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

 



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 Zn→Zn2++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=14⇒pOH=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°=-1.66 V 

Fe3++e-⟶Fe2+,    E°=+0.77 V

Co3++e-⟶Co2+,    E°=+1.81 V

Cr3++3e-⟶Cr(s),    E°=-0.74 V

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 power∝1Reduction potential



Q 17 :

Electrolysis of aqueous CuSO4(0.1 M) 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)

 



Q 19 :

Given at 298 K:

EFe2+/FeΘ=X Volt

EFe3+/FeΘ=Y Volt

The EFe3+/Fe2+Θ in Volt at 298 K is given by:               [2026]

  • 2X-3Y

     

  • 3Y-2X

     

  • 3Y+2X

     

  • Y+X

     

(2)

Fe2++2e-→Fe,  EFe2+/Feo=X,

ΔGFe2+/Feo=-nFEFe2+/Feo=-2FX

Reverse this reaction

Fe→Fe2++2e-;  EFe/Fe2+o=-X,

ΔGFe/Fe2+o=-nFEFe/Fe2+o=-nF(-X)=2FX

Fe3++e-→Fe,  EFe3+/Feo=Y

ΔGFe3+/Feo=-nFEFe3+/Feo=-3FY

Add above two reactions:

Fe3++e-→Fe2+

ΔGFe3+/Fe2+o=ΔGFe3+/Feo+ΔGFe/Fe2+o=-3FY+2FX

-nFEFe3+/Fe2+o=-3FY+2FX

-1FEFe3+/Fe2+o=-3FY+2FX

EFe3+/Fe2+o=3Y-2X