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2551.

Which metal is used as a coating on steel to prevent corrosion:-A. NaB. CaC. KD. Zn

Answer» Correct Answer - D
Discussion
2552.

"Maintenance free" batteries now in use in place of common batteries haveA. Electrode made of lead-lead oxideB. Electrodes made of calcium-containing lead alloyC. Non aqueous solvents as mediumD. Platinum electrodes

Answer» Correct Answer - B
Discussion
2553.

Which substance eliminates bromine from KBr solutionA. `I_(2)`B. `Cl_(2)`C. `HI`D. `SO_(2)`

Answer» Correct Answer - B
Discussion
2554.

A saturated solution in `AgX(K_(sp)=3xx10^(-12))` and `AgY(K_(sp)=10^(-12))` has conductivity `0.4xx10^(-6)Omega^(-1)cm^(-1)`. Given: Limiting molar conductivity of `Ag^+=60Omega^(-1)cm^2mol^(-1)` Limiting molar conductivity of `X^(-)=90Omega^(-1)cm^2mol^(-1)` The conductivity of `Y^(-)` is (in `Omega^(-1)cm^(-1)`):A. `1.45xx10^(-7)`B. `1.45xx10^(-5)`C. `1.45xx10^(-9)`D. None of these

Answer» Correct Answer - A
Discussion
2555.

A saturated solution in `AgX(K_(sp)=3xx10^(-12))` and `AgY(K_(sp)=10^(-12))` has conductivity `0.4xx10^(-6)Omega^(-1)cm^(-1)`. Given: Limiting molar conductivity of `Ag^+=60Omega^(-1)cm^2mol^(-1)` Limiting molar conductivity of `X^(-)=90Omega^(-1)cm^2mol^(-1) The limiting molar conductivity of `Y^(-)` is (in `Omega^(-1)cm^2mol^(-1)`):A. 290B. 2900C. 2.9D. None of these

Answer» Correct Answer - A
Discussion
2556.

Calculate limiting molar conductivity of `CaSO_(4)` given that limiting molar conductivity of calcium and sulphate ions are 119.0 and `160.0" S "cm^(2) mol^(-1)` respectively.

Answer» `Lambda_(CaSO_(4))^(oo)=lambda_(Ca^(2+))^(oo)+lambda_(SO_(4)^(2-))^(oo)=(119.0+160.0)=279 " S "cm^(2) mol^(-1).`
Discussion
2557.

How many Faraday are required to reduce 1 mole of `Cr_(2)O_(7)^(2-)` to `Cr^(3+)` in acid solutions ?A. 2B. 3C. 5D. 6

Answer» Correct Answer - D
(d) The reduction reaction in the acidic medium is :
`Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-) to 2Cr^(3+)+7H_(2)O`
Thus, six Faraday of charge is required.
Discussion
2558.

In the electrolysis of aqueous sodium chloride solution, which of the half cell reaction will occur at anode ?A. `Cl_((aq))^(-)to(1)/(2)Cl_(2)+e^(-)," " E_(cell)^(@)=1.36" volts"`B. `2H_(2)O_((l))to O_(2)+4H^(+)+4e^(-) , E_(cell)^(@)=1.23" volts"`C. `Na_((aq))^(+)+e^(-) toNa_((s)) , " " E_(cell)^(@)=-2.71" volts"`D. `H_((aq))^(+)+e^(-) to (1)/(2)H_(2) , " " E_(cell)^(@)=0.00" volts"`

Answer» Correct Answer - A
(a) It is the correct answer. For details, consult section 3.15 (Electrolysis of aqueous NaCl solution).
Discussion
2559.

In the electrolysis of aqueous sodium chloride solution which of the hall cell reaction will occur at anode?A. `Na_((aq))^(+) + e^(-) to Na_((s)) , E_"cell"^@`=-2.71 VB. `2H_2O_((l)) to O_(2(g)) + 4H_((aq))^(+) + 4e^(-) , E_"cell"^@`=1.23 VC. `H_((aq))^(+) + e^(-) to 1/2H_(2(g)) , E_"cell"^@` =0.00 VD. `Cl_((aq))^(-) to 1/2Cl_(2(g)) + e^(-) ,E_"cell"^@` =1.36 V

Answer» Correct Answer - D
At the anode, the following oxidation reactions are possible :
(d)`Cl_((aq))^(-) to 1//2Cl_(2(g)) + e^(-) , E_"cell"^@`=1.36 V
(b)`2H_2O_((l)) to O_(2(g)) + 4H_((aq))^+ + 4e^(-) , E_"cell"^@`=1.23 V
The reaction at anode with lower value of `E^@` is preferred and therefore , water should get oxidised in preference to `Cl_((aq))^-` However, on account of overpotential of oxygen , reaction (d) is preferred.
Discussion
2560.

In the electrolysis of aqueous sodium chloride solution which of the half cell reaction will occur at anode?

Answer»

(ii) 2H2O(l) → O2(g) + 4H+(aq) + 4e– ; EΘCell = 1.23V
Discussion
2561.

Λ0m is equal to ______________.

Answer»

(ii)  Λ0m(NH4Cl) + Λ0m(NaOH) - Λ0(NaCl)
Discussion
2562.

The cell constant of a conductivity cell _____________.(i) changes with change of electrolyte.(ii) changes with change of concentration of electrolyte.(iii) changes with temperature of electrolyte.(iv) remains constant for a cell.

Answer» (iv) remains constant for a cell.
Discussion
2563.

While charging the lead storage battery:A. `PbSO_(4)` anode is reduced to Pb.B. `PbSO_(4)` cathode is reduced to Pb.C. `PbSO_(4)` cathode is oxidised to Pb.D. `PbSO_(4)` anode is oxidised to `PbO_(2)`

Answer» Correct Answer - A
`PbSO_(4)` (anode) is reduced to Pb. For more details, consult section 3.21.
Discussion
2564.

While charging the lead storage battery:A. `PbSO_(4)` anode is reduced to PbB. `PbSO_(4)` cathode is reduced to PbC. `PbSO_(4)` cathode is oxidised to PbD. `PbSO_(4)` anode is oxidised to `PbO_(2)`

Answer» While charging the lead storage battery the reaction occuring on cell is reversed and `PbSO_(4)(s)` on anode and cathode is converted into Pb and `PbO_(2)` respectively
Hence, opiton (a) is the correct choice
The electrode reactions are as follows
At cathode `PbSO_(4)(s)+2e^(-)rarrPb(s)+SO_(4)^(2-)(aq)` (Reduction)
At anode `PbSO_(4)(s)+2H_(2)OrarrPbO _(2)(s)+SO_(4)^(2-)+4H^(+)+2e^(-)` (Oxidation)
Overall reaction `2PbSO_(4)(s)+2H_(2)OrarrPb(s)+PbO_(2)(s)+4H^(+)(aq)+2SO_(4)^(2-)(aq.)`
Discussion
2565.

While charging the lead storage battery ______________.(i) PbSO4 anode is reduced to Pb.(ii) PbSO4 cathode is reduced to Pb.(iii) PbSO4 cathode is oxidised to Pb.(iv) PbSO4 anode is oxidised to PbO2.

Answer»

(i) PbSO4 anode is reduced to Pb.
Discussion
2566.

While charging the lead storage batteryA. `PbSO_(4)` anode is reduced to Pb.B. `PbSO_(4)` cathode is reduced to Pb.C. `PbSO_(4)` cathode is oxidised to Pb.D. `PbSO_(4)` anode is oxidised to `PbO_(2)`.

Answer» Correct Answer - A
`PbSO_(4)` deposited on anode is reduced to Pb. This electrode now acts as cathode during charging .
Discussion
2567.

In the electrolytic refining of zinc,A. Graphite is at the anodeB. The impure metal is at the cathodeC. The metal ion gets reduced at the anodeD. Acidified zinc sulphate is the electrolyte

Answer» Correct Answer - D
The impure zinc is purified electrolytically. The impure metal is made anode and cathode consists of sheets of pure aluminium. A solution of zinc sulphate acts as an electrolyte. Zinc dissolves from anode and deposits on cathode when electric current is passed.
Discussion
2568.

Copper reduced `NO_(3)^(-)` into `NO` and `NO_(2)` depending upon conc.of `HNO_(3)` in solution. Assuming `[Cu^(2+)] = 0.1M`, and `P_(NO) = P_(NO_(2)) = 10^(-3)` atm and using data answer the following questions: `E_(Cu^(2+)//Cu)^(@) =+ 0.34` volt, at `298K (RT)/(F) (2.303) = 0.06` volt `E_(cell)` for reduction of `NO_(3)^(-) rarr NO` by `Cu(s)` when `[HNO_(3)] = 1M` is [At `t = 298]`A. `10^(1.23)M`B. `10^(0.56)M`C. `10^(0.66)M`D. `10^(0.12)M`

Answer» Correct Answer - C
`(e^(-) +2H^(+) +NO_(3)^(-) rarr NO_(2) +H_(2)O) xx 2`
`Cu rarr Cu^(+2) +2e^(-)`
`4H^(+) +2NO_(3)^(-) +Cu rarr Cu^(+2) +2NO_(2) +2H_(2)O`
`E = 0.79-0.34 -(0.0591)/(2) log.([NO_(2)]^(2)[Cu^(+2)])/([NO_(3)^(-)]^(2)[H^(+)]^(4))`..(ii)
Let `[HNO_(3)] = xM` so `[H^(+)] = [NO_(3)^(-)] = x`
equation (i) & (ii)
`0.62 -(0.0591)/(2) log.(10^(-9))/(x^(10)) = 0.45 -(0.0591)/(2)log.(10^(-7))/(x^(6))`
`0.17 =(0.591)/(6) [-9-10 logx] -(0.0591)/(2)[-7-6logx]`
`0.0518 = 0.0788 logx`
`x = 10^(0.657)M`
Discussion
2569.

Copper reduced `NO_(3)^(-)` into `NO` and `NO_(2)` depending upon conc.of `HNO_(3)` in solution. Assuming `[Cu^(2+)] = 0.1M`, and `P_(NO) = P_(NO_(2)) = 10^(-3)` atm and using data answer the following questions: `E_(Cu^(2+)//Cu)^(@) =+ 0.34` volt, at `298K (RT)/(F) (2.303) = 0.06` volt `E_(cell)` for reduction of `NO_(3)^(-) rarr NO` by `Cu(s)` when `[HNO_(3)] = 1M` is [At `t = 298]`A. `~0.61`B. `~0.71`C. `~0.51`D. `~0.81`

Answer» Correct Answer - B
`(3e^(-) +4H^(+) +NO_(3)^(-) rarr NO +2_(2)O) xx2`
`(Cu rarr Cu^(+2) +2e^(-)) xx 3`
`E = 0.96 -0.34 -(0.0591)/(6) log.([NO]^(2)[Cu^(+2)]^(3))/([NO_(3)^(-)]^(2)[H^(+)]^(8))` ..(i)
since `[HNO_(3)] = 1M` so `[H^(+)] = [NO_(3)^(-)] = 1`
`E = 0.62 -(0.0591)/(6) log (10^(-3))^(2) (0.1)^(3) = 0.70865`
`E = 0.62 -(0.0591)/(6)log (10^(-3))^(2) (0.1)^(3) = 0.70865`
Discussion
2570.

how many grams of cobalt metal will be deposited when solution of cabalt (II) chloride is electrolyzed with a current of 10 amperes for 109 minutes (1Faraday=96,500 C, Atomic mass of Co=59u)A. 4B. 20C. 40D. 0.66

Answer» Correct Answer - B
`w=Zit`
`thereforew=(E_(B)xxIt)/(96500) " "(thereforeE_(B)=(M_(B))/(Z)=(59)/(2))`
`w=(59)/(2)xx(10xx109xx60)/(96500)`
w=19.9
Discussion
2571.

In a metal oxide, there is 20% oxygen by weight. Its equivalent weight isA. 40B. 64C. 72D. 32

Answer» Correct Answer - D
`O_(2)%=20%`
Metal%=80%=`(80)/(20)xx8=32`g of metal.
Discussion
2572.

A cell reaction would be spontaneous if the cell potential and `/_\_rG` are respectively:A. positive and negativeB. negative,negativeC. zero,zeroD. positive,zero

Answer» Correct Answer - a
Discussion
2573.

If the equilibrium constant for the reaction `H^+(aq)+OH^-)(aq)iffH_2O(l)` is `10^13` at certain temperature then what is the `E^(@)` for the reaction, `2H_2O(l)+2e^-)iffH_2(g)+2OH^(-)(aq)`A. 4.74VB. 0.547VC. 4.37VD. 1.09V

Answer» Correct Answer - B
Discussion
2574.

A fuel cell develops an electrical potential from the combustion of butane at 1 bar and 298 K `C_4H_(10)(g)+6.5O_2(g)to4CO_2(g)+5H_2O(l), /_\_rG^(@)=-2746kJ//mol` what is `E^(@)` of a cell?A. 4.74VB. 0.547VC. 4.37VD. 1.09V

Answer» Correct Answer - D
Discussion
2575.

The dissociation constant of a weak acid is `1.6xx10^(-5)` and the molar conductivity at infinite dilution is `380xx10^(-4)` S`m^2mol^(-1)` . If the cell constant is 0.01`m^(-1)` then conductace of 0.1M acid solution is :A. `1.52xx10^(-5)`SB. 1.52 SC. `1.52xx10^(-3)`SD. `1.52xx10^(-4)`S

Answer» Correct Answer - B
Discussion
2576.

What is the % dissociation of `CH_(3)COOH`, if equivalent conductivity at this dilution and at infinite dilution are 148 and 398 S `cm^(2)//mol` respectivelyA. `37.18%`B. `47%`C. `48.78%`D. `10.8%`

Answer» Correct Answer - A
Degree of dissociation, `alpha`
`=("Equivalent conductance at concentration C")/("Equivalent conductance at infinite dilution")`
`=(148)/(398)=0.3718`
i.e., `CH_(3)COOH` is 37.18% dissociated.
Discussion
2577.

For the electrochemicl cell, `M|M^(+)||X^(-)|X E_((M^(+)//M))^(@) = 0.44 V` and `E_((X//X^(-)))^(@) = 0.33 V` From this data one can deduce that :A. `M + X rarr M^(+) + X^(-)` is the spontaneous reactionB. `M^(+) + X^(-) rarr M + X` is the spontaneous reactionC. `E_(cell) = 0.77 V`D. `E_(cell) = -0.77 V`

Answer» Correct Answer - B
`E_(cell) = E_(OP_(M//M^(+)))^(@) + E_(RP_(X//X^(-)))^(@)`
`= -0.44 + 0.33 = -0.11 V` for
`M + X rarr M^(+) + X^(-)`. Thus reaction is non-spontaneous. The spontaneous reaction in `M^(+) + X^(-) rarr M + X, E^(@) = 0.11 V`
Discussion
2578.

The conductivity of a saturated solution of `BaSO_4` is ` 3. 06 xx 10^(-6) "ohm"^(-1) cm^(-1)` and its equivalent conductance is `1.53 "ohm"^(-1) cm^2 equiv^(-1)`. The `K_(sp)` for `BaSO_4` will be .A. ` 4 xx 10^(-12)`B. `2.5xx10^(13)`C. `25xx10^9`D. `10^(-6)`

Answer» Correct Answer - D
` 1.53 = (1000 xx 3 .06 xx 19^(-6))/("Normality")`
Normality `= 2 xx 10^(-3) N`
Molaroty `= (2 xx 10^(-3))/2 =10^(-3) M`
`K_(sp) = 10^(-6) M`.
Discussion
2579.

The conductance `(G)` is the reciprocal ofA. concentrationB. currentC. resistanceD. potential difference

Answer» Correct Answer - C
The reciprocal of resistance offered by a conductor (electrolyte or metal) to the flow of electrocity through it is known conductance `(G)`, the ease with which electricity flows through the conductor.
Condutance `(G) = (1)/("Resistance (R)")`
The `SI` unit of conductance is siements (`ohm^(-1)` or `mho`).
Discussion
2580.

The correct order of equivalent conductance at infinite dilution of LiCl, NaCl and KCl isA. LiClgtNaClgtKClB. KClgtNaClgtLiClC. NaClgtKClgtLiClD. LiClgtKClgtNaCl

Answer» Correct Answer - B
As we go down the group (i.e., from `Li^(+)` to `K^(+)`), the ionic radius increases, degree of solvation decreases and hence effective size decreases resulting into increase in ionic mobility. Hence equivalent conductance at infinite dilution increases in the same order.
Discussion
2581.

For the electrochemicl cell, `M|M^(+)||X^(-)|X E_((M^(+)//M))^(@) = 0.44 V` and `E_((X//X^(-)))^(@) = 0.33 V` From this data one can deduce that :A. M+X `rarr M^(+)+X^(-)` is the spontaneous reactionB. `M^(+)+X^(-)rarrM+X` is the spontaneous reactionC. `E_("cell")=-0.77V`D. `E_("cell")=0.77V`

Answer» Correct Answer - b
Discussion
2582.

The correct order of equivalent conductance at infinite dilution of `LiCl, NaCl` and `KCl` is:A. `LiCI gt NaCI gt KCI`B. `KCI gt NaCI gt LiCI`C. `NaCI gt KCI gt LiCI`D. `LiCI gt KCI gt NaCI`

Answer» Correct Answer - B
In moving down the group (1), as the size of ion increases the solution decreases. `Li^(o+)` smallest ion, is heavily hydrated and due to solution, its size increases and electrical conductance decreases The correct order of electrical conductance is `KCI gt NaCI gt LiCI`.
Discussion
2583.

Under which of the following conditions, conductance conductivity and equivalent conducticity are all equal?A. `1 C C` of the solution contanis `1` eq. of the electrolyteB. `10 C C` of the solution contains `1` eq. of the electrolyteC. `100 C C` of the solution contains `1` eq. of the electrolyteD. `1000 C` of the solution contains `1` eq. of the elelctrolytes

Answer» Correct Answer - A
Imagine `1 C C` of a solution of an electrolyte placed between two large electrode `1 cm` apart. The cross-sectional area of the solution will be `1 cm^(2)`. The conductance of the solution `(G = kappa.l//A)` with evidently be its specific conductance because we are having one `cm` cube of the solution. Further, suppose that `1 CC` of the solution contains `1` gram equivalent of the electrolute dissolved in it. Then, according to the definition of equivalent conductance (the conducting power of all ions produced by one equivalent of the electrolyte in the given solution), the conductance, `Lambda_(eq.)` Thus, in this case
Conducatance (G) = Specific conductance `(kappa)`
= Equivalent conducatance `(Lambda_(eq.))`
Discussion
2584.

Which of the following is an insulator ?A. GraphiteB. AluminiumC. DiamondD. Silicon

Answer» Correct Answer - C
Diamond is an insulator .
Discussion
2585.

Which of the following statement is wrong in the context of molar conductance?A. The solution should contain one mole of the electrolyteB. The distance the electrodes should be `1 cm`C. The area of he electrodes should be large enugh for the solution to touch completelyD. The volume of the solution should be very small.

Answer» Correct Answer - D
Molar conductance is the conductance of a solution between two electrode spaced one centrimeter apart which enclose one mole of the electrolyte between them.
Since the conductivity, `kappa` is the conductance of the cubic centimeter of solution between one centimeter apart, it follows that
`Lambda_(m) = kappaV`
wherer `V`(dilution) is the volume in `cm^(3)` containing one mole of electrolyte.
Discussion
2586.

The units of equivalent conductivity areA. Ohm cmB. Siemens (S)C. `Ohm^(-1) cm^(2)`D. `Ohm^(-1) cm^(-2)`

Answer» Correct Answer - C
Unit of equivalent conductance are `ohm^(-1) cm^(2)` .
Discussion
2587.

The correct order of equivalent conductance at infinite dilution of `LiCl, NaCl` and `KCl` is:A. `LiClgtNaClgtKCl`B. `KClgtNaClgtLiCl`C. `NaClgtKClgtLiCl`D. `LiClgtKClgtNaCl`

Answer» Correct Answer - b
Discussion
2588.

The cell constant is given byA. `kappa/R`B. `A//l`C. `l//A`D. `l//AR`

Answer» Correct Answer - C
The cell constant, which is a constant as far as thed same conducativity cell is used is the ratio of length `(l)`, the distance between the metal plates and area `(A)`, the cross-sectional area of the metal plates. It is usually represented by the symbol `k_("cell")`.
Recall that
Conductivity = conductance `xx` cell constant
Thus
`K_("cell") = kappa//G` or `kappaR`
Discussion
2589.

The correct order of equivalent conductance at infinite dilution of LiCl, NaCl, and KCl isA. `LiCl gt NaCl gt KCl`B. `KCl gt NaCl gt LiCl`C. `NaCl gt KCl gt LiCl`D. `LiCl gt KCl gt NaCl`

Answer» Correct Answer - B
(b) The ionic character decreases as :
`KCl gt NaCl gt LiCl`
Discussion
2590.

Metals have conductivity in the order of `(ohm^(-1) cm^(-1))`A. `10^(4)`B. `10^(5)`C. `10^(-10)`D. `10^(12)`

Answer» Correct Answer - B
Metals are good conductors of electricity and have conductivties in the order `10^(7)(Omegam^(-1))` or `10^(5)(Omegam^(-1))`.
Discussion
2591.

Which of the following solutions of `KCl` has the lowest value of equivalent conductance?A. `0.001 M`B. `0.01 M`C. `0.1 M`D. `1 M`

Answer» Correct Answer - D
Equivalent conductance is inversely proportional to the normality of solution. Since normality is directly related to molarity `(N = n_("factor")M)`, we can say the equivalent conductance is inversely proprotional to the molarity of solution.
Discussion
2592.

Which of the following solutions has the highest equivalent conductance?A. 0.1 M NaClB. 0.050 M NaClC. 0.005 M NaClD. 0.02 M NaCl

Answer» Correct Answer - C
With dilution , the equivalent conductance increases .
Discussion
2593.

Electrolytic conduction differs from metallic conduction in that in the case of electrolytic conductionA. The resistance increases with increasing temperatureB. The resistance decreases with increasing temperatureC. The flow of current does not generate heatD. The resistance is independent of the length of the conductor.

Answer» Correct Answer - B
Electrolytic conduction resistance decreases with increasing temperature.
Discussion
2594.

Electrolytic conduction is due to the movement of :A. electronsB. ionsC. atomsD. electrons as well as ions

Answer» Correct Answer - B
Discussion
2595.

The units of conductivity of the solution areA. Ohm cmB. Siemens (S)C. S cmD. S `cm^(-1)`

Answer» Correct Answer - D
Units of conductivity are ` S cm^(-1)` .
Discussion
2596.

(a) The molar conductivities at infinite dilution of potassium chloride, hydrochloric acid and potassium acetate are 130.1, 379.4 and 95.6 S `cm^(2)mol^(-1)` respectively. Calculate the value of molar conductivity at infinite dilution for acetic acid. (b) If the molar conductivity of given acetic acid solution is 48.5 S `cm^(2)mol^(-1)` at `25^(@)C`, calculate the degree of dissociation of acetic for acetic acid.

Answer» Correct Answer - `wedge^(@)=344.9ohm^(-1)cm^(2)mol^(-1);alpha=0.141`
`wedge^(@)(CH_(3)COOH)=wedge^(@)(CH_(3)COOK)+wedge^(@)(HCl)-wedge^(@)(KCl)=95.6+379.4-130.1=344.9" S "cm^(2)mol^(-1)`
`alpha=(wedge_(m)^(@))/(wedge_(m)^(@))=(48.5)/(344.5)=0.141`
Discussion
2597.

`Omega^(-1) m^(-1)` is the unit ofA. Molar conductivityB. Specific conductivityC. Equivalent conductivityD. Molar conductivity at infinite dilution .

Answer» Correct Answer - B
`Omega^(-1) m^(-1)` is the unit of specific conductivity .
Discussion
2598.

Which of the following solutions has the highest equivalent conductance?A. 0.01 M NaClB. 0.05 M NaClC. 0.005 M NaClD. 0.02 M NaCl

Answer» Correct Answer - C
Discussion
2599.

Electrolytic conduction differs from metallic conduction in the fact in the case of electrolytic conductionA. the resistance increases with increasing temperatureB. the reistance decreases with increasing temperatureC. the flow of current does not generate heatD. the resistance is independent of the length of the conductor .

Answer» Correct Answer - B
Electrolytic conduction in temperature .
Discussion
2600.

Which one of the following solution will have highest conductivity .A. `0.1 M CH_(3) COOH`B. `0.1 M NaCl`C. `0.1 M KNO_(3)`D. `0.1 M HCl`

Answer» Correct Answer - D
NaCl, `KNO_(3) , HCl` are strong electrolytes but the size of `H^(+)` is smallest . Smaller size of the ions , greater is the conductance and hence greater is the conductivity ( k = `G xx` cell const.)
Discussion