Class 12 - Chemistry - Electrochemistry

Question 3.1.

Arrange the following metals in the order in which they displace each other from the solution of their salts.

Al, Cu, Fe, Mg and Zn

Answer:

The following is the order in which the given metals displace each other from the solution of their salts.

Mg, Al, Zn, Fe, Cu

 

Question 3.2.

Given the standard electrode potentials,

K+/K = –2.93V, Ag+/Ag = 0.80V,

Hg2+/Hg = 0.79V Mg2+/Mg = –2.37 V, Cr3+/Cr = – 0.74V

Arrange these metals in their increasing order of reducing power?

Answer:

The lower the reduction potential, the higher is the reducing power.

The given standard electrode potentials increase in the order of

K+/K < Mg2+/Mg < Cr3+/Cr < Hg2+/Hg < Ag+/Ag

Hence, the reducing power of the given metals increases in the following order:

Ag < Hg < Cr < Mg < K

 

Question 3.3.

Depict the galvanic cell in which the reaction

Zn(s) +2Ag+ (aq) →Zn2+ (aq) +2Ag(s) take place. Further show:

(i) Which of the electrode is negatively charged?

(ii) The carriers of the current in the cell.

(iii) Individual reaction at each electrode.

Answer:

The galvanic cell corresponding to the given redox reaction can be represented as:

Zn | Zn2+ (aq) || Ag+ (aq) | Ag

(i)      Zn electrode is negatively charged because at this electrode, Zn oxidizes to Zn2+

and the leaving electrons accumulate on this electrode.

(ii)     Ions are the carriers of current in the cell.

(iii)    The reaction taking place at Zn electrode can be represented as:

Zn(s) à Zn2+ (aq) + 2e-

And the reaction taking place at Ag electrode can be represented as:

Ag+ (aq) + e- àAg(s)

(iv) In aqueous solutions, CuCl2 ionizes to give Cu2+ and Cl ions as:

CuCl2 (aq) à Cu2+ (aq) +2Cl-(aq)

On electrolysis, either of Cu2+ ions or H2O molecules can get reduced at the cathode.

But the reduction potential of Cu2+ is more than that of H2O molecules.

Cu2+ (aq) + 2e- à Cu (aq); E0= +0.34V

H2O (l) +2e- àH2 (g) +2OH- ; E0=-0.83 V

 

Hence, Cu2+ ions are reduced at the cathode and get deposited.

Similarly, at the anode, either of Clor H2O is oxidized. The oxidation potential of H2O is higher than that of Cl.

2Cl-(aq) à Cl2 (g) + 2e- ; E0= -1.36V

2H2O (l) àO2 (g) +4H- + 4e-; E0=-1.23 V

But oxidation of H2O molecules occurs at a lower electrode potential than that of Cl ions because of over-voltage

(extra voltage required to liberate gas). As a result, Cl ions are oxidized at the anode to liberate Cl2 gas.

  

Question 3.4.

Calculate the standard cell potentials of galvanic cell in which the following reactions take place:

(i) 2Cr(s) + 3Cd2+ (aq) → 2Cr3+ (aq) + 3Cd

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

Calculate the ΔrG (-) and equilibrium constant of the reactions

Answer:

  • E0 Cr3+ / Cr = - 0.74 V

E0 Cd2+ / Cd = - 0.40 V

The galvanic cell of the given reaction is depicted as:

Cr(s) |Cr3+ (aq) ||Cd2+ (aq) |Cd(s)

 

Now, the standard cell potential is

Ecell (-) = (E R (-) – EL (-))

= (0.40 – (-0.74))

=+0.34 V

ΔrG (-) = -nFE cell (-)

In the given equation, n =6

F = 96487 C mol−1

Ecell (-) =+0.34 V

Then ΔrG (-) = (−6 × 96487 C mol−1 × 0.34 V)

= −196833.48 CV mol−1

= −196833.48 J mol−1

= −196.83 kJ mol−1

Again,

ΔrG (-) = -RT ln K

  • ΔrG (-) =-2.303 Rt ln K
  • log K = - (ΔrG (-) /2.303 RT)

=-(196.83 x103)/ (2.303 x 8.314 x298)

= 34.496    

 K = antilog (34.496) = 3.13 × 1034

 

  • E0 Fe3+ /Fe2+ =  77 V

E0 Ag+ / Ag = 0.80 V

The galvanic cell of the given reaction is depicted as:

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

        Now, the standard cell potential is

        Ecell (-) = (E R (-) – EL (-))

                  = (0.80-0.77)

                 = 0.03 V

        Here, n = 1.

        Then, ΔrG (-) = -nFE cell (-)

       = (−1 × 96487 C mol−1 × 0.03 V)

      = -2894.61 Jmol-1

     = -2.89 KJ mol-1

Again, ΔrG (-) =-2.303 Rt ln K

  • log K = - (ΔrG (-) /2.303 RT)

=-(2.894.61)/ (2.303 x 8.314 x298)

         = 0.5073

Therefore, K = antilog (0.5073)

= 3.2 (approximately)

 

Question 3.5.

Write the Nernst equation and emf of the following cells at 298 K:

  • Mg(s) |Mg2+(0.001M)||Cu2+(0.0001 M)|Cu(s).
  • Fe(s)|Fe2+(0.001M)||H+(1M)|H2(g)(1bar)| Pt(s)
  • Sn(s)|Sn2+(0.050 M)||H+(0.020 M)|H2(g) (1 bar)|Pt(s)
  • Pt(s)|Br2(l)|Br(0.010 M)||H+(0.030 M)| H2(g) (1 bar)|Pt(s).

Answer:

  • For the given reaction, the Nernst equation can be given as:

Ecell = E cell (-) – (0.0591/n) log [Mg2+]/ [Cu2+]

= {0.34 – (-2.36)} – (0.0591/2) log (.001/.0001)

=2.7 – (0.0591/2) log10

= 2.7 − 0.02955

= 2.67 V (approximately)

  • For the given reaction, the Nernst equation can be given as:

  Ecell = E cell (-) – (0.0591/n) log [Fe2+]/ [H+] 2

= {0-(0.44)} – (0.0591/2) log [0.001]/ [12]

= (0.44 – 0.02955) (-3)

=0.52865 V

=0.53 V (approx.)

  • For the given reaction, the Nernst equation can be given as:

               Ecell = E cell (-) – (0.0591/n) log [Sn2+]/ [H+] 2

             = {0-(0.14)} – (0.0591/2) log (0.050)/ (0.020)2

 

= 0.14 − 0.0295 × log125

= 0.14 − 0.062

= 0.078 V

= 0.08 V (approximately)

  • For the given reaction, the Nernst equation can be given as:

Ecell = E cell (-) – (0.0591/n) log [1]/ [Br-] 2/ [H+] 2

= (0- 1.09) – (0.0591/2) log (1)/ [(0.010)2(0.030)]

=-1.09 -0.02955 x log (1/0.0000009)

=-1.09 – 0.02955 x log (1/9 x 10-8)

=-1.09 -0.02955(0.0453 +7)

=-1.09 – 0.208

=-1.298V

 

 

Question 3.6.

In the button cells widely used in watches and other devices the following reaction takes place:

Zn(s) + Ag2O(s) + H2O (l) → Zn2+ (aq) + 2Ag(s) + 2OH (aq)

Determine ΔrG (-) and E (-) for the reaction.

Answer:

Given - Zn → Zn2 + + 2e - , E0 = 0.76V (anode)

Ag2O + H2O + 2e- →2Ag + 2OH- , E0 = 0.344V (cathode), n = 2

Zn is oxidized and Ag2O is reduced.

Hence, the standard cell potential, E0cell is given as,

E0cell = E0R - E0L

E0cell = (0.344 + 0.76)

Therefore, E0cell = 1.104 V

To calculate the standard Gibb’s free energy, ∆rG0, we use,

rG0 = - nE0F (Equation 1)

= - (2×96487×1.104) J

= - 213043.296 J

Therefore, ∆rG0 = - 2.13×105 J

 

Question 3.7.

Define conductivity and molar conductivity for the solution of an electrolyte.

Discuss their variation with concentration?

Answer:

Conductivity of a solution is defined as the conductance of a solution of 1 cm in length and area of cross-section 1 sq. cm.

The inverse of resistivity is called conductivity or specific conductance. It is represented by the symbol k. If ρ is resistivity, then we can write:

k = (1 / ρ)

The conductivity of a solution at any given concentration is the conductance (G) of one unit volume of solution kept between

two platinum electrodes with the unit area of cross-section and at a distance of unit length.

i.e., G= К (a/l)

Or К .1 = К

(Since a = 1, l = 1)

Conductivity always decreases with a decrease in concentration, both for weak and strong electrolytes.

This is because the number of ions per unit volume that carry the current in a solution decreases with a decrease in concentration.

Molar conductivity:

Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing

1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length.

Ʌm = (К)(A/l)

Now, l = 1 and A = V (volume containing 1 mole of the electrolyte).

Ʌm = (К)(V)

Molar conductivity increases with a decrease in concentration.

This is because the total volume V of the solution containing one mole of the electrolyte increases on dilution.

The variation of Ʌm with √c for strong and weak electrolytes is shown in the following plot:

        Class_12_Chemistry_Electrochemistry_Molar_Conductivity                                                                                                                                                                           

Question 3.8.

The conductivity of 0.20 M solution of KCl at 298 K is 0.0248S cm–1? Calculate its molar conductivity.

Answer:

Given, k = 0.0248 S cm- 1

c = 0.20 M

Molar conductivity, Ʌm = (k x 1000) / (c)

= (0.0248 x1000) / (0.20)

=124 Scm2mol- 1

 

Question 3.9.

The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω.

What is the cell constant if conductivity of 0.001M KCl solution at 298

K is 0.146 × 10–3 S cm–1.

Answer:

Conductivity, κ = 0.146 × 10−3 S cm−1

Resistance, R = 1500 Ω

Cell constant = (κ × R)

= (0.146 × 10−3 × 1500)

= 0.219 cm−1

 

Question 3.10.

The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:

Concentration/M     0.001 0.010 0.020 0.050 0.100

102 × κ/S m–1           1.237 11.85 23.15 55.53 106.74

Calculate Λm for all concentrations and draw a plot between Λm and c½. Find the value of Λ0m.

Answer:

κ = 1.237 × 10−2 S m−1, c = 0.001 M

Then, κ = 1.237 × 10−4 S cm−1, c½ = 0.0316 M1/2

Therefore, Λm = (κ/c)

= [(123.7 x 10-4 Scm-1)/ (0.001 molL-1)] x (1000cm3/L)

 = [(123.7 Scm2mol−1

Given,

κ = 11.85 × 10−2 S m−1, c = 0.010M

Then, κ = 11.85 × 10−4 S cm−1, c½ = 0.1 M1/2

Therefore, Λm = (κ/c)

= [(11.85 x 10-4 S cm-1)/ (0.010 mol L-1)] x (1000 cm3/L)

=118.5Scm2 mol-1

Given,

κ = 55.53 × 10−2 S m−1, c = 0.050 M

Then, κ = 55.53 × 10−4 S cm−1, c1/2 = 0.2236 M1/2

Therefore, κ = (κ/c)

= [(55.53 x 10-4 S cm-1)/ (0.050 mol L-1)] (1000 cm3/L)

= 111.11 S cm2 mol−1

Given,

κ = 106.74 × 10−2 S m−1, c = 0.100 M

Then, κ = 106.74 × 10−4 S cm−1,     c1/2 = 0.3162 M1/2

Therefore, Λm = (κ/c)

  = [(106.74 x10-4) Scm-1/ (0.100) mol L-1]/ [1000 cm3/L)

  = 106.74 S cm2 mol-1

Now the following data,

Since the line interrupts Λm at 124.0 S cm2mol-1, Λm0 =124.0 S cm2mol-1

Class_12_Chemistry_Electrochemistry_Molar_Conductivity_1 

 

Question 3.11.

Conductivity of 0.00241 M acetic acid is 7.896 × 10–5 S cm–1. Calculate its molar conductivity and if Λ0m for acetic acid is 390.5Scm2 mol–1 .

What is its dissociation constant?

Answer:

Given, κ = 7.896 × 10−5 S m−1

c = 0.00241 mol L−1

Then, molar conductivity, Λm = (κ/c)

= (7.896 x10-5 Scm-1)/ (0.00241 molL-1)] (1000 cm3/L)

= 32.76S cm2 mol−1

Λm0 =390.5 Scm2 mol-1

Again,   = 390.5 S cm2 mol-1

α =( Λm/ Λm0)

= (32.76 Scm2 mol-1)/ (390.5 Scm2 mol-1)

= 0.084

Dissociation constant, Ka =(cα2/1- α)

= (0.00241 molL-1) (0.084)2]/ (1-0.084)

=1.86 x 10-5 mol L-1

 

Question 3.12.

How much charge is required for the following reductions:-

(i) 1 mol of Al3+ to Al.

(ii) 1 mol of Cu2+ to Cu.

(iii) 1 mol of MnO4– to Mn2+

Answer:

(i) Al3+ + 3e- → Al

Required charge = 3 F

= (3 × 96487) C = 289461 C

(ii) Cu2+ + 2e- → Cu

Required charge = 2 F

= (2 × 96487) C

= 192974 C

(iii) MnO4- → Mn2+

i.e. Mn7+ +5e- → Mn2+

Required charge

= 5 F

= (5 × 96487) C

= 482435 C

 

 Question 3.13.

How much electricity in terms of Faraday is required to produce:-

(i) 20.0 g of Ca from molten CaCl2

(ii) 40.0 g of Al from molten Al2O3

Answer:

According to the given problem,

Ca2+ + 2e-1 à Ca (40 g)

Electricity required to produce 40 g of calcium = 2 F

Therefore, electricity required to produce 20 g of calcium

= [(2 x 20)/ (40)] F

= 1 F

(ii)     According to the given problem,

Al3+ + 3e- à Al (27g)

Electricity required to produce 27 g of Al = 3 F

Therefore, electricity required to produce 40 g of Al

= [(3x40)/ (27)] F

= 4.44 F

 

 

Question 3.14.

How much electricity is required in coulomb for the oxidation of

(i) 1 mol of H2O to O2.

(ii) 1 mol of FeO to Fe2O3.

Answer:

  • According to the given values,

H2O → H2 + ½ O2

Now, we can write:

O2- → ½ O2 + 2e-

Electricity required for the oxidation of 1 mol of H2O to O2 = 2 F

= (2 × 96487) C

= 192974 C

  • According to the question,

FeO + (1/2) O2 à Fe2O3

                i.e. 2Fe2+ →   3Fe3+ + e-1

For the oxidation of 2 moles of FeO, required charge =2F

For the oxidation of 1 mol of FeO to Fe2O3 = 1 F

= 96487 C

 

 

 

Question 3.15.

A solution of Ni (NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes.

What mass of Ni is deposited at the cathode?

Answer:

Current = 5A

Time = (20 × 60) = 1200 s

Charge = (current) × (time)

= (5 × 1200)

= 6000 C

According to the reaction,

Ni2+ (aq) + 2e- à Ni(s) (58.7 g)

Nickel deposited by (2 × 96487) C = 58.71 g

Therefore, nickel deposited by 6000 C = (58.71 x 6000)/ (2x 96487) g

= 1.825 g

Hence, 1.825 g of nickel will be deposited at the cathode.

 

 

Question 3.16.

Three electrolytic cells A, B, C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series.

A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B.

How long did the current flow? What mass of copper and zinc were deposited?

 

Answer:

According to the reaction:

Ag+ (aq) + e-à Ag(s) (108g)

i.e., 108 g of Ag is deposited by 96487 C.

Therefore, 1.45 g of Ag is deposited by = (96487 X 1.45) / (108) C

= 1295.43 C

Given,

Current = 1.5 A

Time = (1295.43) / (1.5s)

= 863.6 s

= 864 s

= 14.40 min

Again,

Cu2+ (aq) + 2e- à Cu(s) (63.5 g)

i.e., (2 × 96487) C of charge deposit = 63.5 g of Cu

Therefore, 1295.43 C of charge will deposit = [(63.5x1295.43) / (2x96487)] g

= 0.426 g of Cu

Zn2+ (aq) + 2e- à Zn(s) (65.4 g)

i.e., (2 × 96487) C of charge deposit = 65.4 g of Zn

Therefore, 1295.43 C of charge will deposit

= (65.4x1295.43) / (2x96487) g

= 0.439 g of Zn

 

 

Question 3.17.

Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:

  • Fe3+(aq) and I(aq)
  • Ag+ (aq) and Cu(s)
  • Fe3+ (aq) and Br (aq)
  • Ag(s) and Fe 3+ (aq)
  • Br2 (aq) and Fe2+ (aq).

 Answer:

  • Fe3+ (aq) + e- à Fe2+ (aq)] x2;    E0 =+0.77 V

                               2I- à I2(s) + 2e-; E0 =-0.54 V

               ----------------------------------------------------------------------

                2Fe3+ (aq) +2I-(aq) à 2Fe2+ (aq) + I2(s); E0 = +0.23 V

Since E0 for the overall reaction is positive, the reaction between Fe3+ (aq) and I (aq) is feasible.

  • Ag+(aq) + e- à Ag(s) ] x2 ; E0 = +0.80V

Cu(s)  à Cu2+ (aq) +2e- ; E0 = -0.34 V

2Ag+ (aq) + Cu(s)  à 2Ag(s) + Cu2+ (aq); E0 = +0.46

Since for the overall reaction is positive, the reaction between Ag+ (aq) and Cu(s) is feasible.

  • Fe3+ (aq) + e- à Fe2+ (aq)] x2; E0 =+0.77 V

                      2Br- à Br2(s) + 2e-; E0 =-1.09 V

                  --------------------------------------------------------

          2Fe3+ (aq) +2Br-(aq) à 2Fe2+ (aq) + Br2(s); E0 = -0.32 V

 Since E0 for the overall reaction is positive, the reaction between Fe3+ (aq) and Br (aq) is feasible.

  • Ag (s) à Ag+(aq) + e-; E0 = -0.80V

Fe3+ (aq) + e- à Fe2+ (aq);    E0 =+0.77 V

 

Ag(s) + Fe3+ (aq) à Ag+ (aq) +Fe2+ (aq);    E0 =-0.03 V

Since E for the overall reaction is negative, the reaction between Ag (s) and Fe3+ (aq) is not feasible.

  • 2Br- + 2e- à Br2-; E0 =+1.09 V

2Fe3+ (aq) à Fe3+ (aq) + e-] x 2; E0 = -0.77V

Since for the overall reaction is positive, the reaction between Br2 (aq) and Fe2+ (aq) is feasible.

 

 

Question 3.18.

Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO3 with silver electrodes.

(ii) An aqueous solution of AgNO3with platinum electrodes.

(iii) A dilute solution of H2SO4 with platinum electrodes.

(iv)  An aqueous solution of CuCl2 with platinum electrodes.

Answer:

(i) At cathode:

The following reduction reactions compete to take place at the cathode.

Ag+ (aq) + e- à Ag(s); E0 = +0.80V

H+ (aq) + e- à (1/2) H2 (g); E0= 0.00V

The reaction with a higher value of E0 takes place at the cathode. Therefore, deposition of silver will take place at the cathode.

At anode:

The Ag anode is attacked by NO3- ions. Therefore, the silver electrode at the anode dissolves in the solution to form Ag+.

(ii) At cathode:

The following reduction reactions compete to take place at the cathode.

Ag+ (aq) + e- à Ag(s); E0 = +0.80V

H+ (aq) + e- à (1/2) H2 (g); E0= 0.00V

The reaction with a higher value of E0 takes place at the cathode. Therefore, deposition of silver will take place at the cathode.

At anode:

Since Pt electrodes are inert, the anode is not attacked by NO3- ions. Therefore, OH or NO3- ions can be oxidized at the anode.

But OH ions having a lower discharge potential and get preference and decompose to liberate O2.

OH- à OH + e-

4OH- à H2O + O2

  • At the cathode, the following reduction reaction occurs to produce H2

H+ (aq) + e- à (1/2) H2 (g)

At the anode, the following processes are possible.

For dilute sulphuric acid, reaction (i) is preferred to produce O2 gas. But for concentrated sulphuric acid, reaction (ii) occurs.

(iv) At cathode:

The following reduction reactions compete to take place at the cathode.

Cu2+ (aq) + 2e- à Cu(s); E0 = +0.34V

H+ (aq) + e- à (1/2) H2 (g); E0= 0.00V

The reaction with a higher value of E0 takes place at the cathode. Therefore, deposition of copper will take place at the cathode.

At anode:

The following oxidation reactions are possible at the anode.

Cl-(aq) à (1/2) Cl2 (g) +e-1; E0 = 1.36 V

2H2O (l) à O2 (g) + 4H+ (aq) + 4e- ; E0 =+1.23 V

At the anode, the reaction with a lower value of E0 is preferred. But due to the over-potential of oxygen, Cl gets oxidized at the anode to produce Cl2 gas.

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