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

(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

(c) Λ increases and k decreases

(b) 10⁴ Ω^—1 cm² mol^—1

(b) 0.34 V

(d) decrease by 0.0296 V

(b) Ag⁺ and Fe

(b) −110 kJ

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

(c) Pb(s) + SO₄^2—(aq) → PbSO₄(s) + 2e^—

(b) 2 λ⁰ _Al3+ + 3 λ⁰ SO₄^2—

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

In SI units it is expressed as m⁻¹.

If k is conductivity and Λ_m is molar conductivity then,

Λ_m = \(\frac{K}{C}\) x 1000

The SI units of k are S m^-1 and that of c are mol m⁻³.

Hence, the SI units of Λ is S m² mol⁻¹

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) → Cl_2(g)

_{-----------------------------------------------}

Overall reaction : 2K⁺_(l) + 2Cl^—_(l)  →  2K_(s) + Cl_2(g)

• 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.

Given: E⁰_Ni = − 0.25 V, E⁰_Al = −1.66V, Standard cell potential E⁰_cell = ?

Electrode reactions are :

At anode: Al_(s) → Al³⁺_(aq) + 3e^—

At cathode: Ni²⁺_(ag) + 2e^— → Ni_(s)

The standard electrode potential is given by

E⁰_cell = E⁰_cathode − E⁰_anode

∴ E⁰_cell = − 0.25 − (−1.66)

          = − 0.25 + 1.66

          = 1.41 V

The standard cell potential for the reaction is 1.41 V.

(a) Nernst equation for cell potential is,

E_cell = E⁰_cell \(-\frac{2.303RT}{nF}log_{10}\frac{[products]}{[reactants]}\)

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

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.

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)

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.

• 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.

• 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.

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.

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^— → H_2(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)  → Cl_2(g) + 2e^—, E⁰_oxd = − 1.36 V

(ii) Oxidation of water to oxygen gas.

2H₂O_(l) → O_2(g) + 4H⁺_(aq) + 2e^—, E⁰_oxd = − 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) → Cl_2(g) + 2e^—    (Oxidation half reaction at anode)

2H₂O_(l) + 2e^— → H_2(g) + 2OH^—_(aq)  (Reduction half reaction at cathode)

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

2Cl_(aq) + 2H₂O_(l) → Cl_2(g) + H_2(g) + 2OH^—_(aq) (Overall cell reaction)

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

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

Given: E⁰_Co = 1.82 V, E⁰_Mn = − 1.18V

To find:  E⁰_cell 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)

E⁰_cell = E⁰_cathode − E⁰_anode

       = 1.82 − (−1.18)

       = 1.82 + 1.18

       = 3.0 V

E⁰_cell = 3.0 V.

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

ΔGº = − nFE⁰_cell

where ΔGº is standard Gibbs energy change and E⁰_cell is a standard cell potential.

(b) ∴ E⁰_cell = 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 E⁰_cell is independent of the amount of substance and it represents the intensive property.

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

ΔGº = − RT ln K

If E⁰_cell is the standard cell potential (or emf) of the galvanic cell, then

ΔGº = − nFE⁰_cell

By comparing above equations,

ΔGº = − nFE⁰_cell = − RT ln K

∴ nFE⁰_cell = RT ln K

∴ E⁰_cell = \(\frac{RT}{nF}\)ln K

∴ E⁰_cell = \(\frac{2.303RT}{nF}\)log₁₀ K

At 25°C,

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

∴ E⁰_cell = \(\frac{0.0592}{n}\)log₁₀ K

This is a relation between equilibrium constant and E⁰_cell

• 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.

• 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.

Reactions in the cell :

(i) Oxidation at cadmium anode :

Cd_(s) + 2OH^—_(aq) → Cd(OH)_2(s) + 2e^—

(ii) Reduction at NiO_2(S) cathode :

NiO_2(s) + 2H₂O_(l) + 2e^— → Ni(OH)_2(S) + 2OH^—_(aq)

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

Cd_(s) + NiO_2(s) + 2H₂O_(l)  → Cd(OH)_2(s) + Ni(OH)_2(S)

(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.

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) → Cl_2(g)               (secondary process)

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

2Cl^—_(l) → Cl_2(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) → Cl_2(g) + 2e^—   (oxidation half reaction)

2Na⁺_(l) + 2e^— → 2Na_(l)   (reduction half reaction)

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

2 Na⁺_(l) + 2 Cl^—_(l) → 2 Na_(l) + Cl_2(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 Cl_2(g) is nonspontaneous. The electrical energy supplied by the battery forces the reaction to occur.

Given:

Mass of Cu = 2.4 g,

Molar mass of Cu = 63.5 g mol⁻¹

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

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

Cu²⁺_(aq) + 2e^— → Cu_(s)

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

W_Cu = \(\frac{I×t}{96500}\) 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.

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

To find: Standard potential of cell (E⁰_cell)

For the given reaction, n = 1.

E⁰_cell = \(\frac{0.0592}{n}\)log₁₀ K

∴ E⁰_cell = \(\frac{0.0592}{1}\)log₁₀ 1.2 x 10⁶

        = 0.0592 × (6.079) = 0.36 V

The standard cell potential of cell is 0.36 V.

Given :

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

E_Zn2+/Zn = — 0.763 V; [Zn²⁺] = 0.2 M; [H⁺] = 1.6 M; P_H2 = 1.8 atm; E_cell =?

Cell Reaction

Zn_(s) → Zn²⁺_(aq) + 2e^—    (oxidation at anode)

2H⁺_(aq) + 2e^— → H_2(g)     (reduction at cathode)

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

Zn_(s) + 2H⁺_(aq)(1.6M) → Zn²⁺_(aq)(0.02 M) + H_2(g) (1.8 atm)    (overall reaction)

∴ n = 2

E_cell = E⁰_cell − \(\frac{0.0592}{n}log_{10}\frac{[products]}{[reactants]}\)

E_cell = E⁰_cell − \(\frac{0.0592}{n}log_{10}\frac{[Zn^{2+}][H_2]}{[H^+]^2}\)

E⁰_cell = E_H+/H2 − E_Zn2+/Zn

       = 0.0 − (− 0.763)

       = +0.763 V

∴ E_cell = 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 :. E_cell  = 0.7882 V.

Given :

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

E⁰_Zn = - 0.76 V, E⁰_Cr = - 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)

E⁰_cell = E_Cr3+/Cr − E_Zn2+/Zn

       = − 0.74 − (−0.76)

       = 0.02 V

E_cell = E⁰_cell  − \(\frac{0.0592}{n}log_{10}\frac{[Zn^{2+}]^3}{[CR^{3+}]^2}\)

∴ E_cell = 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 : E_cell = 0.03273 V.

(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,

\(\frac{R_{solution}}{l(AC)}=\frac{R_x}{l(BC)}\)

Hence R_(solution) = \frac{l(AC)}{l(BC)}\) × R_x

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

The cell constant is given by

Cell constant = k_KCl X R_(solution)

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

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

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) + SO₄^2—_(aq) → PbSO_4(S) (precipitation)

Pb_(s) + SO₄^2—_(aq) → PbSO_4(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.

PbO_2(s) + 4H⁺_(aq) + 2e^— → Pb²⁺_(aq) + 2H₂O_(l)                (reduction)

Pb_(s) + SO₄^2—_(aq) → PbSO_4(S)                                                          (precipitation)

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

PbO_2(S) + 4H⁺_(aq) + SO₄^2—_(aq) + 2e^—→ PbSO_4(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 :

PbSO_4(s) + 2e^— → Pb_(s) + SO₄^2—_(aq)

Oxidation at the +ve electrode or anode :

PbSO_4(s) + 2H₂O_(l) → PbO_2(s) + 4H⁺_(aq) + SO₄^2—_(aq) + 2e^—

The net reaction during charging is

2PbSO_4(S) + 2H₂O_(l) → Pb_(s) + PbO_2(S) + 4H⁺_(aq) + 2SO₄^2—_(aq)

OR

2PbSO_4(S) + 2H₂O_(l) → Pb_(s) + PbO_2(S) + 2H₂SO_4(aq)

The emf of the accumulator depends only on the concentration of H₂SO_4.

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.

2H_2(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^—.

O_2(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.

2H_2(g) + O_2(g) → 2H₂O_(l) (overall cell reaction)

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^—  → 2NH_3(aq) + H_2(g)  (Reduction half reaction)

2MnO_2(s) + H₂  → Mn₂O_3(s) + H₂O_(l)

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

Zn²⁺_(aq) + 4NH_3(aq) → [Zn(NH₃)₄] ²⁺_(aq)

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^—. ]

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.]

(a) E⁰_Ca = — 2.866 V and E⁰_Cd = — 0.403 V

Since, E⁰_Ca < E⁰_Cd Ca will reduce Cd²⁺_(aq) and reaction would occur.

(b) E⁰_Br = +1.080 V and E⁰_Sn = —0.136

 Since E⁰_Sn < E⁰_Br, reaction will not occur.

(Possible reaction is, Br_2(l) + Sn_(S) → 2Br^—_(aq) + Sn²⁺_(aq)

(c) E⁰_Ag = + 0.799 v and E⁰_Ni = —0.257 v

Since E⁰_Ag > E⁰_Ni, the reaction would not occur.

(Possible reaction is, 2Ag⁺(aq) + Ni_(s) → 2Ag_(s) + Ni²⁺_(aq))