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NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

Page No.36 - Intext Questions

Q2.1: How would you determine the standard electrode potential of the system Mg2+  | Mg?
Ans: The standard electrode potential of Mg2+  | Mg can be measured with respect to the standard hydrogen electrode, represented by Pt(s), H2(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.

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

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

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

Here,NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry for the standard hydrogen electrode is zero.

NCERT Solutions Class 12 Chemistry Chapter 2 - ElectrochemistryNCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry


Q2.2: Can you store copper sulphate solutions in a zinc pot?
Ans: 
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
NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry
Hence, copper sulphate solution cannot be stored in a zinc pot.

Q2.3: Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions
Ans: 
Substances that are stronger oxidising agents than ferrous ions can oxidise ferrous ions.
NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry
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 F2, Cl2, and O2.

Page No.41 - Intext Questions

Q2.4: Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
Ans:

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

= −0.0591 log 1010
= −0.591 V

Q2.5: 
Calculate the emf of the cell in which the following reaction takes place:  Ni(s)+2Ag+ (0.002 M) -> Ni2+ (0.160 M)+2Ag(s) Given that E(-)(cell) = 1.05 V
Ans: Applying Nernst equation we have:
NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry 
= 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

Q2.6: The cell in which the following reactions occurs: 2Fe3+ (aq) + 2I (aq) —> 2Fe2+ (aq) +I2 (s) has E°cell=0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction. 
Ans: 

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry
= −2 × 96487 × 0.236
= −45541.864 J mol−1
= −45.54 kJ mol−1
NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry
= 7.981
Kc = Antilog (7.981)
= 9.57 × 107

Page No.51 - Intext Questions

Q2.7: Why does the conductivity of a solution decrease with dilution?
Ans: 
Conductivity of a solution is the conductance of ions present in a unit volume of the solution. On dilution, the number of ions per unit volume decreases. Hence the conductivity decreases.

Q2.8: suggest a way to determine the NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry value of water.
Ans: 
Λm value of water.
Consider the following chemical equations,
(i) H2O(I) → H+(aq) + OH(aq)
(ii) HCl(aq) → H+(aq) + Cl(aq)
(iii) NaOH(aq) → Na+(aq) + OH(aq)
(iv) NaCl(aq) → Na+(aq) + Cl(aq)
Combining the four equations and using Kohlrausch's law, we get,
Λm(H22O) = Λm(HCl) + Λm(NaOH) − Λm(NaCl)
Hence, if we know the values of Λm for HCl, NaOH and NaCl, we can calculate the value of Λm for water.

Q2.9: 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 λ°(H+)=349.6 S cm2 mol-1 andλ°(HCOO-) = 54.6 S cm2 mol-1 .
Ans:

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry 
= 0.114 (approximately)
Thus, dissociation constant:
NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry


Q2.10: 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: 
= 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
NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry
= 2.25 x 1022 number of electrons
Hence, 2.25 x 1022 number of electrons will flow through the wire.

Q2.11: Suggest a list of metals that are extracted electrolytically.
Ans:
Metals that are on the top of the reactivity series such as sodium, potassium, calcium, lithium, magnesium, aluminium are extracted electrolytically.

Q2.12: Consider the reaction: Cr2O72--+ 14H+ + 6e- -> 2Cr3+ + 7H2O What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr2O72- ? 
Ans: 
The given reaction is as follows:
NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry
Therefore, to reduce 1 mole of NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistrythe required quantity of electricity will be:
= 6 F
= 6 × 96487 C
= 578922 C

Page No.58 - Intext Questions

Q2.13: Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
Ans: 
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:
NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry
When a battery is charged, the reverse of all these reactions takes place.
Hence, on charging, NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistrypresent at the anode and cathode is converted into NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry respectively.

Q2.14: Suggest two materials other than hydrogen that can be used as fuels in fuel cells.
Ans: 
Methane and methanol can be used as fuels in fuel cells.

Q2.15: Explain how rusting of iron is envisaged as setting up of an electrochemical cell.
Ans: 
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,
NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry
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 H2CO3, 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,
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 NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistryi.e., rust.
Hence, the rusting of iron is envisaged as the setting up of an electrochemical cell.

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

Exercises

Q2.1: 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: The following is the order in which the given metals displace each other from the solution of their salts.
Mg, Al, Zn, Fe, Cu

Q2.2: Given the standard electrode potentials,
K+/K = −2.93V, Ag+/Ag = 0.80V,
Hg2+/Hg = 0.79V
Mg2+/Mg = −2.37 V, Cr3+/Cr = − 0.74V
Arrange these metals in their increasing order of reducing power.
Ans: 
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

Q2.3: Depict the galvanic cell in which the reaction Zn(s) + 2Ag+ (aq) → Zn2+ (aq) 2Ag(s) takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.
Ans: The galvanic cell in which the given reaction takes place is depicted as:

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

(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,

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

The reaction taking place at the cathode is given by,

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

Q2.4: Calculate the standard cell potentials of galvanic cells in which the following reactions take place:
(i) 2Cr(s) + 3Cd2+ (aq) → 2Cr3+ (aq) 3Cd
(ii) Fe2+ (aq) Ag+ (aq) → Fe3+ (aq) + Ag(s)
Calculate the ΔrGθ and equilibrium constant of the reactions.
Ans: (i) NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

The galvanic cell of the given reaction is depicted as:

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

Now, the standard cell potential is

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

In the given equation,

n = 6

F = 96487 C mol−1

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry= 0.34 V

Then, NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry= −6 × 96487 C mol−1 × 0.34 V

= −196833.48 CV mol−1

= −196833.48 J mol−1

= −196.83 kJ mol−1

Again,

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry= −RT ln K

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

= 34.496

NCERT Solutions Class 12 Chemistry Chapter 2 - ElectrochemistryK = antilog (34.496)

= 3.13 × 1034

(ii)NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

The galvanic cell of the given reaction is depicted as:

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

Now, the standard cell potential is

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

Here, n = 1.

Then, NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

= −1 × 96487 C mol−1 × 0.03 V

= −2894.61 J mol−1

= −2.89 kJ mol−1

Again, NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

= 0.5073

NCERT Solutions Class 12 Chemistry Chapter 2 - ElectrochemistryK = antilog (0.5073)

= 3.2 (approximately)


Q2.5: Write the Nernst equation and emf of the following cells at 298 K:

(i) Mg(s) | Mg2+ (0.001M) || Cu2+ (0.0001 M) | Cu(s)

(ii) Fe(s) | Fe2+ (0.001M) || H+ (1M)|H2(g)(1bar) | Pt(s)

(iii) Sn(s) | Sn2+ (0.050 M) || H+ (0.020 M) | H2(g) (1 bar) | Pt(s)

(iv) Pt(s) | Br2(l) | Br(0.010 M) || H+ (0.030 M) | H2(g) (1 bar) | Pt(s).

Ans:

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

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

= 2.7 − 0.02955

= 2.67 V (approximately)

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

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

= 0.52865 V

= 0.53 V (approximately)

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

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

= 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:

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

Q2.6: In the button cells widely used in watches and other devices the following reaction takes place:
Zn(s) + Ag2O(s) + H2O(l) → Zn2+(aq) + 2Ag(s) + 2OH(aq)
DetermineNCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry for the reaction.
Ans:

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry
NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

Q2.7: Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Ans: Conductivity : The reciprocal of resistivity is conductivity. Conductivity is the conductance of the electrolytic solution (due to the movement of ions and electrons) of one metre length and unit area of cross section. Mathematically, conductivity, K = 1/p
Conductivity varies with concentration. As the number of ions per cm3 of solution decreases, conductivity decreases.

Molar conductivity : The conductivity of one molar electrolytic solution is termed as molar conductivity. Mathematically,
Molar conductivity (Δm) = k/c,
where c = concemtration (in mol m-3)
K = conductivity (in S m-1)
It should be noted that for strong electrolytes there is only a small increase in Δm value with dilution. However for weak electrolytes, there is a large increase in Δm value with dilution.


Q2.8: The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 Scm−1. Calculate its molar conductivity. 
Ans: Given,
κ = 0.0248 S cm−1
c = 0.20 M
NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

Q2.9: The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10−3 S cm−1.
Ans: Given,
Conductivity, κ = 0.146 × 10−3 S cm−1
Resistance, R = 1500 Ω
∴ Cell constant = κ × R
= 0.146 × 10−3 × 1500
= 0.219 cm−1

Q2.10: The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

Calculate NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistryfor all concentrations and draw a plot between NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistryand c½. Find the value ofNCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry.
Ans: Given,

κ = 1.237 × 10−2 S m−1, c = 0.001 M

Then, κ = 1.237 × 10−4 S cm−1, c½ = 0.0316 M1/2

NCERT Solutions Class 12 Chemistry Chapter 2 - ElectrochemistryNCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

= 123.7 S cm2 mol−1

Given,

κ = 11.85 × 10−2 S m−1, c = 0.010M

Then, κ = 11.85 × 10−4 S cm−1, c½ = 0.1 M1/2

NCERT Solutions Class 12 Chemistry Chapter 2 - ElectrochemistryNCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

= 118.5 S cm2 mol−1

Given,

κ = 23.15 × 10−2 S m−1, c = 0.020 M

Then, κ = 23.15 × 10−4 S cm−1, c1/2 = 0.1414 M1/2

NCERT Solutions Class 12 Chemistry Chapter 2 - ElectrochemistryNCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

= 115.8 S cm2 mol−1

Given,

κ = 55.53 × 10−2 S m−1, c = 0.050 M

Then, κ = 55.53 × 10−4 S cm−1, c1/2 = 0.2236 M1/2

NCERT Solutions Class 12 Chemistry Chapter 2 - ElectrochemistryNCERT Solutions Class 12 Chemistry Chapter 2 - ElectrochemistryNCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

 

= 111.1 1 S cm2 mol−1

Given,

κ = 106.74 × 10−2 S m−1, c = 0.100 M

Then, κ = 106.74 × 10−4 S cm−1, c1/2 = 0.3162 M1/2

NCERT Solutions Class 12 Chemistry Chapter 2 - ElectrochemistryNCERT Solutions Class 12 Chemistry Chapter 2 - ElectrochemistryNCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

= 106.74 S cm2 mol−1

Now, we have the following data:

 

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

Since the line interruptsNCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistryat 124.0 S cm2 mol−1, NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry= 124.0 S cm2 mol−1.

Q2.11: Conductivity of 0.00241 M acetic acid is 7.896 × 10−5 S cm−1. Calculate its molar conductivity and if NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistryfor acetic acid is 390.5 S cm2 mol−1, what is its dissociation constant?
Ans: Given, κ = 7.896 × 10−5 S m−1

c = 0.00241 mol L−1

Then, molar conductivity, NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

= 32.76S cm2 mol−1

Again, NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry= 390.5 S cm2 mol−1

Now, NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

= 0.084

NCERT Solutions Class 12 Chemistry Chapter 2 - ElectrochemistryDissociation constant, NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

= 1.86 × 10−5 mol L−1

Q2.12: How much charge is required for the following reductions:
(i) 1 mol of Al3+ to Al.
(ii) 1 mol of Cu2+ to Cu.
(iii) 1 mol of   Mno4 to MN2+
Ans: 
NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

Q2.13: How much electricity in terms of Faraday is required to produce
(i) 20.0 g of Ca from molten CaCl2.
(ii) 40.0 g of Al from molten Al2O3.
Ans: (i) According to the question,

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

Electricity required to produce 40 g of calcium = 2 F

Therefore, the electricity required to produce 20 g of calciumNCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

= 1 F

(ii) According to the question,

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

Electricity required to produce 27 g of Al = 3 F

Therefore, the electricity required to produce 40 g of Al NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

= 4.44 F


Q2.14: How much electricity is required in coulomb for the oxidation of
(i) 1 mol of H2O to O2.  
(ii) 1 mol of FeO to Fe2O3.

Ans: (i) According to the question,

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

Now, we can write:

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

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

= 2 × 96487 C

= 192974 C

(ii) According to the question,

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

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

= 96487 C

Q2.15: A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?
Ans: Given,

Current = 5A

Time = 20 × 60 = 1200 s

NCERT Solutions Class 12 Chemistry Chapter 2 - ElectrochemistryCharge = current × time

= 5 × 1200  = 6000 C

According to the reaction,

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

Nickel deposited by 2 × 96487 C = 58.71 g

Therefore, nickel deposited by 6000 C NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry = 1.825 g

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


Q2.16: 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 1.45 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: According to the reaction:

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

i.e., 108 g of Ag is deposited by 96487 C.

Therefore, 1.45 g of Ag is deposited by =NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry = 1295.43 C

Given,

Current = 1.5 A

NCERT Solutions Class 12 Chemistry Chapter 2 - ElectrochemistryTime NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

= 863.6 s = 864 s

= 14.40 min

Again,

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

i.e., 2 × 96487 C of charge deposit = 63.5 g of Cu

Therefore, 1295.43 C of charge will deposit NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

= 0.426 g of Cu

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

i.e., 2 × 96487 C of charge deposit = 65.4 g of Zn

Therefore, 1295.43 C of charge will deposit NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

= 0.439 g of Zn

Q2.17: Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:
(i) Fe3+ (aq) and I(aq)                                            
 (ii) Ag+  (aq) and Cu(s)
 (iii) Fe3+  (aq) and Br− (aq)                                      
 (iv) Ag(s) and Fe3+  (aq)
 (v) Br(aq) and Fe2+ (aq).

Ans:

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

SinceNCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistryfor the overall reaction is positive, the reaction between Fe3+ (aq) and I(aq) is feasible.

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

Since NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistryfor the overall reaction is positive, the reaction between Ag+ (aq) and Cu(s) is feasible.

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

SinceNCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry for the overall reaction is negative, the reaction between Fe3+ (aq) and Br(aq) is not feasible.

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

Since NCERT Solutions Class 12 Chemistry Chapter 2 - ElectrochemistryE for the overall reaction is negative, the reaction between Ag (s) and Fe3+ (aq) is not feasible.

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

SinceNCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry for the overall reaction is positive, the reaction between Br2(aq) and Fe2+ (aq) is feasible.


Q2.18: Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO3 with silver electrodes.
(ii) An aqueous solution of AgNOwith platinum electrodes.
(iii) A dilute solution of H2SOwith platinum electrodes.
(iv) An aqueous solution of CuCl2 with platinum electrodes.
Ans: (i) At cathode:

The following reduction reactions compete to take place at the cathode.

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

The reaction with a higher value of NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistrytakes place at the cathode. Therefore, deposition of silver will take place at the cathode.

At anode:

The Ag anode is attacked by NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistryions. 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.

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

The reaction with a higher value of NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistrytakes 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 NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistryions. Therefore, OH or NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistryions can be oxidized at the anode. But OH ions having a lower discharge potential and get preference and decompose to liberate O2.

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

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

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

At the anode, the following processes are possible.

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

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.

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

The reaction with a higher value of NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistrytakes 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.

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

At the anode, the reaction with a lower value of NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistryis preferred. But due to the over-potential of oxygen, Cl gets oxidized at the anode to produce Cl2 gas.

Old NCERT Solution

Q1:NCERT Solutions Class 12 Chemistry Chapter 2 - ElectrochemistryCalculate the potential of a hydrogen electrode in contact with a solution whose pH is 10.
Ans: For hydrogen electrode,   NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry,  it is given that pH = 10

∴[H ] = 10−10 M

Now, using Nernst equation:

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry= NCERT Solutions Class 12 Chemistry Chapter 2 - ElectrochemistryNCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

= −0.0591 log 1010

= −0.591 V


Q2: Calculate the emf of the cell in which the following reaction takes place:

 NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

Given that NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry= 1.05 V
Ans: Applying Nernst equation we have:

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

= 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


Q3: The cell in which the following reactions occurs:

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry has NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry= 0.236 V at 298 K.

Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Ans: Here, n = 2, NCERT Solutions Class 12 Chemistry Chapter 2 - ElectrochemistryT = 298 K

We know that:

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

= −2 × 96487 × 0.236

= −45541.864 J mol−1

= −45.54 kJ mol−1

Again, NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry−2.303RT log Kc

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

= 7.981

Kc = Antilog (7.981)

= 9.57 × 107

Q4: 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:  = 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 NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistrynumber of electrons.

Then,

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

Hence, NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistrynumber of electrons will flow through the wire.

The document NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry is a part of the NEET Course Chemistry Class 12.
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FAQs on NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

1. What is electrochemistry and why is it important?
Ans. Electrochemistry is the branch of chemistry that deals with the study of the relationship between electricity and chemical reactions. It is important because it has applications in various fields such as batteries, corrosion prevention, electroplating, and fuel cells.
2. What is the significance of redox reactions in electrochemistry?
Ans. Redox reactions, which involve the transfer of electrons between two substances, play a crucial role in electrochemistry. These reactions are the basis of electrochemical cells and are responsible for the generation of electricity in batteries.
3. How does an electrochemical cell work?
Ans. An electrochemical cell consists of two electrodes (anode and cathode) immersed in an electrolyte solution. During the cell operation, oxidation occurs at the anode, leading to the release of electrons, which flow through an external circuit to the cathode where reduction takes place.
4. What is the difference between galvanic cells and electrolytic cells?
Ans. Galvanic cells are spontaneous electrochemical cells that convert chemical energy into electrical energy, while electrolytic cells are non-spontaneous cells that require an external source of electrical energy to drive the chemical reaction.
5. How are standard reduction potentials used in electrochemistry?
Ans. Standard reduction potentials are used to determine the feasibility of a redox reaction and to predict the direction of electron flow in an electrochemical cell. These values help in calculating cell potentials and understanding the reactivity of different substances.
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