Q 1 :

One of the commonly used electrodes is calomel electrode. Under which of the following categories calomel electrode come?          [2024]

  • Metal ion – Metal electrodes

     

  • Oxidation – Reduction electrodes

     

  • Metal – Insoluble Salt – Anion electrodes

     

  • Gas – Ion electrodes

     

(3)

Calomel electrode is a metal-metal insoluble salt-anion half cell. It is commonly used as a reference like standard hydrogen electrode.

 



Q 2 :

How can an electrochemical cell be converted into an electrolytic cell?

  • Reversing the flow of ions in salt bridge.

     

  • Applying an external opposite potential greater than Ecell0.

     

  • Exchanging the electrodes at anode and cathode.

     

  • Applying an external opposite potential lower than Ecell0.

     

(2)

When in an electrochemical cell, external voltage greater than Ecell is applied, then direction of flow of charge is reversed. Now external source supplies energy to the cell and cell is now electrolytic cell.

 



Q 3 :

Reduction potential of ions are given below:

     ClO4-                         IO4-                             BrO4-

E°=1.19V         E°=1.65V             E°=1.74V

The correct order of their oxidising power is:                   [2024]

  • ClO4->IO4->BrO4-

     

  • BrO4->IO4->ClO4-

     

  • BrO4->ClO4->IO4-

     

  • IO4->BrO4->ClO4-

     

(2)

              Oxidising power  reduction potential.

              So more is reduction potential of a specie, more is its ability to oxidize the other substance.

              Decreasing oxidising powerBrO4-(E°=1.74V)>IO4-(E°=1.65V)>ClO4-(E°=1.19V)

 



Q 4 :

Match List I with List II.

  List I   List II
  (Cell)   (Use/Property/Reaction)
A. Leclanche cell I. Converts energy of combustion into electrical energy
B. Ni-Cd cell II. Does not involve any ion in solution and is used in hearing aids
C. Fuel cell III. Rechargeable
D. Mercury cell IV. Reaction at anode ZnZn2++2e-

 

Choose the correct answer from the options given below:                    [2024]

  • A–II, B–III, C–IV, D–I

     

  • A–III, B–I, C–IV, D–II

     

  • A–I, B–II, C–III, D–IV

     

  • A–IV, B–III, C–I, D–II

     

(4)

 A. Leclanche cell: It is commonly used dry cell in our clocks. The cell consists of a zinc container that also acts as anode and the cathode is a carbon (graphite) rod surrounded by powdered manganese dioxide and carbon. The space between the electrodes is filled by a moist paste of ammonium chloride and zinc chloride.

      Anode reaction:  Zn(s)Zn2++2e-

      Cathode reaction: MnO2+NH4++e-MnO(OH)+NH3

B. Ni-Cd cell: It is a rechargeable cell. It is more expensive to manufacture than a lead storage battery.

      Cd(s)+2Ni(OH)3(s)CdO(s)+2Ni(OH)2(s)+H2O(l)

C. Fuel cell: These are galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol, etc., directly into electrical energy.

D. Mercury cell:

       Zn(Hg)+HgO(s)ZnO(s)+Hg(l)

 The cell potential is approximately 1.35 V and remains constant during its life as the overall reaction does not involve any ion in solution whose concentration can change during its lifetime.

 



Q 5 :

The standard reduction potentials at 298 K for the following half cells are given below:                           [2024]

Cr2O72-+14H++6e-2Cr3++7H2O,  E0=1.33V

Fe3+(aq)+3e-Fe,  E0=-0.04V

Ni2+(aq)+2e-Ni,  E0=-0.25V

Ag+(aq)+e-Ag,  E0=0.80V

Au3+(aq)+3e-Au,  E0=1.40V

Consider the given electrochemical reactions,

The number of metal(s) which will be oxidized be Cr2O72- in aqueous solution is __________.



(3)

For the above reaction to be spontaneous:

Ereactiono=Ecathodeo-Eanodeo>0

ECr2O72-o-Eanodeo>0

1.33-Eanodeo>0

Eanodeo<1.33V

Thus Fe, Ni and Ag will be oxidized by Cr2O72-.



Q 6 :

FeO42-+2.0 VFe3+0.8 VFe2+-0.5 VFe0

In the above diagram, the standard electrode potentials are given in volts (over the arrow).

The value of  EFeO42-/Fe2+  is                           [2025]

  • 2.1 V

     

  • 1.7 V

     

  • 1.2 V

     

  • 1.4 V

     

(2)

When two reactions are added their Gibbs energy change is added, as Gibbs energy change is an extensive property.

ΔG4=ΔG1+ΔG2

-n4FE4=-n1FE1-n2FE2

-4FE4=-3F×2-1F×0.8

E4=6.84V=1.7 V



Q 7 :

For the given cell

Fe2+(aq)+Ag+(aq)Fe3+(aq)+Ag(s)

The standard cell potential of the above reaction is:

Given:

Ag++e-Ag           Eθ=x V 

Fe2++2e-Fe        Eθ=y V 

Fe3++3e-Fe         Eθ=z V                       [2025]

  • x+2y-3z

     

  • x+2y

     

  • x+y-z

     

  • y-2x

     

(1)

Ag++e-Ag-I  (ΔG1=-nFE1=-1Fx) 

Fe2++2e-Fe-II  (ΔG2=-nFE2=-2Fy) 

Fe3++3e-Fe-III  (ΔG3=-nFE3=-3Fz) 

I+II-III gives the required equation:

Fe2++Ag+Fe3++Ag-IV

ΔG4=ΔG1+ΔG2-ΔG3=-1Fx- 2Fy+3Fz 

-nFE4=-1Fx-2Fy+3Fz 

-IFE4=-1Fx-2Fy+3Fz 

E4=x+2y-3z



Q 8 :

Based on the data given below:

ECr2O72-/Cr3+=1.33 V     ECl2/Cl-=1.36 V

EMnO4-/Mn2+=1.51 V     ECr3+/Cr=-0.74 V

The strongest reducing agent is:                               [2025]

  • Mn2+

     

  • Cl-

     

  • MnO4-

     

  • Cr

     

(4)

Strongest reducing agent is the specie which has highest tendency to get oxidized i.e. the one with highest oxidation potential.

Reduction potential Oxidation potential
ECr2O72-/Cr3+=1.33 V ECr3+/Cr2O72-=-1.33 V
ECl2/Cl-=1.36 V ECl-/Cl2=-1.36 V
EMnO4-/Mn2+=1.51 V EMn2+/MnO4-=-1.51 V
ECr3+/Cr=-0.74 V ECr/Cr3+=0.74 V

 



Q 9 :

The standard reduction potential values of some of the p-block ions are given below. Predict the one with the strongest oxidising capacity.         [2025]

  • ESn4+/Sn2+=+1.15 V

     

  • EPb4+/Pb2+=+1.67 V

     

  • ETl3+/Tl=+1.26 V

     

  • EAl3+/Al=-1.66 V

     

(2)

Specie with highest reduction potential (i.e. highest willingness to get reduced) is strongest oxidizing agent.

 



Q 10 :

The standard electrode potential (M3+/M2+) for V, Cr, Mn and Co are - 0.26 V, - 0.41 V, + 1.57 V and + 1.97 V, respectively. The metal ions which can liberate H2 from a dilute acid are            [2023]
 

  • V2+ and Mn2+

     

  • Mn2+ and Co2+

     

  • Cr2+ and Co2+

     

  • V2+ and Cr2+

     

(4)

The metal ion for which have less value of reduction potential that (EH+/H2=0) can release H2 on reaction with dilute acid. So, V+2 and Cr+2 reduce H+ to H2.



Q 11 :

The standard electrode potential of M+/M in aqueous solution does not depend on           [2023]

  • Sublimation of a solid metal

     

  • Ionisation of a solid metal atom

     

  • Hydration of a gaseous metal ion

     

  • Ionisation of a gaseous metal atom

     

(2)

The standard electrode potential of M+/M in aqueous solution depends on:

• Ionisation of a gaseous metal atom

• Hydration of a gaseous metal ion

• Sublimation of a solid metal



Q 12 :

The reaction

12H2(g)+AgCl(s)H+(aq)+Cl-(aq)+Ag(s)

occurs in which of the given galvanic cells?                        [2023]

  • Pt|H2(g)|HCl(soln)|AgCl(s)|Ag

     

  • Ag|AgCl(s)|KCl(soln)|AgNO3|Ag

     

  • Pt|H2(g)|HCl(soln)|AgNO3(soln)|Ag

     

  • Pt|H2(g)|KCl(soln)|AgCl(s)|Ag

     

(1)



Q 13 :

The standard reduction potentials at 298 K for the following half cells are given below:

NO3-+4H++3e-NO(g)+2H2O     Eθ=0.97 V

V2+(aq)+2e-V                                   Eθ=-1.19 V

Fe3+(aq)+3e-Fe                               Eθ=-0.04 V

Ag+(aq)+e-Ag(s)                              Eθ=0.80 V

Au3+(aq)+3e-Au(s)                          Eθ=1.40 V

The number of metal(s) which will be oxidized by NO3- in aqueous solution is _____________ .          [2023]



(3)

NO3- can oxidise V, Fe and Ag.



Q 14 :

At 298 K, the standard reduction potential for the Cu2+/Cu electrode is 0.34 V.

Given: KspCu(OH)2=1×10-20

Take 2.303RTF=0.059 V

The reduction potential at pH = 14 for the above couple is (-)x×10-2V. The value of x  is __________ .        [2023]



(25)

pH=14pOH=0

[OH-]=1

KspCu(OH)2=[Cu2+][OH-]2=1×10-20

                             [Cu2+]=10-20

ECu2+|Cu=ECu2+|Cu0-0.0592log1[Cu2+]

=0.34-0.0592log1020

= -0.25

=-25×10-2 V



Q 15 :

Consider the following electrochemical cell at 298 K:

Pt | HSnO2-(aq) | Sn(OH)62-(aq) | OH-(aq)  | Bi2O3(s) | Bi(s)

If the reaction quotient at a given time is 106, then the cell EMF (Ecell) is __________×10-1 V (Nearest integer).

Given the standard half-cell reduction potentials as:

EBi2O3/Bi,OH-=-0.44 V and ESn(OH)62-/HSnO2-,OH-=-0.90 V                 [2026]



(4)

Ecell=-0.44-(-0.90)

         =+0.46 V

Applying Nernst equation:

Ecell=Ecell-0.06nlogQ

Ecell=0.46-0.066log106

Ecell=4×10-1

x=4



Q 16 :

Consider the following reduction processes:

Al3++3e-Al(s),    E°=-1.66V 

Fe3++e-Fe2+,    E°=+0.77V

Co3++e-Co2+,    E°=+1.81V

Cr3++3e-Cr(s),    E°=-0.74V

The tendency to act as reducing agent decreases in the order:              [2026]

  • Al>Fe2+>Cr>Co2+

     

  • Al>Cr>Co2+>Fe2+

     

  • Cr>Fe2+>Al>Co2+

     

  • Al>Cr>Fe2+>Co2+

     

(4)

Reducing power1Reduction potential



Q 17 :

Electrolysis of aqueous CuSO4(0.1M) was carried out is two cells I and II. In I, the electrodes are of Cu and is II they were of Pt. As the electrolysis proceeds pH of the electrolyte solution will

  • decrease is II and remain the same in I

     

  • remain the same in both I and II

     

  • increase in both I and II

     

  • increase in I and decrease is II

     

(1)

 



Q 18 :

Saturated solution of KNO3 with agar-agar is used to make 'salt bridge' because:

  • size of K+ is greater than that of NO3-

     

  • velocity of NO3- is greater than that of K+

     

  • velocities of both K+ and NO3- are nearly the same

     

  • both velocities and size of K+ and NO3- ions are same

     

(3)