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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.
| 1551. |
Cell reaction is spontaneous when :A. `E_("red")^(@)` is positiveB. `DeltaG^(@)` is negativeC. `Delta G^(@)` is positiveD. `E_("red")^(@)` is negative |
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Answer» Correct Answer - B (b) `DeltaG^(@)` must be negative for spontaneity of cell. |
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| 1552. |
The rate of a reaction quadruples when temperature charges from 293 K to 312K, Calculate the energy of activation of the reaction assuming that it does not charge with temperature. |
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Answer» According to Arrhenius equation `log ""(k_(2))/(k_(1))=(E_(a))/(2.303)[(1)/(T_(1))-(1)/(T_(2))],` Given `[therefore T_(i) =293 K, T_(2) -T_(1) =20K]` `log ""(4)/(1) =(E_(a))/(2.303 xx(8.314 J mol ^(-1)k^(-1)))xx(20)/(293xx313)` `E_(a)=(0.6021xx2.303xx8.314xx293xx313)/(20) J mol ^(-1)` `=52863mol ^(-1)` `E_(a)=52.863kJ mol ^(-1).` |
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| 1553. |
What is a galvanic cell or a valtaic cell ? Give one example. |
| Answer» Galvanic cell : A device which converts chemical energy into electrical energy by the use of spontaneous redox reaction is called Galvanic cell (or) voltaic cell. Eg: Daniell cell. | |
| 1554. |
The electrode potential of standard hydrogen electrode is assigned a value of ________ |
| Answer» Correct Answer - 0 volt | |
| 1555. |
In cell representation, oxidation half -cell is represented on the _______ |
| Answer» Correct Answer - left-hand side | |
| 1556. |
On increasing temperature, conduction in metallic conductors ______A. increasesB. decreasesC. remains constantD. none of these |
| Answer» Correct Answer - B | |
| 1557. |
The aqueous solution of non-electrolytes contains _______A. atomsB. ionsC. electronsD. molecules |
| Answer» Correct Answer - D | |
| 1558. |
The metal deposited first at the cathode when a solution containing `Fe^(+2), Cu^(+2)` is electrolysed is _______A. FeB. CuC. `Fe^(+3)`D. both 1 and 2 |
| Answer» Correct Answer - B | |
| 1559. |
Which of the following electrolytes exhibits maximum conductivity ?A. 1 M NaClB. 1 M KClC. `1 M Ca(ON_(3))_(2)`D. 1 M `Al_(2) (SO_(4))_(3)` |
| Answer» Correct Answer - D | |
| 1560. |
The position of some metals in the electrochemical series in dectreasing electropositeve character is given as ` Mg gt Al gt Zn gt Cu gt Ag`. What will happen if a copper spoon is used to stir a solution of aluminimum nitrate ?A. The spoon gets coated with aluminiumB. Aan alloy of aluminium and copper is formedC. No reaction occursD. The solution starts turning blue |
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Answer» Correct Answer - C |
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| 1561. |
Zn can displace :A. Mg from its aqueos solutionB. Cu from its aqueos solutionC. Na from its aqueos solutionD. Al from its aqueos solution |
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Answer» Correct Answer - B |
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| 1562. |
Define : (1) Oxidation potential, (2) Reduction potential. |
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Answer» (1) Oxidation potential : It is defined as the difference of electrical potential between metal electrode and the solution around it at equilibrium developed due to oxidation reaction at anode and at constant temperature. (2) Reduction potential : It is defined as the difference of electrical potential between metal electrode and the solution around it at equilibrium developed due to reduction reaction at cathode and at constant temperature. |
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| 1563. |
The answer to each of the following question is a single digit integer, ranging from 0 to 9. if the correct answers to the question number A,B,C and D (say ) are 4,0,9 and 2 respectively. Then the correct correct darkening of bubbles should be as shown on the side.A. In Mg-Al cell, the number of electrons involved in the cell reaction i sB. 0.25 mole of propane is subjected to combustion. If this reaction is used for making a fuel, cell the number of moles of electrons involved in each half cell for this amount of propane will beC. Three litres of 0.5M `K_(2)Cr_(2)O_(7)` solution have to be completely reduced in the acidic medium. The number of faradays of electricity required will beD. For the Mg-Ag cell, how many times the difference between the EMF of the cell and its standard EMF will change if concentration of `Mg^(2+)` ions changed to 0.1M and that of `Ag^(+)` ions is changed from 0.5M to 0.25M. |
| Answer» Correct Answer - A | |
| 1564. |
Consider following half cell reaction and corresponding standard (reduction) electrode potentials. I. `A+e^(-)rarrA:E^(@)=-24V` II. `B^(-)+e^(-)rarrB^(2-):E^(@)=+1.25V` III. `C^(-)+2e^(-)rarrC^(3-):E^(@)=-1.25V` IV. `D+2e^(-)rarrD^(2-):E^(@)=+0.68V` V. ` E+4e^(-)rarrE^(-4):E^(@)=0.38V`. If every ion has concentration 1 M in the cell the largest cell potential at 298K isA. 2.50VB. 1.49VC. 1.06VD. 1.91V. |
| Answer» Correct Answer - A | |
| 1565. |
Consider following half cell reaction and corresponding standard (reduction) electrode potentials. I. `A+e^(-)rarrA:E^(@)=-24V` II. `B^(-)+e^(-)rarrB^(2-):E^(@)=+1.25V` III. `C^(-)+2e^(-)rarrC^(3-):E^(@)=-1.25V` IV. `D+2e^(-)rarrD^(2-):E^(@)=+0.68V` V. ` E+4e^(-)rarrE^(-4):E^(@)=0.38V`. Cell with the largest cell potential isA. `A|A^(-)||B^(2-)|B^(-)`B. `A|A^(-)||B^(2- ),B^(-)|B`C. `Pt|C^(3-),C^(-)|B^(-),B^(2-)| Pt`D. ` Pt|B^(2-) ,B^(-)| |C^(-),C^(3-)|Pt` |
| Answer» Correct Answer - C | |
| 1566. |
The standard oxidation potential of silver electrode is given as `- 0.799 V`. The standard reduction potential of copper electrode is given as 0.33 V. Find out the emf of the cell constituted by these two electrodes. Also write the half -cell reactions involved in the cell and given cell representation |
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Answer» (i) Difference in standard electrode potentials of the electrodes. (ii) Difference between oxidation and reduction potentials (iii) Comparison of standard electrode potentials of the respective metals. (iv) Identification of anode and cathode in the electrochemical cell. (v) Calculation of emf of the cell |
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| 1567. |
What is a standard state of a substance? |
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Answer» The standard state of a substance is that state in which the substance has unit activity or concentration at 25 °C. i.e., For solution having concentration 1 molar, gas at 1 atm, pure liquids or solids are said to be in their standard states. |
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| 1568. |
Define the following terms :(1) Standard electrode potential (2) Standard oxidation potential (3) Standard reduction potential. |
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Answer» (1) Standard electrode potential : It is defined as the difference of electrical potential between metal electrode and the solution around it equilibrium when all the substances involved in the electrode reaction are in their standard states of unit activity or concentration at constant temperature. (2) Standard oxidation potential : It is defined as the difference of electrical potential between metal electrode and the solution around it at equilibrium due to oxidation reaction, when all the substances involved in the oxidation reaction are in their standard states of unit activity or concentration at constant temperature. (3) Standard reduction potential : It is defined as the difference of electrical potential between metal electrode and the solution around it at equilibrium due to reduction reaction, when all the substances involved in the reduction reaction are in their standard states of unit activity or concentration at constant temperature. |
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| 1569. |
What is the standard potential of an electrode according to IUPAC convention? |
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Answer» Standard reduction potential : According to IUPAC convention, the standard potential of an electrode due to reduction reaction at 298 K is taken as the standard reduction potential. In this active mass of the substance has unit value. |
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| 1570. |
What is cell potential or emf of a cell? |
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Answer» Cell potential or emf of a cell : It is defined as the potential difference between two electrodes, responsible for an external flow of electrons from the left hand electrode at higher potential (anode), to the right hand electrode at lower potential (cathode), when connected to form an electrochemical or galvanic cell. Since there is oxidation reaction at left hand electrode (LHE) or anode and reduction reaction at right hand electrode (RHE) or cathode, emf of the galvanic, Ecell, is given by Ecell = (Eoxi)anode + (Ered)cathode Since by IUPAC conventions, generally reduction potentials are used, hence, for the given cell, (∵Eoxi = - Ered) ∴ Ecell = (Ered)cathode - (Ered)anode Similarly, standard emf of the cell, E0cell is given by E0cell = (E0red)cathode - (E0red)anode |
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| 1571. |
In what way fuel cell differs from ordinary galvanic cells? |
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Answer» 1. Fuel cell is a modified galvanic cell in which the thermal energy of combustion reactions is directly converted into electrical energy. 2. In the fuel cell, the reactants are not placed within the cell like ordinary galvanic cells, but they are continuously supplied to the electrodes from outside reservoir. 3. They cannot be recharged unlike ordinary galvanic cell. |
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| 1572. |
What are the applications of the fuel cells? |
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Answer» 1. Fuel cells have been used in the space programme providing electrical energy for a long duration. 2. The fuel cells have been used in automobiles on experimental basis. 3. In case of H2 - O2 fuel cell, used in spacecraft, the water produced is used for drinking for astronauts. 4. The fuel cells using methanol as a fuel for combustion are used in electronic products such as cell phones and laptop computers. 5. The fuel cells have many potential applications as power generators for domestic and industrial uses. |
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| 1573. |
Total charge required to convert three moles of `Mn_(2)O_(4)` to `MnO_(4)^(c-)` in present of alkaline mediumA. `10F`B. `20F`C. `30F`D. 40F` |
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Answer» Correct Answer - c `(Mn^(3//+8))_(3) rarr 3Mn^(6+)+10e^(-)` `:. 10` Faraday charge is required for conversion of `1 mol of Mn_(3)O_(4)` to `MnO_(4)^(2-)`. |
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| 1574. |
The `E_(cell)` for a given cell is `1.2346` and `1.2340V` at `300K ` and `310K`, respectively. Calculate the change in entropy during the cell reaction if the redox change involves three electrons.A. `-17.37 J K^(-1)`B. `+17.37 J K ^(-1)`C. `173.7J K^(-1)`D. `5.79 J K^(-1)` |
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Answer» Correct Answer - a By `DeltaS=(DeltaH-DeltaG)/(T),` `DeltaH=-nF[E-T((delE)/(delT))_(P)]` and `DeltaG=-nEF` `DeltaS=-nF((delE)/(delT))_(p)=3xx96500xx(-(0.0006)/(10))` `=-17.37 J K^(-1)` |
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| 1575. |
The `E^(c-)` for `Cu^(2+)//Cu^(o+),Cu^(o+)//Cu,Cu^(2+)//Cu,` are `0.15V,0.50V` and `0.325V`, respectively. The redox cell showing redox reaction `2Cu^(o+) rarr Cu^(2+)+Cu` is made. `E^(c-)` of this cell reaction and `DeltaG^(c-)` may beA. `E^(c-)=0.175V or E^(c-)=0.350V`B. `n=1 or 2`C. `Delta G^(c-)=-33.775k J`D. All of these |
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Answer» Correct Answer - d `2Cu^(o+) rarr Cu^(2+)+Cu,` Cell `I:Cu^(o+) rarr Cu^(2+)+e)` `Cu^(o+)+e rarr Cu ` `ulbar(2Cu^(o+) rarr Cu^(2+)+Cu)` `E^(c-)=-0.15+0.50` `=+0.35V` `n=1` Cell `II:Cu rarr Cu^(2+)+2e` `2Cu^(o+)+2e rarr2Cu` `ulbar (2Cu^(o+) rarr Cu^(2+)+Cu)` `E^(c-)=-0.325+0.50` `=+0.175V` `n=2` |
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| 1576. |
A curent of `3A` was passed for 1 hour through an electrolyte solution of `A_(x) B_(y)` in water. If `2.977g ` of `A(` atomic weight `106.4)` was deposited at cathode and `B` was a monovalent ion, the formula of electrolyte wasA. `AB_(2)`B. `AB`C. `AB_(3)`D. `AB_(4)` |
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Answer» Correct Answer - d `(W)/(E_(w))=(It)/(96500)(A^(y+)+ye^(_)rarr A)` `(2.977)/(106.4/y)=(3xx1xx60xx60)/(96500)` `:.y=4implies` Electrolyte is `AB_(4)`. |
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| 1577. |
Passage of 5400 C of electricity through an electrolyte deposited 5.954 × 10-3 kg of the metal with atomic mass 106.4. The charge on the metal ion is :(a) + 1 (b) + 2 (c) + 3 (d) + 4 |
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Answer» Option : (a) + 1 |
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| 1578. |
`10800 C` of electricity passed through the electrolyte deposited `2.977g` of metal with atomic mass `106.4 g mol^(-1)`. The charge on the metal cation isA. `+4`B. `+3`C. `+2`D. `+1` |
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Answer» Correct Answer - a `M^(n+)+n e^(-) rarr M` Mole of metal deposited `=(2.977)/(106.4)=0.27mol` `1 mol `of `M` deposited = `n `Faraday `=nxx96500C.` `0.027 mol =nxx965--xx0.027` `:. Nxx96500xx0.027=10800C` `:. n=4` |
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| 1579. |
Predicating the products of electrolysis: What do you expect to be the half reactions in the electrolysis of aqueous `Na_(2)So_(4)` solution. Strategy: Before we look at the electrode reations, we should consider the following facts: (1) Since `Na_(2)So_(2)` does not hydrolyze in water, the `pH`of the solution is close to `7. (2)` The `Na^(+)` ions are not reduced at the cathode and the `SO_(4)^(2-)` ions are not oxidized at the anode. These conclusions are drawn from the electrolysis of water in the presence of sulphuric acid in aqueous sodium chloride solution. |
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Answer» The electrode reaction are Anode: `2H_(2)(l) rarr O_(2)(g) + 4H^(+)(aq.)+4e^(-)` Cathode: `2H_(2)O(l)+2e^(-) rarr H_(2)(g)+2OH^(-)` The overall reaction, obtained by doubling the cathode reaction coefficients and adding the result to the anode reaction, is `6H_(2)O(l) rarr 2H_(2)(g)+O_(2)(g)+4H^(+)(aq.)+4OH^(-)(aq.)` If the `H^(+)` and `OH^(-)` ions ar allowed to mix, then `4H^(+)(aq.)+4OH^(-)(aq.) rarr (4H_(2))(l)` and the overall reaction becomes `2H_(2)O(l) rarr 2H_(2)(g)+O_(2)(g)` |
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| 1580. |
At infinite dilution the ionic condutanc is maximum forA. `F^(-)`B. `Na^(+)`C. `H^(+)`D. `OH^(-)` |
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Answer» Correct Answer - C Since the speed of an ion varies with the potenital applied, it is better to use the term ionic mobility which is defined as the distance travelled by an ion per second under a potential gradient of `1` volt per metre. Potential gradient is given by the potential difference applied at the electrodes divided by the distance between the electrodes. The ionic mobility of `H^(+)` ions is found to be five to ten times that of other ions, expect `OH^(-)` ion. The hydrogen ion, because of its small size has high charge density and is heavily hydrated. Thus, we expect the mobility of `H^(+)` ion to be low rather than high. The high mobility of `H^(+)` ions is hydroxlytic solvents such as `H_(2)O` can be explained by Grouttus type meachanism in which a proton moves raopidly from `H_(3)O^(+)` to a hydrogen-bonded water molecule and is transferred along a series of hydrogen bonded water molecules by a rarrangement of hydrogen bonds (just like a relay race). Amongest negatives ions, `OH^(-)` has the highest ionic mobility. |
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| 1581. |
At cathode, the electrolysis of aqueous `Na_(3)SO_(4)` givesA. `Na`B. `H_(2)`C. `SO_(3)`D. `SO_(2)` |
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Answer» Correct Answer - B Calomel electrode used as a reference electrdrode uses `Hg_(2)Cl_(2)`. |
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| 1582. |
The molar conductance is given by the relation (M `=` concentration in molarity)A. `Lambda_(m) = 1000(k)/(M)`B. `Lambda_(m) = 1000(M)/(k)`C. `Lambda_(m) = 1000 kM`D. `Lambda_(m) = 1000 (k)/(M^(2))` |
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Answer» Correct Answer - A Molar conductivity `(Lambda_(m))` is the conductivity per unit molar concentration and is given by the expression: `Lambda_(m) = (kappa)/(c)` where `c` is in mol `m^(-3)` units if `kappa` is expressed in `S m^(-)` and in mol `cm^(-3)` units it k is expressed in `S cm^(-1)`. Thus molar conductivity is usually expressed in `S m^(2) cm^(-1)` or `S cm^(2) mol^(-1)`. `Lambda_(m)(S m^(2) mol^(-1))= (k(S m^(-1)))/("molarity" (mol L^(-1)) xx 1000 L m^(-3))` `Lambda_(m)(S m^(2) mol^(-1))= (k(S m^(-1)) xx 1000 cm^(3) L^(-1))/("molarity" (mol L^(-1)) xx 1000 L m^(-3))` It may be remebers that, `S m^(2) mol^(-1) = 10000 S cm^(2) mol^(-1)` |
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| 1583. |
The unit of ionic mobility areA. `cm^(2) "volt"^(-1)s^(-1)`B. `cm^(-2) "volt"^(-2)s^(-1)`C. `cm^(-1) "volt"^(-1)`D. `cm^(2) "volt"^(-2)s^(-1)` |
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Answer» Correct Answer - A Mathematically ionic mobility `= ("ionic velocity")/("potential gradient")` `= ("ionic velocity (cm/s)")/("pot, difference (volt)/distance between the electrodes (cm)")` Thus, the unit of ionic mobility is `cm^(2) "volt"^(-1)s^(-1)` |
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| 1584. |
Assertion: `E_(Ag^(+)//Ag)^(@)` increases with increase in concentration of `Ag^(+)` ions. Reason: `E_(Ag^(+)//Ag)` has a positive value.A. Both assertion and reason are true and the reason is the correct explanation of assertion.B. Both assertion and reason are true and the reason is not the correct explanation of assertion.C. Assertion is true but the reason is false.D. Both assertion and reason are false. |
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Answer» Correct Answer - B Correct explanation. `Ag^(+)+e^(-)toAg,E_(Ag^(+)//Ag)+E_(Ag^(+)//Ag)^(@)-(RT)/(nF)"log"(1)/([Ag^(+)])` |
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| 1585. |
In a solution of `CuSO_(4)` how much time will be required to preciitate 2 g copper by 0.5 ampere current?A. 12157.48secB. 102secC. 510secD. 642sec |
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Answer» Correct Answer - A `Cu^(3+)+underset(2"mole 2F")(2e^(-))rarrunderset("1mole" 63.4gm)(Cu)` To deposit 63.5gm of copper, electricity needed `=2xx96500C` To d eposit 2 gm of copper, electricity needed `=2xx(96500)/(63.5)xx2=6078.74C` `Q=6078.74C` `1=0.5` ampere Q=It t(in seconds)`=(Q)/(1)=(6078.74)/(0.5)=12157.48s` |
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| 1586. |
In electroefining of copper a minor percentage of gold accumulates inA. Anode mudB. Cathode mudC. ElectrolyteD. Cathode. |
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Answer» Correct Answer - A We know `W=Z lr therefore t=(W)/(Zl)` `=(2xx96500xx2)/(63.5xx0.5)=12157.68s`. |
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| 1587. |
Which of the following is the use of elec trolysis?A. ElectrorefiningB. ElectroplatingC. Both A and BD. None of the above. |
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Answer» Correct Answer - A See detaisl on electrolysis given in review. |
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| 1588. |
In the cell `Zn|Zn^(2+)||Cu^(2+)|Cu`, the negaitve terminal isA. CuB. `Cu^(2+)`C. `Zn` is cheaper than ironD. `Zn^(2+)` |
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Answer» Correct Answer - C In electrochemical cell anode is taken as negative electrode. |
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| 1589. |
In the cell `Zn|Zn^(2+)||Cu^(2+)|Cu`, the negaitve terminal isA. CuB. `Cu^(2+)`C. `Zn`D. `Zn^(2+)` |
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Answer» Correct Answer - C In an electrochemical cell, anode (Zn) is a negative terminal |
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| 1590. |
For the galvanic cell, `Cu|Cu^(2+)"||"Ag^(+)|Ag`. Which of the following observations is not correct ?A. Cu acts as anode and Ag acts as cathode.B. Ag electrode loses mass and Cu electrode gains mass.C. Reaction at anode, `Cu to Cu^(2+) + 2e^-`D. Copper is more reactive than silver . |
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Answer» Correct Answer - B Cu electrode loses mass and Ag electrode gains mass. |
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| 1591. |
The increase in the molar conductivity of `HCl` with dilution is due toA. Increase in the self ionization of waterB. Hydrolysis of `HCl`C. `Decrease in the self ionization of waterD. Decrease in the interionic forces. |
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Answer» Correct Answer - d Factual statement. |
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| 1592. |
The specific conductance `(k)`of a `0.1M NaOH` solution is `0.0221 S cm^(-1)`. On addition of an equal volume `(V)` of `0.10 M HCI` solution, the value of specific conductance `(k)` falls to `0.0056 S cm^(-1)`. On further addition of same volume `(V)` of `0.10 M HCI`, the value of k rises to `0.017 S cm^(-1)`. calculate equivalent conductivity `(^^)` for `HCI` in `S cm^(2) mol^(-1)`. |
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Answer» Correct Answer - 9 On further adding `V` volume `0.10 M HCI`, concentration of `NaCI` and `HCI` in the final solution are: `[NaCI] = (0.05 xx2)/(3) = (1)/(30)M` `[HCI] = (0.1xx1)/(3) = (1)/(30)M` Also, since specific conductivity is additive: `k = K_(NaCI) +K_(HCI)` `rArr K_(HCI) = 0.017 - 0.0056` `= 0.0114 S cm^(-1)` `rArr ^^_(HCI) = (0.0114 xx 1000)/((1)/(30)) = 342 S cm^(2) mol^(-1)` So ans is `3 +4 +2 = 9` |
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| 1593. |
The units of cell cosntnat areA. `ohm^(-1)`B. `ohm-cm`C. `cm^(-1)`D. `ohm^(-1) cm^(2) eq^(-1)` |
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Answer» Correct Answer - C Cell constant `=(l)/(a)=(cm)/(cm^(2))=cm^(-1)` |
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| 1594. |
The increase in the molar conductivity of `HCl` with dilution is due toA. increase in the self ionisation of waterB. decrease in the self ionisation of waterC. decrease in the interionic forcesD. None of the above |
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Answer» Correct Answer - C Due to increase in volume of water, interionic forces of `HCl` wil decrease. Hence, ionic mobility will increase and consequently will also increase. |
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| 1595. |
For any sparingly soluble salt `[M(NH_(3))_(4)Br_(2)] H_(2)PO_(2)` Given: `lambda_(M(NH_(3))_(4)Br_(2)^(+))^(@) = 400 Sm^(2) - mol^(-1)`. `lambda_(H_(2)PO_(2)^(-))^(@) = 100 S-m^(2) -mol^(1)` Specific resistance of saturated `[M(NH_(3))_(4)Br_(2)] H_(2)PO_(2)` solution is `200 Omega-cm` If solubility product constant of the above salt is `10^(-x)`. What will be the value of `x`. |
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Answer» Correct Answer - 16 `^^_(m)^(oo) = K xx(1000)/(m) xx 10^(-6)` `500 = (1)/(200) xx (1000)/(5) xx 10^(-6)` `S = 10^(-8) mol//L` `K_(sp) = S^(2) = 10^(-16)` |
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| 1596. |
The increase in the molar conductivity of `HCl` with dilution is due toA. increase in the self ionisation of waterB. hydrolysis of HClC. decrease in the self ionisation of waterD. decrease in the interionic, forces. |
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Answer» Correct Answer - D HCl is strong electrolyte so only interionic forces decreases. |
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| 1597. |
Solubility of a sparingly soluble salt S , specific conductance K , and the equivalent conductance `Lambda_(0)` are related asA. `S = (1000 Lambda_(0))/(K)`B. `S = K Lambda_(0)`C. `S = (K)/(1000 Lambda_(0))`D. `S = (1000 K )/(Lambda_(0))` |
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Answer» Correct Answer - D `Lambda = (k xx 1000)/("Molarity")` . For a sparingly soluble salt , `Lambda = Lambda_(0)` and Molarity = Solubility (S) . Hence `S = (k xx 1000)/(Lambda_(0))` |
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| 1598. |
Which one of the following does not represent cell constant correctly ?A. It is the ratio of specific conductance to observed conductanceB. It is the ratio of distance between the electrodes to the area of cross-section of the electrodeC. It is the product of specific resistance and observed resistanceD. It is the product of conductivity and observed resistance . |
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Answer» Correct Answer - C k = `G xx ` cell constant `ne R xx` cell constant . |
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| 1599. |
The incorrect statement isA. Specific conductivity decreases with dilutionB. Equivalent and molar conductivity increase with dilutionC. `Lambda_(oo)` for a weak electrolyte cannot be found by extrapolation of the graph between `Lambda` and concentration of zero concentrationD. Molar conductivity of a strong electrolyte increase with dilution because ionization increases with dilution . |
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Answer» Correct Answer - D Molar conductivity `prop` degree of ionisation . |
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| 1600. |
The dissociation constant of n-butyric acid is `1.6 xx 10^(-5)` and the molar conductivity at infinite dilution is `380 xx 10^(-4) Sm^(2) mol^(-1)`. The specific conductance of the `0.01M` acid solution isA. `1.52 xx 10^(-5) Sm^(-1)`B. `1.52 xx 10^(-2) Sm^(-1)`C. `1.52 xx 10^(-3) Sm^(-1)`D. none |
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Answer» Correct Answer - B `k_(a) = 1.69 xx 10^(-5) = C alpha^(2) = 0.01 alpha^(2) = alpha = 0.04` `alpha =(lambda_(M))/(lambda_(M)^(oo) rArr lambda_(M) = 0.04 xx 380` `lambda_(M) = 15.2 S cm^(2) "mole"^(-1) lambda_(M) = (k xx 1000)/(M)` `15.2 = (k xx 1000)/(0.01)` `k = 1.52 xx 10^(-4) S cm^(-1)` `k = 1.52 xx 10^(-2) S cm^(-1)` |
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