NCERT Solutions for Class 12 Chemistry Chapter 2 Electrochemistry (2025-26)

NCERT Solutions for Class 12 Chemistry Chapter 2 Electrochemistry address the specific challenges of Redox Reactions and Electrochemical Cells. Because the CBSE Class 12th class creates the academic foundation for higher technical education, mastering this chapter is a prerequisite for entry into premier engineering or medical colleges. These Class 12 chemistry electrochemistry NCERT solutions, prepared by subject matter experts, offer a strategic boost to board exam preparation through precision and accuracy.

NCERT Solutions for Class 12 Electrochemistry Download PDF

Electrochemistry serves as the essential bridge between chemical energy and electrical work. In this NCERT Solutions for Class 12 electrochemistry notes chapter 2 overview, the field is defined as the study of chemical changes caused by the passage of an electric current. This branch of science drives modern technology, placing electrochemistry solutions at the heart of energy storage for mobile devices, Electric Vehicles (EVs), and load-leveling systems for renewable energy conversion.

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The ncert solutions class 12 chemistry provided here are built for quick revision and accuracy. Whether Students needs Chemistry Class 12 electrochemistry ncert solutions for daily homework or ncert solutions of electrochemistry class 12 for pre-exam prep, this guide has you covered. Students can download the electrochemistry class 12 ncert solutions pdf for offline access to ensure you are always ready for your chemistry exam.

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Electrochemistry NCERT Chapter 2 Solutions Explanation Steps

In electrochemistry class 12 solutions, the primary focus is the movement of electrons from one element to another. This movement occurs during a redox or oxidation-reduction reaction. When students approach electrochemistry class 12 chemistry solutions, they must follow a logical explanation path: first identifying the anode and cathode based on standard reduction potentials, then formulating the half-cell equations, and finally calculating the net cell EMF. Understanding terms like resistivity, specific conductivity, and molar conductivity is essential for mastering the quantitative aspects of these ncert solutions electrochemistry.

1. How would you determine the standard electrode potential of the system Mg²⁺|Mg?

Ans: A cell with Mg / MgSO₄(1M) as one electrode and standard hydrogen electrode Pt,H₂(1 atm)H⁺(1M) as the second electrode will be set up, and the emf of the cell will be measured along with the direction of deflection in the voltmeter. The direction of deflection indicates that e^- move from the magnesium electrode to the hydrogen electrode, implying that oxidation occurs on the magnesium electrode and reduction occurs on the hydrogen electrode. As a result, the cell can be represented in the following way:

Mg |Mg²⁺(1M) ‖ H⁺(1M)| H₂(1 atm)Pt

E^o_cell = E^o_H⁺/½H2 – E^oMg²⁺ / Mg

put E^o_H⁺/½H2 = 0

∴ E^o_Mg²⁺/Mg = –E^o_cell

2. Can you store copper sulphate solutions in a zinc pot?

Ans: Cu is less reactive than Zn, so Cu is easily shifted from CuSO₄ solution in the following reaction:

Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)

If we express it in relations of emf, it will be like:

Ecell = Ecu²⁺ / Cu⁺ – E^- zn²⁺ / Zn

= 0.34 V – (–0.76 V) = 1.10 V

As, The positive Ecell° value indicates the occurrence of spontaneous reactions. if CuSO₄ is stored in Zn pot, Zn will react will Cu and originality of CuSO₄ will not be maintained.

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

Ans: In the process of oxidation of Fe²⁺ to Fe³⁺, i.e., Fe²⁺ → Fe³⁺ + e^-; the reduction potential value is negative in nature i.e. E^oox = –0.77 V. Only compounds with powerful oxidizing agents and positive reduction potentials larger than 0.77 V may oxidize Fe²⁺ to Fe³⁺, resulting in a positive emf of the cell reaction. This is true for elements like halogens Br₂, Cl₂ and F₂ in the series below Fe³⁺ / Fe²⁺.

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

Ans: For hydrogen electrode, H⁺ + e^- → 1/2H₂

Given: pH = 10

Using Nernst equation,

E_H⁺/½H2 = E^o_H⁺/½H2 – (0.0591/n) log (1/[H⁺])

0 – (0.0591/1) log (1/10^-10)

{pH = 10}
{[H⁺] = 10^-10M} = -0.519×10

= 0.591 V

Hence, the potential of hydrogen electrode in contact with a solution whose pH is 10 is 0.591 V.

5. Calculate the emf of the cell in which the following reaction takes place: Ni_(a) + 2Ag⁺(0.002M) → Ni²⁺(0.160M) + 2Ag_(s) Given that E = (coll)1.05 V.

Ans: Using Nernst equation,

E_cell = E^o_cell - (0.0591/n) log ([Ni²⁺]/[Ag⁺]²)

= 1.05 V – (0.0519/2) log (0.160/(0.002)²)

= 1.05 – (0.0591/2) log(log 4×10⁴)

= 1.05 – (0.0591/2)(4.6021)

= 1.05 – 0.14 V = 0.91 V

Therefore, the emf of the cell is 0.91 V.

6. 0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Ans: 2Fe³⁺(zg) + 2I^-(zq) → 2Fe²⁺(zg) + I₂(g)

For the given cell, the number of transacted electrons i.e. n = 2

Using the given formula;

Δ_rG^o = –FE^o_cell

= –2 × 96500 × 0.236

= –45.55 kJ mol^-1

Also, Δ_rG^o = –2.303RT log K_C

⇒ log K_c = (–Δ_rG^o)/(2.303RT) = (–45.55)/(2.303×8.314×10^-3×298) = 7.983

⇒ K_c = antilog(7.983)

= 9.616×10⁷

7. Why does the conductivity of solution decrease with dilution?

Ans: The conductance of ions contained in a unit volume of solution is called conductivity. The number of ions per unit volume decreases with increase in dilution. As a result, the conductivity drops.

8. Suggest a way to determine the value Λ° of water.

Ans: By using Kohlrausch's law for H₂O, we can calculate Λ_m^°

The Kohlrausch Law says that when dissociation is complete at infinite dilution, each ion adds a definite amount to the electrolyte's equivalent conductance, regardless of the nature of the ion with which it is involved, and the value of equivalent conductance at infinite dilution for any electrolyte is the sum of contributions of its ionic species (cations and anions).

Λ_m^° = Λ_m^°(Hcl) + Λ_m^°(NaOH) – Λ_m^°(NaCl)

The Λ° values of HCl, NaOH and NaCl are known as they are strong electrolytes and dissociates completely. By putting their values in the above equation, we can have value of Λ_m^° for H₂O.

9. The molar conductivity of 0.025 mol L^-1 methanoic acid is 46.1 S cm² mol^-1. Calculate its degree of dissociation and dissociation constant Given λ⁰(H⁺) = 349.6 cm² and λ⁰(HCOO^-) = 54.6 cm² mol^-1

Ans:

Λ^o_m(HCOOH) = λ^o(H⁺) + λ^o(HCOO^-)

= 349.6 + 54.6

= 404.2 S cm² mol^-1

Λ^c_m = 46.1 S cm² mol^-1

∴ α = (Λ^c_m)/(Λ^o_m) = 46.1/404.2 = 0.114

HCOOH ⇌ HCOO^- + H⁺

Initial conc c 0 0

At equi, c(1–∝) c ∝ c ∝

∴ K_a = (cα·cα)/(c(1–α)) = (cα²)/(1–α)

= (0.025×(0.114)²)/(1–0.114) = 3.67×10^-4

10. If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?

Ans: As from the formula,

Q = It

= 0.5×(2×60×60) = 3600C

1F ⇒ 96500C ⇒ 1 mole of e^-1s ∴ 3600C is equivalent to the flow of e^-1s

∴ 6.02×10²³ e^-1s

= (6.02×10²³/96500) × 3600 = 2.246×10²² e^-1s

11. Suggest a list of metals that are extracted electrolytically.

Ans: Na, Ca, Mg, and Al

12. Consider the reaction: CrO₇^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O. What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr₂O₇^2-

Ans: For the above reaction to take place, 1 mol of Cr₂O₇^2- will require 6 F = 6×96500 = 579000C of electricity.

Hence, 579000C of electricity are required for reduction of Cr₂O₇^2- to Cr³⁺

13. Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.

Ans: A lead storage battery has a lead anode, a lead cathode with lead dioxide packed (PbO₂) in it, and a 38% H₂SO₄ solution as the electrolyte. When the battery is in operation, the following reactions occur:

At Anode: Pb_(s) + SO₄^2- → PbSO_4(s) + 2e^-

At cathode: PbO_2(s) + SO₄^2-(ag) + 4H⁺_(aq) + 2e^- → PbSO_4(s) + 2H₂O_(l)

Complete reaction: Pb_(s) + PbO_2(s) + 2H₂SO_4(aq) → 2PbSO_4(s) + 2H₂O_(l)

The reverse process occurs when the battery is charged, in which PbSO₄ deposited on the electrodes is transformed back to Pb and PbO₂, and H₂SO₄ is restored.

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

Ans: Methane and Methanol.

15. Explain how rusting of iron is envisaged as setting up of an electro chemical cell.

Ans: The Water present on the surface of iron dissolves acidic oxides of air like CO₂, SO₂ etc, to form acids which dissociate to give H⁺ ions:

H₂O + CO₂ → H₂CO₃ ⇌ 2H⁺ + CO₃^2-

In the presence of H⁺, iron Fe²⁺ releases e^-1 to form Fe³⁺.

Hence, this acts as anode:

Fe_(s) → Fe²⁺_(aq) + 2e^-

The e^-1 lost then travel through the metal to reach place where these electrons are utilized by H⁺ ions and dissolved oxygen and reduction takes place. Thus acts as cathode:

O_2(g) + 4H⁺_(aq) + 4e^- → 2H₂O_(g)

The overall reaction is:

2Fe_(s) + O_2(g) + 4H⁺_(aq) → 2Fe²⁺_(aq) + 2H₂O_(l)

As a result, an electrochemical cell is established on the surface. The atmospheric oxygen oxidizes ferrous ions to ferric ions, which mix with water to produce hydrated ferric oxide., Fe₂O₃.XH₂O which is known as rust.

NCERT Solutions for Class 12 Electrochemistry Exercise

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

Ans: Mg, Al, Zn, Fe, Cu, Ag

2. Given the standard electrode potentials, K⁺ / K = –2.93 V, Ag⁺ / Ag = 0.80 V

Hg²⁺ / Hg = 0.79 V

Mg²⁺ / Mg = –2.73 V, Cl³⁺ / Cr = 0.74 V

Arrange these metals in their increasing order of reducing power.

Ans: The larger the oxidation potential, the easier it is to oxidize it, and hence the higher the reducing power. As a result, the reducing power of the elements will be in the following ascending sequence: Ag < Hg < Cl < Mg < K.

3. Depict the galvanic cell in which the reaction, 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s) takes place. Further show:

Ans: The set-up will be similar to the one seen below.

[Diagram shows: Ag(s) and Zn(s) electrodes with Ag⁺(aq) and Zn²⁺(aq) solutions connected by a salt bridge and voltmeter V]

(i) Which of the electrode is negatively charged?

Ans: The anode, or zinc electrode, will have a negative charge.

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

Ans: In the external circuit, current will travel from silver to copper.

(iii) Individual reaction at each electrode.

Ans: At anode: Zn_(n) → Zn²⁺_(q) + 2e^-

At cathode: 2Ag⁺_(zq) + 2e^- → 2Ag_(i)

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

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

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

Calculate the Δ_rG^o and equilibrium constant of the reactions

Given

E^o_Cr³⁺,Cr = –0.74 V

E^o_Cr²⁺,Cd = –0.40 V

E^o_Ag⁺,Ag = 0.80 V

E^o_{Fe³⁺,Fe²⁺} = 0.77 V

Ans:

E^ocell = E^ocathode – E^oAnode

= –0.40 V – (–0.74 V) = +0.34 V

Δ_rG^o = –nFE^o_cell

= –6 × 96500 Cmol^-1 × 0.34 V

= –196860 CVmol^-1

= –196860 J mol^-1

= –196.86 kJ mol^-1

–Δ_rG^o = 2.303×8.314×298 log K

196860 = 2.303 × 8.314 × 298 log K

or log K = 34.5014

K = Antilog 34.5014 = 3.172×10³⁴

E^ocell = +0.80 V – 0.77 V = +0.03 V

Δ_rG^o = –nFE^o_cell

= –1 × 96500 CVmol^-1 × (0.03 V)

= –2.895 CVmol^-1 = –2895 J mol^-1

= –2.895 kJ mol^-1

Δ_rG^o = –2.303RT log K

–2895 = –2.303×8.314×298× log K

or log K = 0.5074

or K = Antilog(0.5074) = 3.22

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

(i) Mg_(s)| Mg²⁺(0.001M) || Cu²⁺(0.0001M)Cu_(s)

Ans: Cell reaction:

Mg + Cu²⁺ → Mg²⁺ + Cu(n = 2)

Applying Nernst equation:

E_cell = ECell° - (0.0591/2) log ([Mg²⁺]/[Cu²⁺])

= 1.05 V – (0.0519/2) log (10^-3/10^-4)

= 2.71 – 0.02955 = 2.68 V

(ii) Fe_(s)| Fe²⁺(0.001M) || H⁺(1M)|H_2(g)(1bar)Pt_(S)

Ans: Cell reaction:

Fe + 2H⁺ → Fe²⁺ + H₂(n = 2)

Applying Nernst equation

E_cell = E^o_cell - (0.0591/2) log ([Fe²⁺]/[H⁺]²)

E_cell = 0 - (- 0.44) - (0.0591/2) log (10^-3/(1)²)

= 0.44 – (0.0591/2) × (–3)

= 0.44 + 0.0887 = 0.5287 V

(iii) Sn_(s)|Sn²⁺(0.050M) || H⁺(0.020M) | H_2(g)(1bar)Pt_(S)

Ans: Cell reaction:

Sn + 2H⁺ → Sn²⁺ + H₂(n = 2)

Applying Nernst equation:

E_cell = Ecell - (0.0591/2) log ([sn²⁺]/[H⁺]²)

E_cell = Ecell - (0.0591/2) log (0.05/(0.02)²)

= 0 – (–0.14) – (0.0591/2) log (0.05/(0.02)²)

= 0.14 – (0.0591/2) log 125

= 0.14 – (0.0591/2)(2.0969) = 0.078 V

(iv) Pt_(s)|Br_2(I)|Br^-(0.010M) || H⁺(0.030M)|H_2(g)(1bar)Pt_(S)

Ans: Cell reaction:

2Br^- + 2H⁺ → Br₂ + H₂(n = 2)

Applying Nernst equation:

E_cell = Ecell° - (0.0591/2) log (1/[Br^-]²[H⁺]²)

= (0 – 1.08) – (0.0591/2) log (1/(0.01)²(0.03)²)

= –1.08 – (0.0591/2) log(1.111×10⁷)

= –1.08 – (0.0591/2)(7.0457)

= –1.08 – 0.208 = –1.288 V

As a result, oxidation will take place at the hydrogen electrode, whereas reduction will take place at the Br₂ electrode.

6. In the button cells widely used in watches and other devices the following reaction takes places:

Determine Δ_rG^o and E^o for the reaction.

Given

Zn → Zn²⁺ + 2e^-, E^o = 0.76 V

Ag₂O + H₂O + 2e^- → 2Ag + 2OH^-, E^o = 0.344v

Ans: In the given reaction, Zn is getting oxidized and Ag₂O is being reduced.

E^ocell = 0.344 + 0.76 = 1.104 V

ΔG^o = –nFE^ocell = –2 × 96500 × 1.104 J

ΔG^o = –2.13×10⁵ J

7. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

Ans: The conductivity of a solution is defined as the conductance of a solution with a length of 1 cm and a cross-sectional area of 1 s cm.

When the electrodes are one cm apart and the area of cross-section of the electrodes is large enough that the entire solution is contained between them, the molar conductivity of a solution at a dilution(V) is the conductance of all the ions produced from one mole of the electrolyte dissolved in the solution Vcm³. Λ_m is the most common symbol for conductivity.

The conductivity of a solution (for both strong and weak electrolytes) diminishes as the concentration of the electrolyte drops, i.e. as dilution occurs. This is owing to the fact that when the solution is diluted, the number of ions per unit volume of the solution decreases.

On dilution, the molar conductivity of a solution increases as with decrease in concentration of the electrolyte. This is owing to the fact that when the solution is diluted, the number of ions per unit volume of the solution decreases. The molar conductivity of a solution increases as the electrolyte content drops. This is because dilution increases both the quantity of ions and their mobility. The molar conductivity is known as limiting molar conductivity as the concentration approaches zero.

8. The conductivity of 0.20M solution of KCl at 298 K is 0.0248 S cm^-1. Calculate its molar conductivity.

Ans:

Λ_m = (k × 1000) / Molanity = (0.2485 cm^-1 × 1000 cm³L^-1) / 0.20 moL^-1 = 124 Scm²mol^-1

9. The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500Ω. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10^-3 S cm^-1?

Ans: From the following relation:

Cell constant = Conductivity / Conductance = Conductivity × Resistance

= 0.146 × 10^-3 S cm^-1 × 1500Ω

= 0.218 cm^-1

10. The conductivity of NaCl at 298 K has been determined at different concentrations and the results are given below:

For all concentrations and draw a plot between Λ_m and C^1/2

Find the value of Λ_m^o

Ans: Using the following unit conversion factor, (1S cm^-1)/(100Sm^-1) = 1

Λ^o cm = Intercept of Λm axis = 124.0 S cm² mol^-1, Extrapolation to zero concentration yields this result.

11. Conductivity of 0.00241M acetic acid is 7.896 × 10^-5 Scm^-1. Calculate its. Calculate its molar conductivity. If Λ_m^o for acetic acid is 390.5 S cm² mol^-1, what is its dissociation constant?

Ans: Λ^c_m = (κ × 1000) / Molarity

= (7.896×10^-5 S cm^-1) × 1000 cm³L^-1 / 0.241 molL^-1 = 32.76 S cm² mol^-1

α = (Λ^c_m)/(Λm) = 32.76/390.5 = 8.4×10^-2

k_a = (Cα²)/(1–α) = (0.24 × (8.4×10^-2)²)/(1–0.084) = 1.86×10^-5

12. How much charge is required for the following reductions:

(i) 1 mol of Al³⁺ to Al?

Ans: Aluminum ion have to lose three electrons to be in aluminum elemental state and the reaction is Al³⁺ + 3e^- → Al. The amount of charge needed for reduction of 1 mol will be: Al³⁺ = 3 F = 3×96500C = 289500C.

(ii) 1 mol of Cu²⁺ to Cu?

Ans: cupric ion needs to lose two electrons to be in copper (0) state. The electrode reaction occurring is as follows: Cu²⁺ + 2e^- → Cu

The amount of charge required for conversion of 1 mol of Cu²⁺ = 2 F = 2×96500 = 193000C.

(iii) 1 mol of MnO₄^- to Mn²⁺

Ans: in the compound MnO₄^-, manganese is in +7 oxidation state; which is reduced to +2 oxidation state with the release of five electrons.

MnO₄^- → Mn²⁺

i.e. Mn⁷⁺ + 5e^- → Mn²⁺

The Quantity of charge needed for reduction of 1 mol of MnO^4- to Mn²⁺ = 5 F = 5×96500C = 4825000C.

13. How much electricity in terms of Faraday is required to produce.

(i) 20.0 g of Caa from molten CaCl₂?

Ans: Ca²⁺ + 2e^- → Ca

As There are two electrons transacted in the above reaction, therefore 1 mol of Ca or 40 g of Ca will require = 2 F electricity and 20 g of Ca will require = 1 F of electricity.

(ii) 40.0 g of Al from molten Al₂O₃?

Ans: Al³⁺ + 3e^- → Al

As the number of electrons transacted in the above reduction of aluminum ion is three,

Therefore, 1 mol of Al or 27 g of Al will require = 3 F electricity

And, 40 g of Al will require electricity

= (3/27) × 40 = 4.44F of electricity.

14. How much electricity is required in coulomb for the oxidation of

(i) 1 mol of H₂O to O₂

Ans: The oxidation of water occurs in the following way:

H₂O → H₂ + (1/2)O₂

Or O^2- → (1/2)O₂ + 2e^-

The number of electrons transacting are two. Therefore the amount of electricity needed = 2 F = 2×96500C = 193000C

(ii) 1 mol of FeO to Fe₂O₃

Ans: The oxidation reaction of FeO takes place in the following manner:

FeO + (1/2)O₂ → (1/2)Fe₂O₃

Or Fe²⁺ → Fe³⁺ + e^-

The electron transfer is of one electron unit, therefore the Quantity of electricity essential = 1 F = 96500C

15. A solution of Ni(NO₃)₂ is electrolyzed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

Ans: Firstly, we will find charge in coulombs using the formula: Q = It

Amount of electricity passed (Q) = (5 A) × (20 × 60 sec) = 6000C

The reduction of nickel occurs in the following way:

Ni²⁺ + 2e^- → Ni

Therefore, 2 For 2×96500C is the amount of charge deposit for 1 mol of Ni = 58.7 g

Now,

The 6000C of charge will deposit = (58.7 × 6000)/(2×96500) = 1.825 g of Ni

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 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?

Ans: Given: I = 1.5 A, W = 1.5 g of Ag t = ?, E = 108, n = 1

Applying Faraday's first law of electrolysis,

W = ZIt

Or

W = (E/nF) It

By substituting the values provided in the about formula, t can be calculated;

t = (1.45×96500)/(1.5×108) = 863.73 seconds.

Now for Cu, W₁ = 1.45 g (given) of Ag, E₁ = 108, W₂ = ?

E₂ = 31.75

Applying Faraday's 2^nd law of electrolysis

W₁/W₂ = E₁/E₂

1.45/W₂ = 108/31.75

W₂ = (1.45 × 31.75)/108

Copper deposited = 0.426 g of Cu

likewise, for Zn, W₁ = 1.45 g of Ag, E₁ = 108, W2 = ?, E2 = 32.65

Using formula, W₁/W₂ = E₁/E₂

1.45/W₂ = 108/32.65

W₂ = (1.45 × 32.65)/108 = 0.438 of Zn

Zinc deposited by 1295.6g C is 0.438 g.

17. Predict if the reaction between the following is feasible:

Given standard electrode potentials:

E^o_1/2,I2,I^- = +0.451V

E^o_Cu²⁺,Cu = +0.34V

E^o_1/2Br2,Br^- = +1.09V

E^o_Ag⁺,Ag = +0.80V, E^o_{Fe³⁺,Fe²⁺} = +0.77V

(i) Fe³⁺(aq) and I^-(aq)

Ans: The reaction is feasible if the emf of the cell reaction is positive.

Fe³⁺(aq) + I^-_(aq) → Fe²⁺(aq) + (1/2)I²(g)

i.e., Pt / I² / I^-(aq) ‖ Fe³⁺(aq) | Fe²⁺(aq) | Pt

∴ E^o_cell = E^o_{Fe³⁺/Fe²⁺} – E^o_1/2I2,I²

= 0.77 – 0.54 = 0.23V (Feasibles)

(ii) 2Ag⁺_(aq) + Cu_(s) → 2Ag_(s) + Cu²⁺_(aq)

Ans: 2Ag⁺_(aq) + Cu_(s) → 2Ag_(s) + Cu²⁺_(aq)

i.e., Cu | Cu²⁺(aq) ‖ Ag⁺_(aq) | Ag

∴ E^o_cell = E^o_Ag⁺,Ag – E^o_Cu²⁺,Cu

= 0.80 – 0.34 = 0.46V (Feasible)

(iii) Fe³⁺(aq) + Br^-_(aq) → Fe²⁺_(aq) + (1/2)Br_2(g)

Ans: Fe³⁺(aq) + Br^-_(aq) → Fe²⁺_(aq) + (1/2)Br_2(g)

E^o_cell = 0.77 – 1.09 = –0.32v (Not feasible)

(iv) Ag_(s) + Fe³⁺_(aq) → Ag⁺_(aq) + Fe²⁺(aq)

Ans: Ag_(s) + Fe³⁺_(aq) → Ag⁺_(aq) + Fe²⁺(aq)

E^o_cell = 0.77 – 0.80 = –0.03v(not feasible)

(v) (1/2)Br_2(g) + Fe²⁺(aq) → Br^-_(aq) Fe³⁺(aq)

Ans: (1/2)Br_2(g) + Fe²⁺(aq) → Br^-_(aq) Fe³⁺(aq)

E^o_cell = 1.09 – 0.77 = 0.32V (Feasible)

18. Predict the products of electrolysis of the following:

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

Ans: AgNO₃(s) → Ag⁺(aq) + NO₃^-(aq)

H₂O ⇌ H⁺ + OH^-

At cathode: The discharge potential of Ag⁺ ions is lower than that of H⁺ ions. As a result, Ag ions will be deposited as Ag rather than H⁺ ions.

At anode: When the Ag anode is attacked by NO₃^- ions, the Ag dissolves and forms ions in the solution.

Ag(s) → Ag⁺(aq) + e^-

(ii) An aqueous solution of AgNO₃ with platinum electrodes.

Ans: At cathode: The discharge potential of Ag⁺ ions is lower than that of H⁺ ions. As a result, Ag ions will be deposited as Ag rather than H⁺ ions.

At anode: Because the anode is not attackable, OH^- ions have a lower discharge potential than NO₃^- ions. As a result, OH^- will be evacuated first, followed by NO₃^- ions, which will disintegrate and release O₂.

OH^- → OH + e^-

4OH^- → 2H₂O_(l) + O_2(g)

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

Ans: H₂SO₄ → 2H⁺ + SO₄^2-

H₂O ⇌ H⁺ + OH^-

At cathode: H⁺ + e^- → H

H + H → H_2(g)

At anode: OH^- → OH + e^-

4OH^- → 2H₂O + O_2(g)

Therefore, Hydrogen gas is released at the cathode and O₂ gas at the anode.

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

Ans: CuCl₂ → Cu²⁺ + 2Cl^-

H₂O ⇌ H⁺ + OH^-

At cathode: Cu²⁺ ions will be reduced more than H⁺ ions, resulting in copper deposition at the cathode.

Cu²⁺ + 2e^- → Cu

At anode: Cl^- ions will be released first, followed by OH^- ions that will stay in solution. Cl^- → Cl + e^-

Cl + Cl → Cl_2(g)

Hence, copper will get set down on the cathode and Cl₂ gas will be liberated at the anode.

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