Electrochemistry Question Answers: NCERT Class 12 Chemistry

The standard electrode potential of Mg²⁺| Mg can be measured with respect to the standard hydrogen electrode, represented by Pt_(s), H_2(g)(1 atm) | H⁺_(aq) (1 M).

A cell, consisting of Mg | MgSO4(aq 1 M) as the anode and the standard hydrogen electrode as the cathode, is set up.

Then, the emf of the cell is measured and this measured emf is the standard electrode potential of the magnesium electrode.

E^ø  = E^ø_R  - E^ø_L

Here,E^øR for the standard hydrogen electrode is zero.

∴ E^ø = 0 - E^ø_L

₌ - E^ø_L

Zinc is more reactive than copper. Therefore, zinc can displace copper from its salt solution. If copper sulphate solution is stored in a zinc pot, then zinc will displace copper from the copper sulphate solution.

Zn + CuSO₄  → ZnSO₄ + Cu

Hence, copper sulphate solution cannot be stored in a zinc pot.

Substances that are stronger oxidising agents than ferrous ions can oxidise ferrous ions.

Fe²⁺ → Fe³⁺ + e^-1  ;   E^ø =  - 0.77 V

This implies that the substances having higher reduction potentials than +0.77 V can oxidise ferrous ions to ferric ions.

Three substances that can do so are F_2, Cl₂, and O₂.

For hydrogen electrode, , it is given that pH = 10

∴[H+] = 10 ^{- 10}M

Now, using Nernst equation:

= - 0.0591 log 10¹⁰

= - 0.591 V

Applying Nernst equation we have:

= 1.05 - 0.02955 log 4 × 10⁴

= 1.05 - 0.02955 (log 10000 + log 4)

= 1.05 - 0.02955 (4 + 0.6021)

= 0.914 V

Here, n = 2, E^ø_cell = 0.236V,  T = 298 K

We know that:

The variation of conductivity with dilution can be explained on the basis of number of ions in solution.

In case of weak electrolyte, the number of ions furnished by an electrolyte in solution depends upon the degree of dissociation with dilution. With the increase in dilution, the degree of dissociation increases, as a result the molar conductance increases.

But for strong electrolyte, there is no increase in the number of ions with dilution because strong electrolytes are completely ionized in solution at all concentrations. However, in concentrated solutions of strong electrolytes there are strong forces of attraction between the ions of opposite charges called Interionic Forces. Due to these interionic forces the conducting ability of the ions is less in concentrated solutions. With dilution, the ions becomes far apart from each other and interionic forces decreases. As a result, molar conductance increases with dilution.

C = 0.025 mol L ^{- 1}

Now, degree of dissociation:

Thus, dissociation constant:

I = 0.5 A

t = 2 hours = 2 × 60 × 60 s = 7200 s

Thus, Q = It

= 0.5 A × 7200 s

= 3600 C

We know that 96487C = 6.023 X 10²³ number of electrons.

Then,

Hence, 2.25 X 10²²number of electrons will flow through the wire.

Metals that are on the top of the reactivity series such as sodium, potassium, calcium, lithium, magnesium, aluminium are extracted electrolytically.

The given reaction is as follows:

Cr₂ O₇^2- + 14H⁺ + 6e^- → Cr³⁺ + 8H₂O 

Therefore, to reduce 1 mole of Cr₂ O₇^2-, the required quantity of electricity will be:

= 6 F

= 6 × 96487 C

= 578922 C

A lead storage battery consists of a lead anode, a grid of lead packed with lead oxide (PbO₂) as the cathode, and a 38% solution of sulphuric acid (H₂SO₄) as an electrolyte.

When the battery is in use, the following cell reactions take place:

At anode: Pb_(s) + SO^2-_4(aq)  → PbSO_4(s) + 2e^-

At cathode:

The overall cell reaction is given by,

When a battery is charged, the reverse of all these reactions takes place.

Hence, on charging, present at the anode and cathode is converted into and respectively.

Fuel cells are voltaic cells in which the reactants are continuously supplied to the electrodes. They convert the energy from the combustion of fuels like Hydrogen, CO, Methane directly into electrical energy. Methane, oxygen, carbon monoxide and methanol can be used as fuels in fuel cells.

= 9 [ka²]

Hence, the rate of formation will increase by 9 times.

In the process of corrosion, due to the presence of air and moisture, oxidation takes place at a particular spot of an object made of iron. That spot behaves as the anode. The reaction at the anode is given by,

Fe_(s) → Fe²⁺ _(aq) + 2e^-

Electrons released at the anodic spot move through the metallic object and go to another spot of the object.

There, in the presence of H⁺ ions, the electrons reduce oxygen. This spot behaves as the cathode. These H⁺ ions come either from H₂CO_3, which are formed due to the dissolution of carbon dioxide from air into water or from the dissolution of other acidic oxides from the atmosphere in water.

The reaction corresponding at the cathode is given by,

The overall reaction is:

Also, ferrous ions are further oxidized by atmospheric oxygen to ferric ions. These ferric ions combine with moisture, present in the surroundings, to form hydrated ferric oxide  (Fe₂O₃, x)H₂O    i.e., rust.

Hence, the rusting of iron is envisaged as the setting up of an electrochemical cell.

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

Mg, Al, Zn, Fe, Cu

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

The galvanic cell in which the given reaction takes place is depicted as:

(i) Zn electrode (anode) is negatively charged.

(ii) Ions are carriers of current in the cell and in the external circuit, current will flow from silver to zinc.

(iii) The reaction taking place at the anode is given by,

Zn_(s)  →   Zn²⁺_(aq) + 2e^-

The reaction taking place at the cathode is given by,

Ag⁺_(aq)  + e^-  →  Ag_(s)

(i) E^ø _{Cr³⁺ / Cr}  = - 0.74 V

   E^ø _{Cd²⁺ / Cd}  = - 0.40 V

The galvanic cell of the given reaction is depicted as:

Now, the standard cell potential is

E^ø  = E^ø_R  - E^ø_L

     = -0.40 - (-0.74)

     = + 0.34 V

ΔrG^ø = -nFE^ø_cell

In the given equation,

n = 6

F = 96487 C mol ^{- 1}

E^ø_cell = +0.34 V

Then, ΔrG^ø  = - 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^ø = - RT ln K

= 34.496

K = antilog (34.496)

= 3.13 × 10³⁴

(ii) E^ø _{Fe³⁺ / Fe²⁺}  =  0.77 V

   E^ø _{Ag⁺ / Ag}        =  0.80 V

The galvanic cell of the given reaction is depicted as:

Now, the standard cell potential is

E^ø  = E^ø_R  - E^ø_L

       = 0.80 - 0.77

     = 0.03 L

Here, n = 1.

Then, ΔrG^ø = -nFE^ø_cell

= - 1 × 96487 C mol ^{- 1} × 0.03 V

= - 2894.61 J mol ^{- 1}

= - 2.89 kJ mol ^{- 1}

Again, ΔrG^ø = - 2.303 RT ln K

= 0.5073

K = antilog (0.5073)

= 3.2 (approximately)

(i) For the given reaction, the Nernst equation can be given as:

= 2.7 - 0.02955 = 2.67 V (approximately)

(ii) For the given reaction, the Nernst equation can be given as:

= 0.52865 V = 0.53 V (approximately)

(iii) For the given reaction, the Nernst equation can be given as:

= 0.14 - 0.0295 × log125

= 0.14 - 0.062

= 0.078 V

= 0.08 V (approximately)

(iv) For the given reaction, the Nernst equation can be given as:

E^ø  = 1.104 V

We know that,

ΔrG^ø = -nFE^ø

= - 2 × 96487 × 1.04

= - 213043.296 J

= - 213.04 kJ

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 p is resistivity, then we can write:

k = 1 / p

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., 

(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.

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

A_m = kV

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  A_m with  for strong and weak electrolytes is shown in the following plot:

Given, k = 0.0248 S cm ^{- 1}

c = 0.20 M

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

= 0.0248 x1000 / 0.20

124 Scm² mol ^{- 1}

Given,

Conductivity, k = 0.146 × 10 ^{- 3} S cm ^{- 1}

Resistance, R = 1500 Ω

Cell constant = k × R = 0.146 × 10 ^{- 3} × 1500

= 0.219 cm ^{- 1}

Given,K = 7.896 × 10 ^{- 5} S m ^{- 1}

c = 0.00241 mol L ^{- 1}

Then, molar conductivity, A_m = K/c

= 32.76S cm² mol ^{- 1}

Again, A_m^º = 390.5 S cm² mol ^{- 1}

Now,

= 0.084

Dissociation constant,

= (0.00241 mol L^-1)(0.084)²  / (1-0.084)

= 1.86 × 10 ^{- 5} mol L ^{- 1}

(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

(i) According to the question,

Electricity required to produce 40 g of calcium = 2 F

Therefore, electricity required to produce 20 g of calcium = (2x20) / 40 

= 1 F

(ii) According to the question,

Electricity required to produce 27 g of Al = 3 F

Therefore, electricity required to produce 40 g of Al = (3X40) / 27

= 4.44 F

(i) According to the question,

H₂O  → H₂ + ½ O₂

Now, we can write:

O^2- → ½ O₂ + 2e^-

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

= 2 × 96487 C

= 192974 C

(ii) According to the question,

Fe²⁺  →   Fe³⁺ + e^-1

Electricity required for the oxidation of 1 mol of FeO to Fe₂O₃

= 1 F

= 96487 C

Given,

Current = 5A

Time = 20 × 60 = 1200 s

Charge = current × time

= 5 × 1200

= 6000 C

According to the reaction,

Nickel deposited by 2 × 96487 C = 58.71 g

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

= 1.825 g

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

According to the reaction:

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

= 864/ 60 

= 14.40 min

Again,

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

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

Since E^º for the overall reaction is positive, the reaction between Fe³⁺_(aq) and I ^- _(aq) is feasible.

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

Since Eº for the overall reaction is negative, the reaction between Fe³⁺_(aq) and Br ^- _(aq) is not feasible.

Since Eº for the overall reaction is negative, the reaction between Ag_(s) and Fe³⁺_(aq) is not feasible.

Since for the overall reaction is positive, the reaction between Br_2(aq) and Fe²⁺_(aq) is feasible

(i) At cathode: The following reduction reactions compete to take place at the cathode.

The reaction with a higher value of  E^º 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.

The reaction with a higher value of E^º 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^-  → OH + e^-

4OH^- → 2H₂O +O₂

(iii) At the cathode, the following reduction reaction occurs to produce H₂ gas.

H⁺ _(aq) + e^-  → ½ H_2(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.

The reaction with a higher value of E^º 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.

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