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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 601. |
Assertion `(A):` For a Daniell cell `:` `Zn|Zn^(2+)||Cu^(2+)|Cu` with `E_(cell)=1.1V`, the application of opposite potential greater than `1.1V` results into the flow of electron from cathod to anode.A. If both `(A)` and `(R)` are correct, and `(R)` is the correct explanation of `(A)`.B. If both `(A)` and `(R)` are correct, but `(R)` is not the correct explanation of `(A)`.C. If `(A)` is correct, but `(R)` is incorrect.D. If `(A)` is incorrect, `(R)` is correct. |
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Answer» Correct Answer - b Both `(A)` and `(R)` are correct but `(R)` is not the correct explanation of `(A)`. Refer Figure. Correct explanation `:` On applying external voltage greater than `1.1V` in a Daniell cell, the current flows in the reverse direction `i.e.,` from `Zn` to `Cu(` cathode to anode `)` and electrons flow from `Cu` to `Zn` . `Zn` is deposited at `Zn` electrode and `Cu` dissolves at `Cu` electrode. The reaction is `:` `Zn^(2+)+Cu rarr Zn+Cu^(2+)` |
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| 602. |
Assertion `(A):` For a Daniell cell `:` `Zn|Zn^(2+)||Cu^(2+)|Cu` with `E_(cell)=1.1V`, the application of opposite potential greater than `1.1V` results into the flow of electron from cathod to anode.Reason `(R):` Zn is deposited at anode and Cu is dissolved at cathodeA. If both assertion and reason are correct and reason is correct explanation for assertion.B. If both assertion and reason are correct but reason is not correct explanation for assertion.C. If assertion is correct but reason is incorrect.D. If assertion and reason both are incorrect. |
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Answer» Correct Answer - B (b) Correct explanation. The cell will work in opposite direction. Cu will be oxidised to `Cu^(2+)` ions. The electrons released will move towards anode and will be taken by `Zn^(2+)` ions to form Zn(s). It will be deposited at the anode. |
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| 603. |
The resistance of 0.5 M solution of and electrolyte enclosed enclosed between two platinuim electrodes 1.56 cm apart and having an are of `2.0cm^(3)` was found to be 30 cm. Calculate the molar conductivity of the electrolyte/. |
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Answer» Correct Answer - `0.05 ohm^(-1)cm^(2)mol^(-1)` Step I. Calculation of specific conductance. Specific conductance (k) `=(1)/(R )xx"cell constant" =(1)/((30 ohm))xx((1.5 cm))/((2.0 cm^(2)))=0.025 ohm^(-1)cm^(-1)` Step II. Calculation of molar conductance. Molar conductance `(Lambda_(m))=(k)/(C )=((0.025" ohm"^(-1)"cm"^(-1)))/((0.5" mol cm"^(-3)))=0.05" mol"^(-1)ohm^(-1)cm^(2)` |
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| 604. |
The resistacnce of a conductivity cell filled with 0.01 N KCI at `25^(@)` was found to be 500 `Omega` The specific conductance of 0.01 N KCI at `25^(@)` is `1.41xx10^(-3) Omega^(-) cm^(-1)` The resistance of same cell filled with `0.3 N ZnSO_(4)` at `25^(@)C` was found to be 69 `Omega` Calculate the cell constant equivalent and molar conductivityes of `ZnSO_(4)` solution |
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Answer» Cell constant =0.705 `cm^(-1)` Equivalent conductivity `=34.1 Omega^(-1)cm^(-1)eq^(-1)` Molar conductivity `=68.2 Omega^(-1) cm^(-1) mol^(-1)` |
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| 605. |
When a certain conductance cell was finlled with 0.1 mol `L^(-1)` KCI solution, it had a resistance of 85 ohm at 298K. When the same cell was filled with an aqueous solution of 0.052 mol `L^(-1)` of an electrolyte, the resistance was 96 ohm. Calcualte the molar conductance of the electrolyte at this concentration. (Specific conductance of 0.1 mol `L^(-1)` KCI solution is 1.29xx10^(-2)ohm^(-1)cm^(-1))`. |
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Answer» Correct Answer - `220.2 S cm^(2)mol^(-1)` Step I. Calcualtion of cell constant. Resistance of KCl solution (R )=85 ohm Specific conductance, (k)`=1.29xx10^(-2) ohm^(-1) cm^(-1)` Specific conductance, (k)`=(1)/(R )xx"cell constant"` `:. " " "Cell constant" =kxxR=(1.29xx10^(-2) ohm^(-1)cm^(-1))xx85" ohm" =1.1 cm^(-1)` Step II. Calculation of molar conductance. Resistance of given electrolyte solution, (R )=96 ohm Specific conductance,(k)`=("Cell constant")/(R )=((1.1 "cm"^(-1)))/((96" ohm"))=1.145xx10^(-2)ohm^(-1)cm^(-1)` Concentration, C`=0.052 mol L^(-1)=(0.052" mol")/(1L)=(0.052" mol")/(1000" cm"^(3))=5.2xx10^(-5)" mol cm"^(3)` Molar conductance, `(Lambda_(m))=(k)/(C )=((1.145xx10^(-2) ohm^(-1) cm^(-1)))/((5.2xx10^(-5) mol cm^(-3)))` `=220.2 ohm^(-1) cm^(-1)=220.2 cm^(2) mol^(-1)`. |
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| 606. |
A conductance cell when filled with `0.5 M KCI` solution (conductivity `= 6.67 xx 10^(-3) Omega^(-1) cm^(-1)`) register a resistance of `243 Omega`. Its cell constant is .A. ` 1.62 cm`B. ` 1.62 cm^(-1)`C. ` 1.62 dm^(-1)`D. ` 1.62 m^(-1)` |
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Answer» Correct Answer - B ` K=kR=(6.67 xx 10^(-3) Omega^(-1) cm^(-1)) (243Omega) = 1.62 cm^(-1)`. |
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| 607. |
(a) Calculate the standard free energy change and maximum work obtainable for the reaction. `Zn(s)+Cu^(2+)(aq)hArrCu(s)+Zn^(2+)(aq)` [Given `E_(Zn^(2+)//Zn)^(@)=-0.76V,E_(Cu^(2+)//Cu)^(@)=+0.34V,F=96500" C "mol^(-1)`] (b) also calculate the equilibrium constant for the reaction. |
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Answer» (a) The cell may be represented as `Zn(s)|Zn^(2+)(1M)||Cu^(2+)(1M)|Cu(s)` `E_(cell)^(@)=E_(RHS)^(@)-E_(LHS)^(@)=0.34-(-0.76)=1.10`volt For the given cell reaction, n=2 `DeltaG^(@)=-nFE_(cell)^(@)=-2xx96500" C "mol^(-1)xx1.10V=-212,300CV" "mol^(-1)` Thus, the maximum work that can be obtained from the Daniell cell=212.3 kJ. (b) `DeltaG^(@)=-RT" ln "K_(c)=-2.303RT" log "K_(c)` `therefore-212300=-2.303xx8.314xx298xxlog" "K_(c)" or "logK_(c)=(212300)/(2.303xx8.314xx298)=37.2074` `thereforeK_(c)="Antilog "37.2074=1.6xx10^(37)` |
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| 608. |
`Zn+Cu^(2+)(aq)hArrCu+Zn^(2+)(aq).` Reaction quotient is `Q=([Zn^(2+)])/([Cu^(2+)])` . Variation of `E_(cell)` with log `Q` is of the type with `OA=1.10` `V.E_(cell) ` will be `1.1591V` when A. `[Cu^(2+)]//[Zn^(2+)]=0.01`B. `[Zn^(2+)]//[Cu^(2+)]=0.01`C. `[Zn^(2+)]//[Cu^(2+)]=0.1`D. `[Zn^(2+)]//[Cu^(2+)]=1` |
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Answer» Correct Answer - b `E_(cell)=E^(c-)._(cell)-(0.0591)/(2)log.([Zn^(2+)])/([Cu^(2+)])` From line `OA=E^(c-)._(cell)=1.10V` If `([Zn^(2+)])/([Cu^(2+)])=10^(-2)M,` then `E_(cell)=1.1591V` |
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| 609. |
Tollen reagent is used for the detection of aldehydes. When a solution of `AgNO_(3)` is added to glucose with `NH_(4)OH`, then gluconic acid is formed. `Ag^(o+)+e^(-) rarr Ag," "E^(c-)._(red)=0.8V` `C_(6)H_(12)O_(6)rarr underset(Gluconic aci d)(C_(6)H_(12)O_(7)+)2H^(o+)+2e^(-) , " "E^(c-)._(o x i d ) =-0.05V` `[Ag(NH_(3))_(2)]^(o+)+e^(-) rarr Ag(s)+2NH_(3), " "E^(c-)._(red)=0.337V` `[Use2.303xx(RT)/(F)=0.0592` and `(F)/(RT)=38.92at 298 K ]` When ammonia is added to the solution, `pH` is raised to `11`. Which half cell reaction is affected by `pH` and by how much ?A. `E_("oxi")` will increase by a factor of 0.65 from `E_("oxi")^(@)`B. `E_("oxi")` will decrease by a factor of 0.65 from `E_("cell")^(@)`C. `E_("red")` will increase by a factor of 0.65 from `E_("red")`D. `E_("red")` will decrease by a factor of 0.65 from `E_("red")` |
| Answer» Correct Answer - c | |
| 610. |
Find the equilibrium constant for the reaction `Cu^(2+)+In^(2+)hArrCu^(+)+In^(3+)` Given that `E_("cu^(2+)//Cu^(+))^(@)=0.15V`,? `E_(In^(2+)//In^(+))^(@)=-0.4V`, `E_(In^(3+)//In^(+))^(@)=-0.42V` |
| Answer» Correct Answer - `10^(10)` | |
| 611. |
Tollen reagent is used for the detection of aldehydes. When a solution of `AgNO_(3)` is added to glucose with `NH_(4)OH`, then gluconic acid is formed. `Ag^(o+)+e^(-) rarr Ag," "E^(c-)._(red)=0.8V` `C_(6)H_(12)O_(6)rarr underset(Gluconic aci d)(C_(6)H_(12)O_(7)+)2H^(o+)+2e^(-) , " "E^(c-)._(o x i d ) =-0.05V` `[Ag(NH_(3))_(2)]^(o+)+e^(-) rarr Ag(s)+2NH_(3), " "E^(c-)._(red)=0.337V` `[Use2.303xx(RT)/(F)=0.0592` and `(F)/(RT)=38.92at 298 K ]` Ammonia is always added in this reaction. Which of the followijng must be wrong ?A. `NH_(3)` is a stronger with `Ag^(+)` to form a complexB. `Ag(NH_(3))_(2)^(+)` is a stronger oxidising reagent than `Ag^(+)`C. In the absence of `NH_(3)` silver salt of gluconic acid is formed.D. `NH_(3)` has affected the standard reduction potential of glucose glueconic acid electrode |
| Answer» Correct Answer - d | |
| 612. |
Calculate the equivalent conductivity of 1 M `H_(2)SO_(4)` solution, if its conductivity is `26xx10^(-2)ohm^(-1)cm^(-1)` (Atomic weight of sulphur=32). |
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Answer» Correct Answer - 130 `ohm^(-1)cm^(2)eq^(-1)` 1 M `H_(2)SO_(4)=2N" "H_(2)SO_(4)`. |
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| 613. |
We have taken a saturated solution of `AgBr`, whose `K_(sp)` is `12xx10^(-14).` If `10^(-7)M` of `AgNO_(3)` are added to `1L` of this solutino, find the conductivity `(` specific conductance `)` of the solution in terms of `10^(-7)S m^(-1)` units. Given `:` `lambda^(@)._((Ag^(o+)))=6xx10^(-3)S m^(2) mol^(-1)` `lambda^(@)._((Br^(c-)))=8xx10^(-3)S m^(2)mol^(-1)` `lambda^(@)._((NO_(3)^(C-)))=7XX10^(-3)S m^(2) mol^(-1)` |
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Answer» Solubility of `AgBr` in presence of `10^(-07)` mole of `AgNO_(3)` solution `K_(SP_(AgBr)) = [Ag^(+)][Br^(-)]` `12 xx 10^(-14) = [S + 10^(-7)] xx S` `:. S = 3 xx 10^(-7)M` `:. [Ag^(+)] = [3 xx 10^(-7) + 10^(-7)] = 4 xx 10^(-7) M` `[Br^(-)] = 3 xx 10^(-7) M` `[NO_(3)^(-)] = 3 xx 10^(-7)M` `because kappa = (lambda^(@) xx M)/(1000)` (`lambda` in `S m^(2) mol^(-1) = lambda xx 10^(4) s cm^(2) mol^(-1)`) `:. kappa_(Ag^(+)) = (6 xx 10^(-3) xx 10^(4) xx 4 xx 10^(-7))/(1000)` `= 24 xx 10^(-9) S cm^(-1)` `kappa_(Br^(-)) = (8 xx 10^(-3) xx 10^(4) xx 3 xx 10^(-7))/(1000)` `= 24 xx 10^(-9) S cm^(-1)` `kappa_(NO_(3)^(-)) = (7 xx 10^(-3) xx 10^(4) xx 10^(-7))/(1000)` `= 7 xx 10^(-9) S cm^(-1)` `:. kappa_("total") = kappa_(Ag^(+)) + kappa_(Br^(-)) + kappa_(NO_(3)^(-))` `= 24 xx 10^(-9) + 24 xx 10^(-9) + 7 xx 10^(-9)` `= 55 xx 10^(-9) S cm^(-1) = 55 xx 10^(-7) S cm^(-1)` in the unit of `1 xx 10^(-7) Sm^(-1)`, the conductivity is `55`. |
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| 614. |
Select the equivalent conductivity of `1.0 M H_(2)SO_(4)`, if its conductivity is `0.26 ohm^(-1) cm^(-1)`:A. `130 S cm^(2) eq^(-1)`B. `65 S cm^(2) eq^(-1)`C. `260 S cm^(2) eq^(-1)`D. None of these |
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Answer» Correct Answer - A `Lamda_(eq.)=kappaxx(1000)/(N)" "(because 1.0M H_(2)SO_(4)=2N)` `=(0.26 xx1000)/(2)=130 "S cm"^(2)eq^(-1)` |
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| 615. |
At `25^(@)C`, a conductivity cell was filled with `0.1 M NalL` solution. The conductivity of this solution is `9.2 xx 10^(3) s cm^(-1)` and resistance is `176.6 ohm`. The cross-sectional is `176.6 ohm`. What must have been the distance between the electrodes ?A. `4.5 cm`B. `6.5 cm`C. `9.8 cm`D. `5.9 cm` |
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Answer» Correct Answer - B `kappa = (1)/(R) xx (1)/(a)` or `l = kappa xx R xx a` `= 9.2 xx 10^(-3) xx 176.6 xx 4 = 6.49 cm` |
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| 616. |
9.65 C of electric current is passed through fused anhydrous magnesium chloride. The magnesium metal thus obtained is completely converted into a Grignard reagent. The number of moles of Grignard reagent formed is :A. `5xx10^(-4)`B. `1xx10^(-4)`C. `5XX10^(-5)`D. `1xx10^(-5)` |
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Answer» Correct Answer - C (c ) `Mg^(2+)+underset(2F)2e^(-) to underset(1mol)Mg` `2xx96500" C"` of charge form Mg =1 mol 9.65C of charge form Mg `=((1mol)xx(9.65" C"))/(2xx(96500" C"))=5xx10^(-5)mol` `Mg-=RMgX` Moles of Grignand reagent formed `=5xx10^(-5)mol`. |
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| 617. |
9.65C of electric current is passed through fused anhydrous magnesium chloride. The magnesium metal thus obtained is completely converted into Grignard reagen t. the number of m oles of the original reagent obtained ofA. `5xx10^(-4)`B. `1xx10^(-4)`C. `5xx10^(-5)`D. `1xx10^(-5)` |
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Answer» Correct Answer - C `W=ZxxIxxt` s Where `Z=("equivalent wt.")/(F)=(12)/(96500)` `Ixxt=9.65CthereforeW=(12)/(96500)xx9.65g=12xx10^(-4)g` `25g` of `mg^(2+)=1` mole of `Mg^(2+)` `12xx10^(-4)g` of `mg^(2+)=(1)/(24)xx12xx10^(-4)`mole |
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| 618. |
In the electrolysis of acidulated water, it is desired to obtain 1.12 cc of hydrogen per second under S.T.P. condition. The current to be passed isA. 9.65AB. 19.3AC. 0.965AD. 1.93A |
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Answer» Correct Answer - A No of moles of `H_(2)=(1.12)/(22400)` `therefore` No. of equivalents of hydrogen`=(1.12xx2)/(22400)=10^(-4)` No. of faradays required`=10^(-4)` `therefore`Current to be passed in 1 sec. `=96500xx10^-4` `=9.65`A |
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| 619. |
A current of 0.5 amperes is passed for 30 minute through a voltmeter containing `CUSO_4` Solution. Find the weight of Cu depositedA. 3.18 gB. 0.318 gC. 0.296 gD. 0.150 g |
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Answer» Correct Answer - C `W=ZxxIxxt` `Z=(E)/(96500)=(Mol.wt.//2)/(96500)` [for copper, E=mol. Wt/2] Given, current passed=0.5A time=30mins=`30xx60s` `W=(63.5)/(2)xx(1)/(96500)xx0.5xx30xx60=0.296g`. |
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| 620. |
` 2.5 F` of electricity is passed through a `CuSO_4` solution. The number of gm equivalent of `Cu deposited on anode is .A. ` Zero`B. ` 2.5`C. ` 1.25`D. `5.0` |
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Answer» Correct Answer - B `:. 1 F` obtaine from `1 g` equivalent `:. 2.5 F` obtained from ` 2.5 g` equivalent . |
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| 621. |
When 0.1 Faraday of electricity is passed in aqueous solution of `AlCl_(3)`. The amount of Al deposited on cathode isA. 27gB. 9gC. 0.27gD. 0.9g |
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Answer» Correct Answer - D In `AlCl_(3)` the eq. wt. of `Al=(27)/(3)=9` 1F=9g of Al |
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| 622. |
In an electroplating experiment `m` g of silver is deposited, whe `4` amperes of current flows for `2` mimtes. The amout (in g) of silver deposited by `6` amperes of current flowing for ` 40` seconds will be .A. `4 m`B. `m//4`C. `m//2`D. `2/m` |
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Answer» Correct Answer - C ` m=Zxx 4 xx 120 , M =Z xx 6 xx 40` ` M/m= (6 xx 40)/( 4 xx120) = 1/2 , M=m//2`. |
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| 623. |
In the electroplating of silver over an article made of iron , the electrolyte generally used isA. `AgNO_(3)`B. `AgCl`C. `Ag_(2)SO_(4)`D. `Na[Ag(CN)_(2)]` |
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Answer» Correct Answer - D Complexes are preferred . |
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| 624. |
When electricity is passed through a solution of `AlCl_(3)` and `13.5g` of `Al` is deposited, the number of `Faraday of electricity passed must be `………………….F`.A. `0.50`B. `1. 50`C. `1.00`D. `2.00` |
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Answer» Correct Answer - B Eq. of `Al=(13.5)/(27//3) = 1.5` Thus ` 1.5 F ` is needed. |
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| 625. |
The reaction taking place at the anode when a dilute aqueous solution of `CuSO_(4)` is electrolysed using inert Pt electrodes.A. `Cu to Cu^(2+) + 2 e^(-)`B. `2 SO_(4)^(2-) + 2 H_(2)O to 2 H_(2) SO_(4) + O_(2) + 4e^(-)`C. ` 2 H_(2) O to O_(2) + 4 H^(+) + 4 e^(-)`D. `SO_(4)^(2-) to SO_(2) + O_(2) + 2 e^(-)` |
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Answer» Correct Answer - C At anode (made of Pt) , `H_(2)O` is oxidized more easily than `SO_(4)^(2-)` ions . |
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| 626. |
In the electrolysis of water, one faraday of electrical energy would evolveA. one mole of oxygenB. one g atom of oxygenC. 8 g of oxygenD. 22.4 litres of oxygen |
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Answer» Correct Answer - C Hydrolysis of water: `2H_(2)Oto4H^(+)+4e^(-)+O_(2)` 4 F charge will produce=1 mole `O_(2)=32gm" "O_(2)` 1 F faraday will produce`=(32)/(4)g` gm `O_(2)`. |
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| 627. |
Three faradays of electricity are passed through molten `Al_2O_3` aqueous solution of `CuSO_4` and molten `NaCl` taken in deffernt electrolytic cells. The amout of `Al,Cu` and Na deposited at the cathodes will be in the ration of .A. 1 mol `:` 2 mol `:` 3 molB. 3 mol `:` 2 mol `:` 1 molC. 1 mol `:` 1.5 mol `:` 3 molD. 1.5 mol `:` 2 mol `:` 3 mol |
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Answer» Correct Answer - C `Al^(3+) + 3e^(-) to Al` `Cu^(2+) + 2 e^(-) to Cu` `Na^(+) + e^(-) to Na` Thus, 1 F will deposit `(1)/(3)` mol Al , `(1)/(2)` mol Cu and 1 mol Na , i.e., moles deposited are in the ratio `(1)/(3) : (1)/(2): 1 ` , i.e.,`2 : 3 : 6` or `1 : 1.5 : 3` . |
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| 628. |
Three faradays of electricity are passed through molten `Al_2O_3` aqueous solution of `CuSO_4` and molten `NaCl` taken in deffernt electrolytic cells. The amout of `Al,Cu` and Na deposited at the cathodes will be in the ration of .A. ` 1` mole : 15 ,mole `1` molesB. 1 mole : 2 moles :3 molesC. 1 mole : 2 moles : 3 molesD. 1.5 moles : 2 moles : 3 moles |
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Answer» Correct Answer - A At cathode : `Al^(3+) +3e^- rarr Al` `E_(Al) = ("Atomic mass")/3` At cathode : ` Cu^(2+) +2 e^- rarr Cu` `E_(Cu) = ("Atomic mass")/2` At cathode , `Na^+ + e^- rarr N` ` E_(Na) = ("Atomic mass")/1` for the passage of ` 3F` mole atoms of Al deposited `=1` mole atoms of Cu deposited ` = ( 1xx 3)/2 = 1.5` mole atoms of Na deposited `= 1xx 3 =3`. |
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| 629. |
Three faradays of electricity are passed through molten `Al_2O_3` aqueous solution of `CuSO_4` and molten `NaCl` taken in deffernt electrolytic cells. The amout of `Al,Cu` and Na deposited at the cathodes will be in the ration of .A. 1 mole :2 mole: 3 moleB. 3 mole : 2 mole : 1 moleC. 1 mole : 1.5 mole : 3 moleD. 1.5 mole : 2 mole : 3 mole |
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Answer» Correct Answer - C At cathode: `Al^(3+)+3e^(-)toCu` `E_(Cu)=("Atomic mass")/(3)` At cathode: `Cu^(2+)+2e^(-)toCu` `E_(Cu)=("Atomic mass")/(2)` At cathode: `Na^(+)+e^(-)toNa` `E_(Na)=("Atomic mass")/(1)` For the passage of 3 faraday, mole atoms of Al deposited=1 mole atoms of Cu deposited`=(1xx3)/(2)=1.5` mole atoms of Na deposited `=1xx3=3`. |
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| 630. |
Statement 1: During the electrolysis of water,two faraday of charge will produce a total of 336 litre of gases at S.T.P. at electrodes. Statement 2: Two faraday of charge will produce half mole of `H_(2)` and one fourth mole of `O_(2)` gas.A. Statement 1 is true, statement 2 is true, statement 2 is a correct explanation for statement 1B. Statement 1 is true, statement 2 is true, statement 2 is not a correct explanation for statement 1C. Statement 1 is true, statement 2 is falseD. statement 1 is false,statement 2 is true |
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Answer» Correct Answer - C 2 faraday charge will release 22.4 litre `H_(2) an 11.2` litre `O_(2)` at S.T.P. during the electrolysis of water. `V_(eq)(H_(2))=11.2L` at S.T.P. `V_(eq)(O_(2))=5.6L` at S.T.P. Total volume liberated by 2 faraday cahrge `=2xx11.2+2xx5.6` `=33.6L ` at S.T.P. |
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| 631. |
In the electrolysis of water, one faraday of electrical energy would evovle at STPA. one mole of oxygemB. one g atom of oxygenC. 8g of oxygenD. 22.4litres of oxygen |
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Answer» Correct Answer - C `2H_(2)OrarrO_(2)+4H ^(-)+4e^(-)` 4 faraday produce `O_(2)=32g` 1 faraday produce `O_(2)=(32)/(4)=8g` |
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| 632. |
How much charge (in F) must flow through solution during electrolysis of aq. `Na_(2)SO_(4)` at `STP` to produce `33.6 L` of product gases at `50%` current efficiency: |
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Answer» Correct Answer - 4 `H_(2)O overset(2F)rarr H_(2) +(1)/(2)O_(2)` `V V//2` `(3)/(2)V = 33.6 L` `rArr V = 22.4 L` let passed charge `= Q` `Q xx (50)/(100) = 2F rArr Q = 4F` |
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| 633. |
Electrolysis is a phenomenon where a reaction is carried out by passing electricity through the molten electrolyte or the electrolytic solution in water. In electrolysis, electrolyte first decomposes into anions and cations and thereafter anions undergo oxidation at anode and cations undergo reduction at cathode. for example. XY (molten ) `hArrX^(n+)+Y^(n-)` Oxidation at anode : `Y^(n-)-"ne"^(-)rarrY` Reduction at cathode `X^(n+)+ "ne"^(-)rarrX` In order to predict the electrolytic products accurately an array of substances is arranged in decreasing order tendency of oxidation of the substances or increasing order of standard reducton potential in electrochemical series. the standard electrode potentials of some species (elements,or ions) is given as under. `Cl_(2)(g)+2e^(-)rarr2Cl^(-),E^(@)=1.360V` `Na^(+)+e^(-)rarrNa,E^(@)=-2.71V` `2H_(2)O+e^(-)rarrH_(2)(g)+2OH^(-),E^(@)=-0.83V` `2H^(+)+2e^(-) rarrH_(2),E^(@)=OV` When aqueous solution of 100mL of 1M NaCl is electrolysed using Pt electrodes then answer the following question. What will b e the pH of the resulting solution upon passage of 0.2F chargeA. 13B. 13.301C. 7D. None |
| Answer» Correct Answer - A | |
| 634. |
Electrolysis is a phenomenon where a reaction is carried out by passing electricity through the molten electrolyte or the electrolytic solution in water. In electrolysis, electrolyte first decomposes into anions and cations and thereafter anions undergo oxidation at anode and cations undergo reduction at cathode. for example. XY (molten ) `hArrX^(n+)+Y^(n-)` Oxidation at anode : `Y^(n-)-"ne"^(-)rarrY` Reduction at cathode `X^(n+)+" ne"^(-)rarrX` In order to predict the electrolytic products accurately an array of substances is arranged in decreasing order tendency of oxidation of the substances or increasing order of standard reducton potential in electrochemical series. the standard electrode potentials of some species (elements,or ions) is given as under. `Cl_(2)(g)+2e^(-)rarr2Cl^(-),E^(@)=1.360V` `Na^(+)+e^(-)rarrNa,E^(@)=-2.71V` `2H_(2)O+e^(-)rarrH_(2)(g)+2OH^(-),E^(@)=-0.83V` `2H^(+)+2e^(-) rarrH_(2),E^(@)=OV` When aqueous solution of 100mL of 1M NaCl is electrolysed using Pt electrodes then answer the following question. What will be the electrolytic products of 100 mL 1M aq. solution?A. `H_(2)`B. `Cl_(2)`C. Na and `Cl_(2)`D. `H_(2),Cl_(2)` and `NaOH`. |
| Answer» Correct Answer - D | |
| 635. |
Statement -1 : During the electrolysis of water, two faraday of charge will produce a total of 33.6 litre of gases at STP at electrodes. Statement -2 : In the electrolysis of water, two faraday of charge will produce half mole of `H_(2)` gas and one fourth mole of `O_(2)` gas.A. Statement-1 is true, Statement-2 is true, statement-2 is a correct explanation for statement-1.B. Statement-1 is true , statement-2 is true, statement-2 is not a correct explanation for statement-1.C. Statement-1 is true, statement-2 is false.D. statement-1 is false, statement-2 is true. |
| Answer» Correct Answer - C | |
| 636. |
STATEMENT -1: `E_(cell)^(@)` is negative for electrolytic cell. STATEMENT-2: `/_G^(@)` is +ve for electrolyte cellA. If both the statements are TRUE and STATEMENTS-2 is the correct explantion of STATEMENTS-1B. If both the statements are TRUE but STATEMENTS-2 is NOT the correct explanation of STATEMENTS-1C. If STATEMENTS-1 is TRUE and STATEMENTS-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
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Answer» Correct Answer - A |
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| 637. |
STATEMENT-1: when 2 faraday of charge is passed through 0.1 M `H_2SO_4` (aq) ,11.2 litre `O_2` evolved at STP. STATEMENT-2: Molecular mass of oxygen is 32A. If both the statements are TRUE and STATEMENTS-2 is the correct explantion of STATEMENTS-1B. If both the statements are TRUE but STATEMENTS-2 is NOT the correct explanation of STATEMENTS-2C. If STATEMENTS-1 is TRUE and STATEMENTS-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
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Answer» Correct Answer - B |
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| 638. |
Equivalent conductivity of `BaCl_2,H_2SO_4` and HCl, are `x_1,x_2"and"x_3Scm^(-1)eq^(-1)` at infinite dilution. If conductivity of saturated `BaSo_4` solution is x`Scm^(-1)`, then `K_(sp)` of `BaSO_4` is :A. `(500x)/((x_1+x_2-x_3)^2)`B. `(10^6x^2)/((x_1+x_2-2x_3)^3)`C. `(2.5xx10^5x^2)/((x_1+x_2-2x_3)^2)`D. `(0.25x^2)/((x_1+x_2-2x_3)^2)` |
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Answer» Correct Answer - C |
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| 639. |
Given the following molar conductivity at `25^(@)` C:, HCl,426`Omega^(-1)cm^2mol^(-1)`, NaCl, `126Omega^(-1)cm^2mol^(-1)`, NaC(sodium crotonate), `83Omega^(_1)cm^2mol^(-1)`. What is the dissciation constant of crotonic acid,if the conductivity of a 0.001 M crotonic acid solution is `3.83xx10^(-5)Omega^(-1)cm^(-1)`?A. `10^(-5)`B. `1.11xx10^(-5)`C. `1.11xx10^(-4)`D. 0.01 |
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Answer» Correct Answer - B |
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| 640. |
If same quantity of electricity is passed through `CuCI` and `CuSO_(4)` the ratio of the weights of `Cu` deposited from `CuSO_(4)` and `CuCI` is:-A. `2:1`B. `1:2`C. `1:1`D. `4:1` |
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Answer» Correct Answer - B `W prop E` i.e., `(W_(1))/(W_(2))=(E_(1))/(E_(2))` `(W_(CuCl))/(W_(CuSO_(4)))=(E_(CuSO_(4)))/(E_(CuCl))=(M)/(2M)=(1)/(2)` |
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| 641. |
Using the information in the preceding problem, calculate the solubility product of `AgI` in water at `25^(@)C [E_((Ag +Ag)^(@) = +0.799volt]`:-A. `1.97 xx 10^(-17)`B. `7.91 xx 10^(-17)`C. `1.79 xx 10^(-17)`D. `9.17 xx 10^(-17)` |
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Answer» Correct Answer - B `AgI+e^(-)rarr Ag+ I^(-) E^(@)= -0.151` `{:(E^(@)=-0.95 ,,,E=E^(@)-(0.0551)/(1)log K_(sp)),(-16.074=log_(10)K_(sp),,,10^(-16.074)=K_(sp)):}` `7.91xx10^(-17)=K_(sp)` |
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| 642. |
The cell `Pt(H_(2))(1atm) |H^(+) (pH =?) T^(-) (a=1)AgI(s), Ag` has emf, `E_(298KK) =0`. The electrode potaneial for the reaction `AgI +e^(-) rarr Ag + I^(Theta)` is `-0.151` volt. Calculate the `pH` value:-A. `3.37`B. `5.26`C. `2.56`D. `4.62` |
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Answer» Correct Answer - C `(1)/(2)H_(2)AgI rarr H^(+)+Ag+I^(-) E = 0` `(1)/(2) H_(2)rarr H^(+)+e^(-)` , `E=0.151` `0.151= -(0.059)/(1)log (H^(+))=0.059xxpH` `pH = (0.151)/(0.059) = 2.5` |
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| 643. |
The cell `Pt|H_(2)(g) (1atm)|H^(+), pH = x ||` Normal calomal electrode has EMF of `0.64` volt at `25^(@)C`. The standard reduction potential of normal calomal electrode is `0.28V`. What is the `pH` of solution in anodic compartment. Take `(2.303RT)/(F) = 0.06` at `298K`. |
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Answer» `E_(H^(+)//H_(2))^(@) =- 0.06 log.(1)/([H^(+)]) = - 0.06pH` `rArr 0.64 = E_("cathode") - E_("Anode") = 0.28 -(-0.06pH)` `rArr pH = (0.64 -0.28)/(0.06) =(0.36)/(0.06) =6` |
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| 644. |
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`. |
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Answer» Correct Answer - `pH = 6.61` `E_(cell) = E_(cell)^(@) -(0.0591)/(1)log 10^(-x)` `0.67 =0.28 +(0.0591)/(1) xx x` `x = 6.61` |
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| 645. |
The oxidation potential of a hydrogne electrode at `pH=10` and `p_(H_(2))=1atm` isA. `-0.59V`B. `0.00V`C. `+0.59V`D. `0.059V` |
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Answer» Correct Answer - c `E_(o x i dation)=0.059pH=0.059xx10=0.59V` |
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| 646. |
In Castner `-` Kellner cell for the manufacture of `NaOH(` caustic soda `), Hg` acts as cathode as well as anode. |
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Answer» Correct Answer - T This cell is also called mercurry cathode cell. In this cell, `Hg` flows along the bottom of the cell and acts as cathode in the left hand compartment fitted with graphite anodes in brine `(NaCl)` solution, whereas the right hand compartment is fitted with `Fe` cathodes dippped in very dilute `NaOH` solution and `Hg` acts as anode in this compartment. Thus `Hg` acts as an intermediate electrode by induction, cathode on `LHS` and anode in `RHS.` |
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| 647. |
A hypothetical elecrochemical cell is shown below: `A^(ɵ)|A^(+)(xM)||B^(+)(yM)||B^(o+)` The emf measured is `+0.20 V`. The cell reaction isA. `A+B^(+) to A^(+)+B`B. `A^(+)+B to A+B^(+)`C. `A^(+)+e^(-) to A, B^(+) +e^(-) to B`D. the cell reaction cannot be predicted |
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Answer» Correct Answer - A (a) Electrochemical cell, `A|A^(+) (xM)||B^(+)(yM)|B` EMF of cell is +0.20 V. So, cell reaction is possible. The half-cell reactions are given as follows (i) At negative pole `A to A^(+) +e^(-) " (oxidation)"` (ii) At positive pole `B^(+)+e^(-) to B " (reduction)"` Hence, cell reaction is `A+B^(+) to A^(+) +B, E_(cell)^(@)=+0.20 V` |
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| 648. |
A hypothetical electrochemical cell shown below `overset(" "o+)A|(xM) || B^(+)(yM)|overset(" "o+)B` The e.m.f. measured is `+0.20" V "`. The cell reaction is :A. `A^(+)+e^(-) to A , B^(+)+e^(-) to B`B. The cell reaction cannot be predictedC. `A^(+)+B^(+) to A^(+)+B`D. `A^(+)+B^(+) to A+B^(+)` |
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Answer» Correct Answer - C (c ) The cell reaction is as follows : `A to A^(+)+e^(-)` `B^(+)+e^(-) to B` `A+B^(+) to A^(+)+B` |
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| 649. |
The standard reduction potentials of `Cu^(2+)//Cu` and `Ag^(+)//Ag` electrodes are 0.337 volt and 0.799 volt respectively. Construct a galvanic cell using these electrodes so that its standard e.m.f. is positive. For what concentration of `Ag^(+)`, will the e.m.f. of the cell at `25^(@)C` be zero if the concentration of `Cu^(2+)` is 0.01 M. |
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Answer» Given, `E_(Cu^(2+)//Cu)^(@)=0.337` volt and `E_(Ag^(+)//Ag)^(@)=0.799` volt. The standard emf will be positive if `Cu//Cu^(2+)` is anode and `Ag^(+)//Ag` is cathode. The cell can be represented as : `Cu|Cu^(2+)||Ag^(+)|Ag` The cell reaction is, `cu+2Ag^(+) rarr Cu^(2+)+2Ag` `E_(cell)^(@)=` Oxid. potential of anode + Red. potential of cathode `=-0.337+0.799` `=0.462` volt Applying the Nernst equation, `E_(cell)=E_(cell)^(@)-0.0591/2"log" ([Cu^(2+)])/([Ag^(+)]^(2))` or `"log" ([Cu^(2+)])/([Ag^(+)]^(2))=(0.462xx2)/0.0591=15.6345` `([Cu^(2+)])/([Ag^(+)]^(2))=4.3102xx10^(15)` `[Ag^(+)]^(2)=0.01/(4.3102xx10^(15))` `=0.2320xx10^(-17)` `=2.320xx10^(-18)` `[Ag^(+)]=1.523xx10^(-9) M` |
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| 650. |
The standard reduction potentials of `Cu^(2+)//Cu` and `Ag^(+)//Ag` electrodes are 0.337 volt and 0.799 volt respectively. Construct a galvanic cell using these electrols so that its standard e.m.f. is positive. For what concentration of `Ag^(+)`, will the e.m.f. of the cell at `25^(@)C` be zero if the concentration of `Cu^(2+)` is 0.01 M. |
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Answer» `E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)` =0.799-0.337=0.462 V Applying Nernst equation , `E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Anode])/([Cathode])` Given `" " E_(cell)^(@)=0.462 V , E_(cell)=zero , n=2, [Cu^(2+)]=0.01 M` `0=0.462-(0.0591)/(2)"log"([Cu^(2+)])/([Ag^(+)]^(2))=0.462-0.02955" log"([Cu^(2+)])/([Ag^(+)]^(2)` `"log"([Cu^(2+)])/([Ag^(+)]^(2))=(0.462)/(0.02955)=15.6345` `([Cu^(2+)])/([Ag^(+)]^(2)) ="Antilog" 15.6345=4.3102xx10^(15)` `[Ag^(+)]^(2)=(0.01)/(4.3102xx10^(15))=0.2320xx10^(-17)` `[Ag^(+)]=(0.2320xx10^(-17))^(1//2)=(2.320xx10^(-18))^(1//2)=1.523xx10^(-9) M`. |
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