constant `K=(1)/(beta)impliesK=(1)/(1.7xx10^(7))impliesK=0.588xx10^(-7)impliesK=5.88xx10^(-8)`

Discussion

15.	Calculate the ratio of 4th Bohr's to 2nd Bohr's orbit of He^(+) ion.
Answer» ANSWER :4

Discussion

16.	The type of isomerismexhibited by[ Pt(NH_(3))Cl_(2)] ?
Answer» SOLUTION :The STABILITY constant for the complex `[AG(NH_(3))_(2)]^(+)` is, `beta=1.7xx10^(7)`, overall dissociation constant is the reciprocal of overall stability constant. `K=(1)/(beta)rArr K=(1)/(1.7xx10^(7))rArrK=0.588xx10^(-7)rArr K=5.88xx10^(-8)`

Discussion

17.	Calculate the rate of reaction from the following rate law: -(d[A])/(dt) = k[A]^(1)[B]^(2) when the concentrations of A and B are 0.01 M and 0.02 M respectively and k=5.1xx10^(-3)L^(2)"mol"^(-2)s^(-1).
Answer» Solution :`-(d[A])/(dt)=k[A]^(1)[B]^(2)` SUBSTITUTING the values of k, [A] and [B], we have RATE of reaction `=5.1xx10^(-3) xx (0.01)(0.02)^(2)=5.1xx10^(-3)xx10^(-2)xx4xx10^(-4)` `=20.4 xx 10^(-9) =2.04xx10^(-8)"MOL L"^(-1)s^(-1)`

Discussion

18.	Calculate the rate of flow (volume per second) of a liquid through a capillary tube of diameter 0.20 xx 10^(-3) mand length 1m, eta = 3.00 xx 10^(-5) Nm^(-2)sand pressure gradient = 10 atm.
Answer» SOLUTION :`R=(0.2 xx 10^(-3))/2 = 0.1 xx 10^(-3) m` `=10^(-4) m, p=10 ATM = 10 xx 1.01 xx 10^(5) Pa` `=1.01 xx 10^(6) Pa` We have, `eta = (pi PR^(4)t)/(8VI)` `THEREFORE` rate of flow `=V/t = (pipr^(4))/(8eta l)` `=((22//7) xx (1.01 xx 10^(6))(10^(-4))^(4))/(8 xx (3.0 xx 10^(-3))xx 1)` `=1.32 xx 10^(-8) m^(5) s^(-1)`

Discussion

19.	Calculate the rate constant of a reaction (first order) which is 90% complete in 10 min.
Answer» SOLUTION :Rate CONSTANT. `lambda=2.303/t(LOG)(100)/(100-90)=[2.303/10xx(log)100/10]min^-1=[2.303/10xx(log)10]min^-1` .`=0.2303min^-1`

Discussion

20.	Calculate the radius ratio (r_(+)//r_(-)) and coordination number of Li^(+) and F^(-) ion in LiF crystal structure from the following data : r_(Li^(+)) = 60 pm , r_(F^(-)) = 136 pm
Answer» Solution :`(r_(Li^(+)))/(r_(F^(-)))=(60)/(136)= 0.441` `:.` LiF will have octahedral structure and COORDINATION NUMBER of each ION will be 6.

Discussion

21.	Calculate the radius of the first Bohr orbit of a hydrogen atom. Given that h= 6.63 xx 10^(-27) erg.s, m= 9 xx 10^(-28)g and charge of an electron =4.8 xx 10^(-10) esu
Answer» SOLUTION :We have for hydrogen, Z=1, n=1 `r_(1)= (H^(2))/(4pi^(2) m e^(2))` `= ((6.63 xx 10^(-27))^(2))/(4 xx (3.14)^(2) xx 9 xx 10^(-28) xx (4.8 xx 10^(10))^(2))` `=0.53xx 10^(-8) CM= 0.53 Å`

Discussion

22.	Calculate the radius (in pm) of the largest sphere which fits properly at the centre of the edge of a body centred cubic unit cell. (Given edge length is 100 pm)
Answer» Solution :for bcc `rArr R = sqrt(3)/4 a` `therefore` empty space at edge `a-2R =a -sqrt(3a)/2` `therefore r_("SPHERE") =(a-sqrt(3)/2a)/2 = (2-sqrt(3))/4 XX a` = diameter of sphere =0.067 a, (GIVEN a=100 PM) `=0.067 xx 100 = 6.7` pm

Discussion

23.	Calculate the quantity of ferrous and ferric ions that would be deposited by 1 faraday."(F = 56)"
Answer» SOLUTION :1 mole of ELECTRICITY (i.e., 1 faraday) produces 1 EQ. of matter. Mole of electric charge = 1 faraday (given). `THEREFORE`. No. of equivalent of `Fe^(2+) = 1`. and no. of equivalent of `Fe^(3+) = 1`. `therefore` wt. of `Fe^(2+) = 1 XX (56)/(2)g = 28 g ""` (wt. = eq. `xx` eq. wt.) and wt. of `Fe^(3+) = 1 xx (56)/(3) g = 18.6g`

Discussion

24.	Calculate the quantity of electricity that would be required to reduce 12.3 g of nitrobenzene to aniline if the current efficiency for process is 50 per cent. If the potential drop across the cell is 3 volts, how much energy will be consumed?
Answer» Solution :The reduction reaction may be written as: `C_(6)H_(5)NO_(2)+6H^(+)+6e^(-)toC_(6)H_(5)NH_(2)+3H_(2)O` Thus, 1 mol of `C_(6)H_(5)NO_(2)`, i.e., 123 g require electricity`=6xx96500C=579000C` `therefore12.3`g require electricity=57900 C As the CURRENT EFFECIENCY is 50%, quantity of electricity actually required`=2xx57900C=115800C` Energy consumed=`EMFxx`Qty. of electricity`=3xx115800J=347400J`.

Discussion

25.	Calculate the quantity of electricity needed to reduce one centimole of dichromate in acid medium, to chromic state.
Answer» Solution :The stoichiometric equation is , `Cr_(2)O_(7)^(2-)+14H^(+)+6E^(-) RARR 2Cr^(3+)+7H_(2)O` ONE mole of dichromate`=6` moles of electrons `=` 579000 coul. Quantity of electricity required to reduce one centimole of `Cr_(2)O_(7)^(2-)` to `CR^(3+)=5790` coulomb

Discussion

26.	Calculate the quantity of electricity dellvered by a Daniel celi initially containing 1 L each of 1M Cu^(2+) ion and 1M Zn^(2+) C which is operated until potential drops to 1V. Given E^(@) Zn^(2+)//Zn=-0.76V, E_(Cu^(2+)//Cu)^(@)=+0.34V
Answer» ANSWER :`1.029 XX 10^(5)C`

Discussion

27.	Calculate the proton gain enthalpy of NH_(3) (g) from the following data (in kJ/mole) Delta H_("dissociation")^(@) : H_(2) (g) = 218 , Delta H_("dissociation") : Cl_(2) (g) = 124 Delta H_("formation")^(@) : NH_(3) (g) = -46 , Delta H_(f)^(@) : NH_(4) Cl (s) = -314 Lattice energy of NH_(4) Cl (s) = -683 Ionization energy of H = 1310 Electron affinity of Cl = 348
Answer» SOLUTION :`-718` kJ/mol

Discussion

28.	Calculate the product of uncertainties ofdisplacement and velocity of a moving electron.
Answer» SOLUTION :`5.8xx10^(-5)m^(2)s^(-1)`

Discussion

29.	Calculate the product of uncertainity in position and velocity for an electron of mass 9.1 xx 10^(-31)kg according to Heisenberg uncertainty principle.
Answer» SOLUTION :`5.77 XX 10^(-5) m^2 sec^(-1)`

Discussion

30.	Calculate the product of uncertainity in position and uncertainity in velocity for an electron of mass 9.1xx10^(-31)kg. According to Heisenberg's uncertainty principle. (h=6.6xx10^(-34)kgm^(2)s^(-1),pi=3.14):-
Answer» `5.8xx10^(-5)m^(2)s^(-1)` `5.8xx10^(-5)cm^(2)s^(-1)` `0` `5.8xx10^(-9)m^(2)s^(-1)` SOLUTION :According to heisenberg's uncertainty principle `Deltax xxDeltaV=(h)/(4pim)` `Deltax xxDeltaV=(6.6xx10^(-34)KGM^(2)s^(-1))/(4xx3.14xx9.1xx10^(-31)KG)` `=5.8xx10^(-5)m^(2)s^(-1)`.

Discussion

31.	Calculate the pressure in pascal (Pa) by a 760-mmHg column.
Answer» SOLUTION :101328 PA

Discussion

32.	Calculate the pressure exerted by one mole of CO_2gas at 273 K if the van derWaals constant a = 3.593 dm^6 atm "mol"^(-2). Assume that the volume occupied by CO_2molecules is negligible.
Answer» <P> Solution :APPLY `(p + (a)/(V^2) ) V = RT, V = 22.4 dm^2` 0.9922 atm

Discussion

33.	Calculate the pressure exerted by 5 mole of CO_(2) in one litre vessel at 47^@)C using van der Waal's equation. Also report the pressure of gas if it behaves ideal in nature. Given that a =3.592" atm It"^(2)" mol"^(-2), b=0.0427 It/mol. Also if the volume occupled by CO_(2) molecules is negligible, then calculate the pressure exerted by one mole of CO_(2) gas at 273 K.
Answer» Solution :van DER Waals equation `[P+(n^(2)a)/(V^(2))][V-nb]=NRT` `n_(CO_(2))=5, V=1" litre, "T=320 K, a=3.592, b=0.0427` `therefore [P+25xx(3.592)/(1)][1-5xx0.0427]=5xx0.0821xx320` `therefore P=77.218` atm For ideal behaviour of gas, PV =nRT `therefore P xx1=5xx0.0821xx320` `therefore P=131.36` atm For one mole `[P+(a)/(V^(2))][V-b]=RT` `therefore P =(RT)/(V) -(a)/(V^(2))` `therefore P =(0.0821xx273)/(22.4)-(3.592)/((22.4)^(2))` `therefore P =0.9922` atm The volume occupied by 1 mole at 273 K is 22.4 litre if b is negligible.

Discussion

34.	Calculate the pressure exerted by 10^(23) gas molecules each of mass 10^(22) g in a container of volume one litre. The nms speed is 10^(5) cm sec^(-1).
Answer» SOLUTION :No. of MOLECULES of gas `=10^(23)` mass of one molecules `=10^(-22) g` `V=1" LITRE "=10^(23) cm^(3)` `U_("rms")=10^(5)` cm/sec From kinetic equation `PV =((1)/(3)) mnu^(2)`. `P XX 1 =((1//3)xx10^(23)xx10^(-22)xx(10^(5))^(2))/(10^(3))` `therefore P=-3.3xx10^(7)" dyne cm"^(-2)`

Discussion

35.	Calculate the potential of the following cell reaction at 298K Sn^(4+)(1.50M)+Zn(s)toSn^(2+)(0.50M)+Zn^(2+)(2.0M) The standard potential E^(@) of the cell is 0.89 V. whether the potential of the cell will increase or decrease, if the concentration of Sn^(4+) is increased in the cell? (R=8.314JK^(-1)mol^(-1),F=96500" C "mol^(-1))
Answer» Solution :`E=E_(cell)^(@)-(0.0591)/(N)"log"([Sn^(2+)][Zn^(2+)])/([Sn^(4+)])=0.89-(0.0591)/(2)"log"((0.5)(2.0))/(1.5)=0.895V` if `[Sn^(4+)]` is incrased, the second factor will DECREASE or `E_(cell)` will INCREASE.

Discussion

36.	Calculate the potential of the following cell reaction at 258 K Sn^(4+)(1.50 M)+Zn(s) to Sn^(2+)(0.5 M) +Zn^(2+)(2.0 M) Standard potential of the cell is 0.89 v. s
Answer» Solution :For the reaction : `SN^(4+)(aq)+Zn(s) to Sn^(2+)(aq)+Zn^(2+)(aq)` Accordingto Nernst EQUATION, `E_(cell)^(@)=E^(@)-(0.0591)/(n)"LOG"([Sn^(2+)][Zn^(2+)])/([Sn^(4+)])` `=0.89-(0.0591)/(2)"log"((0.5)(2.0))/((1.50))` `=0.89-(0.0591)/(2)"log "0.667=0.89-0.0295xx(-0.1759)` `=0.89+0.005=0.895" V "`

Discussion

37.	Calculate the potential of the following cell at 298 K Zn//Zn^(2+)(a=0.1)//Cu^(2+)(a=0.01)//Cu E_(Zn^(2+)//Zn)^(@)=-0.762V E_(Cu^(2+)//Cu)^(@)=+0.337V Compare the free energy change for this cell with the free enegy of the cell in the standard state.
Answer» Solution :The overall cell reaction is `Zn+Cu^(2+)(a=0.01)rarrZn^(2+)(a=0.1)+Cu` The cell POTENTIAL GIVEN by nernst equation `E_("cell")=E_("cell")^(@)-(RT)/(2F)ln""(a_(Zn)^(2+)a_(Cu))/(a_(Zn)a_(Cu)^(2+))` `=E_("cell")^(@)-(RT)/(2F)ln""(a_(Zn)^(2+))/(a_(Cu)^(2+))` (Since activity of a pure metal is unity) `E_("cell")^(@)=0.337-(-0.762)=1.099V` `E_("cell")=1.099-(0.0591)/(2)LOG""(0.2)/(0.01)` `=1.099-(0.0591)/(2)log10` `=1.099-0.02956` `=1.0694V` The free energy CHANGE `DeltaG` is given by`DeltaG=-nFE` `DeltaG=-("2 equi/mol")("1.0694 V")("96495 coulomb equiv"^(-1))` `=-206.6" kJ mol"^(-1)` The standard free energy change : `DeltaG^(@)=-("2 equi/mol")(1.099V)("96495 coulomb/equiv")` `=-212.1" kJ/mol."`

Discussion

38.	Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
Answer» SOLUTION :For hydrogen ELECTRODE, `H^(+) + e^(-) to 1/2H_(2)` APPLYING NERNST equation, `E_(H^(+)//1/2H_(2)) = E_(H^(+)//1/2H_(2))^(@) -0.0591/n log 1/([H^(+)])` Substituting the values, we get `E_(H^(+)//1/2H_(2))^(@) =0-(0.0591)/1 log 1/(10^(-10))` [pH = 10 means `[H^(+)]=10^(-10) M]` `=-0.0591 xx 10 = -0.591` V Thus, potential of hydrogen electrode `=-0.591` V

Discussion

39.	Calculate the potential of hydrogen electrode in contact with a solution whose pH=10.
Answer» SOLUTION :`H^(+)+E^(-)rarr1//2H_(2)n=1` `E=E''-(0.0591)/(n)LOG.(1)/([H^(+)])` `E=0-(0.0591)/(1)xxpH` `E=-0.0591xx10v` `E=-0.591//V`

Discussion

40.	Calculate the potential of a silver electrode in a saturated solution of AgBr(K_(sp)=6 times 10^-13) containing, in addition, 0.1 mole per litre KBr. E_(Ag^+,Ag)^@=0.80 volt
Answer» SOLUTION :0.14 VOLT

Discussion

41.	Calculate the potential of a zinc-zinc ion electrode in which the zinc ion activity is 0.001M (E_(Zn^(2+)//Zn)^(@)=-0.76V,R=8.314KJ^(-1)mol^(-1),F=96,500" C "mol^(-1))
Answer» SOLUTION :`E=E^(@)-2.303(RT)/(nF)"log"(1)/(a_(Zn^(2+)))=-0.76-(2.303xx8.314xx298)/(2xx96500)"log"(1)/(10^(-3))=-0.849V`.

Discussion

42.	Calculatethe potential of a half cell having reaction : Ag_2 S (s) + 2e^- iff 2Ag (S) + S^(2-) (aq) in asolution buffered at p^(H) =3 and which is also saturated with0.1 MH_2S (aq) [Given : K_(sp)(Ag_2 S ) = 2 xx 10^(-49) , K_(a1). K_(a2) = 1.1 xx 10^(-21) , E_(As^(+) //Ag)^(@) = 0.8V]
Answer» `-0.19 V ` `0.19 ` ` 1.18 V ` `-0.166 V ` SOLUTION :`K_a = K_(a1). K_(a_2) = ((H^(OPLUS))^(2)(S^(-2)))/([H_2S]) `........... (1) , `K_(sp) = [Ag^(oplus)][S^(-2)] `........ (2) `((1))/((2)) = (K_(a))/(K_(sp))=([H^(oplus)]^(2))/([H_2S]) xx ([S^(2)])/([Ag^(ox)][S^(-2)]), (1.1 xx 10^(-21))/(2 xx 10^(-49)) = (10^(-6) xx 1)/( 0.1 xx [Ag^(oplus)]) implies (Ag^(oplus)) = 4.264 xx 10^(-17) M` ` E_(S^(-2)) .// Ag_2S // Ag =E_(Ag^(+)//Ag)^(0) to (0.859)/(1-1) , E = 0.8 - (0.0591)/(1) log""((1)/(4.264 xx 10^(-17)))` `=0.8p + (0.0591 xx-16.37 ) = (0.8 - 0.9675 ) , E = - 0.166`

Discussion

43.	Calculate the potential ofa Daniel cell, initially containing 1 litre each of 1M Cu^(2+) and 1M Zn^(2+) after a passage of 1 times 10^5 coulombs of charge . E_(Cu^(2+),Cu)^@=+0.34V, E_(Zn^(2+),Zn)^@=-0.76V
Answer» Solution :Moles of ELECTRICITY PASSED `=10^5/96500=1.04F` `therefore` eq. of `Cu^(2+)` removed=1.04 or mole of `Cu^(2+)` removed=0.52 and mole of `Zn^(2+)` PRODUCED= 0.52 Thus, `[Zn^(2+)]`=1.52 M and `[Cu^(2+)]=0.48M` For the Daniel cell, `Cu^(2+)+Zn=Cu + Zn^(2+)` `E=1.10 -0.0591/2 LOG"" 10.52/0.48=1.09V`

Discussion

44.	Calculate the potential of a cell in which hydrogen electrode is immersed in pure water, in a solution with a pH of 3.5 and in a solution with a pH of 10.7.
Answer» SOLUTION :0.425 VOLT

Discussion

45.	Calculate the potential for half cell containing 0.10 M K_(2)Cr_(2)O_(7)(aq),0.20" M "Cr^(3+)(aq) and 1.0xx10^(-4)M H^(+)(aq). The half-cell reaction is Cr_(2)O_(7)^(2-)(aq)+14H^(+)(aq)+6e^(-)to2Cr^(3+)(aq)+7H_(2)O(l) and the standard electrode potential is given as E^(@)=1.33V.
Answer» Solution :`E=E^(@)-(0.0591)/(n)"LOG"([Cr^(3+)]^(2))/([Cr_(2)O_(7)^(2-)][H^(+)]^(14))=1.33-(0.0591)/(6)"log"((0.20)^(2))/((0.10)(10^(-4))^(14))` `=1.33-(0.0591)/(6)log(4xx10^(57))=1.33-(0.0591)/(6)(57+0.6021)=1.33-0.57=0.76V`

Discussion

46.	Calculate the potentialcorresponding to the following cell Pt\|Co^(2+)(2.0 M),Co^(3+)(0.01 M\|\|Cr^(3+)(0.5M) Cr_(2)O_(7)^(2)(4.0 M),H^(+)(1.5 M)\|Pt Given E_(Co^(2+)//Co^(3+))^(2)=-1.82 V, E_(Cr_(2)O_(7)^(2)-//Cr^(3))^(2)=+1.33 V
Answer» `-0.32 v` `+0.32 V` `-0.44 v` `+0.44 V` Solution :`CO^(2+) rarr Co^(3+) +e^(-)xx6` `14 H^(+) +Cr_(2)O_(7)^(2-) +6E^(-) rarr 2Cr^(3+)+7 H_(2) O` `6Co^(2+) +14 H^(+) +Cr_(2) O_(7)^(2-) rarr 6Co^(3+) +2 Cr^(3+) +7 H_(2)O` `E_(cell)^(@)=-1.82 +1.33 V==-0.49` `=- 0.49 -(0.0591)/(6)LOG [10^(-2)]^(6)/([2.0]^(6)[4.0][1.5]^(14))=0.32 V`

Discussion

47.	Calculate the potential (emf) of the cell Cd\|Cd^(2+)(0.10M)\|\|H^(+)(0.20M)\|Pt,H_(2)(0.5atm) (given E^(@) for Cd^(2+)//Cd=-0.403V,R=8.314JK^(-1)mol^(-1),F=96,500" C "mol^(-1))
Answer» Solution :The cell REACTION is: `Cd+2H^(+)(0.20M)toCd^(2+)(0.10M)+H_(2)(0.5atm)` `E_(Cell)^(@)=E_(H^(+),1//2H_(2))^(@)-E_(Cd^(2+),Cd)^(@)=0-(-0.403)=0.403V` ltBrgt Applying NERNST EQUATION to the cell reaction, ltbr `E_(cell)=E_(cell)^(@)-(2.303nRT)/(nF)"LOG"([Cd^(2+)]xxP_(H_(2))^(**))/([H^(+)]^(2))=0.403-(2.303xx8.314xx298)/(2xx96500)"log"(0.1xx0.5)/((0.2)^(2))` `=0.403-0.003=0.400V`

Discussion

48.	Calculate the pOH of a solution at 25^(@)C that contains 1xx10^(-10)M of hydronium ions i.e H_(3)O^(+) :
Answer» `4.00` `9.00` `1.00` `7.00` Solution :`[OH^(-)]=(10^(-14))/([H_(3)O^(+)])=(10^(-14))/(10^(-10))=10^(-4)MOL L^(-1)` `POH = log [OH^(-)]=-log (1xx10^(-4))=4`.

Discussion

49.	Calculate the pOH of a solution at 25^(@)C that contains 1 xx 10^(-10) M of hydronium ions, i.e., H_(3)O^(+)
Answer» <P>`7.000` `4.000` `9.000` `1.000` Solution :Given `H_(3)O^(+)` ion concentration `= 1 xx 10^(-10)` `pH = -LOG[H^(+)], pH = -log[1 xx 10^(-10)]`, `pH = + 10 log 10, pH = 10` We KNOW that, `pH + pOH = 14`....(i) Put the value of pH in EQ. (i) `10 + P^(OH) = 14` `F^(OH) = 4`.

Discussion

50.	Calculate the pH of the following mixtures. Given K_(a) of CH_(3)COOH=2xx10^(-5) and K_(b) of NH_(4)OH=2xx10^(-5) (a) 50mL of 0.10 M NaOH+50mL of 0.10 M HCl. (b) 50mL of 0.10 M NaOH+50mL of 0.10 M CH_(3)COOH (c )=50mLof 0.05M NaOH+50mL of 0.10 M CH_(3)COOH (d) 50mL of 0.10 M NH_(4)OH+50mL of 0.05M HCl (e) 50mL of 0.10 M NH_(4)OH+50mL of 0.10 M HCl. (f) 50mL of 0.05 M NH_(4)OH+50mL of 0.05 M CH_(3)COOH.
Answer» Solution :`{:(,H^(+),+,OH^(-),rarr,H_(2)O),(t=0,5mmol,,5mmol,,-):}` so `pH` of resulting solution `=7`. b `{:(,CH_(3)COOH,+,OH^(-),rarr,CH_(3)COO^(-),+,H_(2)O),(t=0,2.5mmol,,5mmol,,,,-),(,-,,2.5,,2.5,,-):}` `[OH^(-)]=(2.5)/(100)M=2.5xx10^(-2)M` `pOH=3-LOG(2.5)=1.6` `,.pH=12.4` (C) `{:(,CH_(3)COOH,+,OH^(-),rarr,CH(3)COO^(-),+,H_(2)O),(t=0,5mmol,,2.5mmol,,,,),(,2.5,,-,,2.5,,):}` `pH=pK_(a)+log(([CH_(3)COO^(-)])/([CH_(3)COOH]))=pK_(a)=4.7` (d)`{:(,NH_(4)OH,+,H^(+),rarr,NH_(4)^(+),+,H_(2)O),(t=0,5mmol,,2.5mmol,,,,),(,2.5,,-,,2.5,,):}` `pOH=pK_(b)+log(([NH_(4)^(+)])/([NH_(3)]))=4.7` `:.pH=9.3` (e)`{:(,NH_(4)OH,+,H^(+),rarr,NH_(4)^(+),+,H_(2)O),(t=0,5mmol,,5mmol,,0,,),(,0,,0,,5,,):}` `pH=(1)/(2)[14-4.7-log0.05]` `pH=5.3`. (f)`{:(,NH_(4)OH,+,CH_(3)COOH,rarr,CH_(3)COONH_(4),+,H_(2)O),(t=0,2.5mmol,,2.5mmol,,,,),(,-,,-,,2.5,,):}` `pH=7+(1)/(2)pK_(a)-(1)/(2)pK_(b)=7`.

Discussion

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