3195 + Interview Questions in ELECTROCHEMISTRY IN GENERAL AWARENESS Page 1 InterviewSolution

1.	Define molar conductivity of a solution and explain how molar conductivity changes with change in concentration of a solution for a weak and a strong electrolyte.
Answer» For answer, consult section 19.

Discussion

2.	Limiting molar conductivity of `NH_(4)OH` [i.e., `Lambda_(m)^(@)(NH_(4)OH)`] is equal to:A. `Lambda_(m (NH_(4)Cl ))^(@) + Lambda_(m (NaCl))^(@) - Lambda_(m(NaOH))^(@)`B. `Lambda_(m (NaOH))^(@) + Lambda_(m(NaCl))^(@) - Lambda_(m(NH_(4)Cl))^(@)`C. `Lambda_(m(NH_(4)OH))^(@) + Lambda_(m (NH_(4)Cl))^(@) - Lambda_(m(HCl))^(@)`D. `Lambda_(m (NH_(4)Cl))^(@) + Lambda_(m (NaOH))^(@) - Lambda_(m(NaCl))^(@)`
Answer» Correct Answer - D It is a fact

Discussion

3.	Limiting molar conductivity of `NH_(4)OH` [i.e., `Lambda_(m)^(@)(NH_(4)OH)`] is equal to:A. `Lambda_(m)^(@)(NH_(4)Cl)+Lambda_(m)^(@)(Na_(4)Cl)-Lambda_(m)^(@)(NaOH)`B. `Lambda_(m)^(@)(NaOH)+Lambda_(m)^(@)(NaCl)-Lambda_(m)^(@)(NH_(4)Cl)`C. `Lambda_(m)^(@)(NH_(4)OH)+Lambda_(m)^(@)(NH_(4)Cl)-Lambda_(m)^(@)(HCl)`D. `Lambda_(m)^(@)(NH_(4)Cl)+Lambda_(m)^(@)(NaOH)-Lambda_(m)^(@)(NaCl)`
Answer» Correct Answer - D According to Kohlrausch law of independent migration of ions: `Lambda_(m)^(@)(NH_(4)OH) = Lambda_(m)^(@)(NH_(4)^(+))+Lambda_(m)^(@)(OH^(-))` `= Lambda_(m)^(@)(NH_(4)^(+))+Lambda_(m)^(@)(Cl^(-))+Lambda_(m)^(@)(OH^(-))` `= Lambda_(m)^(@)(NH_(4)^(+))-Lambda_(m)^(@)(Cl^(-))-Lambda_(m)^(@)(Na^(+))` `= Lambda_(m)^(@)(NH_(4)Cl)+Lambda_(m)^(@)(NaOH)-Lambda_(m)^(@)(NaCl)`

Discussion

4.	For the cell reaction `Cu^(2+)(C_(1),aq.)+Zn(s)hArrZn^(2+)(C_(2),aq)+Cu(s)` of an electrochemical cell, the change in free energy `(DeltaG)` of a given temperature is a function ofA. `lnC_(1)`B. `lnC_(2)//C_(1)`C. `ln (C_(1) + C_(2))`D. `ln C_(2)`
Answer» Correct Answer - B `E_("cell") = E_("cell")^(@) +(RT)/(nF)"ln"([Cu^(2+)])/([Zn^(2+)])` `= E_("cell")^(@) +(RT)/(nF)"ln"(C_(1))/(C_(2))` Also `DeltaG = -nEF` Thus `DeltaG` is funcation of `"ln"(C_(2))/(C_(1))`.

Discussion

5.	Use of lithium metal as an electrode in high energy density batteries is due to:A. Lithium is highest elementB. Lithium has highest oxidation potentialC. Lithium is quite reactiveD. Lithium does not corrode readily
Answer» Correct Answer - B This is the characteristic of lithium regarding its use in high electron density batteries.

Discussion

6.	The Gibbs energy for the decomposition of `Al_(2)O_(3)` at `500^(@)C` is as follow : `(2)/(3)Al_(2)O_(3) rarr (4)/(3)Al+O_(2), Delta_(r)G= +960 kJ mol^(-1)` The potential difference needed for the electrolytic reduction of aluminium oxide `(Al_(2)O_(3))` at `500^(@)C` isA. `5.0 V`B. `4.5 V`C. `3.0 V`D. `2.5 V`
Answer» Correct Answer - D We are given decomposition reaction: `(2)/(3)Al_(2)O_(3) rarr (4)/(3)Al+O_(2)` The electrolyte `Al_(2)O_(3)` splits into ions as follows: `Al_(2)O_(3) rarr 2Al^(3+)+3O^(2-)` At cathode `2Al^(3+)+6e^(-) rarr 2Al` At anode `3O^(2-) rarr 6e^(-)+(3)/(2)O_(2)` We have `Delta_(r)G^(ɵ) = -nfE_("cell")^(ɵ)` or `E_("cell")^(ɵ) = (Delta_(r)G^(ɵ))/(-nF) = (960 xx 10^(3)J mol^(-1))/(-(6 mol)(96500 C)) = -1.66 V`

Discussion

7.	Based on the data given below, the correct order of reducing power is: `Fe_((aq.))^(3+) + e rarr Fe_((aq.))^(2+), E^(@) = +0.77 V` `Al_((aq.))^(3+) + 3e rarr Al_((s)), E^(@) = -1.66 V` `Br_(2(aq.)) + 2e rarr 2Br_((aq.))^(-), E^(@) = +1.08 V`A. `Br^(-) lt Fe^(2+) lt A1`B. `Fe^(2+) lt Al lt Br^(-)`C. `Al^(-) lt Br^(-) lt Fe^(2+)`D. `Al^(-) lt Fe^(2+) lt Br^(-)`
Answer» Correct Answer - A More is `E_(OP)^(@)`, more is the tendency to get itself oxidised or more is reducing power.

Discussion

8.	Based on the data given below, the correcy order of reducing power is: `Fe_((aq.))^(3+) + e rarr Fe_((aq.))^(2+), E^(@) = +0.77 V` `Al_((aq.))^(3+) + 3e rarr Al_((s)), E^(@) = -1.66 V` `Br_(2(aq.)) + 2e rarr 2Br_((aq.))^(-), E^(@) = +1.08 V`A. `Br^(-) lt Fe^(2+) lt Al`B. `Fe^(2+) lt Al lt Br^(-)`C. `Al lt Br^(-) lt Fe^(2+)`D. `Al lt Fe^(2+) lt Br^(-)`
Answer» Correct Answer - A (a) Electrode potentials in this case are standard oxidation potentials. The standard reduction potentials have opposite signs i.e. `Fe^(2+)(aq) to Fe^(3+)(aq)+e^(-),E^(@)=-0.77" V "` `Al(s) to Al^(3+)(aq)+3e^(-),E^(@)=+1.66" V "` `2Br^(-)(aq) to Br_(2)(aq)+2e^(-),E^(@)=-1.08" V "` Reducing power is in the order : `Br^(-) lt Fe^(2+) lt Al`.

Discussion

9.	Based on the data given below, the correcy order of reducing power is: `Fe_((aq.))^(3+) + e rarr Fe_((aq.))^(2+), E^(@) = +0.77 V` `Al_((aq.))^(3+) + 3e rarr Al_((s)), E^(@) = -1.66 V` `Br_(2(aq.)) + 2e rarr 2Br_((aq.))^(-), E^(@) = +1.08 V`A. `Br^(-) lt Fe^(2+) lt Al`B. `Fe^(2+) lt Al lt Br^-`C. `Al lt Br^(-) lt Fe^(2+)`D. `Al lt Fe^(2+) lt Br^(-)`
Answer» Correct Answer - A Lower the reduction potential, more is the reducing power . `Br^-` lt `Fe^(2+)` lt Al

Discussion

10.	What are elementary reactions ?
Answer» The reactions taking place in one step are called elementary reactions.

Discussion

11.	Which of the following alkali metal ions has the lowest ionic mobility in aqueous solutions?A. `Na^(o+)`B. `F^(c-)`C. `SO_(4)^(2-)`D. `overset(c-)(O)H`
Answer» Correct Answer - D Refer Section `H^(o+)` and `overset(c-)(O)H` ions have highest mobility and mobility of `H^(o+)` is slightly greater than `overset(-)(O)H` ion as explained in Section

Discussion

12.	Which of the following alkali metal ions has the lowest ionic mobility in aqueous solutions?A. `Li^(o+)`B. `Na^(o+)`C. `K^(o+)`D. `Rb^(o+)`
Answer» Correct Answer - A Since solvent is not mentioned so ionic mobility `prop` charge density `( `charge `//` size `)` . `Li^(o+)` has the highest charge density. But in aqueous solution, due to extensive hydration, its mobility becomes lowest amongst the other ions of first group of periodic table.

Discussion

13.	Ammonia released during the reduction at cathode of Leclanche cell combines with zinc ions to formA. `[Zn(NH_(3))_(4)]^(2+)`B. `[Zn(NH_(3))_(2)]^(2+)`C. `[Zn(NH_(3))_(6)]^(2+)`D. `[Zn(NH_(3))_(5)]^(2+)`
Answer» Correct Answer - A Leclanche produces about `1.5V` but drops to about `0.8 V` as the reaction proceeds. This drop is because the reaction products cannot diffuse aways from the electrolyte. If such a cell is left unused for a short time, the voltage may rise back again to about `1.3 V`. The cell apears to be dead, because large excess of `[Zn(NH_(3))_(4)]Cl_(2)` formed crystallizes out around the anode. The electrolyte becomes unable to conduct electricity effectively. A little warming of the cell may restore the voltage again due to the diffusion of the complex.

Discussion

14.	In lead accumulator, anode and cathode are :(a) (Pb + PbO2),Pb (b) Pb, PbO2 (c) PbO2, Pb (d) Pb, (Pb + PbO2)
Answer» Option : (d) Pb, (Pb + PbO₂)

Discussion

15.	What takes place when zinc metal is added to an aqueous solution containing magnesium nitrate and silver nitrate ? 1. Zn is oxidised 2. `Mg^(2+)` is reduced 3. `Ag^(+)` is reduced 4. No reaction takes placeA. 1 and 2 onlyB. 1 and 3 onlyC. 1, 2 and 3 onlyD. 4 only
Answer» Correct Answer - B No reaction takes place

Discussion

16.	Galvanisation is applying a coating ofA. CrB. CuC. ZnD. Pb
Answer» Correct Answer - C Galvanisation is applying a coating of zinc metal to prevent corrosion.

Discussion

17.	In the galvanizing process, iron is coated with zinc. The resulting chemical protection is most similar to that provided when:A. a magnesium bar is connected to an iron pipeB. an iron can is plated with tinC. copper pipes are connected using lead solderD. a copper pipes is covered with epoxy paint
Answer» Correct Answer - A

Discussion

18.	The ionization constant `(K_(a))` of a weak electrolyte is `2.5xx10^(-7)`, while `wedge_(eq)` of its `0.01M` solution is `19.65S cm^(2) eq^(-1)wedge^(@)._(eq)` is `……………….` .
Answer» Correct Answer - `(392 S cm^(2) eq^(-1))` `(392 S cm^(2) eq^(-1))` `K_(a) = c alpha^(2)` `2.5xx 10^(-7)=10^(-2)xxalpha^(2)impliesalpha=5xx10^(-2)` `alpha=(wedge^(c)._(eq))/(wedge^(@)._(eq))` `=5xx10^(-2)=(19.65)/(wedge^(@)._(eq))implieswedge_(eq)^(@)=392 S cm^(2)eq^(-1)`

Discussion

19.	The temperature coefficient of the cell is `((delE)/(delT))_(P)`. Choose the correct statements `(s)`.A. When `((delE)/(delT))_(P)=0` , then `DeltaH=-nFE`B. When `((delE)/(delT))_(P)lt0, ` then `\|nFE\|gt\|DeltaH\|`C. When `((delE)/(delT))_(P)gt0`, then `\|nFE\|lt\|DeltaH\|` Exothermic reactionD. When `((delE)/(delT))_(P)=0,` then `\|DeltaH\|gt\|nFE\|` Endothermic reaction.
Answer» Correct Answer - a,b,c `DeltaG=DeltaH+T((delG)/(delt))_(P)` ` DeltaG=-nFE=DeltaH-nFT((delG)/(delT))_(P)` `implies((delE)/(delT))_(P)=(DeltaH+nFE)/(nF)` Check for `:((delE)/(delT))=0,gtO,lt0`

Discussion

20.	If the temperature coefficient `((delE)/(delT))` is zero for a cell reaction then out of `DeltaS, DeltaH`, and `DeltaG` , the `…………………..` is zero.
Answer» Correct Answer - `(DeltaS)` `(DeltaS)` `DeltaS=nF((dE)/(dT))_(P)` `DeltaS prop ((dE)/(dT))_(P)` So when `((dE)/(dT))_(P)=0,DeltaS=0` `DeltaH=-nF[E_(cell)T((dE)/(dT))_(P),SoDeltaHcancel(=)0` `DeltaG=DeltaH-T DeltaS, ` iF `DeltaS=0,DeltaG=DeltaH cancel(=)0`

Discussion

21.	`13g` of metal `M` is deposited at cathode by passing `0.4F` of electricity . The cathodic reaction is `:` `M^(n+)+n e^(-) rarr M` The formula of metal chloride `(Aw=65)` is `……………………..` .
Answer» Correct Answer - `(MCl_(2))` `(MCl_(2))` `0.4F=0.4Eq of M=(13.0g)/(Ew)` `:. Ew=32.5` Valency `=(Aw)/(Ew)=(65)/(32.5)=2` Formula `=MCl_(2)`

Discussion

22.	`4.5 g` of aluminium (at mass `27 u`) is deposited at cathode from `Al^(3+)` solution by a certain quantity of electric charge. The volume of hydrogen gas produced at `STP` from `H^(+)` ions in solution by the same quantity of electric charge will be:A. 44.8 LB. 22.4 LC. 11.2 LD. 5.6 L
Answer» Correct Answer - D (d) Equivalent mass of `Al =27//3=9` No. of equiv. of Al `=("Mass of Al deposited")/("Equivalent mass")` `=((4.5 g))/((9g" equiv"))=0.5" equiv"` No. of `"equiv"`. of `H_(2)` evolved`=0.5 "equiv"`. Volume of `H_(2)` evolved`=((0.5" equiv"))/((1.0" equiv"))xx11.2L` =5.6 L at S.T.P.

Discussion

23.	On passing 0.1 F of electricity through aluminium chloride, the amount of aluminium meta deposited on cathode is (Atomic weight of `Al=27`)A. `0.27 g`B. `0.3 g`C. `0.9 g`D. `2.7 g`
Answer» Correct Answer - C

Discussion

24.	When 9.65 coulomb of electricity is passed through a solution of silver nitrate (Atomic mass of Ag = 108 g `mol^(-1)`, the amount of silver deposited is :A. `10.8` mgB. `5.4` mgC. `16.2` mgD. `21.2` mg
Answer» Correct Answer - A `w = (E xxi xxt)/(96500) = (108 xx 9.65)/(96500) = 0.0108` g = 10.8 mg .

Discussion

25.	1 mole of Al is deposited by X coulomb of electricity passing through aluminium nitrate solution . The number of mole of silver deposited by X coulomb of electricity from silver nitrate solution isA. 3B. 4C. 2D. 1
Answer» Correct Answer - A X coulomb deposits 1 mole Al or 3 eq. of Al and thus it will deposit 3 mole or 3 eq. of Ag . Ag is monovalent , `[ because (W)/(E)` (for Ag) = `(W)/(E)` (for Al)]

Discussion

26.	If an iron rod is dipped in `CuSO_(4)` solutionA. Blue colour of the solution turns greenB. Brown layer is deposited on iron rodC. No change occurs in the colour of the solutionD. Blue colour of the solution vanishes
Answer» Correct Answer - B Brown layer is deposited o iron rod because Cu has greater reduction potential than that of `Fe^(2+)`.

Discussion

27.	The emf in V of Danneil cell containing 0.1 M `ZnSO_(4)` and 0.01 M `CuSO_(4)` solutions their respective electrodes is `E_(Cu^(2+) \| Cu)^(@) = 0.34`V and `E_(Zn^(2+) \| Zn)^(@) = -0.76 ` VA. 1.10 VB. 1.16 VC. 1.13 VD. 1.07 V
Answer» Correct Answer - D The Daniel cell is represented as `Zn \| Zn ^(2+) \|\|Cu^(2+) \| Cu` and cell reactions `Zn + Cu^(2+) to Zn^(2+) + Cu` `E_(cell)^(@) = E_(RHS)^(@) - E_(LHS)^(@) = 0.34 - (-0.76) = 1.1 V , n = 2 ` A/C to Nernst equation `E_(cell) = E_(cell)^(@) - (0.0592)/(n) log ([Zn^(2+)])/([Cu^(2+)]) because [Zn (s) ] = [ Cu(s) ] =1 ` `= 1.1 V - (0.0592)/(2) "log" (0.1)/(0.01)` `= 1.1 V - (0.0592)/(2) "log" 10^(-1)` =`1.1 V - (0.0592)/(2) = 1.07` V

Discussion

28.	When 9.65 coulomb of electricity is passed through a solution of silver nitrate (Atomic mass of Ag = 108 g `mol^(-1)`, the amount of silver deposited is :A. 10.8 gmB. 5.4 gmC. 16.2 gmD. 21.2 gm
Answer» Correct Answer - A `W_(Ag)=(E_(Ag)xxQ)/(96500)=(108xx9.65)/(96500)` `=1.08xx10^(-2)gm=10.8mg`.

Discussion

29.	Three faradays electricity was passed through an aqueous solution of irono (II) romide. The weight of iron metal (at. At.=56) deposited at the cathode (in gm) isA. 56B. 84C. 112D. 168
Answer» Correct Answer - B `Fe^(2+)+2e^(-)toFe,E_(Fe)=(56)/(2)=28` `W_(Fe)=E_(Fe)xx"Number of faraday"=28xx3=84gm`

Discussion

30.	Blue colour of `CuSO_(4)` solution is discharge slowly when an iron rod is dipped into it. Why?
Answer» Zinc lies below copper in the electrochemical series and can therefore, release electrons to the `Cu^(2+)` ions. As a result, Zn is oxidised while `Cu^(2+)` ions are reduced. Since the concentration of `Cu^(2+)` ions in the solution decreases, the blue colour of the solution is slowly discharged. `Zn(s) +Zn^(2+)(aq) to Zn^(2+)(aq) +Cu(s)`

Discussion

31.	At equilibrium :A. `E_("cell")^(@)=0,DeltaG^(@)=0`B. `E_("cell")^(@)=0,DeltaG=0`C. both are correctD. none is correct
Answer» Correct Answer - B

Discussion

32.	Assertion:When a copper wire is dipped in silver nitrate solution, there is no change in the colour of the solution. Reason : Copper cannot displace silver from its salt solution.A. If both assertion and reason are true and reason is the correct explanation of assertion .B. If both assertion and reason are true but reason is not the correct explanation of assertion .C. If assertion is true but reason is false .D. If both assertion and reason are false .
Answer» Correct Answer - D When a copper wire is dipped in silver nitrate solution it turns blue. Cu is more reactive than Ag hence displaces silver from its salt solution.

Discussion

33.	From the electrochemical series, it can be concluded that :A. `Zn^(2+)` will libreate `H_(2)` from 1 M HClB. Ag metal reacts spontaneously with `Zn^(2+)`C. Zn metal will liberate `H_(2)` from 1M HClD. Ag metal will liberate `H_(2)` from `1M HCl
Answer» Correct Answer - C

Discussion

34.	A solution of Cu(II) sulphate is reacted with KCl and KI. In which case will the `Cu^(2+)` be reduced to `Cu(+)` ?A. In both the casesB. When reacted with KClC. When reacted with KID. In both the cases but in presence of `H^(+)`
Answer» Correct Answer - C

Discussion

35.	1. Give the name for the iron coated with the Zinc.2. Can we store Cl2 gas in copper vessel? Give reason.(E°(Cu2+/Cu) = + 0.34 V E°(Cl2/Cl-) = +1 -36 V)3. The equilibrium constant of a reaction which is difficult to calculate can be calculated by knowing the standard potential of the cell in which reaction takes place. Give the mathematical expression relating equilibrium constant and cell potential
Answer» 1. Galvanised iron 2. No. Since the reduction potential of chlorine is greater than that of copper,chlorine reacts with copper to form cuprous chloride. E°_(Cl²/Cl^-) > (E°_(Cu²⁺/Cu)) 3. E°_cell = \(\frac{2.303RT}{nF}logK_c\) where E°_cell is the standard cell potential (E°_cell = E°_cathode – E°_anode and K_c is the equilibrium constant of the cell reaction.

35.

1. Give the name for the iron coated with the Zinc.2. Can we store Cl2 gas in copper vessel? Give reason.(E°(Cu2+/Cu) = + 0.34 V E°(Cl2/Cl-) = +1 -36 V)3. The equilibrium constant of a reaction which is difficult to calculate can be calculated by knowing the standard potential of the cell in which reaction takes place. Give the mathematical expression relating equilibrium constant and cell potential

Answer»

1. Galvanised iron

2. No. Since the reduction potential of chlorine is greater than that of copper,chlorine reacts with copper to form cuprous chloride.

E°_(Cl²/Cl^-) > (E°_(Cu²⁺/Cu))

3. E°_cell = \(\frac{2.303RT}{nF}logK_c\)

where E°_cell is the standard cell potential (E°_cell = E°_cathode – E°_anode and K_c is the equilibrium constant of the cell reaction.

Discussion

36.	Why is it not possible to measure the single electrode potential?
Answer» Oxidation of reduction cannot take place alone. Moreover, it is arelative tendency and can be measured with respect to a reference electrode only.

Discussion

37.	Define electrochemical series. or Wrtie two applications of electrochemical series.
Answer» The arrangment of various electrodes in the decreasing or increasing order of their standard reduction potential is called electrochemcial series are: (i) To compare oxidizing or reducing powers of different elements. ltbr. (ii) to predict the spontaneity of any redox reaction.

Discussion

38.	Given that the standard electrode potential `(E^(@))` of metals are: `K^(+)//K=-2.93V,Ag^(+)//Ag=0.80V," "Cu^(2+)//Cu=0.34V,Mg^(2+)//Mg=-2.37V`, `Cr^(3+)//Cr=-0.74V,Fe^(2+)//Fe=-0.44V` Arrange these metals in an increasing order of their reducing power.
Answer» Higher the oxidation potential more easily it is oxidised and hence stronger is the reducing power. Therefore order is `Ag lt Cu lt Fe lt Mg lt K`.

Discussion

39.	The values of some of the standard electrode potential are `E^(@)_(Ag^(+)//Ag)=0.80 V.E^(@)(Hg_(21+)^(2+)//Hg)=0.79 VE^(@)_(Cu^(+2)//Cu)=0.34V` What is the sequence of deposition of metals on the cathode ?
Answer» Consider the `E^(@)` values.

Discussion

40.	If the standard electrode poten tial of `Cu^(2+)//Cu` electrode is 0.34V. What is the electrode potential of 0.01 M concentration of `Cu^(2+)`?A. 0.399VB. 0.281VC. 0.222VD. 0.176V
Answer» Correct Answer - B `Cu^(2+)2e^(-)rarrCu` `E=E^(@)-(0.0591)/(n)"log")(1)/([M^(n+)])` `E_(cu^(2+_(//Cu)))-(0.0591)/(2)"log"(1)/(10^(2))`

Discussion

41.	What is the value of `E^(@)` cell in the following reaction? `Cr\|Cr^(3+) (0.1 M)\|\|Fe^(2+) (0.01 M)\|Fe` Given, `E_(Cr^(3+)//Cr)^(@)=-0.74 V, E_(Fe^(2+)//Fe)^(@)=-0.44 V`A. `+ 0.2606 V`B. `0.5212 V`C. `+ 01303 V`D. `-0.2606 V`
Answer» Correct Answer - A The cell reaction is `2Cr+3Fe^(2+) rarr 2Cr^(3+)+3Fe(n=6)` from Nernst equation, `E_("cell")=E_("cell")^(@)-0.059/n "log" ([Cr^(3+)]^(2))/([Fe^(2+)]^(3))` `=[-0.44+0.74]- 0.059/6 xx"log" ((0.1)^(2))/((0.01)^(3))` `=+0.2606 V`

Discussion

42.	Calculate the standard electrode potential of the `Ni^(2+)//Ni` electrode , if the cell potential potential of the cell, `Ni//N^(2+)(0.01 M)//Cu is 0.59" V ". "Given" E_(Cu^(2+)//Cu)^(@)=+0.34 " V "`
Answer» Correct Answer - `-0.2205 V` Cell reaction : `Ni(s)+Cu^(2+)(0.1" M ") to Ni^(2+)(0.01" M ")+Cu(s)` `E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Ni^(2+)])/([Cu^(2+)])` `0.59=E_(cell)^(@)-(0.0591)/(n)"log"(0.01)/(0.1)=E_(cell)^(@)-0.02955" log "(10^(-1))=E_(cell)^(@)+0.02955` `E_(cell)^(@)=0.59-0.02955=0.5605` Now, `" " E_(cell)^(@)=E_((Cu^(2+)//Cu))^(@)-E_((Ni^(2+)//Ni))^(@)` `:. E_((Ni^(2+)//Ni))^(@)=E_((Cu^(2+)//Cu))^(@)-E_(cell)^(@)=0.34-0.5605=0.2205" V "`

Discussion

43.	Calculate the e.m.f. of the cell, `Cr//Cr^(3+)(0.1 M) \|\| Fe^(2+)(0.01 M)//Fe` `"Given" : E_(Cr^(3+)//Cr)^(@)=-0.75" V ", E_(Fe^(2+)//Fe)^(@)=-0.45" V "` `"Cell reaction" : 2Cr(s)+3Fe^(2+)(aq) to 2Cr^(3+)(aq)+3Fe(s)` `{"Hint". E_(cell)=E_(cell)^(@)-(0.0591V)/(6)"log"([Cr^(3+)]^(2))/([Fe^(2+)]^(3)}`
Answer» Correct Answer - `+0.2606 V` Cell reaction : `2Cr(s)+3Fe^(2+)(0.01" M") to 2Cr^(3+)(0.1" M")+3Fe(s)` `E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Cr^(3+)]^(2))/([Fe^(2+)]^(3))=[(-0.45)-(-0.75)]-(0.0591)/(6)"log"((0.1)^(2))/((0.1)^(3))` `=0.30-(0.0591)/(6)log10^(4)=0.30-(0.0591xx4)/(6)=0.30-0.0394=+0.2606" V "`

Discussion

44.	Consider the following cell reaction: `2Fe(2)+O_(2)+4H^(+)(aq)to2Fe^(2+)(aq)+2H_(2)O(l),E^(@)=1.67V` at `[Fe^(2+)]=10^(-3)M,P(O_(2))=0.1` atm and pH=3, the cell potential at `25^(@)C` is,A. 1.47B. 1.77C. 1.87D. 1.57
Answer» Correct Answer - D Applying nernst equation to the given reaction, `E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Fe^(2+)]^(2))/(P_(O_(2))xx[H^(+)]^(4))` Fot the given reaction, n=4 and pH=3 means `[H^(+)]=10^(-3)M` `thereforeE_(cell)=1.67-(0.0591)/(2)"log"((10^(-3))^(2))/(0.1xx10^(-3))^(4)` `=1.67-(0.0591)/(4)log10^(7)=1.67-0.10=1.57V`

Discussion

45.	Statement-1: If an aqueous solution of NaCl is electrolysed the product obtained at the cathode is `H_(2)` gas notn Na. Statement-2: Gases are liberated faster than the metals during the electrolysis of an electrolyte.A. If both (A) and (R) are correct, and (R) is the correct explanation of (A).B. If both (A) and (R) are correct, but (R) is not the correct explanation of (A).C. If (A) is correct, but (R) is incorrectD. If (A) is incorrect, but (R) is correct.
Answer» Correct Answer - C

Discussion

46.	Write a note on Nickel-Cadmium (NICAD) cell.
Answer» (1) Nickel-Cadminum (NICAD) cell is a secondary dry cell. (2) It is rechargable, hence it is a reversible cell. (3) It consists of a cadmium electrode in contact with an alkali and acts as anode while nickel (IV) oxide, NiO₂ in contact with an alkali acts as cathode. The alkali used is moist paste of KOH. (4) Reactions in the cell : (i) Oxidation at cadmium anode : Cd_(s) + 2OH^-_(aq) → Cd(OH)_2(s) + 2e^- (ii) Reduction at NiO_2(s) cathode : NiO_2(s) + 2H₂O_(I) + 2e^- → Ni(OH)_2(s) + 2OH^-_(aq) The overall cell reaction is the combination of above two reactions. Cd_(s) + NiO_2(s) + 2H₂O_(I) → Cd(OH)_2(s) + Ni(OH)_2(s) (5) Since the net cell reaction doesn’t involve any electrolytes but solids, the voltage is independent of the concentration of alkali electrolyte. (6) The cell potential is about 1.4 V. (7) This cell has longer life than other dry cells.

46.

Write a note on Nickel-Cadmium (NICAD) cell.

Answer»

(1) Nickel-Cadminum (NICAD) cell is a secondary dry cell.

(2) It is rechargable, hence it is a reversible cell.

(3) It consists of a cadmium electrode in contact with an alkali and acts as anode while nickel (IV) oxide, NiO₂ in contact with an alkali acts as cathode. The alkali used is moist paste of KOH.

(4) Reactions in the cell :

(i) Oxidation at cadmium anode :

Cd_(s) + 2OH^-_(aq) → Cd(OH)_2(s) + 2e^-

(ii) Reduction at NiO_2(s) cathode :

NiO_2(s) + 2H₂O_(I) + 2e^- → Ni(OH)_2(s) + 2OH^-_(aq)

The overall cell reaction is the combination of above two reactions.

Cd_(s) + NiO_2(s) + 2H₂O_(I) → Cd(OH)_2(s) + Ni(OH)_2(s)

(5) Since the net cell reaction doesn’t involve any electrolytes but solids, the voltage is independent of the concentration of alkali electrolyte.

(6) The cell potential is about 1.4 V.

(7) This cell has longer life than other dry cells.

Discussion

47.	Assertion: In the electrolysis of aqueous NaCl, Na is preferentially discharged at the mercury cathode forming sodium amalgam. Reason: It is due to the fact that hydrogen has high overvoltage at mercury cathode.A. If both assertion and reason are true, and reason is the true explanation of the assertionB. if both assertion and reason are true, but reason is not the true explanation of the assertion.C. if assertion is true, but reason is false.D. If both assertion and reason are false.
Answer» Correct Answer - A R is the correct explanation of A.

Discussion

48.	Cadmium amalgam is prepared by electrolysis of a sodium of `CdCl_2` using a mercury cathode. How long should a current of 4 A be passed in order to prepare 10% by mass Cd in Cd-Hg amalgam on cathode of 4.5 g Hg? (atomic mass of Cd=112)A. 400secB. 215.40 secC. 861.6 secD. 430.8 sec
Answer» Correct Answer - B

Discussion

49.	The number of electrons delivered at the cathode during electrolysis by a current of `1` ampere in 60 seconds is (charger on electron `= 1.60 xx 10^(-19)C`)A. `6 xx 10^(23)`B. `6 xx 10^(20)`C. `3.75 xx 10^(20)`D. `7.48 xx 10^(23)`
Answer» Correct Answer - C Number of electrons `= ("Total charge delivered")/("charge on single electron")` `= (1A)(60 s)/(1.60 xx 10^(-19) C//e^(-)), (Q=It)` `= 37.5 xx 10^(19)` electrons `= 3.75 xx 10^(20)` electrons

Discussion

50.	The number of electrons delivered at the cathode during electrolysis by a current of `1` ampere in 60 seconds is (charger on electron `= 1.60 xx 10^(-19)C`)A. `7.48xx 10^23`B. `6xx10^23`C. `6xx10^20`D. `3.75xx10^20`
Answer» Correct Answer - D `(W)/(E)=(1xx60)/(96500)` `=(6)/(9650)=`no. of mole `e^(-)` no. of `e^(-)=(6)/(9650)xx6.02xx10^(23)` `=3.75xx10^(20)`

Discussion

Explore topic-wise InterviewSolutions in .

Define molar conductivity of a solution and explain how molar conductivity changes with change in concentration of a solution for a weak and a strong electrolyte.

For the cell reaction `Cu^(2+)(C_(1),aq.)+Zn(s)hArrZn^(2+)(C_(2),aq)+Cu(s)` of an electrochemical cell, the change in free energy `(DeltaG)` of a given temperature is a function ofA. `lnC_(1)`B. `lnC_(2)//C_(1)`C. `ln (C_(1) + C_(2))`D. `ln C_(2)`

Use of lithium metal as an electrode in high energy density batteries is due to:A. Lithium is highest elementB. Lithium has highest oxidation potentialC. Lithium is quite reactiveD. Lithium does not corrode readily

What are elementary reactions ?

Which of the following alkali metal ions has the lowest ionic mobility in aqueous solutions?A. `Na^(o+)`B. `F^(c-)`C. `SO_(4)^(2-)`D. `overset(c-)(O)H`

Which of the following alkali metal ions has the lowest ionic mobility in aqueous solutions?A. `Li^(o+)`B. `Na^(o+)`C. `K^(o+)`D. `Rb^(o+)`

Ammonia released during the reduction at cathode of Leclanche cell combines with zinc ions to formA. `[Zn(NH_(3))_(4)]^(2+)`B. `[Zn(NH_(3))_(2)]^(2+)`C. `[Zn(NH_(3))_(6)]^(2+)`D. `[Zn(NH_(3))_(5)]^(2+)`

In lead accumulator, anode and cathode are :(a) (Pb + PbO2),Pb (b) Pb, PbO2 (c) PbO2, Pb (d) Pb, (Pb + PbO2)

What takes place when zinc metal is added to an aqueous solution containing magnesium nitrate and silver nitrate ? 1. Zn is oxidised 2. `Mg^(2+)` is reduced 3. `Ag^(+)` is reduced 4. No reaction takes placeA. 1 and 2 onlyB. 1 and 3 onlyC. 1, 2 and 3 onlyD. 4 only

Galvanisation is applying a coating ofA. CrB. CuC. ZnD. Pb

In the galvanizing process, iron is coated with zinc. The resulting chemical protection is most similar to that provided when:A. a magnesium bar is connected to an iron pipeB. an iron can is plated with tinC. copper pipes are connected using lead solderD. a copper pipes is covered with epoxy paint

The ionization constant `(K_(a))` of a weak electrolyte is `2.5xx10^(-7)`, while `wedge_(eq)` of its `0.01M` solution is `19.65S cm^(2) eq^(-1)wedge^(@)._(eq)` is `……………….` .

If the temperature coefficient `((delE)/(delT))` is zero for a cell reaction then out of `DeltaS, DeltaH`, and `DeltaG` , the `…………………..` is zero.

`13g` of metal `M` is deposited at cathode by passing `0.4F` of electricity . The cathodic reaction is `:` `M^(n+)+n e^(-) rarr M` The formula of metal chloride `(Aw=65)` is `……………………..` .

`4.5 g` of aluminium (at mass `27 u`) is deposited at cathode from `Al^(3+)` solution by a certain quantity of electric charge. The volume of hydrogen gas produced at `STP` from `H^(+)` ions in solution by the same quantity of electric charge will be:A. 44.8 LB. 22.4 LC. 11.2 LD. 5.6 L

On passing 0.1 F of electricity through aluminium chloride, the amount of aluminium meta deposited on cathode is (Atomic weight of `Al=27`)A. `0.27 g`B. `0.3 g`C. `0.9 g`D. `2.7 g`

When 9.65 coulomb of electricity is passed through a solution of silver nitrate (Atomic mass of Ag = 108 g `mol^(-1)`, the amount of silver deposited is :A. `10.8` mgB. `5.4` mgC. `16.2` mgD. `21.2` mg

1 mole of Al is deposited by X coulomb of electricity passing through aluminium nitrate solution . The number of mole of silver deposited by X coulomb of electricity from silver nitrate solution isA. 3B. 4C. 2D. 1

If an iron rod is dipped in `CuSO_(4)` solutionA. Blue colour of the solution turns greenB. Brown layer is deposited on iron rodC. No change occurs in the colour of the solutionD. Blue colour of the solution vanishes

The emf in V of Danneil cell containing 0.1 M `ZnSO_(4)` and 0.01 M `CuSO_(4)` solutions their respective electrodes is `E_(Cu^(2+) | Cu)^(@) = 0.34`V and `E_(Zn^(2+) | Zn)^(@) = -0.76 ` VA. 1.10 VB. 1.16 VC. 1.13 VD. 1.07 V

When 9.65 coulomb of electricity is passed through a solution of silver nitrate (Atomic mass of Ag = 108 g `mol^(-1)`, the amount of silver deposited is :A. 10.8 gmB. 5.4 gmC. 16.2 gmD. 21.2 gm

Three faradays electricity was passed through an aqueous solution of irono (II) romide. The weight of iron metal (at. At.=56) deposited at the cathode (in gm) isA. 56B. 84C. 112D. 168

Blue colour of `CuSO_(4)` solution is discharge slowly when an iron rod is dipped into it. Why?

At equilibrium :A. `E_("cell")^(@)=0,DeltaG^(@)=0`B. `E_("cell")^(@)=0,DeltaG=0`C. both are correctD. none is correct

From the electrochemical series, it can be concluded that :A. `Zn^(2+)` will libreate `H_(2)` from 1 M HClB. Ag metal reacts spontaneously with `Zn^(2+)`C. Zn metal will liberate `H_(2)` from 1M HClD. Ag metal will liberate `H_(2)` from `1M HCl

A solution of Cu(II) sulphate is reacted with KCl and KI. In which case will the `Cu^(2+)` be reduced to `Cu(+)` ?A. In both the casesB. When reacted with KClC. When reacted with KID. In both the cases but in presence of `H^(+)`

Why is it not possible to measure the single electrode potential?

Define electrochemical series. or Wrtie two applications of electrochemical series.

Given that the standard electrode potential `(E^(@))` of metals are: `K^(+)//K=-2.93V,Ag^(+)//Ag=0.80V," "Cu^(2+)//Cu=0.34V,Mg^(2+)//Mg=-2.37V`, `Cr^(3+)//Cr=-0.74V,Fe^(2+)//Fe=-0.44V` Arrange these metals in an increasing order of their reducing power.

The values of some of the standard electrode potential are `E^(@)_(Ag^(+)//Ag)=0.80 V.E^(@)(Hg_(21+)^(2+)//Hg)=0.79 VE^(@)_(Cu^(+2)//Cu)=0.34V` What is the sequence of deposition of metals on the cathode ?

If the standard electrode poten tial of `Cu^(2+)//Cu` electrode is 0.34V. What is the electrode potential of 0.01 M concentration of `Cu^(2+)`?A. 0.399VB. 0.281VC. 0.222VD. 0.176V

What is the value of `E^(@)` cell in the following reaction? `Cr|Cr^(3+) (0.1 M)||Fe^(2+) (0.01 M)|Fe` Given, `E_(Cr^(3+)//Cr)^(@)=-0.74 V, E_(Fe^(2+)//Fe)^(@)=-0.44 V`A. `+ 0.2606 V`B. `0.5212 V`C. `+ 01303 V`D. `-0.2606 V`

Calculate the standard electrode potential of the `Ni^(2+)//Ni` electrode , if the cell potential potential of the cell, `Ni//N^(2+)(0.01 M)//Cu is 0.59" V ". "Given" E_(Cu^(2+)//Cu)^(@)=+0.34 " V "`

Calculate the e.m.f. of the cell, `Cr//Cr^(3+)(0.1 M) || Fe^(2+)(0.01 M)//Fe` `"Given" : E_(Cr^(3+)//Cr)^(@)=-0.75" V ", E_(Fe^(2+)//Fe)^(@)=-0.45" V "` `"Cell reaction" : 2Cr(s)+3Fe^(2+)(aq) to 2Cr^(3+)(aq)+3Fe(s)` `{"Hint". E_(cell)=E_(cell)^(@)-(0.0591V)/(6)"log"([Cr^(3+)]^(2))/([Fe^(2+)]^(3)}`

Consider the following cell reaction: `2Fe(2)+O_(2)+4H^(+)(aq)to2Fe^(2+)(aq)+2H_(2)O(l),E^(@)=1.67V` at `[Fe^(2+)]=10^(-3)M,P(O_(2))=0.1` atm and pH=3, the cell potential at `25^(@)C` is,A. 1.47B. 1.77C. 1.87D. 1.57

Write a note on Nickel-Cadmium (NICAD) cell.

Cadmium amalgam is prepared by electrolysis of a sodium of `CdCl_2` using a mercury cathode. How long should a current of 4 A be passed in order to prepare 10% by mass Cd in Cd-Hg amalgam on cathode of 4.5 g Hg? (atomic mass of Cd=112)A. 400secB. 215.40 secC. 861.6 secD. 430.8 sec

The number of electrons delivered at the cathode during electrolysis by a current of `1` ampere in 60 seconds is (charger on electron `= 1.60 xx 10^(-19)C`)A. `6 xx 10^(23)`B. `6 xx 10^(20)`C. `3.75 xx 10^(20)`D. `7.48 xx 10^(23)`

The number of electrons delivered at the cathode during electrolysis by a current of `1` ampere in 60 seconds is (charger on electron `= 1.60 xx 10^(-19)C`)A. `7.48xx 10^23`B. `6xx10^23`C. `6xx10^20`D. `3.75xx10^20`