NCERT Solutions for Class 12 Chemistry Chapter 2 – Electrochemistry

Intext Questions with Solutions of Class 12 Chemistry Chapter 2 – Electrochemistry

2.1

Ans – The cell’s electromagnetic field (EMF) can be measured with the angle of deflection using a voltmeter. A cell can be placed with Mg/MgSO₄ (1 M) as the primary electrode and the conventional hydrogen electrode Pt, H, (1 atm) H⁺/(l M) as the secondary electrode. The shift in direction indicates that e^-1 s travel from the magnesium electrode to the hydrogen electrode, meaning that reduction occurs on the hydrogen electrode. In contrast, oxidation occurs on the magnesium electrode. The structure of the cell can therefore be shown in the following manner:

2.2

Ans – Cu is readily removed from the CuSO₄ mixture in the subsequent process because it is less sensitive than zinc:
Zn (s) + CuSO₄ (aq) → ZnSO₄ (aq) + Cu (s)

When expressed in terms of emf, it will be as follows:
 E⁰_cell = E⁰_Cu2+/Cu − E⁰_Zn2+/Zn
= 0.34V − (−0.76V) = 1.10V

The presence of involuntary responses is indicated by a positive E⁰_cell measurement. The uniqueness of CuSO₄ will be lost if it is kept in a Zn container because Zinc will interact with copper.

2.3

Ans – The oxidation value of Fe²⁺ ions to Fe³⁺ ions can be depicted below:
Fe²⁺ → ³⁺ + e^– ; E⁰_OX = – 0·77 V 
Particular compounds that are positioned over iron in an electrochemical sequence or that can oxidize Fe²⁺ ions to Fe³⁺ ions that can take up electrons generated during oxidation are capable of doing so. Meanwhile, in the acidic medium, the substances include Cl₂(g), Br₂(g) & Cr₂O₂⁻⁷ ions.

2.4

Ans – For a hydrogen electrode,

It is given that pH = 10.
Using Nernst equation we get,

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

2.5

Ans – Using Nernst formula,

2.6

Ans –

2.7

Ans – The quantity of ions available per unit volume is correlated with a solution’s conductance. When they are diluted, the solution’s particular conductive property or equivalent conductivity also lowers.

2.8

Ans – We can figure out Λm^∘ for water by applying Kohl Rausch’s law.
Irrespective of the type of ion concerned, the Kohlrausch Law states that every ion contributes a particular quantity to the electrolyte’s comparable conductivity when splitting takes place at endless dilution. The quantity of corresponding conductance at infinite dilution for every electrolyte is the combined value of the individual contributions of its ionic constituents (cations followed by anions).
Λm^∘ = Λm^∘ (HCl) + Λm^∘ (NaOH) − Λm^∘ (NaCl)

Since HCl, NaOH, and NaCl are powerful electrolytes and fully split up, their Λ^∘ constants are well-known. The magnitude of Λm^∘ for water molecules may be obtained by entering the corresponding value in the formula mentioned earlier.

2.9

Ans –

2.10

Ans –

2.11

Ans – Electrolysis can be used to obtain extremely reactive elements with significant -ve E° coefficients that can also function as potent reducing substances. Electrolytic reduction is the name of the method that is used. For example, elements like magnesium, sodium, calcium, potassium, etc.

2.12

Ans – For the mentioned reaction to occur, 1 mol of Cr₂O₇²⁻ requires 6 F = 6 × 96500 = 579000 C of electricity.
Therefore, 579000 C of electricity are necessary for the reduction of Cr₂O₇²⁻ to Cr³⁺

2.13

Ans – An anode made of lead, a cathode of a network of lead filled with lead dioxide (PbO₂), and an electrolyte of 38% water in a solution of H₂SO₄ make up a lead storage cell. The following effect occurs while the battery is being used:

The opposite process occurs when the battery is charged, meaning that PbSO₄ that has been accumulated on the conducting surfaces is changed back into Pb and PbO₂, while H₂SO₄ is produced.

2.14

Ans – Methane (CH₄) & methanol (CH₃OH) can be implemented as fuel source as a perfect alternate to hydrogen components in the fuel storage cells.

2.15

Ans – The water available on the above layer of the surface in the iron gets dissolved in the acidic oxides of air including CO₂, and SO₂ to produce acids that spits apart to offer H⁺ ions:
H₂O + CO₂ → H₂CO₃ ⇌ 2H⁺ + CO₃²⁻

Iron Fe²⁺ emits e⁻¹ to produce Fe³⁺ when H⁺ is present. Thus, this serves as an anode:
Fe (s) → Fe²⁺ (aq) + 2e⁻

After that, the e⁻¹ dissipated moves across the metal to a location where reduction occurs and H⁺ ions use these electrons together with oxygen that is dispersed serves as a cathode as a result:
O₂ (g) + 4H⁺ (aq) + 4e⁻ → 2H₂O (g)
Overall reaction will be:
2Fe (s) + O₂ (g) + 4H⁺ (aq) → 2Fe²⁺ (aq) + 2H₂O (l)

Consequently, the outer layer develops an electrochemical cell. Rust, or hydrated ferric oxide, is created when water combines with ferrous ions that have been oxidized by oxygen in the atmosphere to ferric ions (Fe₂O₃.XH₂O).

Exercise Questions with Solutions of Class 12 Chemistry Chapter 2 – Electrochemistry

2.1

Ans – Mg, Al, Zn, Fe, Cu

2.2

Ans – The reducing power increases with the oxidation potential because it can be readily oxidized. Ag < Hg < Cr < Mg < K will experience the rising order of diminishing energy as a result.

2.3

Ans – (i) There will be a negative charge on the anode, which is the zinc electrode.

(ii) Current will flow from silver to copper in the external circuit.

(iii) At anode: Zn (s) → Zn²⁺ (aq) + 2e^–
At cathode: 2Ag⁺ (aq) + 2e^– → 2Ag (s)

2.4

Ans – (i) Calculation of E⁰_cell

Calculation of ∆G^o

Calculation of Equilibrium Constant K_c

(ii) Calculation of E⁰_cell

Calculation of ∆G^o

Calculation of Equilibrium Constant K_c

2.5

Ans – (i) Cell reaction:

(ii) Cell reaction:

(iii) Cell reaction:

(iv) Cell reaction:

2.6

Ans – We know
Zn → Zn²⁺ + 2e⁻, E^∘ = 0.76 V
Ag₂O + H₂O + 2e⁻ → 2Ag + 2OH⁻, E^∘ = 0.344 V
In the specified reaction, Zn undergoes oxidation whereas Ag₂O undergoes reduction.
E^∘_cell = 0.344 + 0.76 = 1.104 V
ΔG^∘ = nFE^∘_cell = −2 × 96500 × 1.104 J
ΔG^∘ = −2.13 × 10⁵ J

2.7

Ans – When an agent has a dimension of 1 cm & a cross-sectional measurement of 1 cm, its conductivity is determined as its conductance. 

The molar conductivity of a fluid at a dilution (V) is the conductance of every ion generated from a single mole of the electrolyte present in the solution when the electrodes are spaced one centimetre apart and the area of their cross-section is sufficient to contain the whole substance between them Vcm3. The majority of highly incorporated indicators for conductivity is Λm.

As the electrolyte saturation decreases, or as dispersion takes place, a solution’s conductivity decreases (for equally weak as well as strong electrolytes). The reason for this is that the quantity of ions per cubic inch of the mixture falls as it becomes dissolved. The molar conductivity of a mixture rises as the quantity of the electrolyte decreases upon diluting. This happens because there is a lower number of ions per cubic inch of the substance when it is dissolved. As a solution’s electrolyte saturation decreases, its molar conductivity rises. The reason for this is that dilution raises the number of ions and their adaptability. As the quantity of concentration inches closer to zero, the molar conductivity is depicted as the restricting molar conductivity.

2.8

Ans –

2.9

Ans –

2.10

Ans – Using the unit conversion factor shown below,

Λ^∘cm = Intercept of Λ_maxis =124.0 S cm² mol⁻¹
The extrapolation to zero concentration produces this outcome.

2.11

Ans –

2.12

Ans – (i) Al³⁺ + 3e → Al is the process that occurs when aluminium ions lose 3 electrons to reach the elemental form of aluminium. For a decrease of 1mol, the required energy is Al³⁺ = 3F = 3 × 96500C = 289500C.

(ii) For it to exist in the copper (0) region, a cupric ion must shed 2 electrons. The following is the electrode response that is taking place: Cu²⁺ + 2e⁻ → Cu. The amount of energy needed to convert one mol of Cu²⁺ is equal to 2F = 2 × 96500 = 193000C.

(iii) 5 electrons are released to decrease manganese from its +7 oxidation condition to a +2 oxidation stage in the molecule MnO₄.
MnO₄ → Mn²⁺ i.e. Mn⁷⁺ + 5e⁻ → Mn²⁺
One mol of MnO₄ must be reduced to Mn²⁺ with an electrical charge value of 5F = 5 × 96500C = 4825000C.

2.13

Ans – (i) Ca²⁺ + 2e⁻ → Ca
One mol of calcium, or 40g of calcium, will call for = 2F electricity since the process above involves a pair of electrons, and 20g of calcium will need = 1F energy.

(ii) Al³⁺ + 3e⁻ → Al
Since there are 3 electrons involved in the decrease of the aluminium ion previously mentioned,
Thus, 27g of Al or one mole of Al will need 3F of power.
Additionally, 40g of aluminium will take 4.44F of power (3/27) times 40.

2.14

Ans – (i) The following reaction happen when water oxidises:

Two electrons are involved in the transaction. Thus, the required amount of electricity is equal to 2 F = 2 × 96500 C = 193000 C

(ii) The following reaction is how the oxidation process of FeO happen:

The electron transfer involves a single electron unit; hence, the required quantity of electricity = 1 F = 96500C

2.15

Ans – The amount of electricity that was passed = (5A) x (20 x 60 sec) = 6000C

2.16

Ans – Given

2.17

Ans – A positive emf indicates the feasibility of the cell reaction.

2.18

Ans – (i) AgNO₃ (s) → Ag⁺ (aq) + NO₃^– (aq)
H₂O ⇆ H⁺ + OH^– 

At the cathode, Ag⁺ ions have a smaller release value than H⁺ ions. Consequently, instead of settling down as H⁺ ions, Ag ions will be implanted as Ag.

At the anode: Ag melts and creates ions in the mixture once NO₃^– ions hit the Ag anode.
Ag (s) → Ag⁺ (aq) + e^– 

(ii) At the cathode, Ag⁺ ions have a smaller release value than H⁺ ions. Consequently, instead of settling down as H⁺ ions, Ag ions will be formed as Ag.

At anode: OH ions have a smaller discharging energy than NO₃⁻ ions due to the anode’s inability to be attacked. Consequently, NO₃⁻ ions will dissolve and emit O₂ once OH has been eliminated initially.
OH⁻ → OH + e⁻
4OH → 2H2O (l) + O₂ (g)

(iii) H₂SO₄ → 2H⁺ + SO₄²⁻
H₂O ⇌ H + OH⁻

At cathode: H⁺ + e⁻ → H
H + H → H₂ (g)

At anode: OH⁻ → OH + e⁻
4OH → 2H₂O + O₂ (g)
Thus, hydrogen gas is released at the cathode and oxygen gas at the anode.

(iv) CuCl₂ → Cu²⁺ + 2Cl⁻
H₂O ⇌ H + OH⁻
Copper metal will be deposited at the cathode as an outcome of Cu²⁺ ions diminishing to greater H⁺ ions.
Cu²⁺ + 2e⁻ → Cu
At the anode, OH^– ions that remain in reaction will be liberated after Cl^– ions.
Cl^– → Cl + e⁻
Cl + Cl → Cl₂ (g)
As a result, Cl₂ vapor will be released at the anode & copper metal will settle on the cathode.

Related Study Resources of Chapter 2 – Electrochemistry

Students can use the links below to get extra study materials for Class 12 Chemistry Chapter 2: Electrochemistry

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1	Class 12 Chemistry Chapter 2 Electrochemistry – Important Questions

NCERT Solutions for Class 12 Electrochemistry

NCERT solutions for students preparing for their Class 12 board exams are very essential. Electrochemistry Class 12 NCERT Solutions provide step-by-step explanations for textbook problems, which make it easy for students to understand complex concepts. Students can develop problem-solving skills and increase confidence in solving questions in the exams by using NCERT solutions.
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The Electrochemistry Class 12 NCERT Solutions deals with Nernst equation, electrochemical cells and Gibbs energy in electrochemical reactions. This helps the students to explain complex problems in a simple way.
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