Solutions-Class-12-Chemistry-Chapter-5-Electrochemistry-Maharashtra Board


Maharashtra Board-Class-12-Chemistry-Chapter-5


Question 1. Choose the most correct option.

(i) Two solutions have the ratio of their concentrations 0.4 and ratio of their conductivities 0.216. The ratio of their molar conductivities will be

(a) 0.54

(b) 11.574

(c) 0.0864

(d) 1.852

Answer :

(a) 0.54

Explanation :

\(\frac{K_1}{K_2}\) = 0.216, \(\frac{C_1}{C_2}\) = 0.4

We know

Λm = \(\frac{K}{C}\) × 100

Λm1 = \(\frac{K_1}{C_1}\) × 100, Λm2 = \(\frac{K_2}{C_2}\) × 100

∴ \(\frac{Λm_1}{Λm_2}\) = \(\frac{K_1/C_1}{K_2/C_2}\)

= \(\frac{K_1/K_2}{C_1/C_2}\) = \(\frac{0.216}{0.4}\) = 0.54

(ii) On diluting the solution of an electrolyte

(a) both Λ and k increase

(b) both Λ and k decrease

(c) Λ increases and k decreases

(d) Λ decreases and k increases

Answer :

(c) Λ increases and k decreases

(iii) 1 S m2 mol1 is eual to

(a) 10-4 S m2 mol1

(b) 104 Ω1 cm2 mol1

(c) 10-2 S cm2 mol1

(d) 102 Ω1 cm2 mol1

Answer :

(b) 104 Ω1 cm2 mol1

(iv) The standard potential of the cell in which the following reaction occurs

H2 (g,1atm) + Cu2+(1M) 2H+(1M) + Cu(s), (E0Cu = 0.34V) is

(a) −0.34 V

(b) 0.34 V

(c) 0.17 V

(d) −0.17 V

Answer :

(b) 0.34 V

(v) For the cell, Pb (s)|Pb2+(1M)||Ag+(1M)|Ag (s), if concentraton of an ion in the anode compartment is increased by a factor of 10, the emf of the cell will

(a) increase by 10 V

(b) increase by 0.0296 V

(c) decrease by 10 V

(d) decrease by 0.0296 V

Answer :

(d) decrease by 0.0296 V

(vi) Consider the half reactions with standard potentials

         (i) Ag+ (aq) + e Ag (s) E0 = 0.8V

         (ii) I2 (s) + 2e  2I (aq) E0 = 0.53V

         (iii) Pb2+ (aq) + 2e  Pb (s) E0 = − 0.13V

         (iv) Fe2+ (aq) + 2e  Fe (s) E0 = − 0.44V

The strongest oxidising and reducing agents respectively are

(a) Ag and Fe2+

(b) Ag+ and Fe

(c) Pb2+ and I

(d) I2 and Fe2+

Answer :

(b) Ag+ and Fe

(vii) For the reaction

Ni(s) + Cu2+ (1M) Ni2+(1M) + Cu (s),

E0 cell = 0.57V ΔG0 of the reaction is

(a) 110 kJ

(b) −110 kJ

(c) 55 kJ

(d) −55 kJ

Answer :

(b) 110 kJ

(viii) Which of the following is not correct?

(a) Gibbs energy is an extensive property

(b) Electrode potential or cell potential is an intensive property.

(c) Electrical work = − ΔG

(d) If half reaction is multiplied by a numerical factor, the corresponding E0 value is also multiplied by the same factor.

Answer :

(d) If half reaction is multiplied by a numerical factor, the corresponding E0 value is also multiplied by the same factor.

(ix) The oxidation reaction that takes place in lead storage battery during discharge is

(a) Pb2+(aq) + SO42(aq) PbSO4(s)

(b) PbSO4(s) + 2H2O(l) PbO2(s) + 4H+(aq) + SO42(aq) + 2e

(c) Pb(s) + SO42(aq) PbSO4(s) + 2e

(d) PbSO4(s) + 2e Pb(s) + SO42(aq)

Answer :

(c) Pb(s) + SO42(aq) → PbSO4(s) + 2e

(x) Which of the following expressions represent molar conductivity of Al2(SO4)3?

(a) 3 λ0 Al3+ + 2 λ0 SO42

(b) 2 λ0 Al3+ + 3 λ0 SO42

(c) 1/3 λ0 Al3+ + 1/2 λ0 SO42

(d) λ0 Al3+ + λ0 SO42

Answer :

(b) 2 λ0 Al3+ + 3 λ0 SO42

Question 2. Answer the following in one or two sentences.

(i) What is a cell constant ?

Answer :

For a given cell, the ratio of separation (l) between the two electrodes divided by the area of cross-section (a) of the electrode is called the cell constant.

Cell constant = b = \(\frac{l}{a}\) cm1

In SI units it is expressed as m1.

(ii) Write the relationship between conductivity and molar conductivity and hence unit of molar conductivity.

Answer :

If k is conductivity and Λm is molar conductivity then,

Λm = \(\frac{K}{C}\) x 1000

The SI units of k are S m-1 and that of c are mol m3.

Hence, the SI units of Λ is S m2 mol1

(iii) Write the electrode reactions during electrolysis of molten KCl.

Answer :

KCl(molten) → K+(l) + Cl+(l)

Reaction at cathode :

2K+(l) + 2e → 2K(s) Reduction

Reaction at anode :

2Cl(l) → 2Cl(g) + 2e

2Cl(g) → Cl2(g)


Overall reaction : 2K+(l) + 2Cl(l)  →  2K(s) + Cl2(g)

(iv) Write any two functions of salt bridge.

Answer :

  • It provides electrical contact between two solutions and thereby completes the electrical circuit.
  • It prevents the mixing of two solutions.
  • It maintains electrical neutrality in both the solutions by the transfer of ions.

(v) What is standard cell potential for the reaction

3Ni(s) + 2Al3+(1M) 3Ni2+(1M) + 2Al(s) if E0Ni = − 0.25 V and E0Al = −1.66V?

Answer :

Given: E0Ni = − 0.25 V, E0Al = −1.66V, Standard cell potential E0cell = ?

Electrode reactions are :

At anode: Al(s) → Al3+(aq) + 3e

At cathode: Ni2+(ag) + 2e → Ni(s)

The standard electrode potential is given by

E0cell = E0cathode − E0anode

∴ E0cell = − 0.25 − (−1.66)

          = − 0.25 + 1.66

          = 1.41 V

The standard cell potential for the reaction is 1.41 V.

(vi) Write Nerst equation. What part of it represents the correction factor for nonstandard state conditions ?

Answer :

(a) Nernst equation for cell potential is,

Ecell = E0cell \(-\frac{2.303RT}{nF}log_{10}\frac{[products]}{[reactants]}\)

(b) The part of equation namely, \(\frac{2.303RT}{nF}log_{10}\frac{[products]}{[reactants]}\) represents the correction factor for nonstandard state conditions.

(vii) Under what conditions the cell potential is called standard cell potential ?

Answer :

The cell potential measured under the standard conditions is called standard cell potential. The standard conditions chosen are 1 M concentration of a solution, 1 atm pressure for gases, solids and liquids in pure form and 25C.

(viii) Formulate a cell from the following electrode reactions :

Au3+(aq) + 3e Au(s)

Mg(s) Mg2+(aq) + 2e

Answer :

An electrochemical cell from above electrode reactions is,

Au3+(aq) + 3e → Au(s)     (Cathode reduction reaction)

Mg(s) → Mg2+(aq) + 2e—   (Anode oxidation reaction)

Cell formula :

Mg(s) | Mg2+(aq) || Au3+(aq) | Au(s)

(ix) How many electrons would have a total charge of 1 coulomb ?

Answer :

Given : 1 Faraday = charge on 1 mol of electrons = 6.022 x 1023 electrons and

1 Faraday = 96500 C

∴  96500 C = 6.022 x 1023 electrons

∴ 1 C ≡ \(\frac{6.022×10^{23}}{96500}\) = 6.24 x 1018 electrons

Number of electrons = 6.24 x 1018.

(x) What is the significance of the single vertical line and double vertical line in the formulation galvanic cell.

Answer :

  • A single vertical line placed between two phases in the galvanic cell represents the phase boundary. It indicates direct contact between them.
  • A double vertical line placed between two solutions indicates that they are connected by salt bridge.

Question 3. Answer the following in brief

(i) Explain the effect of dilution of solution on conductivity ?

Answer :

  • The conductivity of a solution is determined by the presence of ions, with more ions resulting in higher conductance. The specific conductance of an electrolytic solution is the conductance per unit volume.
  • As the concentration of the electrolyte decreases or the dilution increases, the conductivity of the solution decreases. This is because the concentration of the solution decreases, resulting in a decrease in the number of current-carrying ions per unit volume. Molar conductivity increases with dilution.

(ii) What is a salt bridge ?

Answer :

A salt bridge is a U-shaped glass tube containing a saturated solution of a strong electrolyte, like KCl, NH4NO3, Na2SO4 in a solidified agar-agar gel.

A hot saturated solution of these electrolytes in 5% agar solution is filled in the U-shaped tube and allowed it to cool and solidify forming a gel.

It is used to connect two half cells or electrodes forming a galvanic or voltaic cell.

(iii) Write electrode reactions for the electrolysis of aqueous NaCl.

Answer :

Reduction half-reaction at cathode:

At cathode, two reduction reactions compete.

(i) Reduction of sodium ions.

Na+(aq) + e → Na(s), E0 = 2.71V

(ii) Reduction of water to hydrogen gas.

2H2O(l) + 2e → H2(g) + 2OH(aq), E0 = 0.83 V

The standard potential for the reduction of water is higher than that for the reduction of Na+. Hence, water has a much greater tendency to get reduced than the Na+ ion. Therefore, reduction of water is the cathode reaction when the aqueous NaCl is electrolysed.

Oxidation half-reaction at anode:

At anode, there will be competition between oxidation of Cl ion to Cl2 gas as in the case of molten NaCl and the oxidation of water to O2 gas.

(i) Oxidation of Cl ions to chlorine gas

2Cl(aq)  → Cl2(g) + 2e, E0oxd = 1.36 V

(ii) Oxidation of water to oxygen gas.

2H2O(l) → O2(g) + 4H+(aq) + 2e, E0oxd = 0.4 V

The standard electrode potential for the oxidation of water is greater than that of Cl ion or water has a greater tendency to undergo oxidation. Hence, an anode half-reaction would be oxidation of water. However, experiments have shown that the gas produced at the anode is Cl2 and not O2. This suggests that anode reaction is oxidation of Cl to Cl2 gas. This is because of overvoltage.

Net cell reaction:

The net cell reaction is the sum of two electrode reactions.

2Cl(aq) → Cl2(g) + 2e    (Oxidation half reaction at anode)

2H2O(l) + 2e → H2(g) + 2OH(aq)  (Reduction half reaction at cathode)


2Cl(aq) + 2H2O(l) → Cl2(g) + H2(g) + 2OH(aq) (Overall cell reaction)

Since Na+ and OH are left in the solution, they form NaOH(aq)

(iv) How many moles of electrons are passed when 0.8 ampere current is passed for 1 hour through molten CaCl2 ?

Answer :

Given : I= 0.8A; t = 1 x 60 x 60 = 3600 s

Number of moles of electrons = ?

Quantity of electricity passed = I(A) x t(s) = 0.8 × 3600 = 2880 C

1 Faraday = 1 mol electrons

1 Faraday 96500 C

∴ 96500 C = 1 mol electrons

∴ 2880 C = \(\frac{2880}{96500}\) = 0.02984 ≈ 0.03 mol electrons

No. of moles of electrons passed through molten CaCl2 = 0.03

(v) Construct a galvanic cell from the electrodes Co3+| Co and Mn2+| Mn.

E0Co = 1.82 V, E0Mn = − 1.18V. Calculate E0cell.

Answer :

Given: E0Co = 1.82 V, E0Mn = − 1.18V

To find:  E0cell and cell representation

Electrode reactions are

At anode: 3(Mn(s) → Mn(aq) +2e)

At cathode: 2(Co(aq) + 3e → Co(s))

The galvanic cell is,

Mn(s) | M2+(1M) || Co3+(1M)| Co(s)

E0cell = E0cathode − E0anode

       = 1.82 − (−1.18)

       = 1.82 + 1.18

       = 3.0 V

E0cell = 3.0 V.

(vi) Using the relationship between ΔG0 of cell reaction and the standard potential associated with it, how will you show that the electrical potential is an intensive property ?

Answer :

(a) For an electrochemical cell involving n number of electrons in the overall cell reaction,

ΔGº = − nFE0cell

where ΔGº is standard Gibbs energy change and E0cell is a standard cell potential.

(b) ∴ E0cell = ΔGº/−nF

Since ΔGº changes according to number of moles of electrons involved in the cell reaction, the ratio, ΔGº/-nF remains constant.

(c) Therefore E0cell is independent of the amount of substance and it represents the intensive property.

(vii) Not given in Text Book

(viii) Derive the relationship between standard cell potential and equilibrium constant of cell reaction.

Answer :

The equilibrium constant is related to the standard free energy change ΔGº, as follows,

ΔGº = − RT ln K

If E0cell is the standard cell potential (or emf) of the galvanic cell, then

ΔGº = − nFE0cell

By comparing above equations,

ΔGº = − nFE0cell = − RT ln K

∴ nFE0cell = RT ln K

∴ E0cell = \(\frac{RT}{nF}\)ln K

∴ E0cell = \(\frac{2.303RT}{nF}\)log10 K

At 25°C,

\(\frac{2.303RT}{F}\) = \(\frac{2.303×8.314×298}{96500}\) = 0.0592

∴ E0cell = \(\frac{0.0592}{n}\)log10 K

This is a relation between equilibrium constant and E0cell

(ix) It is impossible to measure the potential of a single electrode. Comment.

Answer :

  • According to Nernst theory, electrode potential is the potential difference between the metal and ionic layer around it at equilibrium, i.e. the potential across the electric double layer.
  • For measuring the single electrode potential, one part of the double layer, that is metallic layer can be connected to the potentiometer but not the ionic layer. Hence, single electrode potential can't be measured experimentally.

(x) Why do the cell potential of lead accumulators decrease when it generates electricity ? How the cell potential can be increased ?

Answer :

  • The cell potential depends on sulphuric acid concentration (density). As the cell operates to generate current, H2SO4 is consumed. Its concentration (density) decreases and the cell potential is decreased.
  • During the recharging process by applying external potential slightly greater than 2 V, H2SO4 is regenerated. As a result, its concentration (density) increases and in turn, the cell potential increases.

(xi) Write the electrode reactions and net cell reaction in NICAD battery.

Answer :

Reactions in the cell :

(i) Oxidation at cadmium anode :

Cd(s) + 2OH(aq) → Cd(OH)2(s) + 2e

(ii) Reduction at NiO2(S) cathode :

NiO2(s) + 2H2O(l) + 2e → Ni(OH)2(S) + 2OH(aq)

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

Cd(s) + NiO2(s) + 2H2O(l)  → Cd(OH)2(s) + Ni(OH)2(S)

Question 4. Answer the following :

(i) What is Kohrausch law of independent migration of ions? How is it useful in obtaining molar conductivity at zero concentration of a weak electrolyte ? Explain with an example.

Answer :

(a) Kohlrausch's law : This states that at infinite dilution of the solution, each ion of an electrolyte migrates independently of its co-ions and contributes independently to the total molar conductivity of the electrolyte, irrespective of the nature of other ions present in the solution.

(b) Both cation and anion contribute to molar conductivity of the electrolyte at zero concentration and thus A0 is the sum of molar conductivity of cation and that of the anion at zero concentration.

Thus, Λ0 = n+λ0+ + n λ0

where λ+ and λ+ are molar conductivities of cation and anion, respectively, n+ and n_ are the number of moles of cation and anion specified in the chemical formula of the electrolyte.

(c) Determination of molar conductivity of weak electrolyte at zero concentration:

The theory is particularly useful in calculating A0 values of weak electrolytes from those of strong electrolytes.

For example, Λ0 of acetic acid can be calculated by knowing those of HCl, NaCl and CH3COONa as described below:

Because Λ0 values of strong electrolytes, HCl, CH3COONa and NaCl, can be determined by extrapolation method, the ^o of acetic acid can be obtained.

(ii) Explain electrolysis of molten NaCl.

Answer :

Construction of cell : The electrolytic cell consists of a container in which fused NaCl is placed. Two graphite electrodes are immersed in it. They are connected by metallic wires to a source of direct current that is battery. This is shown in Fig.

Reactions occurring in the cell : Fused NaCl contains Na+ and Cl ions which are freely mobile. When potential is applied, cathode attracts Na+ ions and anode attracts Cl ions.

As these are charged particles, their migration results in an electric current. When these ions reach the respective electrodes, they are discharged according to the following reactions.

Oxidation half reaction at anode :

Cl ions migrate to anode. Each Cl ion, that reaches anode, gives one electron

to anode. It oxidises to neutral Cl atom in the primary process. Two Cl atoms then combine to form chlorine gas in the secondary process.

2Cl(l) → Cl(g) + Cl (g) + 2e(primary process)

Cl(g) + Cl(g) → Cl2(g)               (secondary process)


2Cl(l) → Cl2(g) + 2e—   (overall oxidation)

The battery sucks electrons so produced at the anode and pushes them to cathode

through a wire in an external circuit. The battery thus serves as an electron pump. The electrons from the battery enter into solution through cathode and leave the solution through anode.

Reduction half reaction at cathode : The electrons supplied by the battery are used in cathodic reduction. Each Na+ ion, that reaches cathode accepts an electron from the cathode and reduces to metallic sodium.

Na+(l) + e → Na(l)

Net cell reaction

The net cell reaction is the sum of two electrode reactions.

2Cl(l) → Cl2(g) + 2e   (oxidation half reaction)

2Na+(l) + 2e → 2Na(l)   (reduction half reaction)


2 Na+(l) + 2 Cl(l) → 2 Na(l) + Cl2(g)   (overall cell reaction)

 Results of electrolysis of molten NaCl :

  • A pale green Cl2 gas is released at anode.
  • A molten silvery-white sodium is formed at cathode.

Decomposition of NaCl into metallic sodium and Cl2(g) is nonspontaneous. The electrical energy supplied by the battery forces the reaction to occur.

(iii) What current strength in amperes will be required to produce 2.4g of Cu from CuSO4 solution in 1 hour ? Molar mass of Cu = 63.5 g mol-1.

Answer :


Mass of Cu = 2.4 g,

Molar mass of Cu = 63.5 g mol1

1 hours = 1 x 60 x 60s = 3600s

To find: Current strength (I) (in amperes)

Cu2+(aq) + 2e → Cu(s)

Mole ratio = \(\frac{mole\,of\,Cu}{mole\,of\,electron}\) =\(\frac{1}{2}\)

WCu = \(\frac{I×t}{96500}\) x mole ratio x Molar mass of Cu

∴ I = \(\frac{96500×W_{Cu}}{tx mole\,ratio × Molar\,mass\,of\,Cu}\)

     = \(\frac{96500×2.4}{3600×\frac{1}{2}×63.5}\) = 2.026 A

Required current strength = I = 2.026 A.

(iv) Equilibrium constant of the reaction, 2Cu+(aq) Cu2+(aq) + Cu(s) is 1.2 × 106. What is the standard potential of the cell in which the reaction takes place ?

Answer :

Given: Equilibrium constant of the reaction (K) = 1.2 x 106.

To find: Standard potential of cell (E0cell)

For the given reaction, n = 1.

E0cell = \(\frac{0.0592}{n}\)log10 K

∴ E0cell = \(\frac{0.0592}{1}\)log10 1.2 x 106

        = 0.0592 × (6.079) = 0.36 V

The standard cell potential of cell is 0.36 V.

(v) Calculate emf of the cell Zn(s) | Zn2+(0.2M) || H+(1.6M) | H2 (g 1.8 atm) | Pt at 250C.

Answer :

Given :

Zn(s) | Zn2+(0.2M) || H+(1.6M) | H2 (g 1.8 atm) | Pt

EZn2+/Zn = — 0.763 V; [Zn2+] = 0.2 M; [H+] = 1.6 M; PH2 = 1.8 atm; Ecell =?

Cell Reaction

Zn(s) → Zn2+(aq) + 2e    (oxidation at anode)

2H+(aq) + 2e → H2(g)     (reduction at cathode)


Zn(s) + 2H+(aq)(1.6M) → Zn2+(aq)(0.02 M) + H2(g) (1.8 atm)    (overall reaction)

∴ n = 2

Ecell = E0cell − \(\frac{0.0592}{n}log_{10}\frac{[products]}{[reactants]}\)

Ecell = E0cell − \(\frac{0.0592}{n}log_{10}\frac{[Zn^{2+}][H_2]}{[H^+]^2}\)

E0cell = EH+/H2 − EZn2+/Zn

       = 0.0 − (− 0.763)

       = +0.763 V

∴ Ecell = 0.763 − \(\frac{0.0592}{2}log_{10}\frac{(0.2)(1.8)}{(1.6)^2}\)

           = 0.763 − 0.0296 x log10 0.1406

            = 0.763 − 0.0296 × (− 0.8521)

            = 0.763 + 0.02522

             = 0.7882 V

Answer :. Ecell  = 0.7882 V.

(vi) Calculate emf of the following cell at 250C.

Zn(s) | Zn2+(0.08M) || Cr3+(0.1M) |Cr E0Zn = - 0.76 V, E0Cr = - 0.74 V

Answer :

Given :

Zn(s) | Zn2+(0.08M) || Cr3+(0.1M) | Cr

E0Zn = - 0.76 V, E0Cr = - 0.74 V

Cell reaction :

[Zn(s) → Zn2+(0.08M) + 2e] x 3 (oxidation at anode)

[Cr3+(0.1) + 3e→ Cr(s)] x 2 (reduction at cathode)


3Zn(s) + 2Cr3+(0.1) → 3Zn2+(0.08M) + 2Cr(s) (n=6) (overall reaction)

E0cell = ECr3+/Cr − EZn2+/Zn

       = − 0.74 − (−0.76)

       = 0.02 V

Ecell = E0cell  − \(\frac{0.0592}{n}log_{10}\frac{[Zn^{2+}]^3}{[CR^{3+}]^2}\)

∴ Ecell = 0.02 − \(\frac{0.0592}{6}log_{10}\frac{(0.08)^3}{(0.1)^2}\)

         = 0.02 − 9.867 x 10-3 x .7903 

        = 0.02 − 9.867 x 10-3 x (−1.2907)

         = 0.02 + 0.01273

         = 0.03273 V

Answer : Ecell = 0.03273 V.

(vii) What is a cell constant ? What are its units? How is it determined experimentally ?

Answer :

(a) For a given cell, the ratio of separation (l) between the two electrodes divided by the area of cross-section (a) of the electrode is called the cell constant.

Cell constant = b = \(\frac{l}{a}\) cm-1

(b) In SI units it is expressed as m-1.

(c) The cell constant is determined using the 1 M, 0.1 M or 0.01 M KCl solutions. The conductivity of KCl solution is well tabulated at various temperatures.

The resistance of KCl solution is measured by Wheatstone bridge as shown in the figure.

AB is the uniform wire. Rx is the variable known resistance placed in one arm of Wheatstone bridge. The conductivity cell containing KCl solution of unknown resistance is placed in the other arm of Wheatstone bridge.

D is a current detector. F is the sliding contact that moves along AB. A.C. represents the source of alternating current.

The sliding contact is moved along AB until no current flows. The detector D shows no deflection. The null point is, thus, obtained at C.

According to Wheatstone bridge principle,


Hence R(solution) = \frac{l(AC)}{l(BC)}\) × Rx

By measuring lengths AC and BC and knowing Rx , resistance of KCl solution can be calculated.

The cell constant is given by

Cell constant = kKCl X R(solution)

The conductivity of KCl solution is known. The cell constant, thus, can be calculated.

(viii) How will you calculate the moles of electrons passed and mass of the substance produced during electrolysis of a salt solution using reaction stoichiometry.

Answer :

Calculation of moles of electrons passed : The charge carried by one mole of electrons is referred to as one faraday (F). If total charge passed is Q C,

then moles of electrons passed = \(\frac{Q(C)}{F(C\,mole)}\) mol.

Calculation of mass of product : Mass, W of product formed is given by,

W = moles of product x molar mass of product (M)

   =  \(\frac{Q}{96500}\) x mole ratio x M

   = \(\frac{I×t}{96500}\)  x mole ratio x M

(ix) Write the electrode reactions when lead storage cell generates electricity. What are the anode and cathode and the electrode reactions during its recharging?

Answer :

1) Cell reactions when lead storage cell generates electricity (discharging):

(a) Oxidation at anode (-):

When the cell provides current, spongy lead is oxidised to Pb2+ ions and negative charge accumulates on lead plates. The Pb2+ ions so formed combine with SO4 ions from H2SO4 to form insoluble PbSO4. The net oxidation is the sum of these two processes.

Pb(s) → Pb2+(aq) + 2e(oxidation)

Pb2+(aq) + SO42(aq) → PbSO4(S) (precipitation)

Pb(s) + SO42(aq) → PbSO4(S) + 2e..... (1) (overall oxidation)

(b) Reduction at cathode (+):

The electrons produced at the anode travel through external circuit and re-enter the cell at the cathode. At cathode, PbO2 is reduced to Pb2+ ions in presence of H+ ions. Pb2+ ions formed combine with SO4 ions from H2SO4 to form insoluble PbSO4 that gets coated on the electrode.

PbO2(s) + 4H+(aq) + 2e → Pb2+(aq) + 2H2O(l)                (reduction)

Pb(s) + SO42(aq) → PbSO4(S)                                                          (precipitation)


PbO2(S) + 4H+(aq) + SO42(aq) + 2e→ PbSO4(S) + 2H2O(l) (overall reduction)

2) Recharging of the cell : When the discharged battery is connected to external electric source and a higher external potential is applied the cell reaction gets reversed generating H2SO4.

Reduction at the — ve electrode or cathode :

PbSO4(s) + 2e → Pb(s) + SO42(aq)

Oxidation at the +ve electrode or anode :

PbSO4(s) + 2H2O(l) → PbO2(s) + 4H+(aq) + SO42(aq) + 2e

The net reaction during charging is

2PbSO4(S) + 2H2O(l) → Pb(s) + PbO2(S) + 4H+(aq) + 2SO42(aq)


2PbSO4(S) + 2H2O(l) → Pb(s) + PbO2(S) + 2H2SO4(aq)

The emf of the accumulator depends only on the concentration of H2SO4.

(x) What are anode and cathode of H2-O2 fuel cell ? Name the electrolyte used in it. Write electrode reactions and net cell reaction taking place in the fuel cell.

Answer :

In fuel cell the anode and cathode are porous electrodes with suitable catalyst like finely divided platinum.

The electrolyte used is hot aqueous KOH solution in which porous anode and cathode carbon rods are immersed.

H2 is continuously bubbled through anode while O2 gas is bubbled through cathode.

Working (cell reactions) :

(i) Oxidation at anode : At anode, hydrogen gas is oxidised to H2O.

2H2(g) + 40H(aq) → 4H2O(l) + 4e—  (oxidation half reaction)

(ii) Reduction at cathode : The electrons released at anode travel to cathode through external circuit and reduce oxygen gas to OH.

O2(g) + 2H2O(l) + 4e → 4OH(aq) (reduction half reaction)

(iii) Net cell reaction : Addition of both the above reactions at anode and cathode gives a net cell reaction.

2H2(g) + O2(g) → 2H2O(l) (overall cell reaction)

(xi) What are anode and cathode for Leclanche' dry cell ? Write electrode reactions and overall cell reaction when it generates electricity.

Answer :

A dry cell has zinc vessel as anode and graphite rod as cathode and moist paste of ZnCl2, MnO2, NH4Cl as electrolytes.

At anode :

Zn(s) → Zn2+(0.08M) + 2e  (Oxidation half reaction)

At graphite (c) cathode :

2NH+4(e) + 2e  → 2NH3(aq) + H2(g)  (Reduction half reaction)

2MnO2(s) + H2  → Mn2O3(s) + H2O(l)

There is a side reaction inside the cell, between Zn2+ ions and aqueous NH3.

Zn2+(aq) + 4NH3(aq) → [Zn(NH3)4] 2+(aq)

(xii) Identify oxidising agents and arrange them in order of increasing strength under standard state conditions. The standard potentials are given in parenthesis.

Al (-1.66V), Al3+(-1.66V), Cl2 (1.36V), Cd2+(-0.4V), Fe(-0.44V), I2(0.54V), Br (1.09V).

Answer :

The oxidising agents are I2, Br and Cl2. The increasing strength is

I2(0.54V) < Br (1.09V) < Cl2 (1.36V)

[Note : Actually Br2 acts as an oxidising agent but not Br. ]

(xiii) Which of the following species are reducing agents? Arrange them in order of increasing strength under standard state conditions. The standard potentials are given in parenthesis.

K (-2.93V), Br2(1.09V), Mg(-2.36V), Ce3+(1.61V), Ti2+(-0.37V), Ag+(0.8 V), Ni (-0.23V).

Answer :

Lower the standard reduction potential, higher is reducing power. The reducing agents are Ni, Mg and K. Their increasing strength is,

Ni (-0.23V) < Ti2+(-0.37V) < Mg(-2.36V) < K (-2.93V),

[Note : Cations don't act as reducing agent since they are already in oxidised state.]

(xiv) Predict whether the following reactions would occur spontaneously under standard state conditions.

(a) Ca(s) + Cd2+(aq) Ca2+(aq) + Cd(s)

(b) 2Br(s) + Sn2+(aq) Br2(l) + Sn(s)

(c) 2Ag(s) + Ni2+(aq) 2Ag+(aq) + Ni(s) (use information of Table 5.1)

Answer :

(a) E0Ca = — 2.866 V and E0Cd = — 0.403 V

Since, E0Ca < E0Cd Ca will reduce Cd2+(aq) and reaction would occur.

(b) E0Br = +1.080 V and E0Sn = —0.136

 Since E0Sn < E0Br, reaction will not occur.

(Possible reaction is, Br2(l) + Sn(S) → 2Br(aq) + Sn2+(aq)

(c) E0Ag = + 0.799 v and E0Ni = —0.257 v

Since E0Ag > E0Ni, the reaction would not occur.

(Possible reaction is, 2Ag+(aq) + Ni(s) → 2Ag(s) + Ni2+(aq))


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