ELECTROCHEMISTRY 12th

 

                                                     YASH CLASSES   

                                             ELECTROCHEMISTRY

                                NCERT INTEXT UNSOLVED QUESTIONS

 Q.2.1. How would you determine the standard electrode potential of Mg²⁺ | Mg?

 Ans. We will set up a cell consisting of Mg I MgSO₄ (1 M) as one electrode (by dipping a magnesium wire in 1M MgSO₄, solution) and standard hydrogen electrode Pt, H₂ (1 atm) H1 (1 M) as the second electrode and measure the EMF of the cell and also note the direction of deflection in the voltmeter. The direction of deflection shows that electrons flow from magnesium electrode to hydrogen electrode, i.e., oxidation takes place on magnesium electrode and reduction on hydrogen electrode. Hence, the cell may be represented as:    

     Mg²⁺ | Mg (1 M) || H⁺ (1 M) | H₂, (1 atm), Pt

             E⁰_cell = E⁰_H+, _1/2H2 - E⁰_Mg 2+, _Mg

    Put E⁰_H+, _1/2H = 0, therefore E⁰_Mg 2+, _Mg = -E⁰_cell

 [ where M = molarity]

Q.2.2. Can you store copper sulphate solution in a zinc pot?

 Ans. Zinc is more reactive than copper. Hence, it displaces copper from copper sulphate solution as follows:

    Zn (s) + CuSO₄ (aq)→ ZnSO₄ (aq) + Cu (s) Thus, zinc reacts with CuSO₄ solution. Hence, we cannot store copper sulphate solution in a zinc pot.

Alternatively. E _Zn2+, _Zn = - 0.76 V,

  E _Ca2+, _Ca= 0.34 V To check whether zinc reacts with CuSO₄, solution, le., whether the following reaction takes place or not Zn (s) + CuSO₄ (aq) → ZnSO₄ (aq) + Cu (s), find the EMF of this cell reaction.

  The cell may be represented as: Zn | Zn²⁺|| Cu²⁺ | Ca

        E⁰_cell = E _Ca2+, _Ca- E _Zn2+, _Zn = 0.34 V-(-0.76 V) = 1.10 V

        E⁰_cell is positive, the reaction takes place and we cannot store.

Q. 2.3. Consult the table of the standard electrode potentials and suggest three substances that can oxidize ferrous ions under suitable conditions.

Ans. Oxidation of ferrous ions means that the following reaction should occur:

            Fe²⁺ → Fe³⁺ + e^-,       E⁰_ox = -0.77V

Only those substances can oxidize Fe²⁺ to Fe³ which are stronger oxidizing agents and have positive reduction potentials greater than 0.77 V so that EMF of the cell reaction is positive. This is so for elements lying below Fe³⁺/Fe²⁺ in the electrochemical series, e.g., Br₂, Cl₂ and F₂.     

Q.2.4. Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

 

Ans. For hydrogen electrode, H⁺+ e^-→ [1/2]H₂ Applying Nernst equation. E_H+, _1/2H = E⁰_H+, _1/2H - (0.0591 /n) log [1/ _H+]

                  = 0 – [0.0591/1]log [1/ ₁₀-10] 

{pH= 10 means[H⁺] = 10^-10 M} = - 0.0591 x 10 = -0.0591V

 

Q. 2.5. Calculate the emf of the cell in which the reaction

takes following place

      Ni(s)+2 Ag⁺²(0.002 M) →N⁺² (0.160M) + 2Ag(s)

 

Ans. Applying to the given cell reaction,

      E _cell = E⁰_cell – [0.059/2] log [ Ni ⁺²/Ag ⁺]

              = 1.05 – [0.059/2]log [0.160/(0.002)²]

 

Q. 2.6. The cell in which following reaction occurs –

         2 Fe ⁺³(aq)+2 I^-(aq) → 2 Fe⁺² (aq) + I₂(s) , has

          E⁰_cell = 0.236V at 298 K. calculate the standard Gibbs

      Energy and equilibrium constant of the cell.

     Ans.  ∆G⁰= -nFE⁰, = -2x96500x0.236 J

            = -45548 J = -45.548 kJ.

         K = Antilog [n E ⁰ / 0.054] = Antilog [2x0.236/0.059]

             = 10 ⁸ . 

 

Q. 2.7. Why does the conductivity of a solution decrease with dilution?

Ans-

K= Antilog (7.983) =9.616 x 10⁷ dilution, the number of ions per unit volume decreases. Hence, the conductivity decreases.

 

  Q.2.8 Suggest the method to determine the ꓥ⁰_m value

            of water.

 Ans-      ꓥ⁰_m (H₂O) = λ⁰ _H⁺ + λ⁰ _{o H}^-.

                                 = ꓥ⁰_m (HCl) + ꓥ⁰_m (NaOH) + ꓥ⁰_m (NaCl)

       

  Q.2.9. The molar conductivity of 0.025 mol L^-1. Methanoic

 acid is 46.1 S cm² mol^-1. Calculate its degree of dissociation

   constant. Given λ⁰ _H⁺ = 349.6 S cm² mol^-1.   and

     λ⁰ _HCOO^- = 54.6 S cm² mol^-1.

  Sol.- ꓥ⁰_m (_Hcoo ^-) = λ⁰ _H⁺ + λ⁰_HCOO^-.

                             = 349.6 + 54.6 = 404.2

                     α = [ ꓥ^c_m /ꓥ⁰_m] = 46.1/ 404.2= 0.114

         K_α = [ C α²]/ [1 – α] = [0.025 x0.114 x0.114]/[1 – 0.114]

                            = 3.67 x 10 ^-4.

Q. 2.10. If a current of 0.5 Ampere flows through a metallic

 wire for 2hour then how many electrons would flow

 through the wire.

 Ans- Q = ne =It, n = number of electrons, e= charge of electron

  n = It /e = [0.5 x 2 x 60x 60]/ [1.6 x 10 ^-19] = 2.25 x 10 ²².

 

 Q.2.11 Suggest a list of metals that are extracted electrolytically.

  Ans.- The metals like Na, K, Mg, Al, Ca etc.

        i.e., reactive metals Can be extracted electrolytically.

 

Q.2.12. discovery what is the quantity of electricity in coulomb's needed to reduce one mole of

                  Cr₂O^2-+ 14H → 2Cr³⁺ + 8H₂O       

Ans. 1 mol Cr₂O^2- requires 6moles electron

   For reduction.

  Required charge = 6 x 96500 C

                                  = 57900 C

Q.2.13. write the chemistry of recharging the lead storage battery, highlighting all materials that are involved during the recharging?

Ans. A lead storge battery consists of anode of lead packed

with lead dioxide PbO₂ and 38% solution acid as electrolyte

when the battery is in the use, the following reactions take

 place:

    At anode:          Pb (s) + SO₄^2- → PbSO₄(s) + 2e^-.

    At cathode:     PbO₂ (s) + SO₄^2- (aq) + 4H⁺ (aq) + 2e^- →  PbSO₄(s) + 2H₂O.

Over all reaction: Pb (s) + PbO₂ (s) + 2H₂SO₄(aq) →  2PbSO₄(s) + 2H₂O

On charging the battery, the reverse reaction takes place, i.e., PbSO₄ deposited on the electrodes is converted back into Pb and PbO₂ and H₂SO₄ is regenerated. Reverse of reaction

(i) will be reduction and hence will take place at cathode. Reverse of reaction

(ii)will be oxidation and hence will take place at anode.

 

Q. 2.14 Suggest two materials other than hydrogen that can be used as fuels in fuel cells.

Ans. Methane and methanol.

 

Q.2.15 Explain how rusting of iron is envisaged as setting up of an electrochemical cell.

 Ans. The water layer present on the surface of iron (especially in the rainy season) dissolves acidic oxides of the air like CO₂, SO₂ etc. to form acids which dissociate to give H⁺ ions

         H₂O + CO₂→ H₂CO₃ + 2H⁺ + CO^3-

         In the presence of H⁺ ions, iron starts losing electrons at some spot to form ferrous ions, i.e., its oxidation takes place. Hence, this spot acts as the anode:

           Fe (s) → Fe²⁺ (aq) + 2 e^-.

 The electrons thus released move through the metal to reach another spot where H+ ions and the dissolved oxygen take up these electrons and reduction reaction takes place. Hence, this spot acts as the cathode:

         O₂ (g) + 4 H⁺ (aq) + 4 e^- →2 H₂O (1)

The overall reaction is: 2 Fe (s) + O₂(g) + 4H⁺ (aq) → 2Fe²⁺ (aq) + 2 H₂O (1)

Thus, an electrochemical cell is set up on the surface.

Ferrous ions are further oxidized by the atmospheric oxygen to ferric ions which combine with water molecules to form hydrated ferric oxide, Fe₂O₃.x H₂O, which is rust.