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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.
| 351. |
The reference calomel electrode is made from which of the following ?A. `ZnCI_(2)`B. `CuSO_(4)`C. `Hg_(2)CI_(2)`D. `HgCI_(2)` |
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Answer» Correct Answer - C Calomel electrode `Hg_(2)Cl_(2)` is used as reference electrode. |
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| 352. |
Normal aluminimum electrode cupled with normal hydrogen electrode gives an emf of ` 1.66 V`. So the standard electrode potential of aluminimu is ,A. `-1.66 V`B. `+1.66 V`C. `-0.83 V`D. `+0.83 V` |
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Answer» Correct Answer - B `E_("cell") = E_(OP)^(@) + E_(RP_(H))^(@)` `:. 1.66 = E_(OP_(Al))^(@) + 0` |
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| 353. |
Normal aluminimum electrode cupled with normal hydrogen electrode gives an emf of ` 1.66 V`. So the standard electrode potential of aluminimu is ,A. ` -1. 66 V`B. ` + 1.66 V`C. ` - 0. 83 V`D. ` + 0. 83 V` |
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Answer» Correct Answer - A Al is above hydrogen in the electrochemical secies, therefore `Al^(3+)` has lesser reuction tendecy as compared with `H^(+)`. Hence, hydrogen electrode acts as anode when copled with aluminum electrode. ` E_("cell")^@=E_(H+//H_2)^@ -D_(Al^(3+)//Al)^@` ` :. 1.66V=0.0V- E_(Al^(3+)//Al)^@` ` E_(Al^(3+)//Al) =- 1.66 V`. |
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| 354. |
Match the List-I with List-II `{:(,"List-I (Electrode)",,"List -II (Type)"),(1,"Calomel",(A),"Reference"),(2,"Glass",(B),"Redox"),(3,"Hydrogen",(C),"Membrane"),(4,"Quinhydrone",(D),"Gas"),(,,,):}`A. `1-A, 2-C, 3-D, 4-B`B. `1-B, 2-A, 3-D, 4-C`C. `1-C, 2-B, 30A, 4-D`D. `1 - D, 2-A, 3-C, 4-B` |
| Answer» Correct Answer - A | |
| 355. |
The reference calomel electrode is made from which of the following ?A. `PbO_(2) - PbSO_(4)` mixtureB. `HgCl_(2)`C. `Hg_(2)Cl_(2)`D. `ZnCl_(2)` |
| Answer» Correct Answer - C | |
| 356. |
CALOMEL ELECTRODEA. `1^(0)` reference electrodeB. `2^(0)` reference electrodeC. platinum electrodeD. mercury electrode |
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Answer» Correct Answer - B It is secondary reference electrode . |
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| 357. |
Normal aluminium electrode coupled with normal hydrogen electrode gives an emf of 1.66 volts . So the standard electrode potential of aluminium isA. `-1.66` VB. `+ 1.66` VC. `-0.83` VD. `+ 0.83` V |
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Answer» Correct Answer - A Al undergoes oxidation w.r.t. N.H.E. Hence , its standard electrode potential must be negative . |
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| 358. |
In an aqueous solution how does specific conductivity of electrolytes change with additon of water? |
| Answer» The specific conductivity or conductivity (k) of electrolytes decrease with addition of water or upon dilution because the number of ions per unit volume decrease. For more details consult Section 3.19. | |
| 359. |
The emf of standard H-electrode depends onA. conc. , of HClB. pressure of `H_(2)`C. temperatureD. all |
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Answer» Correct Answer - D All conditions are responsible for SHE. |
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| 360. |
In an aqueous solution how does specific conductivity of electrolytes change with addition of water? |
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Answer» Conductivity decreases because number of ions per unit volume decreases. |
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| 361. |
Which reference electrode is used to measure the electrode potnetial of other electrodes?A. Calomel electrodeB. Ag|Ag Cl electrodeC. `Hg|Hg_(2) Cl_(2) - KCl` electrodeD. `H^(+) | H_(2) `, Pt electrode |
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Answer» Correct Answer - B SHE is primary electrode . |
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| 362. |
Which reference electrode is used to measure the electrode potential of other electrodes? |
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Answer» Standard hydrogen electrode is the reference electrode whose electrode potential is taken to be zero. The electrode potential of other electrodes is measured with respect to it. |
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| 363. |
Consider the cell: `Cu|Cu^(2+)||Cl^(-)|Cl_(2),Pt` Write the reaction that occur at anode and cathode. |
| Answer» Anode: `CutoCu^(2+)+2e^(-)` (Oxidation): Cathode: `Cl_(2)+2e^(-)to2Cl^(-)`(Reduction). | |
| 364. |
Consider a cell given belowCu|Cu2+|| Cl-|Cl2,PtWrite the reactions that occur at anode and cathode |
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Answer» Anode : Cu → Cu2+ + 2e– Cathode : Cl2 + 2e– → 2Cl– Cu is anode as it is getting oxidised. Cl2 is cathode as it is getting reduced. |
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| 365. |
An electrode is prepared by dipping a silver strip into a solution saturateed with `AgACN` and contanining `0.10 M SCN^(-)`. The e.m.f. of voltaic cell constructed by connecting this, as the cathode, to the standard hydrogen half-cell as anode was found to be `0.45 V`. What is the solubility product of `AgSCN`. (Given `E_(Ag^(+)//Ag)^(@) = 0.80 V`) |
| Answer» Correct Answer - `1.169 xx 10^(-7) ;` | |
| 366. |
Select the incorrect statements about dry cell:A. It is also called Leclanche cellB. It is alos called Daniell cellC. Electrolyte used is moist paste of `NH_(4)Cl and ZnCl_(2)`D. Cathodic process is: `2MnO_(2) (s) + 2NH_(4)^(+)(aq) + 2e^(-) rarr Mn_(2)O_(3)(s) + 2NH_(3) (g) + H_(2)O(l)` |
| Answer» Correct Answer - B | |
| 367. |
`DeltaG^(c-)` or the reaction is `,` `4Al+3O_(2)+6H_(2)O+4overset(c-)(O)H rarr 4 Al(OH)_(4)^(c-)` `E^(c-)._(cell)=2.73V` `Delta_(f)G^(c-)._((overset(c)(O)H))=-157k J mol^(-1)` `Delta_(f)G^(c-)._((overset(c-)(O)H))=-237k J mol^(-1)`A. `-3.16xx10^(3)kJ "mol"^(-1)`B. `0.079xx10^(3)kJ "mol"^(-1)`C. `-0.263xx10^(3)kJ "mol"^(-1)`D. `+0.263xx10^(3)kJ "mol"^(-1)` |
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Answer» Correct Answer - a `DeltaG^(@)=-nE^(@)F` `=-12xx2.73xx96500` `=-3.16xx10^(3)kJ` |
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| 368. |
Dipping iron article into a strongly alkaline solution of sodium phosphate :A. does not affect the articleB. forms `Fe_(2)O_(3). X H_(2)O` on the surfaceC. forms iron phosphate filmD. forms ferric hydroxide |
| Answer» Correct Answer - C | |
| 369. |
Calculate `pH` of the half cell `:` `Pt,H_(2)(1 atm)|H_(2)SO_(4)" "E^(c-)=-0.3V` |
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Answer» `E_(2H^(o+)|H_(2))=-0.059pH` `pH=(-0.3)/(-0.059)~~(0.3)/(0.06)~~5` |
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| 370. |
For the redox process `Zn (s) + Cu^(2+) hArr Zn^(2+) + Cu(s) E_("cell")^(@) = + 1.10V` which graph correctly represents `E_("cell")` (Y-axis) as a function of log `([Zn^(2+)])/([Cu^(2+)])` (X-axis) ?A. B. C. D. |
| Answer» Correct Answer - B | |
| 371. |
For hydrogen oxygen fuel cell with reaction `2H_(2)(g)+O_(2)(g) rarr 2 H_(2)O(l)` `DeltaG_(f)^(c-)(H_(2)O)=-237.2kJ mol^(-1)`. Hence, `EMF` of the fuel cell isA. `+2.4 6V`B. `-2.46V`C. `+1.23V`D. `-1.23V` |
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Answer» Correct Answer - c `DeltaG^(@)=-nE^(@)F` `E^(@)=(DeltaG^(@))/(-nF)` `=(-237.2xx10^(3))/(-2xx96500)=1.229V` `=1.23V` |
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| 372. |
A graph is plotted between `E_(cell)` and `log .([Zn^(2+)])/([Cu^(2+)])` . The curve is linear with intercept on `E_(cell)` axis equals to `1.10V`. Calculate `E_(cell)` for the cell. `Zn(s)||Zn^(2+)(0.1M)||Cu^(2+)(0.01M)|Cu` |
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Answer» Cell reaction is `: Zn+Cu^(2+) rarr Zn^(2+)+Cu` `E_(cell)=E^(c-)._(cell)-(0.059)/(2) log[(Zn^(2+))/(Cu^(2+))]" "(i)` `(y=c+mx)" "(ii)` Equation `(i)` represent a straight line like Eq. `(ii)` . `:` Intercept `(c)=E^(c-)._(cell)=1.10V` `:. E_(cell)=1.10V-(0.059)/(2)log .(0.1)/(0.01)` `=1.10V-(0.059)/(2)log 10` `=1.10-0.03" "(` tAKE `0.059~~0.06)` `=1.07V` |
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| 373. |
The standard potential of a cell using the reaction is `1.12`. The heat of the reaction is `-504.2kJ mol ^(-1)` at `25^(@)C`. Calculate the entropy change. |
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Answer» `2Ni(s)+4H_(2)Orarr 2Ni(OH)_(2)+4H^(o+)+4e^(-)` `4H^(o+)+O_(2)+4e^(-)rarr2H_(2)O` `DeltaG^(c-)=-nFE^(c-)=-4xx96500xx1.12=-432320J` `DeltaG^(c-)=DeltaH^(c-)-TDeltaS^(c-)` `:. DeltaS^(c-)=(DeltaH^(c-)-DeltaG^(c-))/(T)` `=-(504.2xx10^(3)J-(-432320))/(298)` `=-241.2JK^(-1)mol^(-1)` |
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| 374. |
Given `:` `NO_(3)^(c-) rarrNO_(2)(` acidic medium `), " "E^(c-)=0.8V` `NO_(3)^(c-) rarr NO_(3)OH(` acidic medium`)" "E^(c-)=0.74V` At what `pH` the above two half reactions will have same `EMF` values ? Assume the concentration of all the species to be unity. `(` Take `0.059~~0.06)` |
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Answer» Balance both equations in acidic medium `NO_(3)^(c-)+e^(-)+2H^(o+)rarrNO_(2)+H_(2)O " "(i)` NO_(3)+6e^(c-)+7H^(o+)rarr NH_(2)OH+2H_(2)O" "(ii)` `E_(1)=E_(1)^(c-) -0.06 log .([NO_(2)])/([NO_(3)^(c-)][H^(o+)]^(2)) [[NO_(2)]=[NO_(3)^(c-)]1M]` `=0.8-0.06log[H^(o+)]^(2)=0.8-0.06xx[-logH^(o+)]` `=0.08-0.06xx2pH` `E_(1)=0.8-0.12pH" "(iii)` Similarly, `E_(2)` for reaction `(ii),` `E_(2)=E_(2)^(c-)-(0.06)/(6)log.([NH_(2)OH])/([NO_(3)^(c-)][H^(o+)]^(7))[[NH_(2)OH]=[NO_(3)^(c-)]=1M]` `=0.74-0.01log [H^(o+)]^(-7)` `=0.74-0.01xx7(-log H^(o+))` `=0.74-0.07pH " "(iv)` Since `E_(1)=E_(2),[`equate Eqs. `(iii)` and `(iv)]` `:.0.8-0.12pH=0.74-0.07pH` `pH=(0.06)/(0.05)=1.2` |
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| 375. |
An aqueous solution containing one mole per litre of each `Cu(NO_(3))_(2), AgNO_(3),Hg(NO_(3))_(2)` is being electrolysed using inert electrodes. The values of standard electrdoe potential in volts (reduction potential) are `Ag|Ag^(+)=+0.80 2Hg|Hg^(2+)=+0.79` `Cu|Cu^(2+)=+0.34 Mg|Mg^(2+)=-2.37` With increasing voltage, the sequence of deposition of metals on cathode will beA. `Ag,Hg,Cu,Mg`B. `Mg,Cu,Hg,Ag`C. `Ag,Hg,Cu`D. `Cu,Hg,Ag` |
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Answer» Correct Answer - C Higher the reduction potential easier to reduce. From an aqueous situation Mg cannot be obtained by electrolysis. Therefore, correct order of depositions as voltage is increased is Ag,Hg, Cu. |
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| 376. |
An electric current is passed through two electrolytic cells connected in series one containing aqueous `AgNO_(3)` solution while the other containing aqueous `H_(2)SO_(4)`. The volume of oxygen that would be liberated at `25^(@)C` and 750 mm pressure from `H_(2)SO_(4)` if 1 mole of `Ag^(+)` ions are deposited from `AgNO_(3)` solution.A. 6.2 LB. 7.2 LC. 8.0 LD. 10.0 L |
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Answer» Correct Answer - A (a) Since Ag is monovalent , `1" mol Ag" -=1" Eq Ag"-=1" Eq "O_(2)` Weight of `O_(2)` liberated =8 g. `T=298" K", P=(750)/(760) atm` `PV=nRT=(W)/(M)RT` `VO_(2)=(WRT)/(MP)` `=((8g)xx(0.0821" L atm " K^(-1) mol^(-1))xx(298K))/((32" g mol"^(-1))xx(0.987" atm"))=6.2" L"` |
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| 377. |
STATEMENT-1 : Using Kohlrausch's law of independent migration of ions, it is possible to calculate Λ for any electrolyte from the λ° of individual ions. STATEMENT-2 : Limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. STATEMENT-3 : When concentration approaches zero, molar conductivity reaches the lowest limit. (1) T T F (2) T T T (3) F T T (4) F F T |
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Answer» The Correct option is (1) T T F |
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| 378. |
STATEMENT-1 : Corrosion of iron is essentially an electrochemical phenomenon. STATEMENT-2 : Corrosion reaction at anode : 2Fe(s)→ 2Fe3+ +6e-STATEMENT-3 : Corrosion reaction at cathode :O2(g) +4H+(aq)+4e- →2H2O(l)(1) T T T (2) T F T (3) T F F (4) F F T |
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Answer» Correct option (2) T F T Explanation: Facts about corrosion Statement 2 : 2Fe(s)→ 2Fe2+ +4e- |
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| 379. |
Assertion Copper rod turns colourless solution of zinc sulphate to light blue.Reason Zinc reduces copper (III) ions to Cu.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. Both A and R are false. |
| Answer» Correct Answer - D | |
| 380. |
Corrosion of iron is essentially an electrochemical phenomenon where the cell reactions areA. Fe is oxidised to `Fe^(2+)` and dissolved oxygen in water is reduced to `OH^(-)`B. (b) Fe is oxidised to `Fe^(3+)` and `H_(2)O` is reduced to `O_(2)^(2-)`C. Fe is oxidised to `Fe^(2+)` and `H_(2)O` is reduced to `O_(2)^(-)`D. Fe is oxidised to `Fe^(2+)` and `H_(2)O` is reduced to `O_(2)` |
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Answer» Correct Answer - A (a) Al anode : `Fe to Fe^(2+)+2e^(-)xx2`. `("Al cathode" : O_(2)+2H_(2)O+4e^(-) to 4OH^(-))/("Overall reaction" , 2Fe+O_(2)+2H_(2)O to 2Fe(OH)_(2))` `:.` It is the correct option. |
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| 381. |
When a zinc rod is suspended in copper sulphate solutionA. the temperature of the solution risesB. the temperature of the solution fallsC. the temperature of the solution remains unchangedD. copper is deposited on zinc rod thereby increasing its weight . |
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Answer» Correct Answer - A A redox reaction is exothermic . |
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| 382. |
Construct a cell using given electrodes at `298K` and also calculate its standart `EMF`. Given ` :` `E^(c-)._(Zn|Zn^(2+))=0.76V` `E^(c-)._(Cu^(2+)|Cu)=0.34V` |
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Answer» Note that `E^(c-)._(Zn|Zn^(2+))` is the standard oxidation potential valuse while that of `E^(c-)._(Cu^(2+)|Cu)` is standard reduction potential value. Therfore, `E^(c-).-(Zn^(2+)|Zn)(` standard reductio potential `)` value is `-0.76v`. Comparing the two standard reduction potential values. `:. E^(c-)._((Cu^(2+)|Cu))gtE^(c-)._((Zn^(2+)|Zn))` Hence, copper electrode will act as cathode and Zinc electrode will act as anode. The cell representation of the cell is `:` `Zn(s)|Zn^(2+)(1.0M)||Cu^(2+)(1.0M)|Cu(s)` `E^(c-)._(cell)=(E^(c-)._(reduction))_(cathode)-(E^(c-)._(reduction ))_(anode)` `=0.34-(-0.76)=1.1V` |
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| 383. |
Which one of the following reactions cannot be used to set u p an electrochemical cell?A. `Fe+2H^(+)rarrFe^(2+)+H_(2)`B. `Mg+Sn^(2+)rarrMg^(2+)+Sn`C. `Cl_(2)+2KBrrarr2KCl+Br_(2)`D. `AgNO_(3)+NaClrarrAgCl+NaNO_(3)`. |
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Answer» Correct Answer - D The reaction `AgNO_(3)NaClrarrAgCl+NaNO_(3)` Cannot be used to set up an electrochemical cell s it is not a redox reaction. |
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| 384. |
Which one of following statements is wrong about in electrochemical cell (ECC) and an electrolytic cell (ELC)?A. GC produces electricity , ELC consumes electricity .B. GC uses a salt bridge/ porous pot , ELC does not .C. Anode of GC is negative while anode of ELC is positiveD. In both GC and ELC , the redox reaction is spontaneous . |
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Answer» Correct Answer - D In GC only redox reaction is spontaneous . |
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| 385. |
Can a solution of `CuSO_(4)` be stored in `(a)` zinc `(Zn)` pot, `(b)` silver `(Ag)` pot ? |
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Answer» `a.` From Table `3.1` , the standard reduction potential of `Zn^(2+)|Zn` is `-0.76V` and standard oxidation potential will be `0.76 V`, whereas the standard reduction potential of `Cu^(2+)|Cu` is `0.34V` and standard oxidation potential will be `-0.34 V`. Therefore ,` E^(c-)._(o x i dation(Zn|Zn^(2+))gtE^(c-)._(o x i d ation(Cu|Cu^(2+)).` Hence, `Zn` is more reactive than `Cu.` Hence, it displaces `Cu` from `CuSO_(4)` solution as follows `:` `Zn(s)+CuSO_(4)(aq)rarrZnSO_(4)(aq)+Cu(s)` or `Zn(s)+Cu^(2+)(aq)rarr Zn^(2+)(aq)+Cu(s)` Thus, `Zn` reacts with `CuSO_(4)` solution. Hence,` CuSO_(4)` solution cannot be stored in `Zn` pot. Alternatively Find `E^(c-)._(cell)` . If it is positive, it means cell reaction will occur, and one cannot store the solution in the pot. If the standard oxidation potential of the metal behaving as pot is greater than the standard oxidation otential of the metal consisting of solution ,then `E^(c-)._(cell)` will be positive and one cannot store the solution in the pot. `:. E^(c-)._(cell)=(E^(c-)._(reduction))_(cathode)-(E^(c-)._(reduction))_(anode)` `=E^(c-).Cu^(2+)|Cu)-E^(c-)._(Zn^(2+)|Zn)` `=0.34-(-0.76)=1.10V` Therefore, a solution of `CuSO_(4)` cannot be stored in `Zn` pot. `b.` From Table `3.1`, the standard reduction potential of `Ag^(o+)|Ag` is `0.80V` and standard reduction potential of `Cu^(2+)|Cu` is `0.34 V` . `E^(c-)._(Ag^(o+)|Ag)gtE^(c-)._(Cu^(2+)|Cu)` or `E^(c-)._(Ag|Ag^(o+))ltE^(c-)._(Cu|Cu^(2+))` Since the standard oxidatioin potential of the metal `(Ag)` behaving as pot is less than the standard oxidation potential of the metal consisting of solution `(CuSO_(4))`, the `E^(c-)._(cell)` will be negative, and one can store the solution `(CuSO_(4))` in the pot of `Ag`. `E^(c-)._(cell)=(E^(c-)._(reduction))_(cathode)-(E^(c-)._(reduction))_(anode)` `=E^(c-)._(Cu^(2+)|Cu)-E^(c-)._(Ag^(o+)|Ag)` `=0.34-0.80=-0.46V` Therefore, one can store `CuSO_(4)` in `Ag` pot. |
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| 386. |
Which one of following statements is wrong about in electrochemical cell (ECC) and an electrolytic cell (ELC)?A. ECC produces electricity ELC consumes electricity.B. ECC uses a salt bridge/porous pot, ELC does notC. Anode of ECC is negative while anode of ELC is positiveD. In both ECC and the redox reaction is spontaneous. |
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Answer» Correct Answer - D In ECC, redox reaction in spontaneous while is ELC, redox reaction is non spontaneous. |
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| 387. |
The measured e.m.f. at `25^(@)C` for the cell reaction, `Zn(s) +XCu^(2+)._((eq))(1.0M) rarr Cu(s) ._((aq))(0.1M)` is 1.3 volt, calculate `E^(@)` for the cell reaction. |
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Answer» Using nernst equation `("at" 298K),E_(cell) = E_(cell)^(@) - (0.0591)/(2) log.([Zn^(2+)(aq)])/([Cu^(2+)(aq)])` Here, `E_(cell) = 1.3 V, [Cu^(2+)(aq)] = 1.0 M, [Zn^(2+)(aq)] = 0.1M, C_(cell) = ?` Substituting the values, `1.3 V = E_(cell)^(@) - (0.0591V)/(2) log.(0.1)/(1.0)` `1.3 = E_(cell)^(@) - 0.02955 V log 10^(-1)` `1.3 V = E_(cell)^(@) + 0.02955 V log 10` `E_(cell)^(@) = 1.3 V - 0.02955 V = 1.27 V` |
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| 388. |
`CuSO_(4)` is not stored in aluminium bottles becauseA. Cu gets oxidisedB. Cu gets reducedC. Al gets reducedD. `CuSO_(4)` gets decomposed |
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Answer» Correct Answer - B Al is more reaction than `Cu` i.e., `Cu^(2+)` gets reduced . |
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| 389. |
Cell reaction for the cell `Zn|Zn^(2+)(1.0M)||Cd^(2+)(1.0M)|Cd` is given byA. `CdrarrCd^(2+)+2e^(-)`B. `Zn^(2+)rarrZn-2e^(-)`C. `Cd+Zn^(2+)rarrZn+Cd^(2+)`D. `Zn+Cd^(2+)rarrZn^(2+)+Cd`. |
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Answer» Correct Answer - D The two half cell reaction for the given cells are: `Zn(s)rarrZn^(2+)+(aq)+2e^(-)` `Cd^(2+)(aq)+2e^(-)rarrCd(s)` New reaction: `Zn(s)+Cd^(2+)(aq)rarrZn^(2+)(aq)+Cd(s)` |
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| 390. |
The reduction potential of the two half cell reaction (occuring in an electrochemical cell) are `PbSO_(4)(s)+2e^(-)rarrPb(s)+SO_(4)^(2-)(aq)(E^(@)=-0.31V)` `Ag^(+)(aq)+e^(-)rarrAg(s)(E^(@)=0.80V)` The fessible reaction will beA. `Pb + SO_(4)^(2-) + 2 Ag^(+) (aq) to 2 Ag (s) + PbSO_(4)`B. `PbSO_(4) + 2 Ag^(+) (aq) to Pb + SO_(4)^(2-) + 2 Ag(s)`C. `Pb + SO_(4)^(2-) + Ag (s) to Ag^(+) (aq) + PbSO_(4)`D. `PbSO_(4) + Ag (s) to Ag^(+) (aq) + Pb + SO_(4)^(2-)` |
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Answer» Correct Answer - A For EMF to be positive , oxidation should occur on lead electrode . |
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| 391. |
The reduction potential of the two half cell reactions (occuring in an electrochemical cell) are `PbSO_(4)+ 2e^(-)rarrPb+SO_(4)^(2-) (E^(@)=-0.31V)` `Ag^(+)(aq)+e^(-)rarrAg(s)(E^(@)=+0.80V)` The fessible reaction will beA. `Pb^(2+)+2Agrarr2Ag^(+)+Pb`B. `Pb^(2+)+H_(2)rarr2H^(+)+Pb`C. `2H^(+)+2Agrarr2Ag^(+)+H_(2)`D. `2Ag^(+)+PbrarrPb^(2+)+2Ag` |
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Answer» Correct Answer - D For the cell reaction to be spontaneous `E_("cell")` should be positive. Therefore, two half cell reaction are `Ag^(+)+e^(-)rarrAg (E^(@)=+0.80V)` `PbrarrPb^(2+)+2e^(-)(E^(@)=+0.13V)` and net reaction is `2Ag^(+)+Pbrarr2Ag+Pb^(2+)` |
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| 392. |
The reduction potential of the two half cell reaction (occuring in an electrochemical cell) are `PbSO_(4)(s)+2e^(-)rarrPb(s)+SO_(4)^(2-)(aq)(E^(@)=-0.31V)` `Ag^(+)(aq)+e^(-)rarrAg(s)(E^(@)=0.80V)` The fessible reaction will beA. `Pb(s)+SO_(4)^(2-)(aq)+2Ag^(+)+(aq)rarr2Ag(s)+PbSO_(4)(s)`B. `PbSO_(4)(s)+2Ag^(+)(aq)rarrPb(s)+SO_(4)^(2-)(aq)+2Ag(s)`C. `Pb(s)+SO_(4)^(2-)(s)+Ag(s)rarrAg^(+)(aq)+PbSO_(4)(s)`D. `PbSO_(4)(s)+Ag(s)rarrAg^(+)(aq)+Pb(s)+SO_(4)^(2-)(aq)` |
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Answer» Correct Answer - A Higher the reduction potential easier to reduce, Therefore two h alf cell reactions are: `Pb(s)+SO_(4)^(2-)(aq)rarrPbSO_(4)(s)+2e^(-)` `Ag^(+)(aq)+e^(-)rarrAg(s)xx2` `Pb (s)+SO_(4)^(2-) (aq)+2Ag^(+)(aq)rarrPbSO_(4)(S)+Ag(s)` |
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| 393. |
Consider the following half reactions `:` `PbO_(2)(s)+4H^(o+)(aq)+SO_(4)^(2-)+2e^(-)rarrPbSO_(4)9s)+2H_(2)O" "E^(c-)=+1.70V` `PbSO_(4)(s)+2e^(-)rarr Pb(s)+SO_(4)^(2-)(aq)" "E^(c-)=-0.31V` `a.` Calculate the value of `E^(c-)` for the cell. `b.` Calculate the voltage generated by the cell if `[H^(o+)]=0.10M` and `[SO_(4)^(2-)]=2.0M` `c.` What voltage is generated by the cell when it is at chemical equilibrium ? |
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Answer» `a. E^(c-)._(cell)=E^(c-)._(Cathode)-E^(c-)._(Anode )` `=1.70-(-0.31)=2.01V` `PbO_(2)+4H^(o+)+SO_(4)^(2-)+2e^(-)rarr PbSO_(4)+2H_(2)O` `Pb+SO_(4)^(2-)rarr PbSO_(4)` `ulbar(PbO_(2)+Pb+4H^(o+)+2SO_(4)^(2-)rarr 2PbSO_(4)+2H_(2)O)` `b.` `E_(cell)=E^(c-)._(cell)-(0.059)/(2)log.(1)/([H^(o+)][SO_(4)^(2-)]^(2))` `=2.01-(0.059)/(2)log.(1)/((0.14)^(4)(2.0)^(2))` `c.` When cell is at equilibrium , `E_(cell)=0` |
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| 394. |
Explain the terms with suitable exapmples. Average rate of a reaction |
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Answer» Average rate of a reaction: The charge in the concentratio of any one of the reactants or products per unit time is called average rate of a reaction. 1) `2HI _((g))to H_(2(g))+I_(2(g))` Rate of reaction `=-1/2 (Delta[HI])/(Deltat)=(Delta[H_(2)])/(Deltat)=(Delta[I_(2)])/(Deltat)` 2) ` Hg_((l))+Cl_(2(g)) to HgCl_(2(g))` Rate of reaction `=(Delta[Hg])/(Deltat)=-(Delta[Cl_(2)])/(Deltat)=(Delta[Hg Cl_(2)])/(Deltat)` 3) `5 Br_((aq))^(-)+BrO_(3(aq))^(-)+6H_((aq))^(+)to 3 Br_(2(aq))+3H_(2)O_((l))` Given reaction is `5 Br_((aq))^(-) +BrO_(3(aq))^(-) +6H_((aq))^(+)to 3 Br _(2(aq))+ 3H_(2) O_((l))` Rate of reaction `=-1/5(Delta[Br^(-)])/(Deltat)=(-Delta[BrO_(3)^(-)])/(Deltat)=(-1)/(6)(Delta[H^(+)])/(Deltat)=(-1)/(3)(Delta[Br_(2)])/(Deltat)=1/3 (Delta[H_(2)O])/(Deltat)` |
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| 395. |
Assertion `(A) : (H_(2)+O_(2))` fuel cell gives a constant voltages throughout its life. Reason `(R): `In this fuel cell, `H_(2)` reacts with `overset(c-)(Oh)H` ions, yet the over all `[overset(c-)(O)H]` does not change.A. If both `(A)` and `(R)` are correct, and `(R)` is the correct explanation of `(A)`.B. If both `(A)` and `(R)` are correct, but `(R)` is not the correct explanation of `(A)`.C. If `(A)` is correct, but `(R)` is incorrect.D. If `(A)` is incorrect, `(R)` is correct. |
| Answer» Correct Answer - a | |
| 396. |
Statement-I: `H_(2)+O_(2)` fuel cell gives a constant voltage throughout its life. Because Statement-II: In this fuel cell, `H_(2)` reacts with `OH^(-)` ions yet the overall concentration of `OH^(-)` ions does not change.A. If both assertion and reason are correct and reason is correct explanation for assertion.B. If both assertion and reason are correct but reason is not correct explanation for assertion.C. If assertion is correct but reason is incorrect.D. If assertion and reason both are incorrect. |
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Answer» Correct Answer - A (a) Reason is the correct explanation for assertion. |
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| 397. |
Consider the argument, “On the basis of position in the electrochemical series, Al does not displace hydrogen from water and acids1. Do you agree with the argument ? 2. Substantiate your answer. |
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Answer» 1. No. 2. Metals which are above hydrogen in the electro chemical series can displace H2 from acids, i.e., elements having lower reduction potential than hydrogen, can liberate H2 from acids and water. Al displaces hydrogen from water and acids. |
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| 398. |
More electropositive elements haveA. positive reduction potentialB. tendency to gain electronsC. negative reduction potentialD. negative oxidation potential |
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Answer» Correct Answer - C `E_(PO)^(@) = +ve` thus `E_(RP)^(@) = -ve` |
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| 399. |
Which of the following is the most electropositive elementA. CarbonB. CalciumC. ChlorineD. Potassium |
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Answer» Correct Answer - D Potassium is more electropositive element, because it is the only alkali metal among the given element. |
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| 400. |
The electrode reactions that takes place at the anode of `CH_(4)-O_(2)` fuel cell is :A. `2O_(2)+8H^(+)+8e^(-) rarr 4H_(2)O`B. `CH_(4)+2H_(2)O rarr CO_(2)+8H^(+)+8e^(-)`C. `CH_(4)+2O_(2) rarr O_(2)+2H_(2)O`D. `2H^(+)+2e^(-) rarr H_(2)` |
| Answer» Correct Answer - B | |