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.

The central dogmaof molecular genetics statesthatthe geneticinformationflows from ………. .

Answer»

AMINO acids Protein DNA
DNA Carbohydrates Proteins
DNA RNA Proteins
DNARNA Carbohydrates

ANSWER :C
Discussion
2.

The central dogma of molecular genetics states that the genetic information flows from

Answer»


DNA, CARBOHYDRATES, PROTEINS
DNA, RNA, Proteins
DNA,RNA,Carbonhydrates

Solution :Amino ACIDS, Protein, DNA
Discussion
3.

The central carbon atom of a free radical contains

Answer»

6 electrons
7 electrons
8 electrons
10 electrons

Solution :The central carbon atom of a FREE radical contains seven electrons. Free radicals are paramagnetic DUE to the PRESENCE of an unpaired electron and are formed by HOMOLYSIS of covalent bonds either by heat or by light.
Discussion
4.

The central atom does not assume sp^(2) hybridisation in

Answer»

`PCl_(3)`
`SO_(3)`
`BF_(3)`
`NO_(3)^(-)`

Solution :In `BF_(3),NO_(3)^(-)` and `SO_(3)`, the central atom assumes `SP^(2)` hybridisation WHEREAS in `PCl_(3)` P-atom involves `sp^(3)` hybridisation.
Discussion
5.

The celll reaction for the given cell is spontaneous if: Pt|Cl_(2)(P_(1) atm)| Cl (1M) | Cl_(2)(P_(2) atm)|Pt

Answer»

`P_(1) GT P_(2)`
`P_(1) LT P_(2)`
`P_(1) = P_(2)`
`P_(2)=1` atm

Answer :B
Discussion
6.

The cells having membrane bound nucleus are called:

Answer»

Eukaryotic
Prokaryotic
Plant TISSUE cell
Animal tissue cell

Answer :A
Discussion
7.

The cell, Zn|Zn^(2+)(1M)||Cu^(2+)(1M)|Cu(E_(cell)^(o)=1.10V) was allowed to be completely discharged at 298K. The relative concentration of Zn^(2+) to Cu^(2+)[([Zn^(2+)])/([Cu^(2+)])] is

Answer»

Antilog (24.08)
37.3
`10^(37.3)`
`9.65xx10^(4)`

Solution :`E_(cell)=E_(cell)^(o)-(0.0591)/(N)log" "Q`
where, `Q=([ZN^(2+)])/([Cu^(2+)])`
For COMPLETE discharge `E_(cell)=0`
So, `E_(Cell)^(0)=(0.0591)/(2)"log"([Zn^(2+)])/([Cu^(2+)])`
`implies|([Zn^(2+)])/([Cu^(2+)])|=10^(37.3)`
Discussion
8.

The cell, Zn|Zn^(2+)(1M)||Cu^(2+)(1M)|Cu,(E_(cell)=1.10V), was allowed to be completely discharged at 298K. The realtive concentration of Zn^(2+) to Cu^(2+),((Zn^(2+)])/([Cu^(2+)])) is

Answer»

`9.65xx10^(4)`
antilog 24.08
`37.3`
`10^(37.3)`

Solution :The cell reaction is: `Zn+CU^(2+)toZn^(2+)+Cu` when the cell is COMPLETELY discharged, `E_(cell)^(@)=(0.0591)/(2)"LOG"([Zn^(2+)])/([Cu^(2+)])`
i.E., `1.10=(0.591)/(2)"log"([Zn^(2+)])/([Cu^(2+)])`
or `([Zn^(2+)])/([Cu^(2+)])=37.3` or `([Zn^(2+)])/([Cu^(2+)])=10^(37.3)`
Discussion
9.

The cell Zn | Zn^(2+)(1 M)||Cu^(2+)(1 M)|Cu E_("cell")^(@)=1.10 Vwas allowed to be copletely discharged at 298 K The relative concentration of (Zn^(2+))/(Cu^(2+)) is

Answer»

antilog (24.08)
37.2
`10^(37.2)`
`9.65xx10^(4)`

Solution :Celliscompletelydischarged it meanswhenequilibriumgetsestablished `E_(cell) =0`
`ZN|Zn^(2+) (1M)||CU^(2+)(1 M)|Cu`
`k_(eq)=[(Zn^(2+))]/[(Cu^(2+))]`
`log k_("eq") =(E_("cell")^(@))/(0.591)=(2xxx1.1)/(0.591)=37.2`
`k_("eq") =10^(37.2)`
Discussion
10.

The cell , Zn//Zn^(2+) (1M) //Cu^(2+) (1M) // Cu (E_("Cell")^@ = 1.10V),was allowed to be completely discharged at 298 K. The relative concentration of Zn^(2+) to Cu^(2+) is

Answer»

37.3
`10^(37.3)`
`9.65 XX 10^4`
ANTILOG (24.04)

ANSWER :B
Discussion
11.

The cell Zn//Zn^(2+) (1 M) ||Cu^(2+)(1M)//Cu(E_("cell")^(0) = 1.10 V)was allowed to be completely discharged at 298 K. The relative concentration of Zn^(2+) to Cu^(2+) (([Zn^(2+)])/([Cu^(2+)])) is 10^(x). The value of x is : (take (2.303 RT)/(F) = 0.059 round off your answer up to one decimal)

Answer»


ANSWER :37
Discussion
12.

The cell voltage depends on _______.

Answer»

NATURE of the electrodes
concentration of the electrolytes
temperature
all of the above

Answer :D
Discussion
13.

The cell voltage of a galvanic cell becomes zero after using for some tme because

Answer»

All the FREE ELECTRONS are used up
Oxidationpotential of the two half cell become different
REDUCTION potential of the two half cells become equal
Oxidation potentials of two half cells become equal in magnitude but opposite in sign.

Solution :Since `E_("red")prop[M^(n+)]`
`THEREFORE` when current is DRAWN the conc. Of eec trolytic solution changes and hence the reduction potential.
Discussion
14.

The cell reaction of the galvanic cell Cu_((s))|Cu_((aq))^(2+)||Hg_((aq))^(2+)|Hg_((l)) is

Answer»

`Hg+CU^(2+)toHg^(2+)+Cu`
`Hg+Cu^(2+)TOCU^(2+)+Hg^(+)`
`Cu+HgtoCuHg`
`Cu+Hg^(2+)toCu^(2+)+Hg`

Solution :`underset("anode OXIDATION")(Cu_((s))|Cu_((AG))^(2+))||underset("cathode reduction")(Hg_((Ag))^(2+)|Hg_((l)))`
Discussion
15.

The cell reaction Zn(s) + Cu^(+2)rarr Zn^(+2) + Cu(s) is best represented as :

Answer»

`Cuabs(Cu^(+2))ABS(ZN^(+2))Zn`
`Znabs(Zn^(+2))abs(Cu^(+2))Cu`
`Cu^(+2)abs(Cu)abs(Zn)Zn^(+2)`
`Ptabs(Zn^(+2))abs(PT)Cu^(+2)`

Answer :B
Discussion
16.

The cell that was used as the primary source of electrical energy on the appolo moon flights is

Answer»

VOLTAIC cell
Mercury cell
Hydrogen-oxygen fuel cell
Lead STORAGE battery

Solution :It ALSO to provide pure water.
Discussion
17.

The cell reaction ,Zn + Cu^(2+)rarrZn^(2+) +Cu is best represented by :

Answer»

`CU//Cu^(2+)||ZN^(2+)//Zn`
`Zn//Zn^(2+)||Cu^(2+)//Cu`
`Cu^(2+)//Cu||Zn//Zn^(2+)`
`PI//Zn^(2+)||PI//Cu^(2+)`

ANSWER :B
Discussion
18.

The cell reaction of the galvanic cell Cu(s)|Cu^(2+)(aq)||Hg^(2+)(aq)|Hg(l) is

Answer»

`Hg+CU^(2+)rarrHg^(2+)+cu`
`Hg+Cu^(2+)rarrCu^(+)+Hg^(+)`
`Cu+HgrarrCuG`
`Cu+Hg^(2+)rarrCu^(2+)Hg`

SOLUTION :Galvaniccell is `Cu(s) |Cu^(2+) (AQ)||Hg^(2+) (aq) |Hg`
in theabove cell oxidatin of COPPER and redcution of mercury takes place its cell reactionis writtenas
`Cu(s) +Hg^(2+) (aq) rarr Cu^(2+) (aq) +Hg`
Discussion
19.

The cell reaction of a cell is Mg+Cu^(2+) to Mg^(2+)+Cu (Given E_(Mg^(2+)//Mg) = -2.37 V , E_(Cu^(2+)//Cu) = 0.337 V) The e.m.f of the cell will be _________.

Answer»

`-2.7 V`
2.7 V
`-2.03 V`
`+2.03 V`

ANSWER :B
Discussion
20.

The cell reaction of the galvanic cell, Cu(s) // Cu^(2+)(aq) // Hg^(2+) (aq) // Hg (l)is

Answer»

`HG + CU^(2+) to Hg^(2+) + Cu`
`Hg + Cu^(2+) to Hg^(+)+ Cu^(+)`
`Hg + Cu^(+) to CuHg`
`Cu + Hg^(2+) to Cu^(2+) + Hg `

ANSWER :D
Discussion
21.

The cell reaction of the cell fuel cell used in the space program is

Answer»

`C(s)+O_(2)(g) rarr CO_(2)(g)`
`CH_(4)(g)+O_(2)(g) rarr CO_(2)+H_(2)(l)`
`2H_(2)(g)+O_(2)(g) rarr 2H_(2)O(l)`
`2H_(2)O(l) rarr 2H_(2)(g)+O_(2)(g)`

Solution :The hydrogen-oxygen fuel cell, which has many applications, is USED in spacecraft to supplement the energy obtained from solar cells.
Liquid `H_(2)` is carried on board a s a propellant. The bolied-off `H_(2)` vapor that ordinarily WOULD be lost is used in a fuel cellto generate electrical power.
Hydrogen (the fuel is supplied to the anode compartment. Oxygen (the oxidizer) is fed into the cathode compartment. The diffusion rates of the gases into the cell are carefully regulated for maximum efficiency. Oxygen is reduced at the cathode, which consists of porous carbon impregnated with a finely divided `Pt` o `Pd` CATALYST.
Cathode
`O_(2)(g)+2H_(2)O(l)+4e^(-)overset("Catalyzing") rarr 4OH^(-)(aq.)`
The `OH^(-)` ions migrate through the electrolyte (an aqueous solution of a base) to the anode. The anode is also porous cabon containing a small amount of catalyst (`Pt, Ag`, or `CoO`). Here `H_(2)` is OXIDIZED to `H_(2)O`.
Anode
`H_(2)(g)+2OH^(-)(aq.) rarr 2H_(2)O(l)+2e^(-)`
The net reaction obtained from the two half-reactions is `2H_(2)(g)+O_(2)(g) rarr 2H_(2)O(l)`
The pure water produced by the cell is drunk by the astronauts.
Discussion
22.

The cell reaction for the given cell is spontaneous if :Pt(H_2)|H^=(1M)||H^+(1M)|Pt(H_2)

Answer»

`P_1 GT P_2`
`P_1 LT P_2`
`P_1=P_2`
`P_1=1 ATM`

ANSWER :A
Discussion
23.

The cell reaction for the given cell is spontaneous if :Pt | Cl_2|CI^(-) (1M)||CI^(-) (1M)|Cl_2| Pt

Answer»

`P_1 GT P_2`
`P_1 LT P_2`
`P_1=P_2`
`P_2=1` atm

Answer :B
Discussion
24.

The cellreactionfor agalvantcell isH_(2(g)) +C1_(2(g))to 2HC1_((aq)) Constructthe cell andalsocalculatestandardemf of cell E_(C1)^(@) = 1.36 V

Answer»

Solution :Cellfor the GIVENREACTION
`Pt, H_(2(g))(1 ATM) | underset(1 M) (HC1_((aq))) | C1_(2(g))( 1 atm),Pt , E_(" cell ")^(@)= 1.36 V`
Discussion
25.

The cell reaction for the given cell is :Pt(H_2)|pH=2||pH=3|Pt(H_2)

Answer»

SPONTANEOUS
Non-spontaneous
In EQUILIBRIUM
EITHER of these

Answer :B
Discussion
26.

The cell, Pt|H_(2)" (1 atm)"||H^(+) (pH=x)| normal calomel electrode, has an emf of 0.67 volt as 25^(@)C. Calculate the pH of the solution. The oxidation potential of calomel electrode on the hydrogen scale is -0.28 volt.

Answer»


ANSWER :PH = 6.61
Discussion
27.

The cell reaction Cu+2Ag^(+)toCu^(+2)+Ag is best represented by

Answer»

`Cu_((s))|Cu_((AQ))^(+2)||Ag_((aq))^(+)|AG`
`Pr|CU^(+2)||Ag_((aq))^(+)|Ag_((s))`
`Cu^(+2)|Cu||Pt|Ag`
NONE of the above representation

Answer :A
Discussion
28.

The cell Pt, H_(2) (1atm) H^(+) (pH =x)| Normal calomel Electrode has an EMF of 0.67V at 25^(@)C .Calculate the pH of the solution. The oxidation potential of the calomel electrode on hydrogen scale is -0.28V.

Answer»


Solution :`E_(CELL) = E_(cell)^(@) -(0.0591)/(1)log 10^(-x)`
`0.67 =0.28 +(0.0591)/(1) XX x`
`x = 6.61`
Discussion
29.

The cell potential for the unbalanced chemical reaction: Hg_(2)^(2+)+NO_(3)^(-)+H_(3)O^(+) to Hg^(2+) +HNO_(2)+H_(2)O+e^(-) Is measured under standard conditions in the electrochemical sell shown in the diagram were Choose the correct statement for the given cell diagram

Answer»

Compartment (X) has less pH than compartment (Y)
(Y) compartment has ACIDIC solution
Current will FLOW from `X to Y` through internal supply
`TRIANGLEG^(@)` for the above cell reaction is more than 1 at the equilibrium

Answer :B
Discussion
30.

The cell potential of the galvanic cell Zn|Zn^(++)|Ag^(+)|Ag , E^(@)(Zn^(++)|Zn)=-0.76 V, E^(@)(Ag^(+)|Ag)=+0.80 V

Answer»

`+ve`
`-ve`
zero
can't say

Solution :`E_(cell)^(0)=UNDERSET(Ag)E_("CATHODE")^(0)-underset(ZN)E_("ANODE")^(0)=1.56 V`
Discussion
31.

The cell potential for the unbalanced chemical reaction: Hg_(2)^(2+)+NO_(3)^(-)+H_(3)O^(+) to Hg^(2+) +HNO_(2)+H_(2)O+e^(-) Is measured under standard conditions in the electrochemical sell shown in the diagram were Equilibrium constant for cell reaction is :

Answer»

`10^(3//2)`
`e^(3/2)`
`e^(-2/3)`
`10^(2/3)`

SOLUTION :`E^(@)-(0.0591)/(n) log K_(c)`
`log K_(c)=(2 XX 0.02)/(0.06) =2/3`
`=10^(2/3)`
Discussion
32.

The cell potential of the galvanic cell, Zn | Zn^(2+) ||Ag^(+) | Ag, whereE_(Zn^(2-)pin)^(@)= -0.76V,E_(Ag'//Ag)^(@) =+0.8V is

Answer»

A) `-1.56 V`
B) `-0.04 V`
C)`+ 1.56 V`
D)`0.004` V

Answer :C
Discussion
33.

The cell membranes are mainly composed of :

Answer»

FATS
PROTEINS
Phospholipids
Carbohydrates.

Answer :C
Discussion
34.

The cell potential (E_("cell")) of a reaction is related as DeltaG = -nFE_("cell") , where DeltaG represents max. useful electrical work . n=no. of molesof electrons exchanged during the reactino of reversiblecell reaction d(DeltaG) = (DeltaV)dP-(DeltaS).dT at constant pressured(DeltaG) = - (DeltaS).dT :' At constant pressure DeltaG = DeltaG = DeltaH-TDeltaS.......(1) :. DeltaG = DeltaH + T((d(DeltaG))/(dT))_P ......... (B) ((dE_("cell"))/(dT))_(P) is known as temperature coefficient of the e.m.f of the cell When DeltaS increases, temperature coefficient of the emf of cell

Answer»

Constant
Decreases
SUDDENLY decreases
Increases

Solution :`DeltaG = DeltaH -T. DeltaS , DELTA G = DeltaH + T ((d(DeltaG))/(DeltaT))_(P)`
` IMPLIES -DeltaS = ((d(DeltaG))/(dT))_P - DeltaS = ((d(-nFE_("cell")^(0)))/(dt))_(P)- DeltaS = -NF((dE_("cell"))/(dt))_(P)`
`((dE_("cell"))/(dt))_P = (DeltaS)/(nF) implies If DeltaS uarr ` then `((dE_("cell"))/(dt))_(P)uarr`
Discussion
35.

The cell potential E_("cell") is ______

Answer»

`E_("cell")^(@)=E_("CATHODE")^(@)-E_("anode")^(@)`
`E_("cell")^(@)=E_("anode")^(@)-E_("cathode")^(@)`
`E_("cell")^(@)=E_("cell")^(@)-0.0591` LOG K
`E_("cell")-0.0591/n`log K

Answer :D
Discussion
36.

The cell membranes are mainlycomposed of ………. .

Answer»

carbohydrates
proteins
phospholipids
fats

ANSWER :C
Discussion
37.

The cell membranes are following statements is not true

Answer»

Buna-S is a copolymer of BUTADIENE and styrene
Natural rubber is a 1,4-polymer of ISOPRENE
In vulcanization, the FORMATION of sulphur bridges between different chains MAKE rubber harder and stronger
Natural rubber has the TRANS configuration at every double bond

Solution :Natural rubber has cis-configuration at each double bond. Gutta-Percha has trans-configuration at each double bond.
Discussion
38.

The cell in which the following reactions occurs : 2Fe_((aq))^(3+)+2l_((aq))^(-) to 2Fe_((aq))^(2+)+I_(2(S)) has E_(cell)^(Theta)=0.236V at 298 K. calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Answer»

Solution :(a) Cell reaction and calculation of `Delta_(R)G^(Theta)`:

Where, n=2 MOL,
`F=96487" C "mol^(-1)`,
`E_(cell)^(Theta)=0.236V`
`THEREFORE Delta_(r)G^(Theta)=-nFE_(cell)^(Theta)`
`=-(2mol)(96487" C "mol^(-1))(0.236V)`
`=-45542(V)`
`=-45542J`
=-45.542 kJ `~~-45.54kJ`
(b) Calculation of equilibrium constant `K_(c)`:
`therefore Delta_(r)G^(Theta)=-2.303RT" log "K_(c)`
`therefore -45542=-2.303xx8.314xx298xxlog" "K_(c)`
`therefore log" "K_(c)=(-45542)/(-2.303xx8.314xx298)`
`therefore log" "K_(c)=+7.9816`
`thereforeK_(c)="Antilog"(7.9816)`
`=9.586xx10^(7)`
`~~9.6xx10^(7)`
Discussion
39.

The cell in which the followingreactionoccurs : 2Fe_(aq)^(3+)+ 2I_(aq)^(-) to2Fe_(aq)^(2+) + I_(2(s)) "has"E_("cell")^(o) = 0.236 V "at" 298 K The equilibriumconstnat of the cell reaction is

Answer»

`6.69xx 10^(-7)`
`9.69 XX 10^(-7)`
`9.69 xx 10^(7)`
`6.69 xx 10^(7)`

ANSWER :C
Discussion
40.

The cell in which the following reaction occurs: 2Fe^(3+)(aq)+2I^(-)(aq)to2Fe^(2+)(aq)+I_(2)(s) has E_(cell)^(@)=0.236V at 298K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Answer»

Solution :`2Fe^(3+)+2e^(-)to2Fe^(2+)` or `2I^(-)toI_(2)+2e^(-)`
`Delta_(r)G^(@)=nFE_(cell)^(@)=-2xx96500xx0.236J=-45.55kJ" "MOL^(-1)`
`Delta_(r)G^(@)=-2.303RT" LOG "K_(c)` or `"log "K_(c)=(Delta_(r)G^(@))/(2.303RT)=(-45.55" kJ "mol^(-1))/(2.303xx8.314xx10^(-3)" kJ "K^(-1)xx298K)=7.983`
`thereforeK_(c)=`ANTILOG `(7.983)=9.616xx10^7`
Discussion
41.

The cell in which the following reaction occurs 2Fe^(3+)(aq)+2I^(-)(aq)to2Fe^(2+)(aq)+I_(2)(s) has E_(cell)^(@)=0.236V at 298K. Calculate the standard gibbs energy of the cell reaction. (Given: 1F=96,500" C "mol^(-1))

Answer»


SOLUTION :For the GIVEN CELL reaction, `n=2,DeltaG^(@)=-nFE_(cell)^(@)`.
Discussion
42.

The cell in which the following reaction occurs 2Fe^(3+)(aq)+2I^(-)(aq)to2Fe^(2+)(aq)++I_(2)(s) has E_(cell)^(0)=0.236V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Answer»

Solution :`n=2`
`DeltaG^(@)=-nFE_("CELL")^(@)=-2xx96500xx0.236J=-45.55kJ//mol`
`DeltaG^(@)=-2.303"RT LOG K"_(c)`
`logK_(c)=(DeltaG^(@))/(-2.303RT)=(45.55xx10^(3))/(2.303xx8.314xx298)=7.983`
`K_(c)="antilog (7.982)"=9.616xx10^(7)`
Discussion
43.

The cell in which electrical energy is used to bring about chemical change is known as ________

Answer»

ELECTROLYTIC cell
galvanic cell
voltaic cell
dynamo

Answer :A
Discussion
44.

The cell in which of the following reaction occurs: 2Fe^(3+)(aq)+2I^(-)(aq) to 2Fe^(2+)(aq)+I_(2)(s)" has "E_("cell")^(0)=0.236V" at "298K Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Answer»

Solution :Data: `E_("cell")^(0)=0.236V, T=298K, N=2, F=96500, R=8.314J//K MOL`
FORMULA 1 . `DeltaG^(0)=-n E_("cell")^(0)F`
`=-2 xx 0.236 xx 96500 =-45.55kJ`
Formula 2. `DeltaG^(0)=-2.303 RRT log K_(C)`
`-45.55=-2.303 xx 8.314 xx 298 xx log K_(C)`
`log k_(c)=(4550)/(2.303 xx 8.314 xx 298)`
`log k_(c)=7.983`
`k_(C)="Antilog "7.983=9.62 xx 10^(7)`
Discussion
45.

The cell for which the cell reaction is H_2 + Cu^(2+) to 2H^(+) + Cu is represented as

Answer»

`Cu//Cu^(2+)//H^(+)//H_(2)`
`H_(2(g))//H^(+) "//"Cu^(2+)//Cu`
`Pt, H_(2)(1 ATM), H^(+)"//"Cu^(2+)//Cu`
`Pt, H_(2)//H_((AQ))^(+) (1atm)"//"Cu^(2+) //Cu`
Discussion
46.

The cell EMF is independent of the concentration of the species of the cell in

Answer»

`Feoverset(c-)(|)FEO(s)|KOH(aq)|NiO(s)|Ni_(2)O_(3)(s) overset(o+)(|)Ni`
`Pt|(H_(2))|HCl|(Cl_(2))Pt`
`Zn|Zn^(2+)||Cu^(2+)|Cu`
`Hg,Hg_(2)Cl_(2)|KCl||AgNO_(3)|AG`

Solution :Cell reaction is `:`

HENCE, `EMF` is independent of `overset(c-)(O)H` ions.
Discussion
47.

The cell emf depends on

Answer»

Temperature
SIZE of cell
[Cathodic electrolyte ]
[Anodic electrolyte ]

SOLUTION :Cell emf DEPENDS on Temperature , Cathodic electrolyte , Anodic electrolyte
Discussion
48.

The cell emf in a cell with cell reaction Cr_2O_7^(2-) + 14H^(+) + 6Fe^(+2) to 2Cr^(+3) + 6Fe^(+3) + 7H_2O could be increased above the standard emf by

Answer»

Increasing `[Cr^(+3)]`
Decreasing `[FE^(+3)]`
Decreasing `[Cr_2O_7^(2-)]`
Decreasing the pH

SOLUTION :`E = E^(0) + (0.0591)/(n) LOG""([Cr_2O_7^(-2)][H]^(14) + [Fe^(+2)]^(6))/([Cr^(+3)][Fe^(+3)]^6)`
Discussion
49.

The cell emf is independent of concentration of the species of the cell in:

Answer»

`Fe|FeO(s)|KOH(aq)|NiO(s)|Ni_(2)O_(3)|Ni`
`Pt|(H_(2))|HCl|Pt(Cl_(2))`
`Zn|Zn^(2+)||CU^(2+)|Cu`
`HG,Hg_(2)Cl_(2)|KCl||AgNO_(3)|AG`

Answer :A
Discussion
50.

The cell edge of a fcc crystal is 100 pm and its density is 10.0 g cm^(-3) . The number of atoms in 100 g of this crystal is:

Answer»

`1 xx 10^(25)`
`2 xx 10^(25)`
` 3 xx 10^(25)`
`4 xx 10^(25)`

Solution :Volume of units cell `= ( 100 xx 10^(-10) cm)^(3)`
`= 10^(-24) cm^(3)`
Density `= 10.0 G cm^(-3)`
Mass of unit cell `= 10^(-24) xx 10`
`= 10 ^(-23) g `
No. of unit cells in 100 g `= ( 100)/( 10^(-23))= 10^(25)`
Since the lattice is fcc and each unit cell has 4 atoms PER unit cell.
No. of atoms `= 4 xx 10^(25)`
Discussion