3195 + Interview Questions in ELECTROCHEMISTRY IN GENERAL AWARENESS Page 61 InterviewSolution

3001.	During the electrolysis of the aqueous solution of copper sulphate using `Pt` electrode, the reaction taking place at anode electrode isA. `Cu^(2+)+2e^(-) rarr Cu`B. `Cu rarr Cu^(2+)+2e^(-)`C. `2H_(2)O rarr 4H^(o+)+O_(2)+4e^(-)`D. `H_(2)O+e^(-) rarr overset(c-)(O)H+1//2H_(2)`
Answer» Correct Answer - c At anode oxidation of `H_(2)O` occurs since the oxidation potential of `H_(2)O` is greater than the oxidation potential os `SO_(4)^(2-)` ion.

Discussion

3002.	On carrying out the electrolysis of acidified water, the volume of hydrogen liberated at `STP` condition is `22.4L`. The volume of oxygen liberated isA. `22.4L`B. `44.8L`C. `11.2L`D. `2.24L`
Answer» Correct Answer - c `1F=1Eq` of `H_(2)=11.2L` `2f=22.4L` `1F=1Eq of O_(2)=5.6L` `2F=11.2L`

Discussion

3003.	What is the ratio of volumes of H2 and O2 liberated during electrolysis of acidified water ?(a) 1 : 2 (b) 2 : 1 (c) 1 : 8 (d) 8 : 1
Answer» Option : (b) 2 : 1

Discussion

3004.	The standard potentials at `298 K` for the following halfreactions are as given ` Zn^(2+) + 2e underset(larr)(rarr) Zn E^o =- 0/ 762 V`. `2 H^+ + 2 Erarr H_2 (g) E^o = 0.000 V` `Cr^(3+) + 3e underset(larr)(rarr) Cr E^o =- 0.740 V` `Fe^(3+) 2e rarr Fe^(2+) E^o =0.772 V` Which of the following is the strongest reducing agent ?A. Zn (s)B. CrC. `H_(2) (g)`D. `Fe^(2+) (aq)`
Answer» Correct Answer - A More negative the standard potential least the reducion tendcy of the ion. The corresponding atom has largest oxidation tendency and thus is a sterong reducing agent. `Zn` is the strongest reucing afent.

Discussion

3005.	For the given cell, `underset(P_(1))(Pt,C1_(2))\|underset(C_(1))(HC1)\|\|underset(C_(2))(HC1)\|underset(P_(2))(C1_(2),Pt)` The emf will be positive when :A. `C_(1) gt C_(2)`B. `C_(2) gt C_(1)`C. `C_(1) = C_(2)`D. None of these
Answer» Correct Answer - A `{:(L.H.S.2Cl^(-)rarrCl_(2)+2e),(R.H.S.2e+Cl^(2)rarrunderset(C_(2))(Cl^(-))):}/(underset(C_(1))(Cl^(-))rarrunderset(C_(2))(Cl^(-)))` Thus, `E_("cell") = -(0.059)/(1)"log"(1)/(C_(1))+(0.059)/(1)"log"(1)/(C_(2))` `= (0.059)/(1)"log"(C_(1))/(C_(2))` Thus, `E_("cell") = +ve` when `C_(1) gt C_(2)`

Discussion

3006.	Nernst equation is `E = E^(@)-(RT)/(nF)lnQ`. If `Q = K_(C)`, then which one is one correct,A. `E = E^(@)`B. `(RT)/(nF)lnQ = E^(@)`C. `E = 0`D. `K_(C) = e^((nE^(@)F)/(RT))`
Answer» Correct Answer - A If `Q = K_(c)`, then `E = 0` Also `E^(@) = (RT)/(nF)lnK_(c)` `K_(c) = e^(nF^(@)F//RT)`

Discussion

3007.	STATEMENT-1: `E_(cell)^(@)`=0` for a chloride ion concentration cell. STATEMENT-2: For this concentration cell where `E_(cell)=(RT)/(nF)In([Cl^-]_(LHS)/([Cl^-]_(RHS)`A. If both the statements are TRUE and STATEMENTS-2 is the correct explantion of STATEMENTS-10B. If both the statements are TRUE but STATEMENTS-2 is NOT the correct explanation of STATEMENTS-12C. If STATEMENTS-1 is TRUE and STATEMENTS-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE
Answer» Correct Answer - B

Discussion

3008.	Redox reaction play a pivotal role in chemistry and biology. The values of standard redox potential `(E^(@))` of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. give below are a set of half-cell reaction (acidic medium) along with their `E^(@)` (V with respect to normal hydrogen electrode) values. using this data obtain the correct explanation to folloing Questions. `I_(2)+2e^(-)to2I^(-)" "E^(0)=0.54` `Cl_(2)+2e^(-)to2Cl^(-)" "E^(0)=1.36` `Mn^(3+)+e^(-)toMn^(2+)" "E^(0)=1.50` `Fe^(3+)+e^(-)toFe^(2+)" "E^(0)=0.77` `O_(2)+4H^(+)+4e^(-)to2H_(2)O" "E^(0)=1.23` Q. Sodium fusion extract, obtained from aniline, on treatment with iron (II) sulphate and `H_(2)SO_(4)` in presence of air gives a prussian blue precipitate. the blue colour is due to the formation ofA. `Fe_(4)[Fe(CN)_(6)]_(3)`B. `Fe_(3)[Fe(CN)_(6)]_(2)`C. `Fe_(4)[Fe(CN)_(6)]_(2)`D. `Fe_(3)[Fe(CN)_(6)]_(3)`
Answer» Correct Answer - A `Na+C+Nto underset("Sodium extract")(NaCN)` `Fe^(2+)+6CN^(-)to[Fe(CN)_(6)]^(4-)` `4Fe^(3+)+3[Fe(CN)_(6)]^(4-)to underset("Prussian blue ppt.")(Fe_(4)[Fe(CN)_(6)]_(3)`.

Discussion

3009.	Number of Faradays required to reduce `3 mol` of `MnO_(4)^(c-)` to `MnO^(2+)` is `……………………….F`
Answer» Correct Answer - `(15F)` `(15F)` `8H^(o+)+5e^(-) +MnO_(4)^(c-) rarr Mn^(2+)+4H_(2)O ` `1 mol` of `MnO_(4)^(c-) -=5e^(-)=5F`. `3 mol `of `MnO_(4)^(C-)=5xx3=15F`

Discussion

3010.	The conductance of a salt solution (AB) measured by two parallel electodes of area `100 cm^2` separated by 10cm was found to be `0.0001Omega^(-1)` . If volume enclosed between two electrode contain 0.1 mole of salt, what is the molar conductivity `(Scm^2mol^(-1))` of salt at same concentration:A. 10B. 0.1C. 1D. None of these
Answer» Correct Answer - B

Discussion

3011.	Equivalent conductivity of `Fe_(2)(SO_(4))_(3)` is relative to molar conductivity by the expression `:`A. `Lambda_(eq)=Lambda_m`B. `Lambda_(eq)=Lambda_m//3`C. `Lambda_(eq)=3Lambda_m`D. `Lambda_(eq)=Lambda_m//6`
Answer» Correct Answer - D

Discussion

3012.	Molar conductivity of a solution of an electrolyte `AB_3` is 150`Scm^2mol^(-1)` . If it ionises as `AB_(3)toA^(3+)+3B^(-)`, its equivalent conductivity will be :A. 150 (in`Scm^2eq^(-1)`)B. 75 (in `Scm^2eq^(-1)`)C. 50 (in `Scm^2eq^(-1)`)D. 80 (in `Scm^2eq^(-1)`)
Answer» Correct Answer - C

Discussion

3013.	If x is specific resistance (in `Scm^2mol^(-1)`) given by:A. `(1000X)/Y`B. `1000(Y)/(X)`C. `(1000)/(XY)`D. `(XY)/(1000)`
Answer» Correct Answer - C

Discussion

3014.	Pure water does not conduct electricity because it :A. is neutralB. is readily decomposedC. is almost totally unionizedD. has a low boiling point
Answer» Correct Answer - C

Discussion

3015.	The relation among conductance (G), specific conductance (K) and cell constant (l/A) us :A. `G=k(l)/A`B. `G=k(A)/l`C. `Gk=l/A`D. G=kAL
Answer» Correct Answer - B

Discussion

3016.	Given`:` Oxidation`H_(2)O_(2) rarr O_(2)+2H^(o+)+2e^(-)" "E^(c-)=-0.69V,` `2F^(c-)rarr F_(2)+2e^(-)" "E^(c-)=-287V,` Reduction `: H_(2)O_(2)+2H^(o+)+2e^(-) rarr 2H_(2)O" "E^(c-)=1.77V,` 2I^(c-)rarr I_(2)+2e^(-)" "E^(c-)=-0.54V,` Which of the following statements is `//` are correct ?A. `H_(2)O_(2)` behaves as an oxidant for `I_(2)//I^(c-)`B. `H_(2)O_(2)` behaves as an reductant for `I_(2)//I^(c-)`C. `I^(c-)//I_(2)` behaves as an reductant for `H_(2)O_(2)`D. None of these is correct
Answer» Correct Answer - a,c Anode `:2I^(c-)rarrI_(2)+2e^(-)+2e^(-),E_(a(o x i d))^(2c-)=-0.535impliesE_(a(Red))^(c-)=+0.535V` Cathode`:H_(2)O_(2)+2H^(o+)+2e^(-)rarr 2H_(2)O,` `E^(c-)._(cell)=(E^(c-)._(cell)-E^(c-)._(a))_(R)=1.77-0.53=+ve` Note that `:` Anode`:H_(2)O_(2) rarr O_(2)+2H^(o+)+2e^(-),` `E^(c_._(a(o x i d))=-0.69V` Cathode`:I_(2)+2e^(-)rarr 2I, " "E^(c-).c_(red))=+0.535V` E^(c-)._(a(red))=+0.69V` `E^(c-)=E^(c-)._(cell)-E^(c-)._(a)=0.535-0.69=-ve`

Discussion

3017.	To obeserve the effect of concentration on the conductivity, electrolytes of different natures are taken in two vessels `A` and `B,` A contains weak electrolyte, `e.g., NH_(4)OH` and `B` contains strong electrolyte, `e.g., NaCl`. In both containers, the concentration of respective electrolyte is increased and the conductivity observed`:`A. In `A` conductivity increases, in `B` conductivity decreaseB. In `A` conductivity decreases while, in `B` conductivity decreaseC. In both `A` and `B` conductivity increasesD. In both `A` and `B` conductivity decreases
Answer» Correct Answer - c For strong and weak electrolytes, increasing concentration will increase conductance `(G)`, decrease molar conductance `(wedge_(m))` and equivalent conductance `(wedge_(eq))` and inceases conductivity `(k)`.

Discussion

3018.	A solution containing `Na^(o+),NO_(3)^(c-), Cl^(c-)`, and `SO_(4)^(2-)` ions, all at unit concentrations, is electrolyzed between nickel anode and platinum cathode. As the current is passed through the cellA. `Ph` of the cathode increasesB. Oxygen is the major product at anodeC. Nickel is deposited at cathode.D. Chlorine is the major product at anode.
Answer» Correct Answer - a,d `a.` At cathod `:` Reduction of `Na^(o+)` does not occur but reduction of `H_(2)O` occurs to give `overset(c-)(O)H` and `H_(2)(g)` , so `pOH` decreases and `pH` increases. At anode `:` Oxidation of `Cl^(c-)` ions occur to give `Cl_(2)(g)`. Likewise, oxidation of `NO^(c-)._(3)` and `SO_(4)^(2-)` does not occur but oxidation of `H_(2)O` occuts to give `H^(o+)` ions and `H_(2)(g)`. So `pH` decreases at anode.

Discussion

3019.	An electrochemical cell is constructed with an open switch as shown below: when the switch is closed, mass of tin-electrode increases. If `E^(@) (Sn^(2+)//Sn) = 0.14V` and for `E^(@) (X^(n+)//X) =- 0.78V` and initial emf of the cell is `0.65V`, determine n and indicate the direction of electron flow in the external circuit.
Answer» Correct Answer - `n = 3 & X`-electrode to Sn-electrode At cathode: `Sn^(+2) +2e^(-) rarr Sn` At anode: `X rarr X^(+n) +n e^(-)` Cell reaction `nSn^(+2) +2X rarr 2X^(+n) +nSn` `E^(@) =- 0.14 +0.78 = 0.64` `E = E^(@) -(0.0591)/(2n)log.([X^(+n)]^(2))/([S^(+2)]^(n))` `0.65 =0.64 -(0.0591)/(2n) [log 0.01 -log (0.5)^(n)]` `0.01 =- (0.0591)/(2n) [-2 +n log 2]` `0.03384 n = 2 -0.3010n` `0.6394 n = 2` `n = 3.12 =3`

Discussion

3020.	What current strength in ampere will be required to liberate 5 g of iodine from potassium iodide solution in 30 minutes ?A. 1.11 ampB. 2.11 ampC. 4 ampD. 10 amp
Answer» Correct Answer - B `I^(-) to (1)/(2) I_(2) + e^(-) , (1)/(2) xx 254 `g require Q = 96500 C . Hence 10 g will require `Q = (96500)/(127) xx 10 = 7598.4 C ` `I = (Q)/(t) = (7594.4)/(60 xx 60) = 2.11 `A .

Discussion

3021.	Which of the following represents the first law of faradayA. `E=mc^(2)`B. `E=hv`C. `m=ect`D. `PV=nRT`
Answer» Correct Answer - C

Discussion

3022.	Silver is removed electrically from 200 ml of a 0.1 N solution of `AgnO_(3)` by a current of 0.1 ampere. How long will it take to remove half of the silver from the solution.A. 16 secB. 96.5 secC. 100 secD. 10 sec
Answer» Correct Answer - B

Discussion

3023.	From the solution of an electrolyte , one mole of electrons will deposit at cathode ,A. 63.5 gm of CuB. 24 gm of MgC. 11.5 gm of NaD. 9.0 gm of Al
Answer» Correct Answer - D 1 g eq. of Al = 9 g = 1 mole of electrons

Discussion

3024.	On passingle C ampere of electricity through a electrolyte solution for t second, m gram metal deposits on cathode. The equivalent weight E of the metal isA. `E=(Cxxt)/(mxx96500)`B. `E=(Cxxm)/(txx96500)`C. `E=(96500xxm)/(Cxxt)`D. `E=(Cxxtxx96500)/(m)`
Answer» Correct Answer - C

Discussion

3025.	On passing 3 A of electricity for 50 min, 1.8 g of metal deposits. The equivalent mass of metal isA. 20.533B. 25.8C. 19.3D. 30.7.
Answer» Correct Answer - C `E = (m)/(Q) xx 96500 = (1.8)/(3 xx 50 xx 60) xx 96500 = 19.3` .

Discussion

3026.	A current of `40` microampere is passed through a solution of `AgNO_(3)` for `32` minute using `Pt` electrodes. An uniform single atom thick layer of `Ag` is deposited covering `43%` cathode surface. What is the total surface area of cathode if each `Ag` atom covers `5.4 xx 10^(-16) cm^(2)` ?
Answer» Correct Answer - `601.65 cm^(2) ;`

Discussion

3027.	Write net charging and discharging reactions for lead storage battery.
Answer» For lead storage battery : Net charging reaction : 2PbSO_4(s) + 2H₂O_(I)→ Pb_(s)+ PbO_2(s) + 2H₂SO_4(aq) Net discharging reaction : Pb_(s) + PbO_2(s) + 2H₂SO_4(4q) → 2PbSO_4(s) + 2H₂O_(I)

3027.

Write net charging and discharging reactions for lead storage battery.

Answer»

For lead storage battery :

Net charging reaction :

2PbSO_4(s) + 2H₂O_(I)→ Pb_(s)+ PbO_2(s) + 2H₂SO_4(aq)

Net discharging reaction :

Pb_(s) + PbO_2(s) + 2H₂SO_4(4q) → 2PbSO_4(s) + 2H₂O_(I)

Discussion

3028.	Which of the following reactions occur at the cathode during the charging of lead storage battery ?A. `Pb^(2+) + 2e^(-) rarr Pb`B. `Pb^(2+) + SO_(4)^(2-) rarr PbSO_(4)`C. `Pb rarr Pb^(2+) + 2e^(-)`D. `PbSO_(4) + 2H_(2)O rarr PbO_(2) + 4H^(+) + SO_(4)^(2-) + 2e^(-)`
Answer» Correct Answer - D

Discussion

3029.	What type of a battery is lead storage battery? Write the anode and the cathode reactions and the overall reactions occurring in a lead storage battery.
Answer» It is secondry cell. For details, consult section 22

Discussion

3030.	Write the anode and cathode reactions and the overall reaction occurring in a lead storage battery.
Answer» For anwer, consult section 22.

Discussion

3031.	During charging of lead storage bttery, the reaction occurring at the cathode is:-A. `Pb^(2+)+2etoPb`B. `PbtoPb^(2+)+2e`C. `PbSO_(4)+2H_(2)OtoPbO_(2)+4H^(+)+SO_(4)^(2-)+2e^(-)`D. `Pb^(2+)+SO_(4)^(2-)toPbSO_(4)`
Answer» Correct Answer - C In an uncharged condition, in lead storage battery, the reaction occurring at the cathode is, `PbO_(2)+4H^(+)+SO_(4)^(2-)+2e^(-)toPbSO_(4)+2H_(2)O` During charging of lead storage battery, the reaction is reversed at both anode and cathode. The reaction occurring at the cathode is, `PbSO_(4)+2H_(2)OtoPbO_(2)+4H^(+)+SO_(4)^(2-)+2e^(-)`

Discussion

3032.	Find `K_(c)` for the complex`:` `[Ag(NH_(3))_(2)]^(o+)hArrAg^(o+)+2NH_(3)` `E^(c-)._((Ag^(o+)//Ag))=0.8V` and `E^(c-)._{[Ag(NH_(3))_(2)]^(o+)\|Ag\|NH_(3))=0.37V`
Answer» Anode reaction `:` `cancel(Ag) hArrAg^(o+)+cancel(e^(_))` Cathode reacion `:[Ag(NH_(3))_(2)]^(o+)hArrcancel(Ag)+2NH_(3)` Cell reaction`ulbar( :[Ag(NH_(3))_(2)]^(o+)hArrAg^(o+)+2NH_(3))` `E^(c-)._(cell)=(E^(c-)._(reduction))_(c)=(E^(c-)._(reduction))_(a)` `=0.37V-0.8V=-0.43V` `:. log K_(c)=(n_(cell)xxE^(c-)._(cell))/(0.0059)` `=(1xx(-.43V))/(0.59)=-7.22` `K_(c)=Anti l og(-7.22)` `=Antil og(-7-022+1-1)` `=Antil og(bar(8).78)=6xx10^(-8)`

Discussion

3033.	Why does conductivity of metallic conductor decrease with an increase in temperature, wheras in case of electrolytic conductors the conductivity increases with an increase in temperature ?
Answer» With an increase in temperature the degree of disosciation of an electrolyte increases, thus more number of ions is produced and rate of migration of ions towards respective electrodes also increases due to increase in the mobility of ions. Therefore, conductivity increases. However, conductivity decreases with decrease in temperature in case of metallic conductors because as temperature increases the vibration of positive kernels and free electrons increases but there is net increase in the KE of electrons, thereby the speed or flow of electrons is hindered or obstructed by increased movement of positive kernels. The obstruction caused by positive kernels to the flow of electrons is called resistance and due to the increase in resistance with increase in temperature, conductivity of a metal decreases.

Discussion

3034.	Calculate the mass of Ag deposited at cathode when a current of 2 amperes was passed through a solution `AgNO_(3)` for 15 minutes.
Answer» Correct Answer - 2.014

Discussion

3035.	The conductivity of 0.01 `mol//dm^(3)` aqueous acetic acid at 300 K is `19.5 xx 10^(-5) ohm^(-1) cm^(-1)` and the limiting molar conductivity of acetic acid at the same temperature is `390 ohm^(-1) cm^(2) mol^(-1)`. The degree of dissociattion of acetic acid is :A. `0.5`B. `0.05`C. `5 xx 10^(-3)`D. `5 xx 10^(-7)`
Answer» Correct Answer - B

Discussion

3036.	What is meant by limiting molar conductivity?
Answer» It is equal to the sum of the ionic conductivities of the ions at infinite dilution.

Discussion

3037.	Limiting molar conductivity of NaBr isA. `Lambda_m^@NaBr = Lambda_m^@NaCl + Lambda_m^@KBr`B. `Lambda_m^@NaBr = Lambda_m^@ NaCl + Lambda_m^@KBr - Lambda_m^@KCl`C. `Lambda_m^@NaBr=Lambda_m^@NaOH + Lambda_m^@NaBr - Lambda_m^@NaCl`D. `Lambda_m^@NaBr=Lambda_m^@NaCl-Lambda_m^@NaBr`
Answer» Correct Answer - B

Discussion

3038.	In a galvanic cell, the electrons flow from :A. atomic a cathode through the solutionB. cathode to anode through the solutionC. anode to cathode through the external circuitD. cathode to anode through the external circuit
Answer» Correct Answer - C

Discussion

3039.	In the galvanic cell, flow of electrons is fromA. anode to cathode through the solutionB. cathode to anode through the solutionC. anode to cathode through the external circuitD. cathode to anode through the external circuit.
Answer» Correct Answer - C In a Galvanic cell electrons move from anode (-ve electrode) to cathode (+ve electrode) through the external circuit.

Discussion

3040.	A solution of `CuSO_(4)` is electrolysed for 10 minutes with a current of `1.5` amperes. What is the mass of copper deposited at the cathode ?
Answer» `t=600` s charg = current `xx` time `=1.5 A xx 600s= 900C` According to the reaction : A`Cu_((aq))^(2+)+2e^(-) =Cu_((s))` We require 2F or `2xx 96487` C to deposite 1 mol or 63 g of Cu. For 900 C, the mass of Cu deposited `=(63 g mol ^(-1)xx900C) //(2xx96487C mol ^(-1))=0.2938g`

Discussion

3041.	Write an expression to co-relate molar conductivity of the electroyte to the degree of dissociation.
Answer» Degree of dissociation (a) `=(Lambda_(m)^(c ))/(Lambda_(m)^(oo))`

Discussion

3042.	What is the direction of flow of conventional current in a galvanic cell?
Answer» Conventional current flows from cathode to anode. It is opposite to the direction of flow of electrons.

Discussion

3043.	A solution of `CuSO_(4)` is electroysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode ? (Molar mass of `Cu=63.5 g//mol)`
Answer» `Q=ixxt` `=1.5xx10xx60` `=1.5xx600=900C` Reaction ocuring at the cathode is `(Cu^+2)+2e^(-1)rarrCu` 2F=`2xx9500` C deposit Cu =1 mole = 63.5 g 900 C will deposit Cu `=(63.5)/(2xx96500)xx900=0.296 g`

Discussion

3044.	The equivalent conductivity of `0.1 M` weak acid is `100` times less than that at infinite dilution. The degree of dissociation of weak electrolyte at `0.1 M` is.A. 1 00B. 10C. 0.01D. 0.001
Answer» Correct Answer - C `alpha=(^^_(m)^(c ))/(^^_(m)^(o))=(x//100)/(x)=(1)/(100)=0.01`

Discussion

3045.	Write an expression to relate molar conductivity of an electrolyte to its degree of dissociation.
Answer» `alpha=(lamda_(m)^(c))/(lamda_(m)^(0))` where `wedge_(m)^(c)`=molar conductivity at given concentration, c and `wedge_(m)^(0)`=molar conductivity at infinite dilution.

Discussion

3046.	The equivalent conductivity of `0.1 M` weak acid is `100` times less than that at infinite dilution. The degree of dissociation of weak electrolyte at `0.1 M` is.A. ` 100`B. `101`C. `0.01`D. `0.001`
Answer» Correct Answer - C `Lambda_v = (Lambda^@)/(100)` `:. Alpha = (Lambda_v)/(Lambda^0) = (Lambda^0)/(100Lambda^0) =0. 01`.

Discussion

3047.	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.A. `{:("(i)","(ii)","(iii)","(iv)"),(H_2,NO_3^(-) , OH^(-), H_2):}`B. `{:("(i)","(ii)","(iii)","(iv)"),(Ag,H^+, O_2, H_2):}`C. `{:("(i)","(ii)","(iii)","(iv)"),(Ag,H^(+),O_(2),Ag):}`D. `{:("(i)","(ii)","(iii)","(iv)"),("Ag","H"^(+),"Ag"^(+),"O"_(2)):}`
Answer» Correct Answer - C

Discussion

3048.	3.8 of motlesn `SnCI_(2)` is electroysed for some time using insert electrodes 0.238 g of Sn is deposited at cathode No substance is lost during the eletroysis find the ratio of weight of `SnCI_(2)` and `SnCI_(4)` after electroysis (Sn=11)
Answer» `SnCI_(2)rarrSn^(+2)+2CI^(-)` `Sn^(+2)+2e^(-)rarrSn` Mass of Sn deposited =0.238 g m=Z it =ZQ Charge `Q=(m)/(z)=(mxxnxxf)/(At.wt)` `Q=(0.238)/(119)xx2xx96500` Q=386 columb For chrone `2CI^(-)rarrCI_(2)+2e` Mass `m_(CI_(2))=Zit=(71)/(nF)xxQ` `=(71)/(2xx96500)xx386=0.142 g` To calculate weight of `SnCI_(4)` produced `SnCI_(2)+CI_(2)rarrSnCI_(4)` `n_(CI_(2))/(1)=n_(SnCI_(4))/(1)` `rarr m_(SnCI_(4))/(261)=0.002` `rarrM_(SnCI_(4))=0.522 g` To calculate weightr of `SnCI_(2)` consumed `SnCI_(2)+CI_(2)rarrSnCI_(4)` `N_(SnCI_(2))/(1)=n_(CI_(2))/(1)=0.002` `M_(SnCI_(2))=0.002xx190=0.38 g` `m_(SnCI_(2))("remaining")=3.8-0.38-0.98=3.04 g` `"Weight ratio" =(SnCI_(2))/(SnCI_(4))=(3.04)/(0.522)=5.829`

Discussion

3049.	The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm-1 . Calculate the molar conductivity.
Answer» A_M= \(\frac{k\times1000}{M}\) = \(\frac{0.0248\,S\,cm^{-1}\times1000\,cm^3}{0.2\,mol}\) = 124 S cm² mol^-1

3049.

The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm-1 . Calculate the molar conductivity.

Answer»

A_M= \(\frac{k\times1000}{M}\)

= \(\frac{0.0248\,S\,cm^{-1}\times1000\,cm^3}{0.2\,mol}\)

= 124 S cm² mol^-1

Discussion

3050.	When an aqueous solution of `AgNO_3` is electroysed between platinum electrodes, the substances liberted at anode and cathode areA. silver is deposited at cathode and `O_2` is liberated at anodeB. silver is deposited at cathode and `H_2` is liberated at anodeC. hydrogen is liberted at cathode and `O_2` is liberated at anode.D. silver is deposited at cathode and Pt is dissolved in electrolyte.
Answer» Correct Answer - A `{:("At cathode",Ag_((aq))^(+)+e^(-) to Ag_((s))),("At anode:",2OH_((aq))^(-) to 1/2O_(2(g))+H_2O_((l))+2e^(-)):}`

Discussion

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During the electrolysis of the aqueous solution of copper sulphate using `Pt` electrode, the reaction taking place at anode electrode isA. `Cu^(2+)+2e^(-) rarr Cu`B. `Cu rarr Cu^(2+)+2e^(-)`C. `2H_(2)O rarr 4H^(o+)+O_(2)+4e^(-)`D. `H_(2)O+e^(-) rarr overset(c-)(O)H+1//2H_(2)`

On carrying out the electrolysis of acidified water, the volume of hydrogen liberated at `STP` condition is `22.4L`. The volume of oxygen liberated isA. `22.4L`B. `44.8L`C. `11.2L`D. `2.24L`

What is the ratio of volumes of H2 and O2 liberated during electrolysis of acidified water ?(a) 1 : 2 (b) 2 : 1 (c) 1 : 8 (d) 8 : 1

For the given cell, `underset(P_(1))(Pt,C1_(2))|underset(C_(1))(HC1)||underset(C_(2))(HC1)|underset(P_(2))(C1_(2),Pt)` The emf will be positive when :A. `C_(1) gt C_(2)`B. `C_(2) gt C_(1)`C. `C_(1) = C_(2)`D. None of these

Nernst equation is `E = E^(@)-(RT)/(nF)lnQ`. If `Q = K_(C)`, then which one is one correct,A. `E = E^(@)`B. `(RT)/(nF)lnQ = E^(@)`C. `E = 0`D. `K_(C) = e^((nE^(@)F)/(RT))`

Number of Faradays required to reduce `3 mol` of `MnO_(4)^(c-)` to `MnO^(2+)` is `……………………….F`

Equivalent conductivity of `Fe_(2)(SO_(4))_(3)` is relative to molar conductivity by the expression `:`A. `Lambda_(eq)=Lambda_m`B. `Lambda_(eq)=Lambda_m//3`C. `Lambda_(eq)=3Lambda_m`D. `Lambda_(eq)=Lambda_m//6`

Molar conductivity of a solution of an electrolyte `AB_3` is 150`Scm^2mol^(-1)` . If it ionises as `AB_(3)toA^(3+)+3B^(-)`, its equivalent conductivity will be :A. 150 (in`Scm^2eq^(-1)`)B. 75 (in `Scm^2eq^(-1)`)C. 50 (in `Scm^2eq^(-1)`)D. 80 (in `Scm^2eq^(-1)`)

If x is specific resistance (in `Scm^2mol^(-1)`) given by:A. `(1000X)/Y`B. `1000(Y)/(X)`C. `(1000)/(XY)`D. `(XY)/(1000)`

Pure water does not conduct electricity because it :A. is neutralB. is readily decomposedC. is almost totally unionizedD. has a low boiling point

The relation among conductance (G), specific conductance (K) and cell constant (l/A) us :A. `G=k(l)/A`B. `G=k(A)/l`C. `Gk=l/A`D. G=kAL

What current strength in ampere will be required to liberate 5 g of iodine from potassium iodide solution in 30 minutes ?A. 1.11 ampB. 2.11 ampC. 4 ampD. 10 amp

Which of the following represents the first law of faradayA. `E=mc^(2)`B. `E=hv`C. `m=ect`D. `PV=nRT`

Silver is removed electrically from 200 ml of a 0.1 N solution of `AgnO_(3)` by a current of 0.1 ampere. How long will it take to remove half of the silver from the solution.A. 16 secB. 96.5 secC. 100 secD. 10 sec

From the solution of an electrolyte , one mole of electrons will deposit at cathode ,A. 63.5 gm of CuB. 24 gm of MgC. 11.5 gm of NaD. 9.0 gm of Al

On passingle C ampere of electricity through a electrolyte solution for t second, m gram metal deposits on cathode. The equivalent weight E of the metal isA. `E=(Cxxt)/(mxx96500)`B. `E=(Cxxm)/(txx96500)`C. `E=(96500xxm)/(Cxxt)`D. `E=(Cxxtxx96500)/(m)`

On passing 3 A of electricity for 50 min, 1.8 g of metal deposits. The equivalent mass of metal isA. 20.533B. 25.8C. 19.3D. 30.7.

A current of `40` microampere is passed through a solution of `AgNO_(3)` for `32` minute using `Pt` electrodes. An uniform single atom thick layer of `Ag` is deposited covering `43%` cathode surface. What is the total surface area of cathode if each `Ag` atom covers `5.4 xx 10^(-16) cm^(2)` ?

Write net charging and discharging reactions for lead storage battery.

Which of the following reactions occur at the cathode during the charging of lead storage battery ?A. `Pb^(2+) + 2e^(-) rarr Pb`B. `Pb^(2+) + SO_(4)^(2-) rarr PbSO_(4)`C. `Pb rarr Pb^(2+) + 2e^(-)`D. `PbSO_(4) + 2H_(2)O rarr PbO_(2) + 4H^(+) + SO_(4)^(2-) + 2e^(-)`

What type of a battery is lead storage battery? Write the anode and the cathode reactions and the overall reactions occurring in a lead storage battery.

Write the anode and cathode reactions and the overall reaction occurring in a lead storage battery.

During charging of lead storage bttery, the reaction occurring at the cathode is:-A. `Pb^(2+)+2etoPb`B. `PbtoPb^(2+)+2e`C. `PbSO_(4)+2H_(2)OtoPbO_(2)+4H^(+)+SO_(4)^(2-)+2e^(-)`D. `Pb^(2+)+SO_(4)^(2-)toPbSO_(4)`

Find `K_(c)` for the complex`:` `[Ag(NH_(3))_(2)]^(o+)hArrAg^(o+)+2NH_(3)` `E^(c-)._((Ag^(o+)//Ag))=0.8V` and `E^(c-)._{[Ag(NH_(3))_(2)]^(o+)|Ag|NH_(3))=0.37V`

Why does conductivity of metallic conductor decrease with an increase in temperature, wheras in case of electrolytic conductors the conductivity increases with an increase in temperature ?

Calculate the mass of Ag deposited at cathode when a current of 2 amperes was passed through a solution `AgNO_(3)` for 15 minutes.

What is meant by limiting molar conductivity?

Limiting molar conductivity of NaBr isA. `Lambda_m^@NaBr = Lambda_m^@NaCl + Lambda_m^@KBr`B. `Lambda_m^@NaBr = Lambda_m^@ NaCl + Lambda_m^@KBr - Lambda_m^@KCl`C. `Lambda_m^@NaBr=Lambda_m^@NaOH + Lambda_m^@NaBr - Lambda_m^@NaCl`D. `Lambda_m^@NaBr=Lambda_m^@NaCl-Lambda_m^@NaBr`

In a galvanic cell, the electrons flow from :A. atomic a cathode through the solutionB. cathode to anode through the solutionC. anode to cathode through the external circuitD. cathode to anode through the external circuit

In the galvanic cell, flow of electrons is fromA. anode to cathode through the solutionB. cathode to anode through the solutionC. anode to cathode through the external circuitD. cathode to anode through the external circuit.

A solution of `CuSO_(4)` is electrolysed for 10 minutes with a current of `1.5` amperes. What is the mass of copper deposited at the cathode ?

Write an expression to co-relate molar conductivity of the electroyte to the degree of dissociation.

What is the direction of flow of conventional current in a galvanic cell?

A solution of `CuSO_(4)` is electroysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode ? (Molar mass of `Cu=63.5 g//mol)`

The equivalent conductivity of `0.1 M` weak acid is `100` times less than that at infinite dilution. The degree of dissociation of weak electrolyte at `0.1 M` is.A. 1 00B. 10C. 0.01D. 0.001

Write an expression to relate molar conductivity of an electrolyte to its degree of dissociation.

The equivalent conductivity of `0.1 M` weak acid is `100` times less than that at infinite dilution. The degree of dissociation of weak electrolyte at `0.1 M` is.A. ` 100`B. `101`C. `0.01`D. `0.001`

3.8 of motlesn `SnCI_(2)` is electroysed for some time using insert electrodes 0.238 g of Sn is deposited at cathode No substance is lost during the eletroysis find the ratio of weight of `SnCI_(2)` and `SnCI_(4)` after electroysis (Sn=11)

The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm-1 . Calculate the molar conductivity.