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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.
| 2501. |
An aqueous solution of an electrolyteA. conducts electricity without any chemical changeB. conducts electricity with chemical decompositionC. is an insulatorD. all are correct |
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Answer» Correct Answer - B Passage of current through an electrolyte brings in chemical changes . |
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| 2502. |
Water is a non-electrolyte but conducts electricity on dissovling a small amount ofA. `O_(2)`B. SugarC. AcetoneD. NaCl |
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Answer» Correct Answer - D NaCl produces `Na^(+)` and `Cl^(-)` ions . |
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| 2503. |
When an electrolytic solution conducts electricity , current is carried out byA. electronsB. cations and anionsC. neutral atomsD. none |
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Answer» Correct Answer - B Passage of current in electrolytic solution is due to migration of ions towards opposite electrodes . |
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| 2504. |
In the electrolysis of which solution `OH^(-)` ions are discharged in preference to `Cl^(-)` ions?A. dilute NaClB. very dilute NaClC. fused NaClD. solid NaCl |
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Answer» Correct Answer - B In case of very dilute solution of NaCl , electrolysis brings in the following changes . Anode :` 2OH^(-) to H_(2)O + (1)/(2) O_(2) + 2e` Cathode : ` 2H^(+) + 2 e to H_(2)` |
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| 2505. |
The amount of ion dischargedf during electrolysis is directly proportional to :A. ResistanceB. TimeC. CurrentD. Chemical equivalent of the ion |
| Answer» Correct Answer - B::C::D | |
| 2506. |
Specific conductance of 0.1 M NaCl solution is `1.01xx10^(-2) ohm^(-1) cm^(-1)` . Its molar conductance in `ohm^(-1) cm^(2) mol^(-1)` isA. `1.01xx10^2`B. `1.01xx10^3`C. `1.01xx10^4`D. `1.01` |
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Answer» Correct Answer - A `Lambda_m=(Kxx1000)/M` `=(1.01xx10^(-2)xx1000)/0.1 = 1.01 xx10^2 ohm^(-1) cm^2 mol^(-1)` |
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| 2507. |
Calculating molar conductance: The resistance of `0.01 M` solution of an electrolyte was found to be `210 ohm` at `25^(@)C`. Calculate them molar conductance of the colution at `25^(@)C`, if the cell constant is `0.88 cm^(-1)` Strategy: First conductivity in `Sm^(-1)` (SI units) and then divide it by molar concentration in mole `m^(-3)` (SI units) to get molar conductance in `Sm^(2) mol^(-1)`. |
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Answer» `G = (1)/R = (1)/(210)S` `K_("cell") = 0.88 cm(-1)` According to Equation (3.18), we have `kappa = ("conductance")("cell constant")` `((1)/(210)S) (0.88 cm^(-1))` `= 0.00419 Scm^(-1)` `= 0.419 Sm^(-1)` Molar concentration of the solution: `C = 0.01 mol L^(-1) = 0.01 mol ((1)/(1000 m^(3)))^(-1)` `= 10 mol m^(-3)` Thus, `Lambda_(m) = (kappa)/(C) = (0.419 Sm^(-1))/(10 mol m^(-3))` `= 0.0419 Sm^(2)mol^(-1)` Alternatively, use the Equation `Lambda_(m) = (kappa(Sm^(-1)))/("molarity "(molL^(-1))xx(1000 L m^(-3))` `= (0.419 Sm^(-1))/((0.01 mol L^(-1)) xx (1000 L m^(-3))` `= 0.0419 Sm^(2) mol^(-1)`. |
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| 2508. |
On diluting the solution of an electrolyte, (a) both ∧ and κ increase (b) both ∧ and κ decrease (c) ∧ increases and κ decreases(d) ∧ decreases and κ increases |
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Answer» Correct answer is (c) ∧ increases and κ decreases |
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| 2509. |
At a particular temperature, the ratio of molar conductance to specific conductance of 0.01 M NaCl is :A. `10^(5) cm^(3) mol^(-1)`B. `10^(3) cm^(3) mol^(-1)`C. `10 cm^(3) mol^(-1)`D. `10^(5) cm^(2) mol^(-1)` |
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Answer» Correct Answer - C `Lambda_(m)= kappa xx 1000/M` `Lambda_(m)/kappa=1000/0.01=10^(5) cm^(3) mol^(-1)` |
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| 2510. |
Two solutions have the ratio of their concentrations 0.4 and ratio of their conductivities 0.216. The ratio of their molar conductivities will be(a) 0.54 (b) 11.574 (c) 0.0864 (d) 1.852 |
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Answer» Correct answer is (a) 0.54 |
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| 2511. |
Solutions of two electrolytes A and B each having a concentration of 0.2 M have conductivities `2xx10^(-2) and 4xx10^(-4)" S "cm^(-1)` respectively. Which will offer greater resistance to the flow of current and why? |
| Answer» `kappa=Gxx(1)/(a)=(1)/(R)(1)/(a),i.e., kappa prop(1)/(R)`. Hence, B will offer greater resistance. | |
| 2512. |
The molar conductivities of KCl, NaCl and `KNO_(3)` are 152,128 and `111" S cm"^(2) mol^(-1)` respectively. What is the molar conductivity of `NaNO_(3)` ?A. 101 S `cm^(2) mol^(-1)`B. `87 S cm^(2) mol^(-1)`C. `-101 S cm^(2) mol^(-1)`D. `-391 S c^(2) mol^(-1)` |
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Answer» Correct Answer - B `Lambda_(NaNO_(3))^(@)=Lambda_(NaCl)^(@)+Lambda_(KNO_(3))^(@)-Lambda_(KCl)^(@)` `=128+111-152=87 S cm^(2) mol^(-1)` |
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| 2513. |
The conductivity of 0.01 mol `L^(-1)`KCl solution is `1.41xx10^(-3)" S "cm^(-1)`. What is the molar conductivity (`S cm^(2) mol^(-1)`) ?A. 14.1B. 1.41C. 1410D. 141 |
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Answer» Correct Answer - D (d) Molar conductivity`(Lambda_(m))=(1000xxk)/(M)` `=(1000xx(1.41xx10^(-3)" S " cm^(-1)))/((0.01" mol "cm^(-3)))=141" S "cm^(-2)mol^(-1)` |
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| 2514. |
Which will have greater molar conductivity and why? Sol A. 1 mol KCl dissolved in 200 cc of the solution. Sol. B. 1 mol KCl dissolved in 500 cc of the solution. |
| Answer» Solution B will have greater molar conductivity because `wedge_(m)=kappaxxV`. With dilution, `kappa` decreases but V increases much more so that product increases. | |
| 2515. |
Fill in the blanks (i) Equivalent wt. of a substance divided by 96500 gives_____ of the substance (ii) The weight deposited by one coulomb of electricity is called____ of the substance (iii) One faraday is the charge present on ______of electrons (iv) One faraday passed through `CuSO_(4)` sol. deposits_____ of Cu. |
| Answer» (i) electtrochemical equivalent, (ii) Electrochemical equivalent, (iii) one mole, (iv) `(1)/(2)` mole | |
| 2516. |
In each of the following pairs, which will allow greater conduction of electricity and why? (a) Silver wire at `20^(@)C`, Same silver wire at `50^(@)C` (b) NaCl solution at `20^(@)C`, same NaCl solution at `50^(@)C` (c) `NH_(4)OH` solution at `20^(@)C`, Sae `NH_(4)OH` solutio at `50^(@)C` (d) 0.1M acetic acid solution, 1M acetic acid solution. |
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Answer» (a) Silver wire at `20^(@)C` because with increase in temperature, metallic conduction decreases due to vibration of kernels. (b) NaCl solution at `50^(@)C` because in case of a strong electrolyte with increase in temperature, the ionic mobilities increase. (c) `NH_(4)OH` at `50^(@)C` because in case of a weak electrolyte, dissociation increases with increase in temperature. (d). 0.1 M acetic acid because with dilution, dissociation/ionization increases. |
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| 2517. |
The reaction, `1//2 H_(2) (g) + AgCl(s) = H^(+) (aq) + Cl^(-) (aq) + Ag(s)` occurs in the galvanic cell . The anode isA. AgClB. AgC. `H_(2)`D. `H^(+)` |
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Answer» Correct Answer - C Anode is `H_(2) | H^(+)` , Cathode is `Ag^(+) |Ag`. |
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| 2518. |
Thereaction, `1//2Hg_(2)(g)+AgCl(s)=H^(+)(aq)+Cl^(-)(aq)+Ag(s)` occurs in the galvanic cellA. `Ag|AgCl(s)|KCl("soln")|AgNO_(3)("soln")|Ag`B. `Pt|H_(2)(g)|HCl("soln")|AgNO_(3)("soln")|Ag`C. `Pt|H_(2)(g)|HCl("soln")|AgCl(s)|Ag`D. `Pt|H_(2)(g)|KCl("soln")|AgCl(s)|Ag` |
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Answer» Correct Answer - C `H_(2)` undergoes oxidation and `AgClAg^(+))` undergoes reduction. |
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| 2519. |
Red hot carbon will remove oxygen from the oxides `XO` and `Yo` but not from `ZO.Y` will remove oxygen from `XO`. Use this evidence to deduce the order of activity of the three metals `X,Y,` and `Z,` putting the most reactive first.A. X , Y , ZB. Z , Y , XC. Y , X , ZD. Z , X, Y |
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Answer» Correct Answer - B Carbon does not reduce ZO and thus Z is more reactive . Also Y removes oxygen from XO and thus Y is more active than X . Thus activity order is Z , Y , X . |
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| 2520. |
Four colourless salt solutions are placed in separate test tubes and a strip of copper is dipped in each. Which solution finally turns blue ?A. `Pb(NO_(3))_(2)`B. `Zn(NO_(3))_(2)`C. `AgNO_(3)`D. `Cd(NO_(3))_(2)` |
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Answer» Correct Answer - C Ag is above Cu in electrochemical series . Metal placed below replaces other placed above in series . |
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| 2521. |
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 will get coated with aluminiumB. aluminium sulphate is formedC. the solution contain aluminium nitrateD. There is no reaction |
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Answer» Correct Answer - C The metal placed below in electrochemical series displaces metal from its salt solution which is placed below in series . |
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| 2522. |
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 will get coated with aluminiumB. An alloy of copper and aluminium is formedC. The solution becomes blueD. There is no reaction |
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Answer» Correct Answer - D Cu is less reactive than Al. |
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| 2523. |
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 will get coated with aluminiumB. An alloy of aluminium and copper is formedC. The solution becomes blueD. There is no reaction |
| Answer» Correct Answer - D | |
| 2524. |
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 will get coated will aluminium.B. An alloy of copper and aluminium is formed.C. The solution becomes blue.D. There is no reaction. |
| Answer» Correct Answer - D | |
| 2525. |
The position of some metals in the electrochemical series in decreasing electropositive 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 aluminium nitrateA. The spoon will get coated with AlB. An alloy of Cu and Al is formedC. The solution becomes blueD. There is no reaction |
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Answer» Correct Answer - D The metal placed below in electrochemical series does not react with that metal salt solution which metal is placed above in series. |
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| 2526. |
The emf of a Daniell cell at `298 K` is `E_(1)` `Zn|ZnSO_(4)(0.01 M)||CuSO_(4)(1.0M)|Cu` When the concentration of `ZNSO_(4)` is `1.0 M` and that of `CuSO_(4)` is `0.01 M`, the `emf` changed to `E_(2)`. What is the relationship between `E_(1)` and `E(2)` ?A. `E_(1) lt E_(2)`B. `E_(1) gt E_(2)`C. `E_(2)=0neE_(1)`D. `E_(1)=E_(2)` |
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Answer» Correct Answer - B (b) According to Nernst equation. `E_(cell)=E_(cell)^(@)-(0.059)/(n)"log"([Zn^(2+)])/([Cu^(2+)])` `E_(1)=E_(cell)^(@)-(0.059)/(2)"log"(0.01)/(1)` `=E_(cell)^(@)-(0.059)/(2)(-2)=E_(cell)^(@)+0.059` `E_(2)=E_(cell)^(@)-(0.059)/(2)"log"(1)/(0.01)` `E_(2)=E_(cell)^(@)-(0.059)/(2)(2)=E_(cell)^(@)-0.059` Hence, `E_(1) gt E_(2)`. |
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| 2527. |
In the electrochemical cell: `Z|ZnSO_(4)(0.01M)||CuSO_(4)(1.0M)|Cu,` the e.m.f. this daniel cell is `E_(1)`. When the concentration of `ZnSO_(4)` is changed to 1.0 M and that of `CuSO_(4)` is changed to 0.01 M, the e.m.f. changes to `E_(2)`. from the following, which one is the relationship between `E_(1) and E_(2)`? (Given, `(RT)/(F)=0.059`)A. `E_(1)=E_(2)`B. `E_(1) gt E_(2)`C. `E_(1) gt E_(2)`D. `E_(2)=0neE_(2)` |
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Answer» Correct Answer - C The cell reaction is `Zn+CuSO_(4)toZnSO_(4)+Cu,n=2` `E=E_(cell)^(@)-(2.303RT)/(2F)"log"([ZnSO_(4)])/([CuSO_(4)])` In 1 st case, `E_(1)=E_(cell)^(@)-(2.303RT)/(2F)"log"(0.01)/(1)` In 2nd case, `E_(2)=E_(cell)^(@)-(2.303RT)/(2F)"log"(1)/(0.01)` Evidently, `E_(1)gtE_(2)`. |
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| 2528. |
The emf of a Daniell cell at `298 K` is `E_(1)` `Zn|ZnSO_(4)(0.01 M)||CuSO_(4)(1.0M)|Cu` When the concentration of `ZNSO_(4)` is `1.0 M` and that of `CuSO_(4)` is `0.01 M`, the `emf` changed to `E_(2)`. What is the relationship between `E_(1)` and `E(2)` ?A. `E_(2)=0cancel=E_(1)`B. `E_(1)gtE_(2)`C. `E_(1)gtE_(2)`D. `E_(1)=E_(2)`. |
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Answer» Correct Answer - B The cell reaction is `Zn+Cu^(2+)rarrZn^(2+)+Cu` `therefore` Nernst equation is `E_(1)=E^(@)-(0.0591)/(2)"log"([Zn^(2+)])/([Cu^(2+)])` when `[Zn^(2+)]` is increased and `[Cu^(2+)]` is decreased the e.m.f. `E_(2)` will b e less than `E_(1)` or `E_(1)gtE_(2)`. |
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| 2529. |
In the electrochemical cell `Zn|ZnSO_(4)(0.01M)||CuSO_(4)(1.0M)|Cu`, the emf of this Daniel cell is `E_(1)`. When the concentration of `ZnSO_(4)` is changed to 1.0M and that of `CuSO_(4)` changed to 0.01M, the emf changes to `E_(2)`. From the followings, which one is the relationship between `E_(1) and E_(2)` (given,`(RT)/(F)=0.059`)A. `E_(1) lt E_(2)`B. `E_(1)gtE_(2)`C. `E_(2)=0neE_(1)`D. `E_(1)=E_(2)` |
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Answer» Correct Answer - B For cell `Zn|ZnSO_(4)(0.01M)||CuSO_(4)(1M)|Cu` Cell reaction `toZn+Cu^(+2)toZn^(+2)+Cu` `E_(1)=E^(o)-(0.059)/(2)"log"(Zn^(+2))/(Cu^(+2))` `E_(1)=E^(o)-(0.059)/(2)"log"(0.01)/(1)` `=E^(o)-(0.059)/(2)"log"(1)/(100)` . . . .(i) Fore cell `Zn|ZnSO_(4)(1M)||CuSO_($)(0.01M)|Cu` `E_(2)=E^(o)-(0.059)/(2)"log"(1)/(0.01)` `=E^(o)-(0.059)/(2)log100` . . . . . ltBrgt `E_(1)gt E_(2)` |
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| 2530. |
The emf of a Daniel cell at 298 K is `E_(1)`, The cell is `Zn|ZnSO_(4)(0.01M)||CuSO_(4)(1M)|Cu` When the concentration of `ZnSO_(4)` is changed to 1M and that of `CuSO_(4)` to 0.01 M, the emf changes to `E_(2)`, the relationship between `E_(1) and E_(2)` will beA. `E_(1)-E_(2)=0`B. `E_(1) lt E_(2)`C. `E_(1) gt E_(2)`D. `E_(1)=10^(2)E_(2)` |
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Answer» Correct Answer - C `underset(n=2)(Zn)+Cu underset(Q=(Zn^(2+))/(Cu^(2+)))(SO_(4)hArrZn)SO_(4)+Cu` `E_(1)-E_(2)=-(0.059)/(2)(log((0.01)/(1))-log((1)/(0.01))` `=-(0.059)/(2)(log((1)/(100))-log100)` `=(0.059)/(2)(-log100-log100)=+(0.059)/(2)xx2impliesSo,E_(1)gtE_(2s)` |
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| 2531. |
The emf of a Daniel cell at 298K is `E_(1)` `Zn|underset((0.01" "M))(ZnSO_(4))||underset((1.0" M"))(CuSO_(4))|Cu` when the concentration of `ZnSO_(4)` is 1.0 M and that of `CuSO_(4)` is 0.01M the emf changed to `E_(2)` what is the relationship between `E_(1) and E_(2)`.A. `E_(2)=0neE_(1)`B. `E_(1)gtE_(2)`C. `E_(1)ltE_(2)`D. `E_(1)=E_(2)` |
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Answer» Correct Answer - B `E_(1)=E_(o)-(0.0591)/(2)log((100)/(0.01))=E_(o)=(0.0591)/(2)xx4` `thereforeE_(1) gt E_(2)`. |
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| 2532. |
Assertion: For the Daniel cell `Zn|Zn^(2+)||Cu^(2+)|Cu` with `E_(cell)=1.1`V, the application of opposite potential greater than 1.1 V results into flow of electron from cathode to anode. Reason: Zn is deposited at anode and Cu is dissolved at cathode.A. if both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is falseD. If the assertion and reason both are false. |
| Answer» Correct Answer - A | |
| 2533. |
Statement-I: In the Daniel cell, if concentration of `Cu^(2+)` and `Zn^(2+)` ions are doubled the emf of the cell will not change. Because Statement-II: If the concentration of ions in contact with the metals is doubled, the electrode potential is doubled.A. is doubleB. is reduced to halfC. remains sameD. becomes four times |
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Answer» Correct Answer - C Remain same , according to Nernst equation . |
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| 2534. |
Estimate the cell potential of a Daniel cell having `1.0 M Zn^(2+)` and originally having `1.0 M Cu^(2+)` after sufficient ammonia has been added to the cathode compartment to make the `NH_(3)` concentration `2.0 M`. Given, `E_(Zn//Zn^(2+))^(@)` and `E_(Cu//Cu^(2+))^(@)` are `0.76` and `-0.34 V` respectively. Also equilirbrium constant for the `[Cu(NH_(3))_(4)]^(2+)` formation is `1 xx 10^(12)`. |
| Answer» Correct Answer - `0.71 V ;` | |
| 2535. |
The cell constant of a conductivity cell is given by :(a) l × a(b) \(\frac{a}{l}\)(c) \(\frac{1}{l\times a}\)(d) \(\frac{l}{a}\) |
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Answer» Option : (d) \(\frac{l}{a}\) |
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| 2536. |
A device that convers energy of combustion of fueles like hydrogen and methane, directly into electrical energy is known as .A. fuel cellB. electrolytic cellC. dynamoD. Ni-Cd cell |
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Answer» Correct Answer - A (a) Fuel cell is a device that converts energy of combustion of fuels like hydrogen and methane, directly into electrical energy. Electrolytic cell converts electrical energy into chemical energy Dynamo is an electrical generator that produces direct current with the use of a commutator. Ni-Cd cell is a type of rechargeable battery which consist of a cadmium anode and a metal grid containing `NiO_(2)` acting as a cathode . |
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| 2537. |
A device that convers energy of combustion of fueles like hydrogen and methane, directly into electrical energy is known as .A. dynamoB. Ni-Ced cellC. fuel cellD. electrolytic cell |
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Answer» Correct Answer - C For cell . |
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| 2538. |
A device that convers energy of combustion of fueles like hydrogen and methane, directly into electrical energy is known as .A. Ni-Cd cellB. Fuel cellC. Electrolytic cellD. Dynamo |
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Answer» Correct Answer - B Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol etc. directly into electrical energy are called fuel cells. In such cells, reactants arc fed continuously to the electrodes and products arc removed continuously to the electrodes and products arc removed continuously form the electrolyte compartment. |
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| 2539. |
When the electric current is passed through a cell having an electrolyte, the positive ions move towards cathode and negative ions toqards the anode. If the cathode is pulled out of the solution .A. The positive and negative ion will move towards anodeB. The positive and negative ion will move towards anode, the the negative ions will stop movingC. the negative ions will continut to move towarsa anode while positive ions will stop movingD. the positive and negative ions will start moving randomly |
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Answer» Correct Answer - D In the absence of electric feld the ions in the solution move radomly due to thermal energy. |
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| 2540. |
When the electric current is passed through a cell having an electrolyte, the positive ions move towards cathode and negative ions toqards the anode. If the cathode is pulled out of the solution .A. the positive and negative ion will move towards anodeB. the positive ions will state moving towards the anode whil negative ions will stop movingC. the negative ions will continut to move towarsa anode while positive ions will stop movingD. the positive and negative ions will start moving randomly |
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Answer» Correct Answer - D After removing cathode no net charge will but ions move randomly. |
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| 2541. |
A device that converts energy of combustion of fuels like hydrogen and methane, directly into electrical energy is known asA. Electrolytic cellB. DynamoC. Ni-Cd cellD. Fuel cell |
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Answer» Correct Answer - D Fuel cell convert chemical energy of fuel like `H_(2),CH_(4)` into electrical energy. |
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| 2542. |
When electric current is passed through a cell having an electrolyte, the positive ions move towards the cathode and the negative ions towards the anode. If the cathode is pulled out of the solutionA. The positive and negative ions will move towards the anodeB. The positive ions will start moving towards the anode, the negative ions will stop movingC. The negative ions will continue to move towards the anode and the positive ions will stop movingD. The positive and negative ions will start moving randomly |
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Answer» Correct Answer - D In the absence of electric field the ions in the solution move randomly due to thermal energy. |
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| 2543. |
`{:("Electrolyte",^^.^(oo)(Scm^(2)mol^(-1))),(KCl,149.9),(KNO_(3),145.0),(HCl,426.2),(NaOAc,91.0),(NaCl,126.5):}` Calculate `^^_(HOAc)^(oo)` using appropriate molar conductance of the electrolytes listed above at infinite dilution in `H_(2)O` at `25^(@)C`A. 517.2B. 552.7C. 390.7D. 217.5 |
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Answer» Correct Answer - C `Lambda_(HOA C)^(@) = Lambda_(NaOAC)^(@) + Lambda_(HCl)^(@) - Lambda_(NaCl)^(@)` `Lambda_(HOAC)^(@) = 91 + 426.2 - 126.5 cm^(2) mol^(-1)` `Lambda_(HOAC)^(@) = 390.75 cm^(2) mol^(-1)` |
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| 2544. |
Name the electrolyte used in fuel cell and mercury cell. |
| Answer» (i) Concentration aqueous KOH solution. (ii) Moist mercuric oxide (HgO) mixed with KOH. | |
| 2545. |
Consider the electrochemical cell : (APP)Zn (s)/Zn2+ (aq) // Cu2+ (aq)/Cu.It has an electrical potential of 1.1 V when concentration of Zn2+ and Cu2+ ions is unity.State the direction of flow of electrons and also specify if Zinc and Copper are deposited or dissolved at their respective electrodes.When:(a) An external opposite potential of 0.8 V is applied.(b) An external opposite potential of 1.1 V is applied.(c) An external opposite potential of 1.4 V is applied. |
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Answer» (a) Electrons flow from Zn rod to Cu rod. (b) No flow of electrons and current. No change observed at Zinc and Copper electrodes (system is equilibrium). (c) Electrons flow from Cu rod to Zn rod. Zinc is deposited and Copper gets dissolved. |
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| 2546. |
For the reaction :2 Ag+ + 2 Hg → 2 Ag + Hg22+E˚ Ag+/Ag = 0.80 V E˚ Hg22+/Hg = 0.79 V Predict the direction in which the reaction will proceed if :[Ag+] = 10-1mol/h [Hg2+] = 10-3mol/h |
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Answer» Cell reaction is : 2 Ag+ + 2 Hg → 2 Ag + Hg22+ Ecell = E˚cell – 0.0591/2 log [Hg22+]/ [Ag+]2 = (0.80 V -0.79 V) - 0.0591/2 log 10-3/(10-1)2 = 0.01 V- 0.0591/2 (-1) = 0.01 + 0.0295 = 0.0395 V Since Ecell is positive, the reaction will be spontaneous in the forward direction. |
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| 2547. |
Rusting of iron is quicker in saline water than in ordinary water. Why is it so ? |
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Answer» In saline water, NaCl helps water to dissociate into H+ and OH-. Greater the number of H+, quicker will be rusting of iron. |
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| 2548. |
Rusting of `Fe` is quicker in saline water than in ordinary water. |
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Answer» Correct Answer - T The `Na^(o+)` and `Cl^(c-)` present in saline water increase the conductance of the solution in contact with the metal surface. This accelerates the formation of `Fe^(2+)` ions and hence that of rust, `Fe_(2)O_(3).Xh_(2)O`. |
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| 2549. |
Calculate the emf of the cell at 298 K `Sn_((s))|Sn^(2+)(0.05M)||H_((aq))^(+)(0.02M) |H_(2)` 1 atm. Pt Given that `E_(sn^(2+)//Sn)^(0)=-0.144V` |
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Answer» Cell equation `: Sn _((s))+2H_((aq))^(+)to Sn _((aq))^(+)+H_(2(g))` Nernst equation `:E_(cell)=E_(cell)^(0)-(0.0591)/(2)log ""([Sn^(2+)])/([H^(+)]^(2))` EMF of the cell `E_(cell) =[0-(-0.14)]-(0.0591)/(2)log ""([0.05])/([0.02]^(2))` `=0.14-(0.0591)/(2)xx[2.097]=0.14-0.0620=0.078V` `EMF=0.078V` |
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| 2550. |
The conductivity of `0.20M` solution of KCl at 298 K is `0.0248S cm^(-1).` Calculate molar conductance. |
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Answer» Conductivity `(k) =0.0248S cm^(-1)=0.0248 ohm^(-1) cm^(-1)` Molar concentration `[c] =0.20mol L^(-1)` `=([0.2mol])/([1000cm^(3)])=2.0xx10^(-4)mol cm ^(-3)` Molar conductivity `[^^_(m)]=k/c=([0.0248ohm^(-1)cm^(-1)])/([2.0xx10^(-4)mol cm^(-3)])` `=124 ohm^(-1) cm ^(-2)or 1245 mol ^(-1)cm^(2)` Molar condu tivity `=124s mol ^(-1) cm^(2)` |
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