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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.
| 651. |
The oxidation potentials of Zn, Cu, Ag, `H_(2)` and Ni are 0.76, -0.34, 0.80, 0 and 0.25 volt respectively. Which of the following reactions will provide maximum voltage ?A. `Zn+Cu^(2+) rarr Cu + Zn^(2+)`B. `Zn+2Ag^(+) rarr 2Ag+zn^(2+)`C. `H_(2) + Cu^(2+) rarr 2H^(+) + Cu`D. `H_(2)+Ni^(2+) rarr 2H^(+) + Ni` |
| Answer» Correct Answer - B | |
| 652. |
The standard electrode potentials of Zn, Ag and Cu are -0.76, 0.80 and 0.34 volt respectively, then:A. Ag can oxidised Zn and CuB. Ag can reduce `Zn^(2+)` and `Cu^(2+)`C. Zn can reduce `Ag^(+)` and `Cu^(2+)`D. Cu can oxidise Zn and Ag |
| Answer» Correct Answer - C | |
| 653. |
`18.97 g` fused `SnCl_(20` was electrolyzed using inert electrodes. `1.187g Sn` was deposited at cathode. If nothing is obtained during electrolysis, calculate the ration of weight of `SnCl_(2)` and `SnCl_(4)` in fused state after electrolysis. Given`:` Atomic weight of `Sn=118.7, Mw `of `SnCl_(2)=189.7, Mw` of `SnCl_(4)=260.7` |
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Answer» Fused `SnCl_(2)overset(El ectrolysis)rarr Sn^(2+)+2Cl^(c-)` At cathode `: Sn^(2+)+2e^(-)rarr Sn,,,(` Reduction `)` At anode `: 2Cl^(c-)rarr Cl_(2)(g)+2e^(-),,,(` Oxidation `)` Further , `Cl_(2)(g)` formed at anode reacts with left over `SnCl_(2)` to give `SnCl_(4)` . `SnCl_(2) +Cl_(2) rarr SnCl_(4)` During electrolysis`:` Eq of `SncL_(2)` lost `=` Eq of `Cl_(2)` formed `=` Eq of `Sn` formed. `Eq` of `Sn` formed `=(W_(Sn))/(Ew of Sn)` `=(1.187)/(118.7//2) " "Sn^(2+)+2e^(-)rarrSn` =`2xx10^(-2)" "n` factor of `Sn=2` `Eq` of `SnCl_(2)` lost `=Eq` of `Cl_(2)` formed `=Eq `of `SnCl_(4)` formed `=Eq` of `Sn` formed `=2xx10^(-2)` Now total loss is equivalent of `SnCl_(2)` during complete course `=Eq` of `SnCl_(2)` lost during electrolysis is `+ Eq` of `SnCl_(2)` lost during reaction with `Cl_(2)` `=2xx10^(-2)+2xx10^(-2)=4xx10^(-2)` `Eq` of `SnCl(` initially `)=(18.97g)/(189.7//2)=2xx10^(-1)` `Eq` of `SnCl_(2)` left is molten solution `=2xx10^(-1)-4xx10^(-2)=0.16` `Eq` of `SnCl_(4)` formed `=2xx10^(-2)=0.02` `("Weight of SnCl left")/("Weight of SnCl formed ")=("Weight of "SnCl_(2)"left" xx Ew of SnCl_(2))/("Weight of "SnCl_(4) "formed "xx Ew of SnCl_(4))` `=((0.16xx189.7//2))/((0.02xx260.7//2))=5.82` |
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| 654. |
Assertion`(A):` Whne acidified `ZnSO_(4)` solution is electrolyzed between `Zn` electrodes, it is `Zn` that is deposited at the cathode and `H_(2)(g)` is not evolved. Reason `(R):` The electrode potential of `Zn` is more negative than hydrogen as the overpotential for hydrogen evolution in `Zn` is quite large.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 - A | |
| 655. |
`DeltaG` for the reaction `(4)/(3)Al+O_(2)to(2)/(3)Al_(2)O_(3)` is -772kJ `mol^(-1)` of `O_(2)`. Calculate the minimum EMF in volts requird to carry out and electrolysis of `Al_(2)O_(3)` |
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Answer» `AltoAl^(3+)+3e^(-)` `(4)/(3)mol" of "Al=(4)/(3)xx3mol" "e^(-)=4mol" "e^(-)` n=4 `DeltaG=-nFE` `-772xx1000J=-4xx96500xxE" "thereforeE=(772xx1000)/(4xx96500)=2.0V` |
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| 656. |
`DeltaG` for the reaction `:` `(4)/(3) Al+O_(2)rarr (2)/(3)Al_(2)O_(3)` is `-772 kJ mol^(-1)` of `O_(2)`. Calculate the minimum `EMF` in volts required to carry out an electrolysis of `Al_(2)O_(3)` |
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Answer» `Al to Al^(3+)+underset(=4" mol electrons") underset(4//3xx3" mol")(3e^(-))` `DeltaG=-nFE_(cell)` `E_(cell)=-(DeltaG)/(nF)` `DeltaG=772" kJ "mol^(-1)=772xx1000" J "mol^(-1)` `=772xx1000" CV "mol^(-1)` `n=4,=96500" C "mol^(-1)` `E_(cell)=((772xx1000" CV "mol^(-1)))/((4)xx(96500" C "mol^(-1)))=2V` |
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| 657. |
Identify the cathode, anode and calculate cell emf when a cell is constructed with `A (E^(@) = -0.44 V) and B (E^(@) = 0.337 V)` |
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Answer» (i) Comparison of standard electrode potentials of A and B (ii) Relation between standard reduction potential of a metal and its tendency to get oxidised/reduced. (iii) Identification of anode and cathode based on the relative tendency of getting oxidised/reduced. (iv) Calculation of cell emf. |
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| 658. |
The correct representation of Nernst`s equaion is .A. `E_(M^(n+) //M) = E_( M^(n+)//M)^@ + n/( 0.591) log (M^(n+))`B. `E_(M^(n+)) //M) = E_(M^(n+)//M)^@ + n/( 0.591) log (M^(n+))`C. `E_(M^(n+) //M) = E_(M^(n+)//M)^@ + n/( 0.591) log (M^(n+))`D. None of the above |
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Answer» Correct Answer - C ` E=E^@ - (RT)/(nF) In 1/([M^(n+)]) , E=E^@ + (RT)/(nF) In [M^(n+)]` `E=E^@ + ( 2.303 RT)/(nF) log [M^(n+)]` Substituting the value of `R,T (298 K)` and `F` we get `E=E^@ + ( 0.591)/n log (M^(n+))`. |
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| 659. |
The electrical resistance of a column of 0.05 M NaOH solution of diameter 1 cm and length 50 cm is `5.55xx10^(3)`ohm. Calcalate the resistivity, conductivity and molar conductivity. |
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Answer» (i) Calculation of resistivity. Electrical resistance of the solution, `R=5.55xx10^(3)Omega` Area of cross-section of the column `(a)=pir^(2)=3.14xx((1)/(2))^(2)cm^(2)=0.785cm^(2)` Length of the column `(l)=50cm` Applying the formula `R=rho(l)/(a)` `rho=(Rxxa)/(l)=((5.55xx10^(3)Omega)(0.785cm^(2)))/(50cm)=87.135Omegacm` i.e., Resistivity `(rho)=87.135Omega` cm (ii) Calculation of conductivity. Conductivity `(kappa)=(1)/(rho)=(1)/(87.135Omega" cm")=0.01148" S "cm^(-1)` (iii) Calculation of molar conductivity Molar conductivity `(wedge_(m))=(kappaxx1000)/("Molarity")=((0.01148" S "cm^(-1))(1000cm^(3)L^(-1)))/((0.05" mol "L^(-1)))=229.6" S "cm^(2)mol^(-1)` |
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| 660. |
The emf of the cell, `Zn|Zn^(2+) (0.01 M)||Fe^(2+) (0.001 M)|Fe` at 298 K is 0.2905 V then the value of equilibrium constant for the cell reaction is :A. `e^(0.32//0.0295)`B. `10^(0.32//0.0295)`C. `10^(0.26//0.0295)`D. `10^(0.32//0.0591)` |
| Answer» Correct Answer - B | |
| 661. |
The molar conductivity of a solution of a weak acid HX (0.01M) is 10 times smaller than the molar conductivity of a solution of a weak aid HY (0.10M). If `lamda_(x^(-))^(0)=lamda_(y-)^(0),` the difference in their `pK_(a)` values, `pK_(a)(HX)-pK_(a)(HY)`, is (consider degree of ionization of both acids to be ltlt1). |
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Answer» Correct Answer - C `lamda_(x-)^(0)=lamda_(y-)^(0)` `implieslamda_(H^(+))^(o)+lamda_(X^(-))^(o)=lamda_(H^(+))^(o)+lamda_(Y^(-))^(o)implieslamda_(HX)^(o)=lamda_(HY)^(o)` . . . (i) Also `(lamda_(m))/(lamda_(m)^(o)=alpha,` so `lamda_(m)(HX)=lamda_(m)^(o)alpha_(1) and lamda_(m)(HY)=lamda_(m)^(o)alpha_(2)` (where `alpha_(1)` and `alpha_(2)` are degrees of dissociation of HX and HY respectively). Now, given that `lamda_(m)(HY)=10lamda_(m)(HZ)implieslamda_(m)^(o)alpha_(2)=10xxlamda_(m)^(o)alpha_(1)` `alpha_(2)=10alpha_(1)` . . . (ii) `K_(a)=(Calpha^(2))/(1-alpha)`, but `alpha ltlt 1`, therefore, `K_(a)=Calpha^(2)` `implies(K_(a)(HX))/(K_(a)(HY))=(0.01alpha_(1)^(2))/(0.1alpha_(2)^(2))=(0.01)/(0.1)xx((1)/(10))^(2)=(1)/(1000)` `implieslog(K_(a)(HX)-pK_(a)(HY))=-3` `impliespK_(a)(HX)-pK_(a)(HY)=3` |
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| 662. |
The molar conductivity of a solution of a weak acid HX (0.01M) is 10 times smaller than the molar conductivity of a solution of weak acid HY (0.10M). If `lamda_(X^(-))^(@)=lamda_(Y^(-))^(@)`, the difference in their `pK_(a)` values, `pK_(a)(HX)-pK_(a)(HY)` is (consider degree of ionization of both acids to be lt lt1). |
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Answer» Correct Answer - C Suppose `wedge_(m)(HX)` is represented by `wedge_(m_(1))` and `wedge_(m)(HY)` by `wedge_(m_(2))` Then given that `wedge_(m_(1))=(1)/(10)wedge_(m_(2))` `HX hArr H^(+)+X^(-),K_(a)=([H^(+)][X^(-)])/([HX])` Representing `K_a` of HX by `K_(a_(1))`, then as `K_(a)=Calpha^(2)` and `alpha=(wedge_(m))/(wedge_(m)^(@))`, we have `K_(a_(1))=C_(1)((wedge_(m_(1)))/(wedge_(m_(1))^(@)))^(2)` . . . (i) Similarly, `HYhArrH^(+)+Y^(-),K_(a)=([H^(+)][H^(-)])/([HY])` Representing `K_(a)` of HY by `K_(a_(2))`, we have `K_(a_(2))=C_(2)((wedge_(m_(2)))/(wedge_(m_(2))^(@)))^(@)` . . . (ii) As `lamda_(X^(-))^(@)=lamda_(Y^(-))^(@),wedge_(HX)^(@)=wedge_(HY)^(@)` or `wedge_(m_(1))^(@)=wedge_(m_(2))^(@)` From eqns. (i) and (ii) `(K_(a_(1)))/(K_(a_(2)))=(C_(1))/(C_(2))(wedge_(m_(1)))/(wedge_(m_(2)))^(2)=(0.01)/(0.1)((1)/(10))^(2)` ltBrgt `=0.001=10^(-3)` or `logK_(a_(1))=logK_(a_(2))=log^(10^(-3))=-3` or `-logK_(a_(1))-(-logK_(a_(2)))=3` or `pK_(a_(1))-pK_(a_(2))=3` |
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| 663. |
The addition of `Br_(2)` to `NaI` turns the solution violet. |
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Answer» Correct Answer - T In this case, redox reaction occurs and `I^(c-)` ions reduce `Br_(2)` to `Br^(c-)` ions and themselves are oxidized to `I_(2)` to give violet colour. `Br_(2)+2I^(c-) rarr 2Br^(c-)+I_(2)(` violet `)` |
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| 664. |
Based on the following informations arrange four metals, A,B,C and D in order of increasing ability to act as reducing agents : (I) Only C react with 1 M HCl to give `H_(2)`(g) (II) When A is added to solution of the other metal ions, mettalic D is formed but not B or CA. `DgtAgtCgtB`B. `AgtDgtCgtB`C. `BgtDgtAgtC`D. `DgtAgtBgtC` |
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Answer» Correct Answer - D |
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| 665. |
Based on the following information arrange four metals `A,B,C` and `D` in order of decreasing ability to act as reducing agents `:` `(I)` Only A,B, and C react with `1M HCl` to give `H_(2)(g)` `(II)` When C is added to solutions of the other metal ions, metallic B and D are formed `(III)` Metal C dows not reduced `A^(n+)`A. `CgtAgtBgtD`B. `CgtAgtDgtB`C. `AgtCgtDgtB`D. `AgtCgtBgtD` |
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Answer» Correct Answer - D |
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| 666. |
Preidict what will happern if molecular bromine `(Br_(2))` is added to a solution containing `NaCl` and `NaI` at `25^(@)C`. Assume all species are in their standard states.A. `Cl_(2)(g)` is realeasedB. `I_(2)(s)` is obtainedC. Both `Cl_(2)` and `I_(2)` are obtainedD. Neither `Cl_(2)` nor `I_(2)` is obtained |
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Answer» Correct Answer - B To predict what redox reactions will take place, we need to compare the standard reduction potential for the following half reactions: `{:(I_(2)(s)+2e^(-)hArr2I^(-)(aq., 1M)E^(@)=0.53V),(Br_(2)(l)+2e^(-)hArr2Br^(-)(aq., 1M)E^(@)=1.07 V),(Cl_(2)(g,1 "bar")+2e^(-)hArr 2Cl^(-)(aq.,1M)E^(@)=1.36V):}` Since `E_(Br_(2)//Br^(-))^(@)gtE_(I_(2)//I^(-))^(@),Br_(2)` gets reduced by `I^9-)` i.e. `Br_(2)` will oxidized `I^(-)`. Since `E_(Br_(2)//Br^(-))^(@)gtE_(I_(2)//I^(-))^(@)`, `Br_(2)` will not oxidize `Cl^(-)`. Therefore, the only redox reaction that will occur appreciably under standard state conditions is `{:("Oxidation"2I^(-)(aq.,1M)hArrI_(2)(s)+2e^(-)),("Reduction" Br_(2)(l)+2e^(-)hArr 2Br^(-)(aq.,1M)),(bar("Overall" 2I^(-)(aq.,1M)+Br_(2)(l)hArr I_(2)(s)+2Br(aq., 1M))):}` Note that the `Na^(+)` ions are inert and do not enter into the redox reaction. |
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| 667. |
The cell, `Zn|Zn^(2+)(1M)||Cu^(2+)(1M)|Cu,(E_(cell)=1.10V)`, was allowed to be completely discharged at 298K. The realtive concentration of `Zn^(2+)` to `Cu^(2+),((Zn^(2+)])/([Cu^(2+)]))` isA. `9.65xx10^(4)`B. antilog 24.08C. `37.3`D. `10^(37.3)` |
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Answer» Correct Answer - B The cell reaction is: `Zn+Cu^(2+)toZn^(2+)+Cu` when the cell is completely discharged, `E_(cell)^(@)=(0.0591)/(2)"log"([Zn^(2+)])/([Cu^(2+)])` i.e., `1.10=(0.591)/(2)"log"([Zn^(2+)])/([Cu^(2+)])` or `([Zn^(2+)])/([Cu^(2+)])=37.3` or `([Zn^(2+)])/([Cu^(2+)])=10^(37.3)` |
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| 668. |
Consulting Table 3.1, arrange the following species in order of increaing stregth as oxidizing agents: `MnO_(4)^(-)` (in acidic solution) `Sn^(+)`, `Al^(3+)`, `Co^(3+)`, and `Ag^(+)`. Assume all species are in their standard states.A. `Al^(3+) lt Sn^(2+) lt Ag^(+) lt MnO_(4)^(-) gt Co^(3+)`B. `Al^(3+) lt Sn^(2+) lt Ag^(+) lt Co^(3+) gt MnO_(4)^(-)`C. `Al^(3+) ltSn^(2+) lt Co^(3+) lt Ag^(+) gt MnO_(4)^(-)`D. `Al^(3+) lt Co^(2+) lt Sn^(3+) lt Ag^(+) gt MnO_(4)^(-)` |
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Answer» Correct Answer - A Using Table 3.1, we write the half-reactions in the order in which they appear there: `Al^(3+)(aq., 1 M)+3e^(-) underset(larr)(rarr) Al(s) E^(@) = -1.66 V` `Sn^(2+)(aq., 1 M)+2e^(-) underset(larr)(rarr) Sn(s) E^(@) = -0.14 V` `Ag^(2)(aq.,1M)+e^(-)hArr Ag(s)E^(@)=0.80 V` `MnO_(4)^-)(aq.,1M)+8H^(+)(aq.,1M)+5e^(-)hArr Mn^(2-) (aq.,1M)+4H_(2)O(1)E^(@) = 1.51V` `Co^(3+)(aq.,1M)+e^(-)hArr Co^(2+)(aq.,1M)E^(@)=1.82V` Oxidizing agent oxidizes the other and gets reduced. Thus, strength of an oxidizing agent is directly related to its reduc-tion potential i.e. the more positive the standard reducation potential, the greater the tendency of the species to be reduced or the stronger the species as an oxidizing agent. Thus the oxidizing agents increases in strength with the increases of reduction potential as follows: `Al^(3+),Sn^(2+) lt Ag^(+) lt MnO_(4)^(-) lt Co^(3+)` |
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| 669. |
Determining the realtive strengths of oxidizing and reducing agents a. Arrange the following oxidizing agents in order of increasing strength unde standard-state conditions: `Br_(2)(1), Fe^(3+)(aq.), Cr_(2)O_(7)^(2-)(aq.)` b. Arrange the following reducine agents in order of increasing strength under standard state condition: `Al(a), Na(s), Zn(s)` Strategy: Pick our the half reactions in Table 3.1 that involve the given oxidizing or reducing agents and list them, along with their `E^(@)` value increases (i.e. becomes more positive) whereas the strength of a reducing agent increases as the `E^(@)` value decreases (i.e. becomes more negative). |
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Answer» (a). List the half-reation that involve `Br_(2), Fe^(3+)`, and `Cr_(2)O_(7)^(2-)`, in the order in which they occur inTable 3.1 `Cr_(2)O_(7)^(2-)(aq.)+14H^(+)(aq.)+6e^(-)hArr2Cr^(3+)(aq.)+7H_(2)O(1)E^(@)=1.33V` `Br_(2)(1)+2e^(-)hArr2Br^(-)(aq.)E^(@)=1.09V` `Fe^(3)(aq.)+e^(-)hArr Fe^(2+)(aq.)E^(@)=0.77V` We can see that `Cr_(2)O_(7)6(2-)` has the greatest tendency to be reduced (largest `E^(@)`), and `Fe^(3+)` has the least tendency (smallest `E^(@)`). The species that has the freatest tendency to be reduced is the strongest oxidizin agent. So oxidizing strength increases in the `Fe^(3+)ltBr_(2)ltCr_(2)O_(7)^(2-)`: As a shortcut, cimply note that the strength of the oxidizing agents, listed on the left-side of Table 3.1, increases on moving up in the Table. (b) List the half reactions that involve `Al(s), Na(s)` and `Zn(s)` in the order in which they occur in Table 3.1: `Zn^(2+)(aq.) + e^(-) rarr Zn(s) E^(@) = 0.76 V` `Al^(3+)(aq.) + 3e^(-) rarr Al(s) E^(@) = -1.66 V` `Na^(+)(aq.) + e^(-) rarr Na(s) E^(@) = -2.71 V` The last half-reaction has the least tendency to occur in the forward direction (most negative `E^(@)`) and the greatest tendency to occur in the reverse direction. Therefore, `Na` is the strongest reducing agent, and reducing strength increases in the order `Zn lt Al lt Na`. As a shortcut, note that the strength of the reduing agents, listed on the right side of Table 3.1, increases on moving down the table. |
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| 670. |
Which of the following statements is correct for a galvanic cell?A. Reduction occurs at cathodeB. Oxidation occurs at anodeC. Electrons flow from anode to cathodeD. All the statements are correct. |
| Answer» Correct Answer - D | |
| 671. |
On the basis of the given equivalent conductivity `lamda_(oo)(NH_(4)Cl)=130` `lamda_(oo)(OH^(-))=174` `lamda_(oo)(Cl^(-))=66` The value of `lamda_(oo)(NH_(4)OH)` will beA. 304B. 238C. 108D. 64 |
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Answer» Correct Answer - B `NH_(4)Cl+OH^(-)toNH_(4)OH^(-)+Cl^(-)` `130+174tox+66` `304tox+66` `304-66tox` `x=238` |
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| 672. |
In the electrodeposition of Ag , the silver ions areA. reduced to anodeB. reduced to anodeC. oxidised to anodeD. oxidised to cathode |
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Answer» Correct Answer - B `Ag^(+) + e to Ag` |
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| 673. |
Calculate the emf of the following cell: `Cu(s)|Cu^(2+) (aq)||Ag^(+) (aq)| Ag(s)` Given that, `E_(Cu^(2+)//Cu)^(@)=0.34 V, E_(Ag//Ag^(+))^(@)=-0.80 V`A. `+0.46V`B. `+1.14V`C. `+0.57V`D. `-0.46V` |
| Answer» Correct Answer - A | |
| 674. |
The EMF of a cell isA. sum of two oxidation potentialsB. sum of two potentialsC. difference of two electrode potentialsD. None of the above |
| Answer» Correct Answer - C | |
| 675. |
The art of electroplating was given byA. FaradayB. EdisonC. GrahamD. Brugan |
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Answer» Correct Answer - A Faraday discovereed electrolysis . |
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| 676. |
The limiting molar conductivities of HCl, `CH_(3)COONa` and NaCl are respectively 425,90 and 125 mho `cm^(2)mol^(-1)` at `25^(@)C`. The molar conductivity of 0.1 M `CH_(3)COOH` solution is 7.8 mho `cm^(2)mol^(-1)` at the same temperature. The degree of dissociation of 0.1 M acetic acid solution at the same temperature isA. 0.1B. 0.02C. 0.15D. 0.03 |
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Answer» Correct Answer - B `wedge^(@)` for `CH_(3)COOH=lamda_(CH_(3)COO^(-))^(@)+lamda_(Na^(+))^(@)` `lamda_(H^(+))^(@)+lamda_(Cl^(-))^(@)-lamda_(Na^(+))^(@)-lamda_(Cl^(-))^(@)` `=lamda_(CH_(3)COO^(-))^(@)+lamda_(H^(+))^(@)` `=90+425-125=390mho" "cm^(2)mol^(-1)` Degree of dissociation`(alpha)=(wedge_(m)^(@))/(wedge_(m)^(@))=(7.8)/(390)=0.02` |
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| 677. |
The limiting molar conductivities of HCl, `CH_(3)COONa` and NaCl are respectively 425, 90 and 250 mho `cm^(2)mol^(-1)` at `25^(@)C`.t he molar conductivity of 0.1 M `CH_(3)COOH` solution is 7.8 mho `cm^(2)mol^(-1)` at the same temperature. The degree of dissociation of 0.1 M acetic acid solution at the same temperature isA. `0.10`B. `0.02`C. `0.15`D. `0.03` |
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Answer» Correct Answer - B `wedge_(CH_(3)COOH)^(@)=wedge_(CH_(3)COONa)^(@)+wedge_(HCl)^(@)-wedge_(NaCl)^(@)` `=90+425-125=390" mho "cm^(2)mol^(-1)` `alpha=(wedge_(m)^(@))/(wedge_(m)^(@))=(7.8)/(390)=.02` |
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| 678. |
Calculate the emf of the following cell: `Cu(s)|Cu^(2+) (aq)||Ag^(+) (aq)| Ag(s)` Given that, `E_(Cu^(2+)//Cu)^(@)=0.34 V, E_(Ag//Ag^(+))^(@)=-0.80 V`A. 0.046 VB. 0.46 VC. 0.57 VD. `-0.46 V` |
| Answer» Correct Answer - B | |
| 679. |
Which of the following does not promote corrosion ?A. presence of impurityB. presence of moistureC. higher activity of the metalD. higher temperature |
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Answer» Correct Answer - A At high temperature , moisture is not present , so corrosion does not occur . |
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| 680. |
Consider the following reaction, `Zn(s)+Cu^(2+) (0.1 M) rarr Zn^(2+) (1 M)+Cu(s)` above reaction, taking place in a cell, `E_("cell")^(@)` is `1.10 V. E_("cell")` for the cell will be `(2.303 (RT)/(F)=0.0591)`A. `1.80 V`B. `1.07 V`C. `0.82 V`D. `2.14 V` |
| Answer» Correct Answer - B | |
| 681. |
The logarithm of the equilibrium constant of the cell reaction corresponding to the cell `X(s)|X^(2+)(aq)|| Y^(+)|Y(s)` with standard cell potential `E_("cell")^(@)=1.2V` is given byA. 12.5B. 21.5C. 40.5D. 47.2 |
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Answer» Correct Answer - C `E^(@)=(0.0591)/(n)"log"k(n=2)` `therefore "log" k=(1.2xx2)/(0.591)=40.6` |
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| 682. |
The limiting molar conductivities of HCl, `CH_(3)COONa` and NaCl are respectiley 425, 90 and 125 mho `cm^(2) " mol"^(-1)` and `25^(@)C`. The molar conductivity of 0.1M `CH_(3)COCH` solution is 7.8 mho `cm^(2) "mol"^(+1)` at the same temperature isA. 0.1B. 0.02C. 0.15D. 0.03 |
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Answer» Correct Answer - B `^^_(CH_(3)COOH)^(@)=^^_(CH_(3)COONa)^(@)+^^_(HCl)^(@)-^^_(NaCl)^(@)` `=90+425-125=390 mho cm^(2)"mol"^(-1)` `a=(^^_(m)^(@))/(^^_(m)^(@))=(7.8)/(390)=0.02` |
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| 683. |
In electrochemical corrosion of metals, the metal undergoing corrosion:A. acts as anodeB. acts as cathodeC. undergoes reductionD. neutral |
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Answer» Correct Answer - A Metal undergoes oxidation during corrosion and thus act as anode . |
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| 684. |
On the basis of electrochemical theory of aqueous corrosion, the reaction occurring at the cathode isA. `O_(2)(g)+4H^(+) (aq)+4e^(-) rarr 2H_(2)O(l)`B. `H_(2)(g)+2OH^(-)(aq) rarr 2H_(2)O (l) +2e^(-)`C. `Fe^(2+)(aq)+2e^(-) rarr Fe(s)`D. `Fe^(3+) (aq) +e^(-) rarr Fe^(2+) (aq)` |
| Answer» Correct Answer - A | |
| 685. |
A current is passed through two cells connected in series. The first cell contains `X(NO_(3))_(3)(aq)` and the second cell contains `Y(NO_(3))_(2)(aq)`. The relative atomic masses of X and Y are in the ratio 1:2, what is the ratio of the liberated mass of X to that of Y?A. `3:2`B. `1:2`C. `1:3`D. `3:1` |
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Answer» Correct Answer - C if atomic mass of X=a, then that of Y=2a. As eq.wt=atomic mass/valency, Eq. wt. of X=a/3, Eq. wt. of Y=2 a/2=a. `("Mass of "X)/("Mass of "Y)=("Eq. wt. of "X)/("Eq. wt. of Y")=(a//3)/(a)=(1)/(3)=1:3` |
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| 686. |
A current is passed through two cells connected in series. The first cell contain `x(NO_(3))_(2)` (aq) and the second cell contains `y(NO_(3))_(2)(aq)` . The relative atomic masses of x and y are in the ratio of x to that of y?A. `3:2`B. ` 1:2`C. `1:3`D. ` 3:1` |
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Answer» Correct Answer - C If atomic mass of x=a, then that of y=2a `therefore` Eq. mass `=("at. Mass")/("valency")` `therefore` Eq. mass of `x=a//3`. Eq mass of `y=2a//2=a`. `("Mass of x")/("Mass of y")=("Eq. mass of x")/("E q mass of y")` `=(a//3)/(1)=(1)/(3)= 1:3` |
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| 687. |
In electrochemical corrosion of metals, the metal undergoing corrosionA. anodeB. cathodeC. neither anode nor cathodeD. either anode or cathode depending upon its standard reduction potential. |
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Answer» Correct Answer - A Metal undergoing corrosion acts an anode . |
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| 688. |
Given below are half-cell reaction: `Mn^(2+)+2e^(-) rarr Mn,, E^(@) = -1.81 V` `2(Mn^(3+)+e^(-) rarr Mn^(2+)),, E^(@) = +1.51 V` The `E^(@)` for `3Mn^(2+) rarr Mn+2Mn^(3+)` will be:A. `-0.33 V`, the reaction will not occurB. `-0.33 V`, the reaction will occurC. `-2.69 V`, the reaction will not occurD. `-2.69 V`, the reaction will occur |
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Answer» Correct Answer - C To obtain the given reaction, we reverse the second reaction and add it to the first one: `{:(Mn^(2+)+2e^(-) rarr Mn,),(2Mn^(2+)rarr2Mn^(3+)+2e^(-),):}/(3Mn^(2+)rarrMn+2Mn+2Mn^(3+))" "{:(E^(@) = -1.81 V),(E^(@) = -1.51 V):}` Thus, `E^("cell")^(@) = E_(SOP)^(@)+E_(SRP)^(@)` `= (-1.18 V)+(-1.51 V)` `= -2.69 V` Since `DeltaG_("cell")^(@) = -nFE_("cell")^(@)` negative `E_("cell")^(@)` indicates positive `DeltaG_("cell")^(@)` i.e., a non-spontaneous cell reaction. |
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| 689. |
A current is passed through two cells connected in series. The first cell contains `X(NO_(3))_(3(aq))` and the second cell contains `Y(NO_(3))_(2(aq))`. The relative atomic masses of X and Y are in the ratio 1:2. What is the ratio of liberated mass of X to that of Y.A. `3:2`B. `1:2`C. `1:3`D. `3:1` |
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Answer» Correct Answer - C The oxidation states of X and Y are `X^(3+) and Y^(2+)` given atomic masses are in the ratio of 1:2 `therefore` Equivalent masses are in the ratio `(1)/(3):(2)/(2)` or `(1)/(3):1` or `1:3` |
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| 690. |
Which of the following statement is correct?A. `E_("cell)` and `Delta_(r) G` of the cell reaction both are extensive properties.B. `E_("cell")` and` Delta _(r) G` of the cell reaction both are intensive properties .C. `E_("cell")` is an intensive and `Delta_(r) G` of cell reaction an extensive property .D. `E_("cell")` is an extensive and `Delta_(r) G` of the cell reaction is an intensive property . |
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Answer» Correct Answer - C It is a fact . |
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| 691. |
A quantity of electrical charge that brigns about the depositiion of `4.5g Al` from `Al^(3+)` at the cathode will also produce the following volume `(STP)` of `H_(2)(g)` from `H^(o+)` at the cathode.A. ` 11. 2 L`B. ` 44.8 L`C. ` 5.6 L`D. ` 22 .4 L` |
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Answer» Correct Answer - C Eq. of `H_2 =` Eq . Of `Al = ( 4.5)/(27 .3) =0.5` `:. 1 eq .H_2 = 11 . 2 L` ` :. 0 5 eq . H_2 = 5. 6 L` . |
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| 692. |
The emf of a galvanic cell is positive when free energy change of reaction isA. `gt0`B. `lt0`C. `=0`D. no relationship of free energy chagne and e.m.f. |
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Answer» Correct Answer - B `DeltaG=-nFE^(@)`. If Emf is positive (spontaenous reaction) so `DeltaG=-ve` i.e. less than zero. |
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| 693. |
The emf of a galvanic cell is positive when free energy change of reaction isA. `gt 0`B. `lt 0`C. `= 0`D. no relationship of free energy change and emf |
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Answer» Correct Answer - B `DeltaG=-nFE` If `E=+ve`, then `DeltaG=-ve` or `lt 0` |
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| 694. |
An electric current is passed through silver nitrated solution using silver electrodes . ` 10. 79 g` of silver qas found to be deposited on the cathode fi the same amount of electricity is passed through copper sulphate solutin using copper electrodes. the weihgt of copper deposited on teh cathode is .A. ` 6.4 G`B. ` 2.3 g`C. ` 12 .8 g`D. ` 3.2 g` |
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Answer» Correct Answer - D Number of equivalents of silve formed = Number of equivalents of copper formed . In `AgNO_3, AG` is in + 1 oxidation state. In `CuSO_4` Cu is in ` + 2` oxidation state. Equivalent weight `Ag = ( 108)/1 = 108` Equivalent weight `Cu = (63.6)/@ = 31 . 8`. `M_1/M_2 = E_1 /E_2 , ( 10. 97)/(MCu) = (108)/(31.8)` `M_(Cu) = (10. 79 xx 31. 8)/( 108) = 3 2 g`. |
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| 695. |
An electric current is passed through silver nitrated solution using silver electrodes . ` 10. 79 g` of silver qas found to be deposited on the cathode fi the same amount of electricity is passed through copper sulphate solutin using copper electrodes. the weihgt of copper deposited on teh cathode is .A. 6.4 gB. 2.3 gC. 12.8 gD. 3.2 g |
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Answer» Correct Answer - D Number of equivalents of silver formed=Number of equivalents of copper formed In `AgNO_(#)`, Ag is in +1 oxidation state. In `CuSO_(4)`, Cu is in +2 oxidation state. Equivalet weight of `Ag=(108)/(1)=108` Equivalent weight of `Cu=(63.6)/(2)=31.8` `(M_(1))/(M_(2))=(E_(1))/(E_(2)),(10.79)/(M_(Cu))=(108)/(31.8)` `M_(Cu)=(10.79xx31.8)/(108)=3.2gm` |
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| 696. |
When, during electrolysis of a solution of `AgNO_(3) 9650` colombs of charge pass through the electroplating path, the mass of silver deposited on the cathode will be:A. `21.6 g`B. `108 g`C. `1.08g`D. `10.8g` |
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Answer» Correct Answer - D No. of equivalents of Ag deposited `=(Q)/(F)` `=(9650)/(96500) =(1)/(10)` No. of equivalents `= (wt)/(Eq.wt) = (1)/(10)` `:. (wt)/(108) = (1)/(10)` `:. Wt = 10.8 g` |
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| 697. |
The emf of a galvanic cell is positive when free energy change of reaction isA. zeroB. positiveC. negativeD. not definite |
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Answer» Correct Answer - A At equilibriumof the cell reaction , `E_("cell") = 0` . |
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| 698. |
How many moles of platinum will be deposited on the cathode when 0.60 F of electricity is passed through a 1.0 M solution of `Pt^(4+)` ?A. 0.60 molB. 0.15 molC. 0.30 molD. 0.45 mol |
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Answer» Correct Answer - B `Pt^(4+)(aq)+underset(0.6F)underset(4F)4e^(-) to underset(((1mol)xx(0.6F))/((4F))=0.15" mol")underset(1" mol")(Cu(s))` |
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| 699. |
Hydrogen electrode isA. reversible with respect to `H^(+)` ionsB. a secondary reference electrodeC. an indicator electrodeD. irreversible with respect to `H^(+)` ions |
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Answer» Correct Answer - A By definition , SHE is reversible with respect to `H^(+)` ions . The oxidation reaction is = `(1)/(2) H_(2) to H^(+) + e^(-)` |
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| 700. |
When, during electrolysis of a solution of `AgNO_(3) 9650` colombs of charge pass through the electroplating path, the mass of silver deposited on the cathode will be:A. 10.0 gB. 21.6 gC. 108 gD. 1.08 g |
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Answer» Correct Answer - A (a) No. of moles of `Ag=(1" mol")xx((9650" C"))/((96500" C"))` deposited`=0.1" mol"`. Mass of Ag deposited `=0.1xx108` =10.8 g |
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