Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Calculate the emf of the following cell at 298 K: Cr(s)//Cr^(3+) (0.1 m)//Fe^(2+) (0.01 M)//Fe(s) [Given: E_("cell")^(@) = +0.30 V]

Answer»

Solution :The REACTION is:
`2Cr(s) + 3Fe^(2+)(s) (aq) to 3Fe(s) + 2Cr^(3+) (aq)`
In this case number of electrons involved, n=6
Apply the equation,
`E_("CELL") =E_("cell")^(@) -(2.303RT)/(NF) log ([Cr^(3+)]^(2))/([Fe^(2+)]^(3))`
Substituting the values in the above equation, we have
`E_("cell") =0.30 -0.059/6 log [10^(-1)]^(2)/[10^(-3)]^(3) = 0.30 -0.059/6 log 10^(4) = 0.30 -0.04 =0.26 V`
Discussion
2.

Calculate the EMF of the following cell at 25^@C, Pt, H_2 (1 atm) / H^+(0.09 M) | | OH^– (0.075M)/ H_2 (1 atm), Pt (K_w of H_2O = 1 xx 10^(–14))

Answer»


ANSWER :0.3489
Discussion
3.

Calculate the emf of the following cell at 25^(@) C Zn| Zn^(2+) (0.001 M) || H^(+) (0.01 M) | H_(2) (g) (1 bar) | Pt (s) E_(Zn^(2+)//Zn)^(@) = -0.76 V, E_(H^(+)//H_(2))^(@) = 0.00 V

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Solution :The cell reaction is:
`Zn(s) + 2H^(+) (aq) to Zn^(2+) (aq) + H_(2)`, Then n=2
`E = E^(@) - 0.059/2 LOG ([Zn^(2+)])/([H^(+)]^(2)`
`=0.76 - 0.0295 log ([0.001])/([0.01]^(2)) = 0.76 - 0.0295 log 10`
`=(0.76 - 0.0295) "VOLT" = 0.7305` volt
Discussion
4.

Calculate the emf of the following cell at 25^@C H_2(g) (1atm)//H^+ (aq)|| OH^(-) (aq)//O_2(g) (1atm) from the following data: H_2+1/2O_2=H_2O,DeltaG^@=-226.8kJ//mol e

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SOLUTION :The electrodic process are,
`1/2 O_2+H_2O +2e=2OH^-` ……….at cathode (RHS)
`H_2=2H^+ +2e`…………at anode (LHS)
The cell reaction is
`1/2 O_2+ H_2O+ H_2=2H^+ + 2OH^(-)`
for which,
`E_(cell)=E_(cell)^@-.0591/2log[H^+]^2[OH^-]^2`
`E_(cell)=E_(cell)^@-.05921/2 log (K_w)^2`
`E_(cell)=E_(cell)^@+0.8274 (K_w=10^-14)`
Now multiply the given eqn. (II) by 2 and add eqn. (i) we get the cell reaction
`H_2+ 1/2O_2+H_2O=2H^+ +2OH^(-), Delta G^@=-226.8+2 times 76.8`
`=-74.4 kJ`
`=-74400J`
Again we have,
`Delta G^@=-2.303RT logk`
`-74400=-2.303 times 8.314 times 298 times log k`
`log k=13.04`
`therefore E^@=(2.303 RT)/(n F) log k`
`=.591/2times 13.04`
`=0.3853`
SUBSTITUTING the `E^@` values in eqn. (1)
`E_(cell)= 0.3853+0.8274`
`=1.2128V`
Discussion
5.

Calculate the emf of the cells formed by the various combinations of the following standard half cells. Here [M^(n+)]=1 mol L^(-1), since we are considering standard cells. (i) Zn^(2+)(aq)//Zn(s) (ii) Cr^(3+)(aq)//Cr(s) (iii) Cu^(2+)(aq)//Cu(s) (IV) Ni^(2+)(aq)//Ni(s) (v) Co^(2+)(aq)//Co(s) (vi) Ag^(+)(aq)//Ag(s) also calculate the standard potentials of such cells.

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Solution :The `E^(@)` values (reduction potentials) for the given half -cells obtained from Table 8.3 are given below : `{:("Half-cell":Zn^(2+)//Zn, Cr^(3+)//Cr, Cu^(2+)//Cu,Ni^(2+)//Ni, Co^(2+)//Co,Ag^(+)//Ag),(E^(@) "value(v)": -0.76,-0.74,+0.34,-0.25,-0.28,+0.80):}`
The increasing order of `E^(@)` value is :
`Zn^(2+)//Zn lt Cr^(3+)//Cr lt Co^(2+)//Co lt Ni^(2+)//Ni lt Cu^(2+)//Cu lt Ag^(+)//Ag`.
From the DATA it is evdent that
(a) A cell with `Zn^(2+)//Zn` as anode can have all the FIVE other ELECTRODES acting as cathode. The emf of the cells in all the cases are :
(i)`E_((Cr^(3+)//Cr))^(@)-E_((Zn^(2+)//Zn))^(@)=(-0.74)-(-0.76)=0.02 V`
(ii) (ii)`E_((Co^(2+)//Cu))^(@)-E_((Zn^(2+)//Zn))^(@)=(-0.28)-(-0.76)=0.48 V`
(III) `E_((Ni^(2+)//Ni))^(@)-E_((Zn^(2+)//Zn))^(@)=(-0.25)-(-0.76)=0.51 V`
(iv) `E_((Cu^(2+)//Cu))^(@)-E_((Zn^(2+)//Zn))^(@)=(+0.34)-(-0.76) = 1.10 V`
(v) `E_((Ag^(+)//Ag))^(@)-E_((Zn^(2+)//Zn))^(@)=(+0.80)-(-0.76) = 1.56 V`
(b) A cell with `Cr^(3+)//Cr` as anode can have all the remaining four electrodes acting as cathode.
The emf of the cells in all the cases are :
(i)`E_((Co^(2+)//Co))^(@)-E_((Cr^(3+)//Cr))^(@)=(-0.28)-(-0.74)=0.46 V`
(ii)`E_((Ni^(2+)//Ni))^(@)-E_((Cr^(3+)//Cr))^(@)=(-0.25)-(-0.74)=0.49 V`
(iii)`E_((Cu^(2+)//Cu))^(@)-E_((Cr^(3+)//Cr))^(@)=(+0.34)-(-0.74)=1.08 V`
(iv)`E_((Ag^(+)//Ag))^(@)-E_((Cr^(3+)//Cr))^(@)=(+0.80)-(-0.74)=1.54 V`
(c ) A cell with `Co^(2+)//Co` as anode can have all the remaining three electrodes acting as cathode.
The emf of the cells can be calculated as : (i)`E_((Ni^(2+)//Ni))^(@)-E_((Co^(2+)//Co))^(@)=(-0.25)-(-0.28)=0.03 V`
(ii)`E_((Cu^(2+)//Cu))^(@)-E_((Co^(2+)//Co))^(@)=(+0.34)-(-0.28)=0.62 V`
(iii)`E_((Ag^(+)//Ag))^(@)-E_((Co^(2+)//Co))^(@)=(+0.80)-(-0.28)=1.08 V`
(d) A cell with `Ni^(2+)//Ni` as anode can have the remaining two elctrodes acting as the cathode.
The emf of the cells can be calculated as :
(i)`E_((Cu^(2+)//Cu))^(@)-E_((Ni^(2+)//Ni))^(@)=(+0.34)-(-0.25)=0.59 V`
(ii)`E_((Ag^(+)//Ag))^(@)-E_((Ni^(2+)//Ni))^(@)=(+0.80)-(-0.25)=1.05 V`
(e) A cell with `Cu^(2+)..Cu` as anode can have only `Ag^(+)//Ag` as cathode.
(i)`E_((Ag^(+)//Ag))^(@)-E_((Cu^(2+)//Cu))^(@)=(+0.80)-(+0.34)=0.46 V`
Discussion
6.

Calculate the emf of the cell. Zn|Zn^(2+)(0.001M)||Ag^(+)(0.1M)|Ag The standard potential of Ag//Ag^(+) half - cell is +0.80 V and Zn//Zn^(2+) is -0.76V.

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Solution :Step 1 : Write the half-cell reactions of the anode and the cathode.
Then ADD the anode and cathode half reactions to obtain the cell reaction and the value of `E_("cell")^(@)`
`{:("Cathode",:,2Ag^(+)+2e^(-),rarr,2Ag,E^(@)=+0.80),("Anode",:,""Zn,rarr,Zn^(2+)+2e^(-),E^(@)=-0.76V),("Cell",:,Zn+"2Ag"^(+),rarr,Zn^(2+)+2Ag,E^(@)=1.56V):}`
Step 2. K for the cell reaction `=([Zn^(2+)])/([AG^(+)]^(2))`
Substituting the given values in the Nernst equation and solving for `E_("cell")`, we have
`E_("cell")=E_("cell")^(@)-(0.0591)/(n)logK`
`=1.56-(0.0591)/(2)log""([Zn^(2+)])/([Ag^(+)]^(2))`
`=1.56-(0.0591)/(2)log""([10^(-3)])/([10^(-1)]^(2))`
`=1.56-0.02955`
`=1.58955V`
Calculation of Equilibrium constant for the cell reaction
The Nernst equation for a cell is
`E_("cell")=E_("cell")^(@)-(0.0591)/(n)logK`
`"or"logK=(E_("cell")^(@))/(0.0591)`
Discussion
7.

Calculate the emf of the cell Zn//ZnO_(2)^(-), OH_((aq))^(-), HgO //Hg given that E^(@) values of OH^(-), ZnO_(2)^(-)//Zn and OH^(-), HgO //Hg half cells are -1.216V and 0.098 V respectively.

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SOLUTION :HINT : `E^(@)=E_(R)^(@)-E_(L)^(@),E^(@)=1.314V`
Discussion
8.

Calculate the emf of the cell Zn//ZnO_(2), OH^(-)""_(aq),-HgO""//Hg given that E^(@) values of OH^(-), ZnO and OH^(-), HgO""//Hg half cells are -1.216 V and 0.098 V respectively.

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SOLUTION :Given :
`Zn//ZnO_(2)^(-), OH^(-)""_((aq)), HgO""//Hg`
`""E^(@)""_(R)=0.098V, E_(L)=-1.216V`
Formula : `therefore E^(@)""_(CELL)=E^(@)""_(R)-E^(@)""_(L)`
Solution : `E^(@)""_(cell)=0.098-(-1.216)`
`""=0.098+1.216=+1.314`
`E^(@)=+1.314V`.
Discussion
9.

Calculate the emf of the cell : Mg|Mg^(2+)(0.1 M)||Ag^(+)(1.0xx10^(-3) M)|Ag Given that E_(cell)^(@)=3.15" V".

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SOLUTION :`E_(cell)^(@)=E_(cell)^(@)-(0.0591)/(n)"LOG"([MG^(2+)])/([Ag^(+)]^(2))`
`=3.15-(0.0591)/(2)"log"(0.1)/((1.0xx10^(-3))^(2))`
`=3.15-0.02955" log "5`
`=3.15-0.02955xx0.6989`
`=3.15-0.0207=3.13~~3.0`
Discussion
10.

Calculate the emf of the cell in which the following reaction takes place: Ni(s)+2Ag^(+)(0.002M)toNi^(2+)(0.160M)+2Ag(s) Given that E_(cell)^(@)=1.05V

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SOLUTION :Applying NERNST equation to the GIVEN cell REACTION,
`E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Ni^(2+)])/([Ag^(+)]^(2))=1.05V-(0.0591)/(2)"log"(0.160)/((0.002)^(2))=1.05-(0.0591)/(2)log(4XX10^(4))`
`=1.05-(0.0591)/(2)(4.6021)=1.05-0.14V=0.91V`
Discussion
11.

Calculate the emf of the cell in which the following reaction takes place: Ni(s)+ 2Ag^+(0.002M) to Ni^2(0.160 M)+2Ag(s). Given that E_(cell)^0=1.05V

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SOLUTION :`E_(cell)=E_(cell)^0-0.059/nlog[NI^(2+)]/[AG^+]^2`
`=1.05-0.059/2log0.160/(0.002)^2=1.05-0.059/2log(4xx10^4)`
`1.05-0.0295xx4.6021=1.05-0.135=0.914V`
Discussion
12.

Calculate the emf of the cell in which the following reaction takes place: Ni_((S))+2Ag^(+)(0.002M) to Ni^(2+)(0.160M)+2Ag_((S))""[E_(cell)^(Theta)=1.05V]

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Solution :* In given CELL FOLLOWING reaction is possible
`Ni_((S))+2Ag^(+)(0.002M)toNi^(2+)(0.160M)+2Ag_((S))`
Where, n=2, `[Ni^(2+)]=0.160M,[Ag^(+)]=0.002M,E_(cell)^(Theta)=1.05V`.
* Note: Here, solid Ni and Ag is neglected
`THEREFORE E_(cell)=E_(cell)^(Theta)-(0.059)/(n)"log"([Ni^(2+)])/([Ag^(+)]^(2))`
`=1.05-(0.059)/(2)"log"((0.160)/((0.002)^(2)))`
`=1.05-0.0295log(40000)`
`=1.05-0.0295xx4.6021`
`=1.05-0.1358`
=0.9142
`~~0.91V`
Discussion
13.

Calculate the emf of the cell in mV (atleast first two digits must match with correct answer) Ag(s),AgIO_(3)(s)|Ag^(+)(xM),HIO_(3)(1M)||Zn^(+2)(1M)|Zn(s) if K_(sp)=3xx10^(-8) for AgIO_(3) and K_(a)=(1)/(6) for HIO_(3) and E_(cell)^(0) for 2Ag+Zn^(+2)to2Ag^(+)+Zn is -1.56V. log3=0.48)Take(PT)/(F)=0.059) (giving your answer in magnitude only)

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Solution :`K_(a)=(Calphal^(2))/(1-ALPHA)implies(1)/(6)=(alpha^(2))/(1-alpha)impliesalpha=(-1+-sqrt((1)^(2)+4xx6xx1))/(12)=(-1+-sqrt(1+24))/(12)=(1)/(3)`
`therefore[IO_(3)^(-)]=1xx(1)/(3)=(1)/(3)Mimplies[Ag^(+)]=(3xx10^(-8))/((1)/(3))=9xx10^(-8)M`
Now `2Ag+Zn^(+2)overset(2e^(-))to2Ag^(+)+Zn`
gives `E=-1.56+(0.059)/(2)LOG((1)/(9xx10^(-8))^(2))=-1.144V=-1144mV`
Discussion
14.

Calculate the emf of the cell in which the following reaction takes place Ni(s) + 2Ag^(+)(0.002M) to Ni^(2+) (0.160 M)+ 2Ag(s) Given that E_("cell")^(@) = 1.05 V.

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SOLUTION :Applying Nernst equation to the given CELL reaction:
`E_("cell") = E_("cell")^@ - (0.0591)/x LOG (NI^(2+))/([Ag^+]^2)= 1.05 V - (0.0591)/2 log (0.160)/((0.002)^2)`
`= 1.05 - (0.0591)/2 log (4 xx 10^(4)) = 1.05 - (0.0591)/(2) xx 4.602`
`= 1.05 - 0.14 V = 0.91 V`.
Discussion
15.

Calculate the emf of the cell in which the following reaction takes place : Ni(s) + 2Ag^(+) (0.002 M) to Ni^(2+) (0.160 M) + 2Ag (s) Given that: E_("cell")^(@) = 1.05 V

Answer»

Solution :Applying Nernst equation and substituting the values, we GET
`E_("cell") = E_("cell")^(@) -(0.0591)/N log ([Ni^(2+)])/([Ag^(+)])^(2) = 1.05 - 0.0591/2 (4.6021) = 1.05 - 0.14 V = 0.91 V`
Thus emf of the cell = 0.91 V.
Discussion
16.

Calculate the emf of the cell having the cell reaction 2Ag^(+)+Zn iff 2Ag+Zn^(2+) " and " E^(@)""_(cell)=1.56V " at " 25^(@)C when concentration of Zn^(2+)=0.1 M " and " Ag^(+)=10M in the solution. ""["Hint :" E_(cell)=E^(@)""_(cell)-(RT)/(nF)In([Zn^(2+)])/[Ag]^(2)]

Answer»

Solution :Given :
`E^(@)""_(cell)=1.56V" "[Zn^(2+)]`
`""=0.1 M [AG^(+)]=10M`
Formula :
`E_("Cell")=E^(@)""_(cell)-(RT)/(NF)In.([Zn^(2+)])/[Ag^(+)]^(2)`
Solution : `""=1.56-0.02955 LOG 0.001,`
`""=1.56-(-0.08865)`
`""=1.56+0.08865=1.6486V`
`E_("Cell")=1.6486V`.
Discussion
17.

Calculate the EMF of the cell formd fromAg & Ni electrodes E^(@).Ag^(+)//Ag=(+0.799V) E^(@)Ni^(2+)//Ni= -0.250V

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ANSWER : 1.049 V
Discussion
18.

Calculate the emf of the cell having the cell reaction 2Ag^(+)+ZnhArr 2Ag+Zn^(2+) and E_("cell")^(@)=1.56V" at "25^(@)C when concentration of Zn^(2+)=0.1M and Ag^(+)=10M in the solution.

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Solution :HINT : `E_("CELL")=E_("cell")^(@)-(RT)/(NF)ln""([Zn^(2+)])/([AG^(+)]^(2))`
Discussion
19.

Calculate the emf of the cell Cr|Cr^(3+) (0.1 M) ||Fe^(2+) (0.01 M)|Fe ("Given: "E_(Cr^(3+)//Cr)^(@)=- 0.75" volt,"E_(Fe^(2+)//Fe)^(@)=-0.45" volt")

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ANSWER :0.261 VOLT
Discussion
20.

Calculate the EMF of the cell Cu//Cu^(2+)(0.1M)"//"Cl^(-) (0.01)// 1/2 Cl_(2), Pt E^(@) Cu^(2+)//Cu = +0.33V E^(@) 1/2 Cl_(2)//Cl^(-) = +1.36V

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ANSWER :1.073 V
Discussion
21.

Calculate the EMF of the cell for the reaction Mg_((s))+2Ag_((aq))^(+)rarr Mg_((aq))^(2+)+2Ag_((s)). ("Given " :E^(@)Mg^(2+)//Mg=-2.37V, E^(@)Ag^(+)//Ag=0.80V, [Mg^(2+)]=0.001M[Ag^(+)]=0.001M and log10^(5)=5).

Answer»

SOLUTION :`E_("CELL")^(@)=E_("cathode")^(@)-E_("anode")^(@)`
`E_("cell")^(@)=E_(Ag^(+)//Ag)^(@)-E_(Mg^(2+)//Mg)^(@)`
`E_("cell")^(@)=0.80-(-2.37)`
`E_("cell")^(@)=3.17V`
`E_("cell")=E_("cell")^(@)-(0.059)/(n)log""([Mg^(2+)])/([Ag^(+)]^(2))`
`E_("cell")=3.17-(0.059)/(2)log""([0.001])/([0.0001]^(2))`
`E_("cell")=3.17-(0.059)/(2)LOG10^(5)`
`E_("cell")=3.17-(0.059)/(2)xx5`
`E_("cell")=3.022V`
Discussion
22.

Calculate the EMF of the cell containing chromium and cadmium electrodes (Given E_(Cr^(3+)//Cr)^(@)=-0.74V,E_(Cd^(2+)//Cd)^(@)=-0.40V)

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ANSWER :0.34V
Discussion
23.

Calculate the emf of the cell, Cd|Cd^(2+)(0.001M)||Fe^(2+)(0.6M)|Fe at 25^(@)C. The standard reduction potential of Cd//Cd^(2+) and Fe//Fe^(2+) electrodes are -0.403 and -0.441 volt respectively.

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Solution :For the cell as REPRESENTED, `E_(cell)^(@)=E_(Fe^(2+)//Fe)^(@)-E_(Cd^(2+)//Cd)^(@)=-0.441-(0.403)=-0.038V`
Cell reaction: `Cd+Fe^(2+)(0.6M)toCd^(2+)(0.001M)+Fe`
`E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Cd^(2+)])/([Fe^(2+)])=-0.038-(0.0591)/(2)"log"(10^(-3))/(0.6)=-0.038+0.0821=0.0441V`
ALTERNATIVELY, CALCULATE electrode potentials of L.H.S. & R.H.S. ELECTRODES separately. then
`E_(cell)=E_(RHS)-E_(LHS)`.
Discussion
24.

Calculate the emf of given cell {:(Ag|AgNO_(3)||AgNO_(3)|Ag,,E_(Ag^(+)//Ag)=0.80V),(0.001 M, 0.1 M):}

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Solution :The GIVEN cell is concentrated cell. Hence, `E_("cell")^(@)=0`
`E_("cell") = E_("cell")^(@) - 0.0591/n log [("PRODUCT")]/[("Reactant")]`
`E_("cell") = 0-(0.0591)/1 log 10^(-3)/10^(-1) = 0.118` volt
Discussion
25.

Calculate the EMF of cell Ni//Ni^(2+) (0.01M)"//"Cl^(-)0.1M//Cl_(2), Pt E^(@)Ni^(2+)//Ni= -0.250V : E^(@)Cl_(2)//Cl^(-)= +1.360V

Answer»


ANSWER :`+1.73V`
Discussion
26.

Calculate the emf and DeltaGof cell reaction for the following cell at 25 °C : Mg(s) | Mg^(2+) (0.001 M) || Cu^(2+) (0.0001 M) + Cu(s) [E_(Mg^(2+)//Mg)]^(@) = -2.37 V, E_(Cu^(2+)//Cu)^(@) =+0.34 V and 1F = 96500C mol^(-1)

Answer»

Solution :The following reactions take place in the cell:
`Mg(s) to Mg^(2+) (aq) + 2e^(-)`
`Cu^(2+) (aq) + 2e^(-) to Cu(s)`
`Mg(s) + Cu^(2+) (aq) to Mg^(2+) (aq) + Cu(s)`
Thus, n=2
APPLYING Nernst equation and substituting the values, we GET
`E_("cell") = E_(Cu^(2+)//Cu)^(@) -E_(Mg^(2+)//Mg)^(@) -0.0591/2 LOG ([Mg^(2+)])/([Cu^(2+)])`
`=0.34 V - (-2.37 V) - 0.0591/2 log 0.001/0.0001`
`=2.71 V - 0.0591/2 log 10 = 2.71 V - 0.0295 V = 2.6805V`
`DeltaG` can be OBTAINED as under:
`DeltaG = -nFE =-2 xx 2.6805 xx 96500`
`=-517336.5 J mol^(-1) = -517.34 kJ mol^(-1)`
Discussion
27.

Calculate the e.m.f. of cell {:(Pt_(H_(2))),(1atm):}|{:(CH_(3)COOH),(0.1M):}||{:(NH_(4)OH),(0.01M):}|{:(Pt_(H_(2))),(1atm):} (K_(a) for CH_(3)COOH = 1.8 xx 10^(-5), K_(b) for NH_(4)OH = 1.8 xx 10^(-5))

Answer»


ANSWER :`-0.4575` VOLT ;
Discussion
28.

Calculate the electronic velocity in the first Bohr orbit of hydrogen.

Answer»

Solution :For hydrogen, Z=1, n=1
`v_(1)= (2pi E^(2))/(h)`
`=(2 xx (3.14) xx (4.8 xx 10^(-10))^(2))/(6.63 xx 10^(-27))`
`=2.19 xx 10^(8) cm//s`
Discussion
29.

Calculate the electronegativity of fluorine from the following data: E_(H-H)=10.4.2 kcal mol^(-1) E_(F-F)= 36.6 kcal mol^(-1) E_(H-P)= 134.6 kcal mol^(-1) Electronegativity of H= 2.05

Answer»

Solution :ENERGY for 100% COVALENT bond (H-F)= `sqrt(E_(H-H) XX E_(F-F))`
=`sqrt(104.2 xx 36.6)`
=61.75 kcal
Resonance energy `(Delta)=` acutal bond energy- energy for 100% covalent bond
=134.6-61.75
=72.85 kcal
Electronegativity difference= `0.18 sqrt(Delta)`
Electronegativity of F- electronegativity of `H= 0.18 sqrt(Delta)`
Electronegativity of `F= 0.18 xx sqrt(72.85) + 2.05` = 3.586
Discussion
30.

Calculate the electronegativity of fluorine from following data : F_(H-H) =104. 2 k cal mol ^(-1) E _(F-F) =36.6 kcal mol ^(-1) E _(H-F) = 134. 6 kcal mol ^(-1) Electronegativity ofH is 2.05

Answer»

Solution :On Paulling SCALE:
`x _(F) -x_(H) =0.182sqrt(Delta _(H)-F)`
(using B.E. in kcal `mol ^(-1))`
`Delta _(H-F) = E_(H-F) - SQRT( E_(H-H) xx E _(F -F))`
`=13.45 - sqrt( 104.2 xx36.6) = 72.84` kcal
From (i)
`x_(F) -x_(H) = 0.182 sqrt(72.84)+ 1.5534`
`x_(F)=x_(H) + 1.4434=2.05+ 1.5534=3.6034`
Discussion
31.

Calculate the electron gain enthalpy of fluorine atom using the following data. Make Born – Haber's cycle. All the values are in kJ mol^(-1) at 25^(@) C , Delta H_("diss") (F_(2)) = 160 , Delta H_(f)^(0) (NaF(s)) = -571 , I.E. [Na (g) ] = 494 , DeltaH_("vap") [Na(s)] = 101 . Lattice energy of NaF (s) = -894.

Answer»

SOLUTION :`-352 KJ MOL^(-1)`
Discussion
32.

Calculate the electrode potential of the electrode Zn//Zn^(2+) (conc. =0.1M) at 25^(@)C Given that E_(Zn//Zn^(2+))^(@)=0.7618 volt.

Answer»


Solution :STANDARD REDUCTION potential, VIZ, `E_(Zn^(2+)//Zn)^(@)=-0.7618V`.
Discussion
33.

Calculate the Electrode potential of single electrode.Cu^(2+)(0.01M)//Cu" "(E^(@)= +0.337V)

Answer»


ANSWER :0.277V
Discussion
34.

Calculate the electrode potential forunderset(1 atm)((Pt)H_2) | H^(+) (c = 0.1)

Answer»

SOLUTION :`-0.0591 V`
Discussion
35.

Calculate the electrochemical equivalent of the lightest element.

Answer»


ANSWER :`1.045xx10^(-5)` gm/coulomb
Discussion
36.

Calculate the electrode potential of given electrode Pt, Cl_(2) (1.5 "bar") | 2Cl^(-) (0.01 M), Solu E_(Cl_(2)//2Cl^(-))^(@) = 1.36 V tion:

Answer»

Solution :The REACTION of electrode is
`{:(Cl_(2)(g) + 2E^(-), to, 2Cl^(-)),(1.5 "bar",,0.01 M):}`
`E = e^(@) -(0.0591)/n LOG ([Cl^(-)]^(2))/P_(Cl_(2))`
`=1.36 - (0.0591)/2 log (0.01)^(2)/1.5 = 1.483 V`
Discussion
37.

Calculate the electrode potential developed when a silver electrode in dipped in 0.025 M silver nitrate solution at 289K.

Answer»

SOLUTION :`E=E^@+0.0591/N LOG [M^(+n)]`
`E=+0.8+0.0591/1 log (0.025)`
`E=+0.8+0.0591/1 log 2.5 TIMES 10^-2`
`E=+0.70531 V`
Discussion
38.

Calculate the electrode potential at 25^@C of Cr_2O_7^(2–)//Cr^(3+) electrode at pOH = 11 in a solution of 0.01 M both in Cr^(3+) and (Cr_2O_7^(2–)) = 1.33 V. If this electrode is coupled with a hydrogen gaseous electrode at pH = 6, find the emf of the spontaneous cell.

Answer»


ANSWER :1.2973
Discussion
39.

Calculate the packing efficiency in simple cubic lattice.

Answer»

Solution :Packing efficiency of metal crystal for simple CUBIC lattice.
Let a be the edge length of the UNIT cell and r be radius of the sphere.
Volume of the sphere `= (4)/(3) pi r^(3)`

As the spheres at the corners are touching each other, the edge length a = 2r
Volume of the cube `= a^(3) = 8r^(3)`
% of the SPACE occupied by spheres
`= ("Volume of sphere")/("Volume of cube") xx 100`
`= ((4)/(3) pi r^(3) xx 100)/(8r^(3)) = 52.4%`
Packing efficiency of simple cubic lattice is 52.4%
EXAMPLE : Silicon and GERMANIUM (Any one).
Discussion
40.

Calculate the efficiency [percentage of volume occupied and unoccupied] of packing in case of a metal crystal for simple cubic.

Answer»

Solution :For simple cubic `a=2r`.
`"Efficiency "=("Volume OCCUPIED by all spheres")/("Total volume of cube")XX100`
`"Efficiency "=((4)/(3)PIR^(3))/((2r)^(3))xx100 =(4xxpixxr^(3))/(3xx8xxr^(3))xx100=(pi)/(6)xx100=(22)/(7xx6)xx100=(2200)/(42)=52.4%`
Percentage of volume occupied `=52.4%`.
Percentage of volume UNOCCUPIED `=47.6%`.
Discussion
41.

Calculate the efficiency of packing in case of a metal crystal for (i) simple cubic (ii) body-centred cubic (iii) face-centred cubic (with the assumptions that atoms are touching each other).

Answer»

Solution :(i) Simple CUBIC :
Number of spheres in a unit cell `=8xx(1)/(8)=1`
Volume of the sphere `=(4)/(3)pir^(3)"(r is the radius of the sphere)"`
Volume of the cube `=a^(3)=(2r)^(3)=8r^(3)`
Effciency of packing `=(4//3pir^(3))/(8r^(3))=(pi)/(6)=0.524 or 52.4%`.
(ii) Body - centred cubic structure :
Assuming that body centre touches the spheres at the corner.
Body diagonal AD = 4r
FACE diagonal `AC=sqrt(AB^(2)+BC^(2))=sqrt(a^(2)+a^(2))=sqrt2a`
Body diagonal `AD=sqrt(AC^(2)+CD^(2))=sqrt(2a^(2)+a^(2))=sqrt3a`
`sqrt3a=4r or a=(4r)/(sqrt3)`,
Volume of unit cell `=a^(3)=((4r)/(sqrt3))^(3)=(64r^(3))/(3sqrt3)`

Number of spheres in unit cell `=8xx(1)/(8)+1=2`
Volume of two spheres `=2xx(4)/(3)pir^(3)=(8)/(3)pir^(3)`
Efficiency of packing `=(8pir^(3)//3)/(64r^(3)//3sqrt3)=(pisqrt3)/(8)=0.68 or 68%`
(iii) Face - centred cubic :
`AC=sqrt(AB^(2)+BC^(2))=sqrt(a^(2)+a^(2))=sqrt2a`
`therefore""sqrt2a=4r or a=(4)/(sqrt2)xxr`
Volume of the unit cell `=a^(3)=(32)/(sqrt2)r^(3)`
Number of sphere in a unit cell `=8xx(1)/(8)+6xx(1)/(2)=4`
Volume of four spehres `=4xx(4)/(3)pir^(3)=(16)/(3)pir^(3)`
Efficiency of packing `=(16pir^(3)//3)/(32r^(3)//sqrt2)=0.74 or 74%`.
Discussion
42.

Calculate the effective nuclear charge experienced by the 4s electron in potassium atom.

Answer»

SOLUTION :The electronic configuration of K atom is
`K_19 = (1s^2) (2s^2 2p^6) (3s^2 3p^6) 4s^(1)`
Effective nuclear CHARGE (Z*) = Z – S
Z* `=19 - [(0.85xx" No. of electron in "(n-1)^(TH)" shell" + (1.00" total number of electron in the INNER shells")]`
`=19- [ 0.85 xx (8) + (1.00 xx 10)]`
Z* `= 2.20`
Discussion
43.

Calculate the effective neutron capture radius of a nucleus having a cross section of 1.0 barn.

Answer»

Solution :1 barn ` = 10^(-24)CM^2` and area of CIRCLE =`pi r^2`
`5.6 xx 10^(-13) cm`
Discussion
44.

Calculate the ebullioscopic constant for water. The heat of vaporization is 40.685 kJ mol^(-1)

Answer»

`0.512 K KG mol^(-1)`
`1.86 K kg mol^(-1)`
`5.12 K kg mol^(-1)`
`3.56 Kkg mol^(-1)`

SOLUTION :`K_(b)=(RT_(0)^(2)M)/(1000DeltaH_("vapour"))`
`=(8.314xx(373.15)^(2)xx18)/(1000xx40.685)xx10^(-3)=0.512 K kg mol^(-1)`
Discussion
45.

Calculate the EAN of central atom in the following complexes: (i) [Cr(CO)_(6)] (ii) [Fe (eta^(5)-C_(5)H_(5))_(2)]

Answer»


Solution :In `[FE(H^(5)-C_(5)H_(5))_(2)]`
EAN of `Fe= 26-2+2(6)=36`
Discussion
46.

Calculate the E and E^(@) of the cell Ni | Ni^(2+) || Cu^(2+) | Cu from the following half-cell reactions: {:(Ni^(2+) + 2e^(-) rarr Ni,,E^(@) = - 0.25 " volt"),(Cu^(2+) + 2e^(-) rarr Cu,,E^(@) = + 0.34 " volt"):} (Given: [Ni^(2+)] = 1 M " and " [Cu^(2+)] = 10^(-3) M)

Answer»


ANSWER :`E_("CELL")^(@) = 0.59 " VOLT"; E_("cell") = 0.5015` volt
Discussion
47.

Calculate the ebullioscopic constant for water. The heat of vaporisation is 40.685 kJmol^(-1)

Answer»

0.512 K KG `MOL^(-1)`
1.86 K kg `mol^(-1)`
5.12 K kg `mol^(-1)`
3.56 K kg `mol^(-1)`

ANSWER :A
Discussion
48.

Calculate the dmole fraction of ethyl alcohol and water in a solution in which 46 g of ethlyl alcohl and 180g of water are mixed together.

Answer»


Solution :`"No. of moles of ethyl alcohol" (n_(C_(2)H_(5)OH))=("Mass of ethyl alcogol")/("Molar mass")=((46g))/((46g mol^(-1)))=1mol`
`"No. of moles of WATER"(n_(C_(2)H_(5)O))=("Mass of water")/("Molar mass")=((180G))/((18"g mol"^(-1)))=10 mol`
`"MOLE fration of ethyl alcohol "(x_(C_(2)H_(5)OH))=n_(C_(2)H_(5)OH)/(n_(C_(2)H_(5)OH)+n_(H_(2)O))=((1 mol))/((1+10)mol)=0.09`.
Discussion
49.

Calculate the DU of following compounds : (i) C_(6)H_(6)ClBrO, (ii) C_(5)H_(9)N

Answer»


SOLUTION :(i) `D U = (6+1-8//2) = 3`" " (II)`D U = 5+1 -((9-1)/(2)) = 2`
Discussion
50.

Calculate the distance between 111 planes in a crystal of Calculate the distance between 111 planes in a crystal of Ca. the answer is

Answer»

1.61nm
0.610nm
0.321 nm
None of these

Solution :We have, `d=(a)/(SQRT(h^(2)+k^(2)+I^(2))):d_(111)=(0.556)/(sqrt(I^(2)+I^(2)+I^(2)))`
=0.321 nm.
Discussion