94420 + Interview Questions in Chemistry in Class 12

1.	Electrolytic reducation of Al_(2)O_(3) Electrolyte (Al_(2)O_(3)+cryolite) Cathode : fraphite inside Fe -Container Pick out the statement which is wrong if anode is made of nikel instesd of graphoite?
Answer» ni is costly Anode will be affected by the prouct formed Graphite remain UNAFFECTED by prouced `F_(2)` NIKKEL may be affected at HIGH temperature Answer :b

Discussion

2.	Electrolytic reduction is used in the extraction of
Answer» Magnesium Sodium Aluminium GOLD SOLUTION :MG, Na, Al electropositive metals

Discussion

3.	Electrolytic oxidation of H_2SO_4gives rise to the formation of persulphuric acid, H_2S_2O_8.2H_2SO_4 to H_2S_2O_8 + 2H^(+) + 2eIf 4 litres of O_2and 11.2 litres of H_2were produced at NTP, determine the mass of H_2S_2O_8produced.
Answer» SOLUTION :`2H_2SO_4to H_2S_2O_8 + H_2` `2H_2O to 2H_2+ O_2` EQ. of `H_2S_2O_8 = ` eq. of `H_2 ` - eq. of `CO_2` 27.72g

Discussion

4.	Electrolytic process is used for the extraction of
Answer» Alkali metal ALKALINE earth METALS AL All Solution : Alkali metals, alkaline earth metals Al are EXTRACTED by electrolytic process.

Discussion

5.	Electrolytic conduction is due to movement of
Answer» MOLECULES ATOMS ions electrons Answer :C

Discussion

6.	Electrolytic conduction differs from metallic conduction in the fact in the case of electrolytic conduction
Answer» the resistance increases with INCREASING temperature the reistance decreases with increasing temperature the flow of CURRENT does not generate heat theresistance is INDEPENDENT of the length of the conductor . SOLUTION :Electrolytic conduction in temperature .

Discussion

7.	Electrolytic conduction differs from metallic conduction in that in the case of electrolytic conduction
Answer» The RESISTANCE INCREASES with INCREASING temperature The resistance decreases with increasing temperature The flow of current does not generate heat The resistance is INDEPENDENT of the LENGTH of the conductor. Solution :Electrolytic conduction resistance decreases with increasing temperature.

Discussion

8.	Electrolytic cell containing molten nickel chloride and aluminium chloride solutions are arranged in a series. If on passing same current through both the solution, if 18 gm Al is obtained then how much Ni is obtained ? (Atomic mass of Al=27 and Ni=58.5 gm mol^(-1))
Answer» 58.5 gm 29.25 gm 117 gm 5.85 gm Solution :(i) Reaction when Al METAL is formed : `Al^(3+)+3e^(-) to Al_((S))` So, 3 MOL `e^(-) to 1` mol AI So, 27 gm Al required 3 F CURRENT. So for 18 gm Al, amount of electricity `=(18xx3)/(27)=2F` (ii) 2F current is also PASSED through `Ni^(2+)` solution as it is in series. `Ni^(2+)+2e^(-) to NI_((S))`

Discussion

9.	Electrolytic cell is used to convert
Answer» ELECTRICAL energy to chemcial energy electrical energy to MECHANICAL energy mechanical energy to chemical energy chemcial energy to electrical energy Solution :Nonspontaneous REDOX reaction CONVERTS electrical energy Taction to chemical energy.

Discussion

10.	Electrolytic cell and Electrochemical cell or Galvanic cell or Voltaic cell.
Answer» Solution :`{:("ELECTROLYTIC CELL","electrochemical cell (Galvanic cell or Voltaic cell)"),("1. This device is used to bring about a non-spontaneous chemical REACTION by passing an electric current.","1. This device is used to produce elexctrical energy by spontaneous chemical reaction.,"),("2. It is used to bring about a chemical reaction GENERALLY FRO the dissocition (electrolysis) of compounds.","2. It is used to generate electrocity."),("3. In this cell, electrical energy is converted into chemical energy.","3. In this cell, chemical energy is converted into electrical energy."),("4. In this cell, the cathode is negative and anode is positive.","4. In this cell, the cathode is positive and the anode is negative."),("5. Electrolytic cells are irreversible.","5. Electrochemical cells are reversible".):}`

Discussion

11.	{:("Electrolyte:",KCl,KNO_(3),HCl,NaOAc,NaCl),(wedge^(oo)(Scm^(2)mol^(-1)),149.9,145.0,426.2,91.0,126.5):} Calcualte wedge_(HOAc)^(oo) using appropriate molar conductances of the electrolytes listed above at infinite dilution in H_(2)O at 25^(@)C.
Answer» 517.2 552.7 390.7 217.5 Solution :`wedge_(HOAC)^(OO)=wedge_(NAOAC)^(oo)+wedge_(HCL)^(oo)-wedge_(NaCl)^(oo)` ltBrgt `=91.0+426.2-126.5=390.7`

Discussion

12.	{:("Electrolyte","Battery"),((i)NH_(4)Cl+ZnCl_(2)+H_(2)O,"(a) Mercury button cell"),("(ii) Paste of KOH and ZnO","(b) Lithium - ion battery"),("(iii) 38% by mass of "H_(2)SO_(4),"(c ) Leclanche cell"),("(iv) Lithium salt in an organic solvent","(d) Lead storage battery"):}
Answer» `{:(A,B,C,D),(c,a,d,b):}` `{:(A,B,C,D),(d,c,b,a):}` `{:(A,B,C,D),(a,b,c,d):}` `{:(A,B,C,D),(c,a,d,b):}` ANSWER :A

Discussion

13.	Electrolytes when dissolved in water dissociate into ions, because :
Answer» They are unstable The WATER DISSOLVES them The FORCES of repulsion increases The forces of ELECTROSTATIC attraction are BROKEN down by water Answer :C

Discussion

14.	Electrolyte when dissolved in water dissociate into their constituent ions. The degree of dissociation of an electrolyteincreases with
Answer» Increasing concentration of the electrolyte Decreasing concentration of the electrolyte Decreasing temperature Presence of a SUBSTANCE yielding a COMMON ionsely PROPORTIONAL to the concentration of the electrolyte Solution :Because the degre of DISSOCIATION is inversely proportional to the concentration of the electrolyte.

Discussion

15.	Electrolysis rules of Faraday states that mass deposited on electrode is proportional to
Answer» `m oo l^2` `m oo Q` `m oo Q^2` NONE of these SOLUTION :It is faraday's law.

Discussion

16.	Electrolysis temperature is maximum for
Answer» `AsH_(3)` `NH_(3)` `PH_(3)` `SbH_(3)` Solution :`NH_(3)` is most THERMALLY stable hydride. Hence, ELECTROLYSIS TEMPERATURE is MAXIMUM.

Discussion

17.	Electrolyte can conduct electricity because
Answer» ANSWER :D

Discussion

18.	Electrolyte can conduct electricity because ___________.
Answer» their molecules CONTAIN unpaired electrons, which are mobile their molecule contain loosly held electrons which GETS FREE under the influence of VOLTAGE the molecules BREAK up into ions when a voltage is applied the molecules are broken up into ions when the electrolyte is fused or is dissolved in the solvent Answer :D

Discussion

19.	Electrolysis of water with 1 faraday electricity gives
Answer» 1 mole of oxygen 1 gram equivalent of oxygen 1 molecule of oxygen 1 ATOM of oxygen Solution :That's the THUMB rule, one faraday current RELEASES one gram equivalent of the CONCERNED substance.

Discussion

20.	Electrolysis of water with 1 Faraday electricity gives :
Answer» 1 MOLE of OXYGEN 1 gram equilvalent of oxygen 1 molecule of oxygen 1 atom of oxygen ANSWER :B

Discussion

21.	Electrolysis of queous potassium sulphate containing litmus, develop red colour at anode and blue colour at cathode. Why ?
Answer» Solution :During the electrolysis of aqueous `K_2SO_4` , the chemical CHANGES involved are, At anode (oxidation of water) = `2H_2O(L) to O_2 + 4E^(-) + 4H^(+)(aq)` At CATHODE (reduction of water) = `2H_2O (l) + 2e^(-) to H_2(g) + 2OH^(-)(aq)` Solution is ACIDIC around anode and litmus is red. Solution is basic around cathode and litmus is blue.

Discussion

22.	Electrolysis of molten sodium chloride leads to the formation of
Answer» `Na` and `H_(2)` Na and `O_(2)` `H_(2)` and `O_(2)` Na and `Cl_(2)` Solution :`UNDERSET("MOLTEN")(2NaCl) OVERSET("Electrolysis")to underset("Cathode")(2Na)+underset("Anode")(Cl_(2))`

Discussion

23.	Electrolysis of molten NaCl leads to the formation of :
Answer» SODIUM and oxygen sodium and hydrogen hydrogen and oxygen sodium and chlorine Answer :D

Discussion

24.	Electrolysis of KBr (aq) gives Br_(2) at anode but of KF (aq) does not give F_(2) . Give reason for disparity in behaviour.
Answer» Solution :`L_(F_(2)//F^(-))^(@)`has the HIGHEST reduction potential. THEREFORE, `F_(2)`is not liberated. Reduction potential of `O_(2)`is higher than that of `Br_(2)` . Thus, ELECTROLYSIS of KBr (aq) gives `Br_(2)`and not `O_(2)`

Discussion

25.	Electrolysis of fused KCl.MgCl_2.6H_2O gives:
Answer» POTASSIUM only MAGNESIUM only Magnesium and chloride Potassium, magnesium and chlorine Answer :D

Discussion

26.	Electrolysis of dilute aqueous solution of KCl gives?
Answer» `O_2` at anode `Cl_2` at anode K at cathode `H_2` at cathode SOLUTION :Anode : `2Cl^(-) to Cl_2 + 2E^(-)` , Cathode : ` 2H^(+) + 2e^(-) to H_2`

Discussion

27.	Electrolysis of dilute aqueous NaCl solution was carried ut be passing 10 milli ampere current. The time required to liberate 0.01 mol of H_(2) gas at the cathode is (1Faraday=96500 C mol^(-1))
Answer» `9.65xx10^(4)sec` `19.3xx10^(4)sec` `28.95xx10^(4)sec` `38.6xx10^(4)sec` Solution :`NaCl+aq to Na^(+)+CL^(-)` `H_(2)OhArrH^(+)+OH^(-)` `H^(+)+e^(-)to(1)/(2)H_(2)` Thus, 0.5 mole of `H_(2)` is liberated by 1F=96500C `therefore0.01V` mole of `H_(2)` will be liberated by CHARGE `=(96500)/(0.5)xx0.01=1930C` `Q=Ixxt` or `t=(Q)/(I)=(1930C)/(10xx10^(-3)A)`

Discussion

28.	Electrolysis of dilute aqueous sodium chloride solution was carried out by passing 10 milliampere current. The time required to liberate 0.01mol of H_2 gas at the cathode is (1 Faraday= 96500 C mol^(-1)).
Answer» `9.65 XX 10^4s` `19.3xx10^4s` `28.95 xx10^4s` `38.6 xx 10^4 s` ANSWER :B

Discussion

29.	Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H_(2) gas at the cathode is (1 Faraday=96500 C mol^(-1))
Answer» `9.65xx10^(4)`sec `19.3xx10^(4)`sec `28.95xx10^(4)sec` `38.6xx10^(4)`sec Solution :Faraday's LAW Equivalent of `H_2` PRODUCED `=(1xxt(sec))/(9)` `0.01xx2=(10xx10^(-3)t)/(96500)=96500xx2=t` `19.3xx10^(4)sec=1`.

Discussion

30.	Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 rnilli ampere current. The time required to liberate 0.01 mole of H_(2) gas at the cathode is (1 Faraday =96500 C mol^(-1) )
Answer» `9.65 xx10^(4)SEC` `19.3 XX10^(4)sec` `28. 95xx10^(4)sec` `38.6 xx10^(4)sec` Solution :`H_(2)O hArrH^() OH^()` `H^() * c ^() * 1/2 H_(2)` `therefore0.5` mole of `H_(2)` is liberated by 1 `F = 96500C 0.01 ` mole of `H_(2)` will beliberated by `=(96500)/(0.5)0.01""=1930C` `Q=1xx1` `t * (Q)/(T)* (1930)/(10 * 10^(3)A) 19.3* 10^(4)sec`

Discussion

31.	Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H_(2) gas at the cathode is (1 F = 96500 C "mol"^(-1))
Answer» `9.56xx10^(4)SEC` `19.3xx10^(4)sec` `28.95xx10^(4)sec` `38.6xx10^(4)sec` Solution :Number of EQUIVALENTS of HYDROGEN = 0.02

Discussion

32.	Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberated 0.01 mole of H_(2) gas at cathode is: ("1 faraday =96500"C mol"^(-1)).
Answer» `9.65xx 10^(4)"sec"` `19.3 xx 10^(4)"sec"` `28.95 xx 10^(4) sec` `38.6 xx 10^(4)" sec"` SOLUTION :Mass of 0.01 MOL of `H_(2)=0.02g` `W=(LT E)/(96500)` `0.02=(10 xx 10^(-3) xx t xx 1)/(96500)` `t=19.3 xx 10^(4)" sec"`.

Discussion

33.	Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli-ampere current. The time required to liberate 0.01 mole of H_(2) gas at the cathode is : (1 faraday = 96500 C mol^(-1))
Answer» `9.65 XX 10^(4)` SEC `19.3 xx 10^(4)` sec `28.95 xx 10^(4)` sec `38.6 xx 10^(4)` sec Solution :`(W)/(E) = (i xx T)/(96500)` `0.01 xx 2 = (10 xx 10^(-3) xx t)/(96500)` ` t = 19.3 xx 10^(4)` sec

Discussion

34.	Electrolysis of concentrated aqueous solution of an alkali salt of fatty acid gives :
Answer» Alkanes Alkenes Alkynes Alcohols. Answer :A

Discussion

35.	Electrolysis of cold concentrated aqueous solution of potassium succinate yields
Answer» Ethane Ethyne Ethene Ethane-1, 2-diol Solution :`UNDERSET(CH_2COOK)OVERSET(CH_2COOK)\|+2H_2Ooverset"Electrolyis"to underset"ANODE"underset(CH_2)underset(\|\|)CH_2+2CO_2 +2KOH+underset"CATHODE"(H_2)`

Discussion

36.	Electrolysis of concentrated solution of potassium acetate gives :
Answer» ETHANE ETHYLENE BUTANE ACETYLENE ANSWER :A

Discussion

37.	Electrolysis of cold conc. Aqueous solution of potassium succinate yields
Answer» Ethane Ethene Ethyne Ethane-1-2-diol Answer :B

Discussion

38.	Electrolysis of cold concentrated aqueous solution of potassiumsuccinate yields:
Answer» ETHANE Ethyne Propene Ethane-1,2-diol Answer :C

Discussion

39.	Electrolysis of brine gives
Answer» `Cl_(2)` `H_(2)` NaOH `Na+H_(2) + Cl_(2)` Solution :`2NaCl RARR 2Na^(+) + 2Cl^(-)`( Ionisation ) , Anode `: 2Cl rarr Cl_(2) + 2e^(-)` ( Oxidation) Cathode `: H_(2) O overset( 2e) ( rarr) H_(2) + 2OH^(-) `( REDUCTION) `2Na^(+)+2OH^(-) rarr 2NaOH`

Discussion

40.	Electrolysis of aqueous solution of Na_(2)SO_(4) is carried out in presence of graphite electrode.
Answer» Na is OBTAINED at cathode. `O_(2)` gas is obtained at cathode. `H_(2)` gas is obtained at ANODE. `O_(2)` gas is obtained at anode. Solution :`Na_(2)SO_(4) to 2Na_((aq))^(+)+SO_(4(aq))^(2-)` `H_(2)` gas is obtained at cathode on reduction of water, no reduction of `Na^(+)` is OBSERVED on cathode. `2H_(2)O_((L))+2e^(-) to H_(2(g))+2OH_((aq))^(-)` `O_(2)` is obtained at anode on OXIDATION of water, no oxidation is observed on `SO_(4)^(2-)`. `H_(2)O_((l)) to (1)/(2)O_(2(g))+2H_((aq))^(+)+2e^(-)`.

Discussion

41.	Electrolysis of aqueous HCl solution produces
Answer» `H_(2)` GAS at the anode `H_(2)` gas at the cathode `Cl_(2)` gas at the cathode `Cl_(2)` and `O_(2)` gases both at the anode Solution :`2H^(+) + 2 e to H_(2)`

Discussion

42.	Electrolysis of aqueous HCI solution produces
Answer» Solution :`2H^(+)to2e^(-)H_(2(G))` at CATHODE.

Discussion

43.	Electrolysis of aqueous alkali metal chloride does not liberatemetal . Why ?
Answer» SOLUTION :In aqueous solutions, both sodium cation and water MOLECULES are present around cathode. Water is PREFERENTIALLY reduced to liberate HYDROGEN, since metal cation is stable.

Discussion

44.	Electrolysis of aqueous copper sulphatesolution liberateswhich of the following gasesat the anodes ?
Answer» `SO_(3)^(-)` `SO_2` `O_2` `H_2` ANSWER :C

Discussion

45.	Electrolysis of aqueous HF produces O_2 at anode but not F_2 Explain.
Answer» Solution :In aqueous solution , fluoride `(F^(-))` is very stable Electrolysis of aqueous fluoride liberates oxygen , due to the preferential OXIDATION of WATER . `2H_(2)Oto4e^(-)+4H^(+)+O_(2)`

Discussion

46.	Electrolysis of aq solution of brine gives
Answer» `Cl_(2)` `H_(2)` NaBr KI Solution :`2NaCl rarr 2NA^(+) + 2Cl^(-)` at CATHODE `: 2H_(2) O + 2e^(-) rarr 2OH^(-) + H_(2) UARR` `2Na^(+) + 2HO^(-) rarr 2NaOH` at anode `: 2Cl^(-) rarr CL^(2) uarr`

Discussion

47.	Electrolysis of an aqueous solution of sodium chloride produces____at the cathode and ____at the anode.
Answer» SOLUTION :`H_(2),Cl_(2)`

Discussion

48.	Electrolysis of an aqueous solution of copper sulphate using platinum electrodes produces____ at the cathode and ___at the anode.
Answer» SOLUTION :`CU,O_(2)`

Discussion

49.	Electrolysis of an aqueous solution of AgNO_(3) with silver electrodes produces ul((i)) at cathode while ul((ii)) ions are dissolved from anode. When Pt electrodes are used ul((iii)) is produced at anode and ul((iv)) at cathode.
Answer» `{:("(i)", "(ii)","(III)","(IV)"),(H_(2), NO_(3)^(-) , OH^(-) , H_(2)):}` `{:("(i)", "(ii)","(iii)","(iv)"),(Ag, H^(+) , O_(2) , H_(2)):}` `{:("(i)", "(ii)","(iii)","(iv)"),("Ag", "Ag"^(+) , O_(2) , Ag):}` `{:("(i)", "(ii)","(iii)","(iv)"),(Ag, H^(+), "Ag"^(+), O_(2)):}` Answer :C

Discussion

50.	Electrolysis of an aqueous solution of AgNO_3 with silver electrodes produce __(i)___ at cathode while ___(ii)___ ions are dissolved from anode. When Pt electrodes are used ____(iii)___ is produced at anode ___(iv)___ at cathode.
Answer» `{:("(i)","(II)","(III)","(iv)"),(H_2,NO_3^(-) , OH^(-), H_2):}` `{:("(i)","(ii)","(iii)","(iv)"),(Ag,H^+, O_2, H_2):}` `{:("(i)","(ii)","(iii)","(iv)"),(Ag,H^(+),O_(2),Ag):}` `{:("(i)","(ii)","(iii)","(iv)"),("Ag","H"^(+),"Ag"^(+),"O"_(2)):}` Answer :C

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Electrolytic reducation of Al_(2)O_(3) Electrolyte (Al_(2)O_(3)+cryolite) Cathode : fraphite inside Fe -Container Pick out the statement which is wrong if anode is made of nikel instesd of graphoite?

Electrolytic reduction is used in the extraction of

Electrolytic oxidation of H_2SO_4gives rise to the formation of persulphuric acid, H_2S_2O_8.2H_2SO_4 to H_2S_2O_8 + 2H^(+) + 2eIf 4 litres of O_2and 11.2 litres of H_2were produced at NTP, determine the mass of H_2S_2O_8produced.

Electrolytic process is used for the extraction of

Electrolytic conduction is due to movement of

Electrolytic conduction differs from metallic conduction in the fact in the case of electrolytic conduction

Electrolytic conduction differs from metallic conduction in that in the case of electrolytic conduction

Electrolytic cell containing molten nickel chloride and aluminium chloride solutions are arranged in a series. If on passing same current through both the solution, if 18 gm Al is obtained then how much Ni is obtained ? (Atomic mass of Al=27 and Ni=58.5 gm mol^(-1))

Electrolytic cell is used to convert

Electrolytic cell and Electrochemical cell or Galvanic cell or Voltaic cell.

{:("Electrolyte:",KCl,KNO_(3),HCl,NaOAc,NaCl),(wedge^(oo)(Scm^(2)mol^(-1)),149.9,145.0,426.2,91.0,126.5):} Calcualte wedge_(HOAc)^(oo) using appropriate molar conductances of the electrolytes listed above at infinite dilution in H_(2)O at 25^(@)C.

{:("Electrolyte","Battery"),((i)NH_(4)Cl+ZnCl_(2)+H_(2)O,"(a) Mercury button cell"),("(ii) Paste of KOH and ZnO","(b) Lithium - ion battery"),("(iii) 38% by mass of "H_(2)SO_(4),"(c ) Leclanche cell"),("(iv) Lithium salt in an organic solvent","(d) Lead storage battery"):}

Electrolytes when dissolved in water dissociate into ions, because :

Electrolyte when dissolved in water dissociate into their constituent ions. The degree of dissociation of an electrolyteincreases with

Electrolysis rules of Faraday states that mass deposited on electrode is proportional to

Electrolysis temperature is maximum for

Electrolyte can conduct electricity because

Electrolyte can conduct electricity because ___________.

Electrolysis of water with 1 faraday electricity gives

Electrolysis of water with 1 Faraday electricity gives :

Electrolysis of queous potassium sulphate containing litmus, develop red colour at anode and blue colour at cathode. Why ?

Electrolysis of molten sodium chloride leads to the formation of

Electrolysis of molten NaCl leads to the formation of :

Electrolysis of KBr (aq) gives Br_(2) at anode but of KF (aq) does not give F_(2) . Give reason for disparity in behaviour.

Electrolysis of fused KCl.MgCl_2.6H_2O gives:

Electrolysis of dilute aqueous solution of KCl gives?

Electrolysis of dilute aqueous NaCl solution was carried ut be passing 10 milli ampere current. The time required to liberate 0.01 mol of H_(2) gas at the cathode is (1Faraday=96500 C mol^(-1))

Electrolysis of dilute aqueous sodium chloride solution was carried out by passing 10 milliampere current. The time required to liberate 0.01mol of H_2 gas at the cathode is (1 Faraday= 96500 C mol^(-1)).

Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H_(2) gas at the cathode is (1 Faraday=96500 C mol^(-1))

Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 rnilli ampere current. The time required to liberate 0.01 mole of H_(2) gas at the cathode is (1 Faraday =96500 C mol^(-1) )

Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H_(2) gas at the cathode is (1 F = 96500 C "mol"^(-1))

Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberated 0.01 mole of H_(2) gas at cathode is: ("1 faraday =96500"C mol"^(-1)).

Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli-ampere current. The time required to liberate 0.01 mole of H_(2) gas at the cathode is : (1 faraday = 96500 C mol^(-1))

Electrolysis of concentrated aqueous solution of an alkali salt of fatty acid gives :

Electrolysis of cold concentrated aqueous solution of potassium succinate yields

Electrolysis of concentrated solution of potassium acetate gives :

Electrolysis of cold conc. Aqueous solution of potassium succinate yields

Electrolysis of cold concentrated aqueous solution of potassiumsuccinate yields:

Electrolysis of brine gives

Electrolysis of aqueous solution of Na_(2)SO_(4) is carried out in presence of graphite electrode.

Electrolysis of aqueous HCl solution produces

Electrolysis of aqueous HCI solution produces

Electrolysis of aqueous alkali metal chloride does not liberatemetal . Why ?

Electrolysis of aqueous copper sulphatesolution liberateswhich of the following gasesat the anodes ?

Electrolysis of aqueous HF produces O_2 at anode but not F_2 Explain.

Electrolysis of aq solution of brine gives

Electrolysis of an aqueous solution of sodium chloride produces____at the cathode and ____at the anode.

Electrolysis of an aqueous solution of copper sulphate using platinum electrodes produces____ at the cathode and ___at the anode.

Electrolysis of an aqueous solution of AgNO_(3) with silver electrodes produces ul((i)) at cathode while ul((ii)) ions are dissolved from anode. When Pt electrodes are used ul((iii)) is produced at anode and ul((iv)) at cathode.

Electrolysis of an aqueous solution of AgNO_3 with silver electrodes produce __(i)___ at cathode while ___(ii)___ ions are dissolved from anode. When Pt electrodes are used ____(iii)___ is produced at anode ___(iv)___ at cathode.

Electrolysis of an aqueous solution of sodium chloride producesat the cathode and at the anode.

Electrolysis of an aqueous solution of copper sulphate using platinum electrodes produces__ at the cathode and _at the anode.

Electrolysis of an aqueous solution of AgNO_3 with silver electrodes produce (i)_ at cathode while _(ii)_ ions are dissolved from anode. When Pt electrodes are used __(iii)_ is produced at anode _(iv)_ at cathode.