

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.
551. |
The number of electron involved when one faraday of electricity is passed through an electrolytic solution is:A. 96500B. `8 xx 10^(6)`C. `12 xx 10^(16)`D. `6 xx 10^(23)` |
Answer» Correct Answer - D | |
552. |
The number of electron involved when one faraday of electricity is passed through an electrolytic solution is:A. `12 xx 10^(46)`B. `96500`C. `6.023 xx 10^(23)`D. `22.4 xx 10^(23)` |
Answer» Correct Answer - C 1 Faraday = `6.023 xx 10^(23)` electrons |
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553. |
When 0.2 Faraday of electricity is passed through an electrolytic solution, the number of electrons involved are :(a) 96500 (b) 1.603 × 10-19 (c) 1.2046 × 1023(d) 12 × 106 |
Answer» Option : (c) 1.2046 × 1023 |
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554. |
A metal-insoluble salt electrode consists ofA. crystals of an insoluble salt coated with a metalB. a piece of metal placed in a solution containing a sparingly soluble saltC. a piece of metal coated with one of its insoluble salts.D. a metal fused with an insoluble salt at high temeperature |
Answer» Correct Answer - A It is the most common type of electrode where a metal is in equilibrium with a solution containing its ions. Second one is a redox electrode, thirds one is metal-insoluble salt-anion electrode and the fourth one is a gas electrode. |
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555. |
The emf of a galvanic cell constituted with the electrodes `Zn^(2+)//Zn(E^(@)=-0.76" V")` and `Fe^(2+)//Fe(E^(@)=-0.41" V")` is :A. `-0.35" V"`B. `+1.17" V"`C. `+0.35" V"`D. `-1.17" V"` |
Answer» Correct Answer - C (c ) `E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)` `=(-0.41)-(-0.76)=0.35" V"` |
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556. |
A galvanic cell is composed of two hydrogen electrodes one of which is a standard one. In which of the following solutions should the other electrode be immersed to get maximum emf ?A. 0.1 M HClB. `0.1 M CH_(3)COOH`C. `0.1 M H_(3)PO_(4)`D. `0.1 M H_(2)SO_(4)` |
Answer» Correct Answer - C (d) `0.1 M H_(3)PO_(3)` will liberate maximum `H^(+)` ions in solution. |
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557. |
Given that `E^o` values of `Ag^+//Ag, K^+//K, Mg^(2+)//Mg` and `Cr^(3+) //Cr` are `0.08 V, -2 .93 V, -2 37 V` and ` -0. 74 V` respectively. Therefore the order for the reducing power of the metal is .A. ` Ag gt Cr gt Mg gt K`B. ` Ag lt Cr lt Mg lt K`C. ` Ag gt Cr gt K gt Mg`D. ` Cr gt Ag gt Mg gt K` |
Answer» Correct Answer - B The greater the standard reduction potential, the greater tendency to reducing power of metal. K has greater value `( - 2. 93)` so `K gt Mg gt Cr gtAg` . |
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558. |
Given that `E^o` values of `Ag^+//Ag, K^+//K, Mg^(2+)//Mg` and `Cr^(3+) //Cr` are `0.08 V, -2 .93 V, -2 37 V` and ` -0. 74 V` respectively. Therefore the order for the reducing power of the metal is .A. `Ag gtCr gtMg gt K`B. `Ag gtCr gtMg gt K`C. `Ag gt Cr gt K gt Mg`D. `Cr gt Ag gt Mg gt K` |
Answer» Correct Answer - B More the negativ e `E^@` value , larger the reducing power of the metal. |
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559. |
Which of the following represents the electrodes of the first kind ?A. `Zn^(2+)|Zn`B. `Fe^(3+), Fe^(2+)|Pt`C. `AgCl|Ag|Cl^(-)`D. `Pt|Cl_(2)|Cl^(-)` |
Answer» Correct Answer - B It is a metal-insoluble salt-anion electrode. It consists of a metal `(Hg)` in contact with a saturated solution of its sparingly soluble salt `(HgCl_(2)`, calomel) and another solubel salt `(KCl)` having common anions `(Cl^(-))`. The saturated calomel electrode is often used as a reference electrode (secondary reference) in place of hydrogen elec-trode which is inconvenient and difficult to prepare. As a reference electrode, saturated calomel electrode is reversible and has a fixed potential. Mercurous chloride (calomel) is reduced to mearcury, when saturated calomel electrode `(SCE)` is used as cathode accord-ing to the equation: `Hg_(2)Cl_(2)+2e^(-)hArr 2Hg(1)+2Cl^(-)(aq.)` It can be represented as `Hg_(2)Cl_(2)|Hg|Cl^(-)`. The half-cell potential for saturated calomel electrode is `0.2682 V`. |
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560. |
Cell potential is the difference in `"______"` between the two electrodes of a galvanic cell that cause charge to flow.A. Potential energyB. Kinetic energyC. internal energyD. Enthalpy |
Answer» Correct Answer - A Work is needed to move electrons in a wire or tow move ions through a solution to an electrode. The sitution is analogous to pumping water form one point to anoter. Work must be expended to pump the water. The analogy may be extended. Water moves spontaneously form a point at high pressure to point at low pressure. thus, a pressure differnce is required. The work expended in moving water through a pipe depends on the volume of water and the pressure differnece. The situ-ations with electicity is similar. An electric change moves spontaneously from a point at high electric potential (high electrical pressure) to a point at low electric potential (low electrical pressure). The work needed to move an electric chage through a conductor depends on the total charge noved and the potential difference i.e. the difference in elec-trical potential (electrical pressure) between two points. The potential energy difference (or the potential difference is most conveniently decribed as potential, which is the potential energy per unit chartge. The `SI` unit of charge is the coulomb `(C)`, the `SI` unit of potential enerfy is the joule `(J)` and the `SI` unit of potential is the Volt `(V)`. The relationship between the three quantities is given by the equation. `1 V = 1(J)/(C)` or `1 J = 1 C xx 1 V` where `1 C` is the amount of achge transferred when a current of `1` ampere `(A)` flows for `1` second `(s)`. [Since `1` watt (`W)` is `1 JS^(-1)`, the current passing through the buulb in `1s` is `1 C`.] When `1 C` of charge moves between two electrodes that differ in electrical potential by `1V, 1 J` of energy is released by the cell and can be used to do electrical work. Remember, the electrical work expended in moving a charge through a conductor is Electrical work = charge `xx` potential difference |
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561. |
Salt bridge is used toA. connect two half cellsB. maintain electrical neutrallyC. reduced junction potentialD. all |
Answer» Correct Answer - D These are the functions of salt bridge . |
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562. |
`KCl` cannot be used as a salt bridege for the cell `Cu(s)|CuSO_(4)(aq.)||AgNo_(3)(aq.)|Ag(s)` becauseA. `Cl_(2)` gas is evolvedB. `CuCl_(2)` gets precipitatedC. `SO_(2)` gas is evolvedD. `AgCl` gets precipitated |
Answer» Correct Answer - C In this notation, the anode (or oxidation half-cell) is always written first on the lect white the cathode (or reduction half-cell) is always written last on the rifht. The other components appear in the order in which we would encounter them in moving form the anode to the cathode. The two electrodes are electrically connected by means of a slat bridge, denoted by two by vertical bars. The cell terminals are at the extreme ends in this cell notation. and a single cerical bar indicates a phase boundary-sat between a solid terminal and the electrode solution. for example, the zinc electode is a solild while the `Zn^(2+)` ions (form `ZnSO_(4)`) are in solution. thus, we draw a line between `Zn` and `Zn^(2+)` to show the phase boundries. Note that there is also a line between the `ZnSO_(4)` solution and the `KCl` solution in the salt bridge because these two solutions are not mixed physically and therefore constitute two separate phases. |
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563. |
What is the number of coulombs required for the conversion of one mole of `MnO_(4)^(-)` to one mole of `Mn^(2+)` ?A. `5 xx 96500`B. `3 xx 96500`C. 96500D. 9650 |
Answer» Correct Answer - A | |
564. |
What is the difference between galvanic cell and electrolytic cell ?A. In galvanic cell, electrical energy is profuced while in electrolytic cell electrical energy is consumedB. In galvanic cell, anode is (-) ve while in electrolytic cell anode is (+) veC. In galvanic cell, cathode is (+) ve while in electrolytic cell anode is (-) veD. All are correct |
Answer» Correct Answer - D | |
565. |
In the class, teacher wrote some ions on the black board. It is given below.1. Which among the ions, gain electron more easily?2. Is it true that the reduction potential of hydrogen is less than that of Na and Mg?3. Justify your answer. |
Answer» 1. Mg2+ 2. No. The reduction potential of hydrogen is greater than that of both Na and Mg. 3. In electrochemical series, sodium and magnesium lie above hydrogen. Standard reduction potential of hydrogen electrode is taken as zero volt. |
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566. |
KCl cannot be used as a salt bridge for all the cell Cu(s) | `CuSO_(4) (aq) | | AgNO_(3) (aq) |Ag(s)` becauseA. `CuCl_(2)` gets precipitatedB. `Cl_(2)` gas is evolvedC. `AgCl` gets precipitatedD. None of the above . |
Answer» Correct Answer - C KCl reacts with `AgNO_(3)` to form a ppt. of AgCl. |
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567. |
Assertion `(A):` A saturated solution of `KCl` is used in making salt bridge. Reason `(R):` Ionic mobilities of `K^(o+)` and `Cl^(c-)` are comparable.A. 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. |
Answer» Correct Answer - A (a) Reason is the correct explanation for assertion. |
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568. |
Assertion `(A):` A saturated solution of `KCl` is used in making salt bridge. Reason `(R):` Ionic mobilities of `K^(+)` and `Cl^(-)` are comparable.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. |
Answer» Correct Answer - A Both `(A)` and `(R)` and correct and `(R)` is the correct reason For `(A)`. As a referrence `:wedge_(mK^(o+))^(@)=73.52S cm^(2) mol^(-1)` and `wedge_(m Cl^(c-)_^(@)=76.34 S cm^(2) mol^(-1)` |
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569. |
Given that standard electrode potentials`(E^(@))` of metals are : `K^(+)//K=-2.93" V",Ag^(+)//Ag=0.80" V",Cu^(2+)//Cu=0.34" V", Mg^(2+)//Mg=-2.37" V"` `Sn^(2+)(aq) to Sn^(4+)(aq)+2e^(-),E^(@)=-0.15" V"` Construct the redox reaction equation from the two half-reactions and calculate the cell potential from the standard potentials and predict if the reaction is reactant or product favoured. |
Answer» Please note that less the `E^(@)` value of an electrode, more is its reducing power. The increasing order of reducing power is : `Ag^(+)//Ag lt Cu^(2+)//Cu lt Fe^(2+)//Fe lt Mg^(2+)//Mg lt K^(+) lt K`. |
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570. |
How many coulombs of electricity are required for the reduction of 1 mole of `MnO_(4)^(-)` to `Mn^(2+)` ?A. 96500 CB. `1.93 xx 10^(5)C`C. `4.83 xx 10^(5)C`D. `9.65 xx 10^(6)C` |
Answer» Correct Answer - C | |
571. |
What is the cell reaction occurring in Daneill cell (galvanic cell) ?A. `Cu(s)+ZnSO_(4)(aq.) rarr CuSO_(4)(aq.) + Zn(s)`B. `Zn(s)+CuSO_(4)(aq.) rarr Cu(s)+ZnSO_(4)(aq.)`C. `Ni(s)+ZnSO_(4)(aq.) rarr NiSO_(4)(aq.)+Zn(s)`D. `2Na(s)+CdSO_(4)(aq.) rarr Na_(2)SO_(4)(aq.)+Cd(s)` |
Answer» Correct Answer - B | |
572. |
In the experiment set up for the measurement of EMF of a half cell using a reference electrode and a salt bridge. When the salt bridge is removed, the voltageA. Does not changeB. Decreases to half the valueC. Increase to maximumD. Drops to zero |
Answer» Correct Answer - D It connect two solutions and complete the circuit. |
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573. |
In an experimental set-up for the measurement of potential of a half-cell using the reference electrode and a salt bridge, the salt bridge was suddenly removed.1. What happens to the voltage when the salt bridge is removed?2. Justify your answer.3. Write any two examples of inert electrolytes used for the construction of salt bridge. |
Answer» 1. Voltage ceases. 2. When the salt bridge is removed the migration of ions between the half cells is hindered which results in the accumulation of charge near the electrodes. This decreases the cell voltage. 3. KCl, NH4NO3 |
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574. |
In an experiment setup for the measurement of `EMF` of a half cell using a reference electrode and a salt bridge, when the salt bridge is removes, the voltageA. Does not changeB. Increase to maximumC. Decreases to half the valueD. Drops to zero |
Answer» Correct Answer - d Factual statement. |
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575. |
Statement -1 : `KCl` and `NH_(4)Cl` cannot be used in salt bridge of a cell containing `Ag^(+), Hg_(2)^(2+)` and `Tl^(+)` ions. Statement -2 : Cell will be destroyed due to precipitation of metal chlorides.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 - A | |
576. |
The standard electrode potential a `Ag^(+)//Ag` is +0.80 V and of `Cu^(2+)//Cu` is +0.34 V. These electrodes are connected through a salt bridge and if :A. copper electrode acts as cathode, then `E_(cell)^(@)` is +0.46 voltB. silver electrode acts as anode, then `E_(cell)^(@)` is + 0.46 voltC. copper electrode acts as cathode, then `E_(cell)^(@)` is - 0.34 voltD. silver electrode acts as anode, then `E_(cell)^(@)` is + 1.14 volt |
Answer» Correct Answer - C | |
577. |
In an experiment setup for the measurement of `EMF` of a half cell using a reference electrode and a salt bridge, when the salt bridge is removes, the voltageA. does not changeB. increases to maximumC. decreases half the valueD. drops to zero |
Answer» Correct Answer - D | |
578. |
A salt bridge may contain:A. a saturated solution of KCl and agar-agarB. a saturated solution of `KNO_3` and agar-agarC. a saturated solution of `NH_4NO_3` and agar-agarD. all of these |
Answer» Correct Answer - D |
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579. |
Select the equivalent conductivity of `1.0 M H_(2)SO_(4)`, if its conductivity is `0.26 ohm^(-1) cm^(-1)`: |
Answer» Correct Answer - `130 ohm^(-1)cm^(2)(g eq.)^(-1)` Conductivity or specific conductance (k)`=26xx10^(-2)" ohm^(-1) cm^(-1)` Molarity of `H_(2)SO_(4)` solution (M)=1M Normality of `H_(2)SO_(4)` solution (N)`=1xx2=2N " " (because H_(2)SO_(4) "is a disbasic acid")` `Lambda_(E)=(kxx1000)/(N)=((26xx10^(-2)" ohm"^(-1)cm^(-1))xx1000cm^(3))/((2.0"g eq")^(-1))=130" ohm"^(-1)cm^(2)("g eq")^(-1)`. |
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580. |
For a spontaneous reaction `DeltaG^(@)`, Equilibrium constant (K) and `E_(cell)` will be respectively.A. `-ve gt 1 gt +ve`B. `+ve gt 1 gt -ve`C. `-ve lt 1 lt -ve`D. `-ve gt 1 gt -ve`. |
Answer» Correct Answer - A (a) is the correct answer. |
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581. |
What volume of `0.1N FeSO_(4)` can be oxidized by a current of 2 ampere hours ?A. `0.746 L`B. `7.46L`C. `1.482L`D. `0.373L` |
Answer» Correct Answer - a `1F=96500C=1Eq of Fe^(2+)` `2xx36003600C-=(2xx3600)/(96500)-=0.0746Eq` `:. Vxx 0.1 -=0.0746` `V=0.746` |
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582. |
The conductance of solution of an electrolyte is equal to that of its specific conductance. The cell constant of the conductivity cell is equal toA. ResistanceB. FaradayC. ZeroD. Unity |
Answer» Correct Answer - D | |
583. |
The specific conductance of 0.05 N solution of an electrolyte at 298 K is 0.002 S cm^(-1). Calculate the equivalent conductance. |
Answer» The specific conductance, `k =0.002 " S " cm^(-1)=0.002 " oh "m^(-1) cm^(-1)` Equivalent concentration (C ) =0.05 N =0.05 g eq. `L^(-1)` `=(0.05 g" eq".)/(1 L)=(0.05 g" eq".)/(1000 cm^(3))=0.05xx10^(-3) f" eq". cm^(-3)` `"Equivalent conductance", (Lambda_(E))=(k)/(C)=((0.002 " oh "m^(-1) cm^(-1)))/((0.05xx10^(-3)g " eq ". cm^(-3)))` `=40 " oh "m^(-1) cm^(2)(g " eq".)^(-1)`. |
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584. |
The resistance of `1N` solution of acetic acid is `250ohm`, when measured in a cell of cell constant `1.15 cm^(-1)`. The equivalent conductance `(` in `ohm^(-1) cm^(2)eq^(-1))` of `1N` acetic acid isA. `4.6 ohm^(-1) cm^(2) eq^(-1)`B. `9.2 ohm^(-1) cm^(2) eq^(-1)`C. `18.4 ohm^(-1) cm^(2) eq^(-1)`D. `0.023 ohm^(-1) cm^(2) eq^(-1)` |
Answer» Correct Answer - A `kappa=Cxxl/A=1/250xx1.15=4.6xx10^(-3) ohm^(-1) cm^(-1)` `Lambda_(e) =kappaxx1000/N =4.6xx10^(-3) xx1000/1=4.6 ohm^(-1) cm^(2) eq^(-1)` |
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585. |
The resistance of `1N` solution of acetic acid is `250ohm`, when measured in a cell of cell constant `1.15 cm^(-1)`. The equivalent conductance `(` in `ohm^(-1) cm^(2)eq^(-1))` of `1N` acetic acid isA. `4.6`B. `9.2`C. `18.4`D. `0.023` |
Answer» Correct Answer - a `k=(1)/(R)xxG=GG^(**)=(1)/(250)xx1.15=0.0046S cm^(-1)` `wedge_(eq)=(kxx1000)/(N)=(0.0046xx1000)/(1)=4.6ohm^(-1)cm^(2)eq^(-1)` |
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586. |
Assertion `(A) : ` The ration of specific conductivity to the observed conductance does not depend upon the concentration of the solution taken in the conductivity cell. Reason `(R) :` Specific conductivity decreases with dilution whereas observed conductance increases with the dilution.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 incorrectD. If (A) is incorrect, but (R) is correct. |
Answer» Correct Answer - B | |
587. |
The resistance of `1N` solution of acetic acid is `250ohm`, when measured in a cell of cell constant `1.15 cm^(-1)`. The equivalent conductance `(` in `ohm^(-1) cm^(2)eq^(-1))` of `1N` acetic acid isA. `46.0`B. `9.2`C. `18.4`D. `2.3` |
Answer» Correct Answer - A `k=GxxG^(**)=(1)/(T)xxG^(**)` `=(1)/(250)xx1.15` `wedge_(eq)=(kxx1000)/(N)=(1.15xx1000)/(250xx0.1)=46S cm^(2) eq^(-1)` |
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588. |
A weak electrolyte having the limiting equivalent conductance of 400 S `cm^(2)g`. `"equivalent"^(-1)` at 298 K is 2% ionized in its 0.1N solution. The resistance of this solution (in ohms) in an electrolytic cell of cell constant 0.4 `cm^(-1)` at this temperature is:-A. 200B. 300C. 400D. 500 |
Answer» Correct Answer - D `alpha=(wedge_(m))/_(wedge_(oo))` `wedge_(oo)=400S" "cm^(2)g" "eq^(-1)` `wedge_(m)=wedge_(oo)xxalpha=400xx(2)/(100)=8` `wedge_(m)=(kxx1000)/(C)` `k=(1)/(R)xx`cell constant `wedge_(m)=(1)/(R)xx`cell constant`xx(1000)/(C)` `R=("cell constant")/(wedge_(m)xxC)xx1000` `R=(0.4xx1000)/(8xx0.1)=500ohm` |
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589. |
(A) The cell constant of a cell depends upon the nature of the material of the electrodes. (R) The observed conductance of a solution depends upon the nature of the material of the electrodes.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 incorrectD. If (A) is incorrect, but (R) is correct. |
Answer» Correct Answer - D | |
590. |
The resistance of a `1N` solution of salt is `50 Omega`. Calculate the equivalent conductance of the solution, if the two platinum electrodes in solution are 2.1 cm apart and each having an area of `4.2 cm^(2)`. |
Answer» `k = (1)/(rho) = (1)/(R) ((l)/(A)) = (1)/(50) xx (2.1)/(4.2) = (1)/(100)` and `lambda_(eq) = (k xx 1000)/(N) = (1)/(100) xx (1000)/(1) = 10` | |
591. |
Statement-1: The cell constant of a conductivity cell depends upon the nature of the material of the electrodes. Statement-2: The electrodes of the cell are coated with platinum black to avoid polarisation effects.A. Statement-1 is True, statement-2 is true, statement-2 is a correct explanation of statement-1B. Statement-1 is true, statement-2 is true, statement-2 is not a correct explanation of statement-1.C. Statement-1 is true, statemet-2 is falseD. Statement-1 is false, statement-2 is true |
Answer» Correct Answer - D Correct Statement-1. the cell constant of a conductivity cell depends upon the distance between the electrodes and the area of their cross-section and not on the material of the electrodes. |
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592. |
The resistance of ` aN//10KCI` solution is ` 245 Omega`. Calculate the equivalent conductance of the solution if the electrodes in the cell are `4 cm` apart and each haveing an area of ` 7.0 sq.cm`.A. `23. 32 cm^2 eq^(-1)`B. `23. 23 S m^2 eq^(-1)`C. `2. 332 S cm^2 eq^(-1)`D. None of these |
Answer» Correct Answer - A ` 245 = 1/(Lambda_(eq)) xx 4/7 ( 1000)/(0.1) rArr Lambda_(eq) = 23.32 S cm^2 eq^(-1)`. |
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593. |
(A) The cell constant of a cell depends upon the nature of the material of the electrodes. (R) The observed conductance of a solution depends upon the nature of the material of the electrodes.A. 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. |
Answer» Correct Answer - D (d) Correct assertion : The value of cell constant (l/a) depends upon distance between electrodes and the conductance of the solution. Correct reason : The observed conductance depeds upon the nature of the electrolyte and the concentration of the solution. |
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594. |
Assertion : Copper does not get corroded in the acidic medium. Reason : Free energy for this process is positive.A. 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. |
Answer» Correct Answer - A (a) Reason is the correct explanation for assertion. |
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595. |
The electrical resistance of a column of `0.05 M N aOH ` solution of diameter `1 cm` and length `50 cm` is `5.55 xx 10^(3)ohm`. Calculate its resistivity , conductivity, and molar conductivity. |
Answer» Area `(a)=pi r^(2)=3.14xx((1cm)/(2))^(2) ` `=0.785 cm^(2)` `=0.785xx10^(-4)m^(2)` `l=50 cm=0.5m` `R=(rhol)/(a)` or `rho=(Ra)/(l)=(5.55xx10^(3)ohmxx0.785cm^(2))/(50cm)` `=87.135ohm cm` Conductivity `(k)=(1)/(rho)=((1)/(87.135))S cm^(-1)` `=0.01148 S cm^(-1)` Molar conductivity `(wedge_(m))=(kxx1000)/(M)cm^(3)L^(-1)` `=(0.01148Scm^(-1)xx1000cm^(3)L^(-1))/(0.05mol L^(-1))` `=229.6 S cm^(2) mol^(-1)` Alternatively The value of different quantities in `SI` units `(i.e.,` in terms of `"m"` instead of `"cm")` `rho=(Ra)/(l)=(5.55xx10^(3)ohm xx 0.785 xx 10^(-4)m^(2))/(0.5m)` ` =87.135xx10^(-2)ohm m` `k=(1)/(rho)=(100)/(87.135)ohm m=1.148 S m^(-1)` and `wedge _(m)=(k)/(c)=(1.148Sm^(-1))/(50 mol m^(-3))=229.6xx10^(-4)S m^(2) mol ^(-1)` |
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596. |
A hydrogen gas electrode is mae by dipping platinum wire in a solution of HCl of pH=10 and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would beA. 1.81 VB. 0.059VC. 0.59VD. 0.118V |
Answer» Correct Answer - C `(1)/(2)H_(2)(g)toH^(+)+e^(-)` `E_(O.P.)=E_(O.P.)^(o)-(0.059)/(n)log([H^(+)])/((P_(H_(2)))^(1//2))` `E_(O.P.)=0-(0.059)/(1)"log"(10^(-10))/((1)^(1//2))(pH=10,[H^(+)]=10^(-10)M)` `E_(O.P.)=0.59V` |
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597. |
A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl or `pH = 10` and by passing bydrogen gas around the platinum wire at one atm pressure . The oxidation potential of electrode would be ?A. 0.059 VB. 0.59 VC. 0.118 VD. 1.18 V |
Answer» Correct Answer - B (b) For hydrogent electrode, oxidation half reaction is `underset((1 atm))(H_(2)) to underset(" (At pH 10)")(2H^(+))+2e^(-)` If pH=10 `H^(+)=1xx10^(-pH)=1xx10^(-10)` From Nernst equation, `E_(cell) =E_(cell)^(@)-(0.0591)/(2)log""([H^(+)]^(2))/(p_(H_(2)))` For hydrogen electrode, `E_(cell)^(@)=0` `E_(cell)-(0.0591)/(2) log""((10^(-10))^(2))/(1)` `=+(0.0591xx1)/(2) log""(1)/(10^(-10))` `=0.0591 log 10^(10)=0.0591xx10=0.591 V` |
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598. |
A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl or `pH = 10` and by passing bydrogen gas around the platinum wire at one atm pressure . The oxidation potential of electrode would be ?A. ` 0.59 V`B. ` 0. 11 8 V`C. ` 1.18 V`D. ` 0.059 V` |
Answer» Correct Answer - A `underset (1atm) (H_2)rarr underset(10^(-10))(2H^+ +2^(-)` `E_(H_2//H^+) = 0- (0.059)/9 log. (10^(-10))^(2)/1` `E_(H_2 //H^+) = + 0.59 V`. |
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599. |
A hydrogen gas electrode is made by dipping platinum wire is a solution of HCl of pH=10 and by passing hydrogen gas around the platinum wire at one atom prssure. The oxidation potential of the electrode would beA. 0.059 VB. 0.59 VC. 0.118 VD. 0.18 V |
Answer» Correct Answer - C For hydrogen electrode, reduction reaction is `H^(+)+e^(-)to(1)/(2)H_(2)` `E_(H^(+)//(1)/(2)H_(2))^(@)=E_(H^(+)//(1)/(2)H_(2))^(@)-(0.0591)/(n)"log"(p_(H_(2))^(1//2))/([H^(+)])` `=0-(0.0591)/(n)"log"(1)/(10^(-10))=-0.591` `(pH=10" means "[H^(+)]=10^(-10)M)` `therefore`Oxidation potential`=0.591V` |
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600. |
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 the assertion and reason are true but the reason is ont the correct explanation of assertionB. If both the assertion and reason are true but the reason is not the correct explanation of assertion.C. If the assertion is true but reason is false.D. If assertion is false but reason is rue |
Answer» Correct Answer - c Zn is -ve elvctrode and Cu is `+ ve` electrode in danidel cell. |
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