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.

For the cell Zn|Zn^(2+)||Cu^(2+)|Cu if the concentration of Zn^(2+) and Cu^(2+) ions is doubled, the emf of the cell:

Answer»

Doubles
Reduces of half
Remains same
Becomes zero

Answer :C
Discussion
2.

For the cell: Zn(s)|ZnSO_(4)(aq)||CuSO_(4)(aq)Cu(s), calculate standard cell potential if standard state reduction electrode potentials for Cu^(2+)//Cu and Zn^(2+)//Zn are +0.34V and -0.76V respectively.

Answer»


ANSWER :1.10 V
Discussion
3.

For the cell Zn|Zn^(2+) (C_1)||Zn^(2+) (C_2)|Zn. DeltaG is negative if :

Answer»

`C_1 = C_2`
`C_1 GT C_2 `
`C_2 gt C_1`
NONE of these

Solution :`DeltaG = nFE_("cell")=-nF[(0.059)/(2) log_(10)""(c_2)/(c_1)] c_2 gt c_1 , ` Then `DeltaG=-ve`
Discussion
4.

For the cell Zn|Zn^(+2)||Cu^(+2)|Cu, if the concentration of Zn^(+2) and Cu^(+2) ions is doubled, the emf of the cell

Answer»

doubles
reduces to half
remains same
becomes zero

Solution :Ration of log`[ZN^(2+)]/[CU^(2+)]` remain same.
Discussion
5.

For the cell, Zn(s)abs(Zn^(2+))abs(Cu^(2+))Cu(s), the standard cell voltage, E^(0)""_(cell) is 1.10 V.When a cell using these reagents was prepared in the lab, the measured cell voltage was 0.98One possible explanation for the observed voltage is :

Answer»

there were `2.00` mole of `Cu^(2+)` but only `1.00` mol of `Zn^(2+)`
the Zn electrode had TWICE the surface area as that of the Cu electrode
the `[Zn^(2+)]` concentration was LARGER than the `[Cu^(2+)]` concentration
the volume of the `Zn^(2+)` ION solution was GREATER than the volume of the `Cu^(2+)` ion solution.

Answer :C
Discussion
6.

For the cell Zn(s)|Zn^(2+) (2M)||Cu^(2+) (0.5 M)|Cu(s) (a) Write equation for each half reaction. (b) Calculate the cell potential at 25^@C

Answer»

SOLUTION :(a) Half reaction OXIDATION half reaction:
(i) `Zn(s) to Zn^(2+) (aq) +2e^(-)`
Reduction half reaction:
(II) `Cu^(2+) (aq) +2e^(-) to Cu(s) `
`E_(cell)^@=E_(cathode)-E_(anode)`
`=+0.34-(-0.76)=1.1`
`E_(cell)^@=E_(cell)^@-0.059/2log""(Products)/(Reactants)`
`Zn_((s))+Cu^(2+) (0.5M)to Cu_((s))+Zn^(2+)(2M)`
`E_(cell)^@=1.1-0.059/2 LOG""2/0.5`
`=1.1-0.0295 times 0.6020`
`=1.1 -0.0177`
=1.082V
Discussion
7.

For the cell Zn(s) | Zn^(2+)(2 M) ||Cu^(2+) (0.5 M) | Cu(s) (a) Write the equation for each half-reaction. (b) Calculate the cell potential at 25^(@) C. [Given, E_(Zn^(2+)//Zn)^(@) =-0.76 V, E_(Cu^(2+)//Cu)^(@) = +0.34 C]

Answer»

Solution :`E^(@) =0.34 -(-0.76) = 1.1 V`
`E_("cell") =E_("cell")^(@) -(0.0591)/2 LOG (ZN^(2+))/(Cu^(2+)) =1.1 -0.0591/2 log 2/0.5`
`=1.1 -0.0295 log 4 = 1.1 -0.0295 xx 0.6021 = 1.1 - 0.0177 = 1.0823` V
Discussion
8.

For the cellZn(s) | Zn^(2+) || Cu^(2+) | Cu(s) , the standard cell voltageE_("cell")^(0) is 1.10 V . When a cell using these reagents was prepared in the lab, the measured cell voltage was 0.97 V . One possible explanaiton of the observed voltage is

Answer»

there are 2.00 mol of `Cu^(2+) ` bu only 1.00 mol of `ZN^(2+) `
the Zn ELECTRODE had twicethe surfaceof the Cu electrode
the `[Zn^(2+)]` was LARGER than the `[Cu^(2+)]`
the VOLUME of the `Zn^(2+)` SOLUTION was larger than the volume of the `Cu^(2+)` solution

Solution :`Zn + Cu^(+2) to Cu + Zn^(+2) , E=E^(0) + (0.0591)/(2) log""((Cu^(+2))/(Zn^(+2)))`
`0.98 = 1.1 + (0.0591)/(2) log""([Cu^(+2)])/([Zn^(+2)]) ,` So , ` [Zn^(+2)] gt [Cu^(+2)]^(2)`
Discussion
9.

For the cell Zn |Zn^(2+) | | Cu^(2+) | Cu , if the concentration of Zn^(2+) and Cu^(2+) ions is doubled , the emf of the cell

Answer»

DOUBLES
reduces to HALF
remains same
becomes zero

Solution :`E_("cell") = E_("cell")^(@) + (0.059)/(2) "log" ([Cu^(2+)])/([Zn^(2+)])` , THUS on doubling concentration of both `Cu^(2+) and Zn^(2+)` , there will be no effect on `E_("cell")`.
Discussion
10.

For the cell Zn// Zn^(2+) // // Cu^(2+) // Cu , if the concentration of Zn^(2+) and Cu^(2+) ions is doubled , the emf of the cell

Answer»

DOUBLES
REDUCES to HALF
REMAINS same
remains ZERO
Discussion
11.

For the cell, TI|TI^(+)(0.001M)|Cu^(2+)(0.1M)|Cu,E_(cell) at 25^(@)C is 0.83 V. this can be increased

Answer»

by increasing `[CU^(2+)]`
by increasing `[TI^(+)]`
by decreasing `[Cu^(2+)]`
by decreasing `[TI^(+)]`

Solution :`2TI+Cu^(2+)to2TI^(+)+Cu,E_(cell)-E_(cell)^(@)-(0.0591)/(2)"log"([TI^(+)]^(2))/([Cu^(2+)])`
Discussion
12.

For the cell, TI|TI^(+)(0.001 M)||Cu^(2+)(0.1 M)|Cu(s),E_(cell)^(@) at 25^(@)C is 0.83" V". It can be increased by :

Answer»

INCREASING `[Cu^(2+)]`
increasing `[TI^(+)]`
decreasing `[Cu^(2+)]`
decreasing `[TI^(+)]`

SOLUTION :(a,d) are correct.
Discussion
13.

For the cell ,Tl |Tl^+(0.001M)|| Cu,E_(cell) at 25^@ C is 0.83V.E_(cell) can be increased :

Answer»

By increasing `[CU^(2+)]`
By increasing `[Ti^+]`
By decrasing` [Cu^(2+)]`
None of the above

Answer :A
Discussion
14.

For the cell reaction Zn_((s)) + Cu_("0.1 M")^(2+) to Zn_(0.1 M)^(2+) + Cu_((s)) if the standard EMF of the cell is E^(@), then

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`E GT E^(@)`
`E lt E^(@)`
`E = E^(@)`
`E le E^(@)`

SOLUTION :`E = E^(@) - 0.591 " log" (0.01)/(0.1)`
`E = E^(@) + 0.059` V
Discussion
15.

For the cell reaction, Sn(s)+Pb^(2+)(aq)toSn^(2+)(aq)+Pb(s), E_(Sn^(2+)//Sn)^(@)=-0.136V, E_(Pb^(2+)//Pb)^(@)=-0.126V, Calculate the ratio of concentration of Pb^(2+) to Sn^(2+) ion at which the cell reaction will be reversed?

Answer»

SOLUTION :For the given cell, `E_(cell)^(@)=E_(Pb^(2+)//Pb)^(@)-E_(Sn^(2+)//Sn)^(@)=0.126-(-0.136)=0.01V`
For the given cell REACTION,
`E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Sn^(2+)])/([Pb^(2+)])=0.01-(0.0591)/(2)"log"([Sn^(2+)])/([Pb^(2+)])=0.01+0.0295"log"([Pb^(2+)])/([Sn^(2+)])`
At equilibrium `E_(cell)=0`
`therefore0.01+0.0295"log"([Pb^(2+)])/([Sn^(2+)])=0` or `"log"([Pb^(2+)])/([Sn^(2+)])=-(0.01)/(0.0295)=-0.3390=overline(1).6610`
`THEREFORE([Pb^(2+)])/([Sn^(2+)])="Antilog "overline(1).6610=0.458`
THUS, so long as `([Pb^(2+)])/([Sn^(2+)]) gt0.458`, the cell reaction as given will place. when `([Pb^(2+)])/([Sn^(2+)]) lt 0.458,E_(cell)` will become -ve, i.e., the reaction will be reversed.
Discussion
16.

For the cell reaction X(s)+ 2Y^(+) to X^2 + 2Y k_c has been found to be 10^12. The E_("Cell")^@ is :

Answer»

0.708 V
1.36 V
0.354 V
1.006 V

Answer :C
Discussion
17.

For the cell reaction : Sn (s) + Pb^(2+) (aq) to Sn^(2+) (aq) + Pb (s) E_(Sn^(2+)|Sn)^(@) = -0.140 , E_(Pb^(2+) |Pb)^(@) = -0.126 V Calculate the ratio of concentration of Pb^(2+) to Sn^(2+) ion at which the cell reaction be reversed .

Answer»

Solution :For the cell , `E^(@) = E_(PB^(2+)|Pb)^(@) - E_(Sn^(2+) |Sn)^(@) = -0.126 - (-0.140)`
`= 0.014 V `
Applying Nernst equation
`E = E^(@) - (0.059)/(2) log ([Sn^(2+)])/([Pb^(2+)])`
`= 0.014 + (0.059)/(2) "log" ([Pb^(2+)])/([Sn^(2+)])`
At equilibrium , E = 0
`therefore 0.014 + (0.059)/(2) log ([Pb^(2+)])/([Sn^(2+)]) = 0`
or log `([Pb^(2+)])/([Sn^(2+)]) = - (0.014xx 2)/(0.059) = -0.474`
`therefore ([Pb^(2+)])/([Sn^(2+)]) =` antilog `(-0.474) = 0.336`
Thus ,the cell reaction will occur TILL `([Pb^(2+)])/([Sn^(2+)])` is more than `0.336` V .
When `([Pb^(2+)])/([Sn^(2+)])` BECOMES less than 0.336 V , `E_(cell)` will become negative and reaction will be reversed .
Discussion
18.

For the cell reaction, Ni|Ni^(2+)||Ag^(+)|Ag Calculate the equilibrium constant at 25^(@)C. How much maximum work would be obtained by the operation of this cell ?

Answer»


ANSWER :`3.98xx10^(35)`, MAX. WORK `=202650 J`
Discussion
19.

For the cell reaction Ni (s)|NI^(2+)(aq)||Ag^(+)(aq)|Ag(s), calculate the equilibrium constant at 25^(@)C. How much maximum work would be obtained for the operation of this cell ? (Given E_(Ni^(2+)//Ni)^(@)=-0.25" V " and E_(Ag^(+)//Ag)^(@)=0.80" V ")

Answer»


SOLUTION :`Ni(s)+2Ag^(+)(aq) hArr Ni^(2+)(aq)+2Ag(s)`
`E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)=0.80-(-0.25)=1.05" V "`
`E_(cell)^(@)=(0.0591)/(n)log K_(c )"or""log "K_(c )=(nE^(@)cell)/(0.0591)`
`log K_(c )=(1.05xx2)/(0.0951)=35.6 "or"K_(c )="Antilog "35.6=3.981xx10^(35)`
`(-DeltaG^(@))=Max.work =nFE^(@) cell`
`=2" MOL"xx(96500" C mol"^(-1))xx(1.05" V")=202650" J "=202.65 KJ`
Discussion
20.

For the cell reaction, Mg_((s))+Cu_((aq))^(2+) rarr Cu_((s)) +g^(2+)(aq.) the standard reduction potentials of Mg and Cu are -2.37 V and 0.34 V respectively. The e.m.f. of the cell is

Answer»

2.03V
`-2.03V`
`2.71V`
`-2.71V`

ANSWER :C
Discussion
21.

For the cell reaction, Cu_(C_(2))^(2+)(aq)+Zn(s) to Zn_(C_(1))^(2+)(aq)+Cu(s) The change in free energy (DeltaG) at a given temperature is a function of :

Answer»

In `C_(1)`
In `(c_(2)//c_(1))`
In `(c_(1)+c_(2))`
In `c_(2)`

Solution :(B) `-DeltaG=nFE_(CELL)=nFE_(cell)^(@)+RT" In "([Cu^(2+)])/([Zn^(2+)])`
`-DeltaG` is a function of In `(C_(2)//C_(1))`
Discussion
22.

For the cell reaction Cu^(2+)(C_(1aq))+Zn_((s))+Zn^(2+)(C_(2aq))+Cu_((s)) of an electrochemical cell, the change in free energy at a given temperature is a function of

Answer»

ln `(C_(1))`
ln `(C_(2))`
ln `(C_(1)+C_(2))`
ln `(C_(2)//C_(1))`

Solution :`E_(cell)=E_(cell)^(@)-(RT)/(nF)` In `(C_(2))/(C_(1)) and DELTAG=-nFE_(cell)`
HENCE `DeltaG` is the FUNCTION of `ln ((C_(2))/(C_(1)))`
Discussion
23.

For the cell reaction, 2H_(2)(g)+O_(2)(g)to2H_(2)O(l),Delta_(r)S_(298K)^(@)=-0.32" kJ "K^(-1) What is the value of Delta_(f)H^(@)(H_(2)O)? Given: O_(2)(g)+4H^(+)(aq)+4e^(-)to2H_(2)O(l),E^(@)=1.23V

Answer»

`-189.71 " kJ "mol^(-1)`
`-285.08" kJ "mol^(-1)`
`-379.42" kJ "mol^(-1)`
`-570.16" kJ "mol^(-1)`

Solution :For the given CELL reaction involving FORMATION of 2 MOLES of `H_(2)O`,
`DeltaG^(@)=-nFE_(cell)^(@)=-4xx96500xx1.23J`
`=474780J=-474.78kJ`
(n=4 for the given cell raction)
`DeltaG^(@)=DeltaH^(@)-TDELTAS^(@)`
ORR `DeltaH^(@)=DeltaG^(@)-TDeltaS^(@)`
`=-474.78kJ+298K(-0.32kJK^(-1))`
`=-474.8-95.36kJ=-570.16 kJ`
This is enthalpy change for the formation of 2 moles of `H_(2)O`
`therefore` Enthalpy change for the formation of 1 mole of
`H_(2)O(Delta_(f)H^(@))=-(570.16)/(2)=-285.08kJ`.
Discussion
24.

For the cell reaction 2Fe_((aq))^(3+)+2I_((aq))^(-) to 2Fe_((aq))^(2+)+I_(2(aq)) E_(cell)^(Theta)=0.24V at 298K. The standard Gibbs energy (Delta_(r)G^(Theta)) of the cell reaction is : [Given that Faraday constant F=96500 C mol^(-1)]

Answer»

23.16 KJ `mol^(-1)`
`-46.32` kJ `mol^(-1)`
`-23.16` kJ `mol^(-1)`
`46.32" kJ "mol^(-1)`

SOLUTION :`2Fe_((aq))^(3+)+2I_((aq))^(-) to 2Fe_((aq))^(2+)+I_(2(aq))`
`DELTAG^(@)=-nFE^(@)""` (where,n=2)
`=-2xx96500xx(0.24)`
`=-46320J`
`=-46.32` kJ `mol^(-1)`
Discussion
25.

For the cell reaction ,Cu^(2+)(aq)(C_2) +Zn(s) rarrZn^(2+)(aq)(C_1)+Cu(s), the change in free energy(Delta G) at a given temperature is a function of :

Answer»

`lnC_1`
`LN(C_2//C_1)`
`ln(C_1+C_2)`
`ln C_2`

ANSWER :B
Discussion
26.

For the cell reaction 2Fe^(3+)(aq) + ""^(21-)(aq) to 2Fe^(2+)(aq) + 1_2 (aq) E_("cell")^(@) = 0.24V at 298K.The standard Gibbs energy (Delta, G^@) of the cell reactions is :

Answer»

`-46.32 KJ MOL^(-1)`
`-23.16 KJ mol^(-1)`
`46.32 KJ mol^(-1)`
`23.16 KJ mol^(-1)`

ANSWER :A
Discussion
27.

For the cellreaction 2Fe^(3+)(aq)+21^(-)(aq) rarr 2Fe^(2+)(aq)+1_(2)(aq)E_(cell)^(@)=0.24V at 298 K. The standard Gibbs energy (Delta, G^(@)) of the cell reactions is :

Answer»

`-46.32" KJ "MOL^(-1)`
`-23.16" KJ "mol^(-1)`
`46.32"KJ "mol^(-1)`
`23.16" KJ "mol^(-1)`

ANSWER :A
Discussion
28.

For the cell reaction, 2Cr^(4+)+Coto2Ce^(3+)+Co^(2+)""E_((cell))^(@) is 1.89V and E_(Co//Co^(2+))^(@)=-0.028. If E_(Ce^(4+)//Ce^(3+))^(@)

Answer»

`-1.64V`
`+1.64V`
`-2.08V`
`+2.17V`

SOLUTION :In this cell Co is oxidised and it ACTS as anode and Ce acts as cathode.
`E_(Cell)^(@)=E_("cathode")^(@)-E_("anode")^(@)=1.89=E_(Cell)^(@)-(-0.28)`
`E_(Cell)^(@)=-1.89-0.28=1.61`VOLT.
Discussion
29.

For the cell Pt|H_2(0,4atm)|H^+(pH=1)||H^+(pH=2)|H_2(0.1atm)|Pt The measured potential at 25^(@)C is

Answer»

(-0.1)V
-0.5
(-0.041)
None

Answer :C
Discussion
30.

For the cell prepared from electrode A and B, electrode A:(Cr_2O_7^(2-))/(Cr^(3+)), E_("red")^(0)=+1.33V and electrode B :(Fe^(3+))/(Fe^(2+)), E_("red")^(0)=0.77 V, which of the following statement is not correct ?

Answer»

The electrons will flow B to A (in the OUTER circuit) when connections are made.
The standard e.m.f. of the CELL with be 0.56 V
A will be POSITIVE electrode
None of the above

Solution :A ACTS as cathode and B acts as anode because reduction potential of A is more.
Anode is B `implies` is NEGATIVE electrode.
`implies` electrons flow from B to A .
`implies` e.m.f=(1.33-0.77) V
Discussion
31.

For the cell prepared from electrode A and B , Electrode A : Cr_(2)O_(7)^(2-) | Cr^(3+) , E_("red")^(@) = 1.33 V and Electrode B : Fe^(3+) | Fe^(2+) , E_("red")^(@) = 0.77V . Which of the following statement is correct ?

Answer»

The electrons will flow from B to A when connection are MADE
The emf of the cell will be `-0.56`V
A will be negative electrode
B will be positive electrode

Solution :`E_(RP_(Cr_(2)O_(7)^(2-)|CR^(3+))^(@) = 1.33 V ` and `E_(RP_(Fe^(3+)|Fe^(2+))^(@) = 0.77 V E_(RP_(Fe^(3+) |Fe^(2+))^(@) ` is less thus will oxidise or electron will flow Fe(B) electrode to Cr(A) electrode . Also Fe electrons will be negative . Also
`E_(cell)^(@)= E_(RP_(Cr))^(@) - E_(RP_(Fe))^(@)`
`= 1.33 - 0.77 = + 0.56` V
Discussion
32.

For the cell prepared from electrodA , Cr_2O_7^(2-) |Cr^(3+) , E_("red")^(@) = + 1.33Vand electrode B: Fe^(3+) | Fe^(2+), E_("red")^(@) = 0.77V. Which of the following statement is correct ?

Answer»

The ELECTRONS will flow from B to A when the connection is made
The e.m.f. of the CELL will be 0.56V
A will be POSITIVE ELECTRODE
All of the above

Answer :D
Discussion
33.

For the cell Mg_((s))|Mg_((aq.))^(2+)||Ag_((aq.))^(+)|Ag_((s)), calculate the equilibrium constant at 25^(@)C and the maximum work that can be obtained during operation of cell. (Given, E_(Mg//Mg^(2+))^(@) = +2.37 V and E_(Ag^(+)//Ag)^(@) = +0.80V, R = 8.314 J)

Answer»


ANSWER :`107.457, 3.17 V, 6.118 XX 10^(2)KJ ;`
Discussion
34.

For the cell prepared from electrode A and B: Electrode A:Cr_2O_7^(2-)|Cr^(3+),E_(red)^@ =+1.33 V and Electrode B:Fe^(3+)/Fe^(2+),E_(red)^@=0.77 V.Which of the following statements are correct :

Answer»

THEELECTRONS will flow from B to A when CONNECTION are made
The emf of the cell will be 0.56V
A will be POSITIVE electrode.
All of these

ANSWER :D
Discussion
35.

For the cell: Mg(s)|Mg^(2+)(aq)||Ag^(+)(aq)|Ag(s), calcualte the equilibrium constant of the cell reaction at 25^(@)C and maximum work that can be obtained by operating the cell, E_(Mg^(2+)//Mg)^(@)=-2.37V and E_(Ag^(+)//Ag)^(@)=+0.80V(R=8.31" J "mol^(-1)K^(-1)).

Answer»


Answer :Max. work `DELTAG^(@)=611.81" KJ EQM. Const, K"=1.891xx10^(107)`
Discussion
36.

For the cell Mg(s) |Mg^(2+)(aq)||Ag^(+)(aq)|Ag (s), calculate the equilibrium constant at 25^@C and maximum work that can be obtained during operation of cell. Given : E_(Mg^(2+)|Mg)^@ = + 2.37 V and E_(Ag^(2+)|Ag)^@ = 0.80V

Answer»

Solution :`{:("Oxidation at anode", Mg to Mg^(2+)+2e^(-), (E_("ox")^(@)) = 2.37V),("Reduction at cathode ", Ag^+ + e^(-) to Ag, (E_("red")^@) = 0.80 V):}`
`:. E_("cell")^(@) = (E_("ox")^(@))_("anode") + (E_("red")^(@))_("cathode") = 2.37 + 0.80 = 3.17V`
Overall reaction,
`Mg + 2Ag^(+) to Mg^(2+) +2Ag`
`DeltaG^@ = -NFE^@`
`= -2 XX 96400 xx 3.17`
`= -6.118 xx 10^5J`
We know that `W_("MAX") = DeltaG^@`
`:. W_("max") = +6.118 xx 10^5 J`
Relationship between `DeltaG^@` and `K_(eq)` is
`Delta G = -2.303 RT log K_(eq)`
`DeltaG = -2.303 xx 8.314 xx 298 log K_(eq) "" [ :. 25^@C = 298 K]`
`log K_(eq) = (6.118 xx 10^5)/(2.303 xx 8.314 xx 298) IMPLIES (6.118 xx 10^5)/(5705.84)`
`log K_(eq) = 107.223`
`K_(eq) = "Antilog" (107.223)` .
Discussion
37.

For the cell Mg_((s))abs(Mg^(2+)""_((aq)))abs(Ag^(+)""_((aq)))Ag_((s)), calculate the equilibrium constant at 25^(@)C and maximum work that can be obtained during operation of cell. Given : E_(Mg^(2+)|Mg)^(@)=-237V " and " E_(Ag^(2+)|Ag)^(@)=0.80V

Answer»

Solution :oxidation at anode
`MG rarr mg^(2+)+2e^(-)"...(1)"`
`""(E_("ox")^(@))=2.37V`
Reduction at cathode
`Ag^(+)+e^(-) rarr Ag "...(2)"`
`""(E_("red")^(2))=0.80V`
`therefore E_("CELL")^(@)=(E_("ox")^(@))_("anode")+(E_("red")^(@))_("cathode")`
`""=2.37+0.80`
`""=3.17V`
Overall REACTION
Equation (1) + `2 times " equation (2)" rArr`
`Mg+2Ag^(2+) rarr Mg^(2+)+2Ag`
`""DELTAG^(@)=-nfE^(@)`
`""=-2 times 96500 times 3.17`
`""=611.810J`
`DeltaG^(@)=-6.12 times 10^(5)J`
`W=6.12 times 10^(5)J`
`DeltaG^(@)=-2.803" RT log " K_(c)`
`rArr logK_(c)=(6.12 times 10^(5))/(2.803 times 8.314 times 298)`
`""K_(c)=` Antilog of (107.2).
Discussion
38.

For the cell Mg|Mg^(2+)||Ag^(+)|Ag, calculate the equilibrium constant at 25^(@)C and the maximum work that can be obtained during the operation of the cell. (Given : E_(Mg//Mg^(2+))^(@)=2.37 volt and E_(Ag^(2+)//Ag)^(@)=0.80 volt

Answer»


ANSWER :`K=2.86xx10^(107); W_(max)=6.118xx10^(5) J`
Discussion
39.

For the cell , Hg,Hg_2 Cl_2//Cl (0.1M)//Cl^(-)(0.01M)//Cl_2, Pt E_("cell")^@ is 1.10V , Hence , E_("cell")is

Answer»

1.1591V
`-1.1591V`
1.0409 V
`-1.0409 V`

ANSWER :A
Discussion
40.

For the cell: Cu(10g)|underset((C_(1)))(CuSO_(4))"solution""||"underset((C_(2)))(ZnSO_(4))"solution""|"Zn(10g),E_("cell")=EV E_("cell") for the cell : Cu(20g)|underset((C_(1)))(CuSO_(4))"solution""||"underset((C_(2)))(ZnSO_(4))"solution""|"Zn(20g), is

Answer»

E VOLT
2E volt
`E/2` volt
1.1 volt

Answer :A
Discussion
41.

For the cell given below Zn|Zn^(2+)||Cu^(2+) | Cu, (E_("cell")-E_("cell")^(0))is -0.12 V . It will be when :

Answer»

`[ZN^(2+)]//[CU^(2+)]=10^(2)`
`[Zn^(2+)]//[Cu^(2+)]=10^(-2)`
`[Zn^(2+)]//[Cu^(2+)]=10^(4)`
`[Zn^(2+)]//[Cu^(2+)]=10^(-4)`

Solution :`E = E^(0)-(0.0591)/(2) log""([Zn^(+2)])/([Cu^(+2)]) , (E_("cell") - E_("cell")^(0)) =(-0.86)/(2) log""([Zn^(+2)])/([Cu^(+2)])-0.12 = (-0.06)/(2) log""([Zn^(+2)])/([Cu^(+2)])`
`log""([Zn^(+2)])/([Cu^(+2)]) = (0.12 XX 2)/(0.06) = (0.24)/(0.06) = 4, ([Zn^(+2)])/([Cu^(+2)]) = 10^(4)`
Discussion
42.

For the cell: A|A^(m+)||B^(n+)|B,E_("cell")=-1.1V

Answer»

right ELECTRODE is cathode
the CELL shall not OPERATE
left electrode is cathode
electrons FLOW from left to right in the external circuit

Answer :C
Discussion
43.

For the cell (at 298 K ) Ag_((g)) // AgCl_((s)) // Cl_((aq))^(-) ||AgNO_(3(aq)) // Ag_((s)) Which of the following is correct ?

Answer»

The EMF of the cell is zero when `[Ag^(+)]_("ANODIC " ) = (Ag^(+) )_("cathodic")`
The amount of `AgCl_((s))` precipitate in anodic compartment will decrease with the working of the cell
The concentration of `[Ag^(+)]` is CONSTANT in anodic compartment with the working of cell
`E_("cell") = E_(Ag^(+)//Ag)^(0)-E_(Cr^(-)//AgCl//Ag^(-))^(0) (0.059)/(1) LOG""(1)/([Cl^(-1)]) ` anodic

Solution :`underset("anode")(Ag//AgCl)//KCl_((aq)) // //AgnO_(3(aq)) //underset("cathode")(Ag_((s)) ) , E_("cell") = E_("cathode")^(0) -E_("anodic")^(0)`
`E_("cathode")^(0) = E_(Ag^(+)//Ag)^(0) + (0.059)/(1) log [Ag^(+)] , E_("cell") =E_(O.P)^(0)-(0.059)/(1) log[Ag^(+)] + E_(R.P)^(0) + (0.059)/(1) log""(1)/([Cl^(-)])`
E.M.F of cell is zero when `[Ag^(+)]_("cathode") = [Ag^(+)]_("anode")`
Discussion
44.

For the calomel electrode Hg,Hg_(2)|Cl^(-)(aq), electrode potential measured at different Cl^(-) ion concentration are plotted against log [Cl^(-)]. The variation is correctly represented by the plot.

Answer»




SOLUTION :The half cell reaction of the calomel electrode (written as reduction reaction) is
`Hg_(2)Cl_(2)(s)+2e^(-)to2Hg(l)+2Cl^(-)(AQ)`
`E=E^(@)-(0.0591)/(2)log[CL^(-)]^(2)`
or `E=E^(@)-0.0591log[Cl^(-)]`
THUS, plot of E vs log `[Cl^(-)]` is linear with a negative slope (i.e., E decreases LINEARLY with log `[Cl^(-)]`)
Discussion
45.

For the carbylamine reaction we need hot alc. KOH and

Answer»

any AMINE and CHLOROFORM
chloroform and AG powder
a primary amine and chloroform
a mono alkyl amine and trichloromethane

Solution :`1^(@)` Amine and chloroform.
Discussion
46.

For the carbylamine reaction we need hot alcoholic KOH and :

Answer»

Any AMINE and chloroform
Chloroform and SILVER powder
A PRIMARY amine and an ALKYL halide
Any monoalkyl amine and TRICHLORO methane

Answer :D
Discussion
47.

For the bimolecular gaseour reaction: 2R(g) to Products, the fraction of molecules having sufficient energy for effective collision is 2.06 xx 10^(-9) at 127^(@) C. The activation energy for reaction is about (ln 2.06 = 0.727, ln 10 = 2.303) .

Answer»

16 kJ/mole
16 kcal/mole
8 kcal/mole
80 kcal/moile

ANSWER :B
Discussion
48.

For the bleaching of hair, the substance used is :

Answer»

`SO_2`
BLEACHING POWDER
`H_2O_2`
`O_3`

ANSWER :C
Discussion
49.

For the allotropic change represented by the equation C (graphite) toC (diamond), DeltaH = 1.9 kJ. If 6 g of diamond and 6 g of graphite are separately burnt to yield CO_(2), the heat liberated in first case is

Answer»

LESS than in the SECOND CASE by 1.9 KJ
more than in the second case by 11.4 kJ
more than in the second case by 0.95 kJ
less than in the second case by 11.4 kJ

Answer :C
Discussion
50.

For the allotropic change represented by equation C(diamond) rarr C(graphite), the enthalpy change is DeltaH=-1.89 kJ. If 6g of diamond and 6g of graphite are separately burnt to yield carbon dioxide, the heat liberated in the first case is

Answer»

Less than in the second case by 1.89 kJ
More than in the second case by 1.89 kJ
Less than in the second case by 11.34 kJ
More than in the second case by 0.945 kJ

Solution :`C_(("GRAPHITE"))rarrC_(("diamond")), DeltaH=-1.89 kJ`
`C_(("graphite"))+O_(2)rarrCO_(2), DeltaH=-DeltaH_(1)`
`C_(("diamond"))+O_(2)rarrCO_(2), DeltaH=-DeltaH_(2)`
`(-DeltaH_(1))-(-DeltaH_(2))=1.89 kJ` for 12 G `C_("Diamond")rarrC_("graphite")`
for combustion of 6g, `C_("Diamond")rarrC_("graphite") =-1.89//2=-0.945 kJ`.
Discussion