Electrochemistry Class 12 Important Questions With Answers

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Electrochemistry class 12 important questions with answers from CBSE 2020 to 2009. Class 12th Chemistry important questions in chapter 3 Electrochemistry for free.

The most important questions for annual examination from chapter 3 Electrochemistry are given here for free . The additional questions for practice from XII class CBSE exam are collected from various sources. It includes questions asked in previous year exams, questions provided by Kendriya Vidyalaya KV schools, DoE, Delhi government school, Navodaya Vidyalaya samiti NVS and other government and private schools in India and abroad that are affiliated to CBSE, New Delhi. Class 12 Chemistry Electrochemistry have different set of questions. The questions includes 1,2,3,4 & 5 mark questions as per the latest CBSE curriculum for the current session.

Electrochemistry Class 12 Important Questions with Answers CBSE

2020 Electrochemistry Class 12 Questions With Answer

2019 Electrochemistry Class 12 Questions With Answer

2018 Electrochemistry Class 12 Questions With Answer

2017 Electrochemistry Class 12 Questions With Answer

What is the effect of catalyst on:
(i) Gibbs energy (ΔG) and
(ii) activation energy of a reaction? (Delhi 2017 & All India 2017)
Answer:
(i) There will be no effect of catalyst on Gibbs .energy.
(ii) The catalyst provides an alternative pathway by decreasing the activation energy of a reaction

Calculate the degree of dissociation (a) of acetic acid if its molar conductivity (Λm) is 39.05 S cm2 mol-1.
Given: λ°(H+) = 349.6 S cm2 mol-1 and λ°(CH3COO) = 40.9 S cm2 mol-1 (Delhi 2017)
Answer:
Λ°m(HAC) = λ°H+ + λ°AC
= λ°CH3COOH = λ°H+ + λ°CH3COO
= 349.6 S cm2 mol-1 + 40.9 S cm2 mol-1
= 390.5 S cm2 mol-1

Write the name of the cell which is generally used in hearing aids. Write the reactions taking place at the anode and the cathode of this cell. (All India 2017)
Answer:
Mercury cells are used in hearing aids.
Reaction at anode:
Zn (Hg) + 2OH → ZnO (s) + H2O + 2e
Reaction at cathode:
HgO + H2O + 2e → Hg (l) + 2OH

Write the name of the cell which is generally used in transistors. Write the reactions taking place at the anode and the cathode of this cell. (All India 2017)
Answer:
Leclanche cells (Dry cell) is used in transistors.
Reaction at Anode:
Zn(s) → Zn2+ + 2e
At Cathode:
MnO2 + NH+4 + e → MnO(OH) + NH3

Write the name of the cell which is generally used in inverters. Write the reactions taking place at the anode and the cathode of this cell. (Delhi 2017)
Answer:
Lead storage battery is used in inverters.
At Anode:
Pb(s) + SO24(aq) → PbSO4 (s) + 2e
At Cathode:
PbO2(s) + SO24(aq) + 4H+ (aq) + 2e
PbSO4 (s) + 2H2O

(i) Calculate the mass of Ag deposited at cathode when a current of 2 amperes was passed through a solution of AgNO3 for 15 minutes.
[Given: Molar mass of Ag = 108 g mol-1 1F = 96,500 C mol-1)
(ii) Define fuel cell. (Delhi 2017)
Answer:
(i) Q = I × t …(Charge = Current ∝ Time)
. = 2 × 15 × 60 = 1800 C
∵ 96500 C deposit Ag = 108 g
∴ 1800 C deposit Ag = 10896500 × 1800
= 2.0145 g
(ii) Cells that convert the energy of combustion of fuels like hydrogen, methanol, methane, etc directly into electrical energy are called fuel cells.

(a) The cell in which the following reaction occurs:
2Fe3+ (aq) + 2I(aq) → 2Fe2+ (aq) + I2(s)
has Eo Cell  = 0.236 V at 298 K. Calculate the standard Gibbs energy of the cell reaction. (Given: 1F = 96,500 C mol-1)
(b) How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours? (Given: 1F = 96,500 C mol-1) (All India 2017)
Answer:
(a) 2Fe3+ (aq) + 2I (aq) → 2Fe2+ (aq) + I2 (s)
For the given reaction, n = 2, E° = 0.236 V
Using formula
ΔG° = -nF E°cell
= -2 × 96,500 C mol-1 × 0.236 V
∴ ΔG° =-45.55 kj mol-1

(b) Given:
I = 0.5 A
t = 2 hrs. = 2 × 60 ×60 s = 7,200 s
Q = I × t = 0.5 × 7200 = 3,600 C
96,500 C electricity flows to produce
= 6.022 × 1023 electrons
∴ 1 C electricity flows to produce

Why an electrochemical cell stop working after some time? The reduction potential of an electrode depends upon the concentration of solution with which it is in contact. (All India 2017)
Answer:
As the cell works, the concentration of reactants decrease. Then according to Le chatelier’s pri¬nciple it will shift the equilibrium in backward direction. On the other hand if the concentration is more on the reactant side then it will shift the equilibrium in forward direction. When cell works concentration in anodic compartment in cathodic compartment decreases and hence E° cathode will decrease. Now EMF of cell is
E0cell = E0cathode – E0anode
A decrease in E°cathode and a corresponding increase in E°anode will mean that EMF of the cell will decrease and will ultimately become zero i.e., cell slops working after some time.

2016 Electrochemistry Class 12 Questions With Answer

From the given cells: Lead storage cell, Mercury cell, Fuel cell and Dry cell
Answer the following:
(i) Which cell is used in hearing aids?
(ii) Which cell was used in Apollo Space Programme?
(iii) Which cell is used in automobiles and inverters?
(iv) Which cell does not have long life?(Delhi 2016)
Answer:
(i) Mercury cell is used in hearing aids.
(ii) Fuel cell was used in the Apollo Space Programme.
(iii) Lead storage cell is used in automobiles and inverters.
(iv) Dry cell does not have a long life.

Calculate e.m.f. of the following cell at 298 K: 2Cr(s) + 3Fe2+ (0.1 M) → 2Cr3+ (0.01 M) + 3 Fe(s)
Given: E0(Cr3+| Cr) = -0.74 V E0(Fe2+ | Fe) = -0.44 V (Delhi 2016)
Answer:
Cell reaction: 2Cr(s) + 3Fe2+ (0.1 M) → 2Cr3+ (0.01 M) + 3Fe(s)
Given: E0(Cr3+/Cr) = -0.74
E0(Fe2+/Fe) = -0.44 V

2015 Electrochemistry Class 12 Questions With Answer

1. (a) Following reactions occur at cathode during the electrolysis of aqueous silver chloride solution :
Ag+(aq) + e → Ag(s) E° = +0.80 V
H+(aq) + e12H2(g) E° = 0.00 V
On the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode and why?
(b) Define limiting molar conductivity. Why conductivity of an electrolyte solution decreases with the decrease in concentration? (Delhi 2015)
Answer:
(a) At the cathode Ag+ (aq) + e → Ag(s)
reaction is feasible, because Ag+ ion has higher reduction potential i.e. higher E° value.
(b) Limiting molar conductivity or the molar conductivity of solution at infinite dilution is the sum of molar conductivity cations and anions.
Conductivity of an electrolyte solution decreases on dilution because number of ions per unit volume decreases.

2. Calculate the time to deposit 1.27 g of copper at cathode when a current of 2A was passed through the solution of CuSO4.
(Molar mass of Cu = 63.5 g mol-1,1 F = 96500 C mol-1) (All India 2015)

Electrochemistry Class 12 Important Questions With Answers

3. Calculate emf of the following cell at 25°C :
Fe | Fe2+ (0.001 M) || H+ (0.01 M) | H2(g) (1 bar) | Pt(s)
E0(Fe2+ | Fe) = -0.44 V E0(H+ | H2) = 0.00V (Delhi 2015)
Answer:
Fe | Fe2+ (0.001 M) || H+ (0.01 M) | H2(g) (1 bar) | Pt(s)

4. Conductivity of 2.5 × 10-4 M methanoic acid is 5.25 × 10-5 S cm-1. Calculate its molar conductivity and degree of dissociation.
Given : λ0(H+) = 349.5 Scm2 mol-1
and λ0(HCOO) = 50.5 Scm2 mol-1. (All India 2015)
Answer:
Concentration is 2.5 x 10-4 M
K = 5.25 × 10-5 Scm-1.
Λcm=K×1000 Concentration 

5. Calculate the time to deposit 1.17 g of Ni at cathode when a current of 5 A was passed through the solution of Ni(NO3)2.
(Molar mass of Ni = 58.5 g mol-1, 1 F = 96500 C mol-1).
Answer:

Electrochemistry Class 12 Important Questions With Answers

6. (a) Following reactions occur at cathode during the electrolysis of aqueous copper (II) chloride solution:
Cu2+(aq) + 2 e-2 ———–> Cu(s)           E° = + 0.34 V
H+(aq) + e-1  ————-> 1/2H2(g)       E° = 0.00 V
On the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode and why ?
(b) State Kohlrausch law of independent migration of ions. Write its one application.
Answer: (a) At cathode Cu2+ + 2e-1 ———-> Cu(s)
The above reaction is feasible at cathode because E°Cu2+/Cu = + 0.34, because reduction potential of Cu2+ is higher than H+ as E0H+/H2= 0.
(b) Kohlrausch law: It states that the limiting molar conductivity of an electrolyte is equal to the sum of contribution of cations as well as anions. Application: It helps in calculating A0 (limiting molar conductivities) of weak electrolytes.

7. Conductivity of 2.5 X 10-4M methanoic acid is 5.25 X 10-5 S cm-1 Calculate its molar conductivity and degree of dissociation.
Given: A°(H+) = 349.5 S cm2 mol-1 and A°(HCOO ) = 50.5 S cm2
Answer:

Electrochemistry Class 12 Important Questions With Answers

8. The conductivity of 0.1 mol L-1 solution of NaCl is 1.06 X 10-2 S cm-1. Calculate its molar conductivity and degree of dissociation (a).
Given Z°(Na+) = 50.1 S cm2mol-1 and 7°(C1-) = 76.5 S cm2 mol-1.

Electrochemistry Class 12 Important Questions With Answers

2014 Electrochemistry Class 12 Questions With Answer

State Kohlrausch law of independent migration of ions. Why does the conductivity of a solution decrease with dilution? (All India 2014)
Answer:
Kohlrausch law of independent migration of ions: The limiting molar conductivity of an electrolyte (i.e. molar conductivity at infinite dilution) is the sum of the limiting ionic conductivities of the cation and the anion each multiplied with the number of ions present in one formula unit of the electrolyte
Λ°m for AxBy = xλ°+ + yλ°
For acetic acid Λ° (CH3COOH) = λ°CH3COO + λ°H+
Λ°(CH3COOH) = Λ° (CH3COOK) + Λ° (HCl) – Λ° (KCl)

Define the following terms :
(i) Fuel cell
(ii) Limiting molar conductivity (Λ°m) (All India 2014)

(iii) Molar conductivity (Λm)
(iv) Secondary batteries (All India 2014)

Answer:
(i) Fuel cells : These cells are the devices which convert the energy produced during combustion of fuels like H2, CH4, etc. directly into electrical energy.
(ii) The molar conductivity of a solution at infinite dilution is called limiting molar conductivity and is represented by the symbol Λ°m.

(iii) Molar conductivity: Molar conductivity of a solution at a given concentration is the conductance of the volume ‘V’ of a solution containing one mole of electrolyte kept between two electrodes with area of cross section ‘A’ and distance of unit length. It is represented by Λm (lamda).
Λm = KAl I = 1 and A = V
∴ Λm = KV Unit = S cm2 mol-1
(iv) Secondary batteries : Those batteries which can be recharged by passing an electric current through them and can be used again and again are called secondary batteries.

Define the following terms :
(i) Rate constant (k)
(ii) Activation energy (Ea) (Comptt. Delhi 2014)
Answer:
(i) Rate constant (k): It is a proportionality constant and is equal to the rate of reaction when the molar concentration of each of the reactants is unity.
(ii) Activation energy (Ea): The minimum extra amount of energy absorbed by the reactant molecules to form the activated complex is called activation energy.

Set up Nemst equation for the standard dry cell. Using this equation show that the voltage of a dry cell has to decrease with use. (Comptt. All India 2014)
Answer:
Cell reaction of a dry cell can be represented as
Zn + Hg2+ → Zn2+ + Hg (n = 2)
Nemst equation
Ecell = E°cell0.059 log([Zn2+]/[Hg2+])
The voltage of dry cell has to decrease because the concentration of electrolyte decreases in the reactions.

(a) Calculate ΔrG0 for the reaction
Mg (s) + Cu2+ (aq) → Mg2+ (aq) + Cu (s)
Given : E0cell = + 2.71 V, 1 F = 96500 C mol-1
(b) Name the type of cell which was used in Apollo space programme for providing electrical power. (All India 2014)
Answer:
(a) ΔrG0 = – nFE0
= -2 × 96500 × 2.71 (∵ n = 2)
= -523,030 J mol-1 = -523.03 KJ mol-1
(b) Fuel cell was used in Apollo space programme for providing electrical power.

The resistance of 0.01 M NaCl solution at 25° C is 200 Ω. The cell constant of the conductivity cell used is unity. Calculate the molar conductivity of the solution.(Comptt. All India 2014)
Answer:
For 0.01 M NaCl solution,
R = 200 Ω, cell constant is unity.
∴ Conductivity (K) =  Cell constant  Resistance 
= 1200 = 0.005 Sm-1
Concentration of solution = 0.01 M = 0.01 mol L-1
= 0.01 × 103 mol m-3 = 10 mol m-3
Molar conductivity = KCm=0.00510
5 × 10-4 Sm2 mol-1

2013 Electrochemistry Class 12 Questions With Answer

The conductivity of 0.20 M solution of KCl at 298 K is 0.025 S cm-1. Calculate its molar conductivity. (Delhi 2013)
Answer:
Molar conductivity Λm = 1000×κM
Given : K = 0.025 S cm-1, M = 0.20 M
Hence, Λm = 0.025×10000.20 ∴ Λm = 125 S cm2 mol-1

The standard electrode potential (E°) for Daniel cell is +1.1 V. Calculate the ΔG° for the reaction
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
(1 F = 96500 C mol-1). (All India 2013)
Answer:
We know, ΔG° = -nF E°cell
Given : E°cell = 1.1 volt
∴ ΔG° = -2 × 96500 C mol-1 × 1.1 volt
= -212300 CV mol-1
= -212300 J mol-1 = -212.3 KJ mol-1

The conductivity of 0.001 M acetic acid is 4 × 10-5 S/cm. Calculate the dissociation constant of acetic acid, if molar conductivity at infinite dilution for acetic acid is 390 S cm2/mol. (Comptt. Delhi 2013)
Answer:
Given : K = 4 × 10-5 S/cm, M = 0.001 M
Λ°m = 390 S cm2/mol, k = ?
Using the formula

Electrochemistry Class 12 Important Questions With Answers

Calculate the emf of the following cell at 298 K: Fe(s) | Fe2+ (0.001 M) || H+ (1M) | H2(g) (1 bar), Pt(s) (Given E°cell = +0.44V) (Delhi 2013)
Answer:
As Fe + 2H+ → Fe2+ + H2 (n = 2)
According to Nernst equation

      

(a) State and explain Kohlrausch law.
(b) How much electricity in terms of Faradays is required to produce 20 g of calcium from molten CaCl2? (Comptt. Delhi 2013)
Answer:
(a) Corrosion is an electrochemical phenomenon. At a particular spot of iron j oxidation it takes place and the spot behaves j as anode and the reaction will be: i
At anode :
2Fe(s) → 2Fe2+ + 4e
Electrons released above move through the metal to another spot and reduce oxygen in presence of H+ and the spot behaves as cathode with the reaction.
At cathode :
O2(g) + 4H+ (aq) + 4e → 2H2O(I)
Overall reaction :
2Fe(s) + O2(g) + H+(aq) → 2Fe2+(aq) + 2H2O (l)
The ferrous ions are further oxidised by atmospheric oxygen to form hydrated ferric oxide (Fe2O3 xH2O) as rust.

(b) Ca22+ + 2e → Ca
Thus, 1 mole of Ca i.e. 40 g of Ca requires electricity = 2F
∴ 20 g of Ca will require electricity = 1F

2012 Electrochemistry Class 12 Questions With Answer

Express the relation among cell constant, resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of a solution related to its conductivity? (All India 2012)
Answer:
K=(1/R)(L/A) =Q(L/A)
where Q is conductance and R is Resistance;
L/A is cell constant;
K is specific conductivity
∴ Λm = K×1000/M S cm2 mol-1

The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm2 mol-1. Calculate the conductivity of this solution. (All India 2012)
Answer:

C = 1.5 M, Λm = 138.9 S cm2 mol-1
Λm = K×1000/C
∴K = Λm×C/1000=138.9×1.5/1000 = 0.20835 S cm-1

A zinc rod is dipped in 0.1 M solution of ZnSO4. The salt is 95% dissociated at this dilution at 298 K. Calculate the electrode potential.
[ E°Zn2+ /Zn = – 0.76 V] (Comptt. Delhi 2012)
Answer:
The electode reaction is given as
Zn+2 + 2e → Zn
Using Nernest Equation

Electrochemistry Class 12 Important Questions With Answers

Write the reactions taking place at cathode and anode in lead storage battery when the battery is in use. What happens on charging the battery ? (Comptt. All India 2012)
Answer:
At Anode: Pb + SO4-2 → PbSO4 + 2e
at Cathode : PbO2 + SO4-2 + 4H+ + 2e → PbSO4 + 2H2O
On charging the battery, the reaction is reversed and PbSO4 on anode and cathode is converted into Pb and PbO2 respectively.

The electrical resistance of a column of 0.05 M NaOH solution of diameter 1 cm and length 50 cm is 5.55 × 103 ohm. Calculate its resistivity, conductivity and molar conductivity.(All India 2012)
Answer:
A = πr2 = 3.14 × (0.5)2 = 0.785 cm2, l = 50 cm

Electrochemistry Class 12 Important Questions With Answers

A voltaic cell is set up at 25°C with the following half cells :
Al/Al3+ (0.001 M) and Ni/Ni2+ (0.50 M)
Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.
E0Ni2+/Ni=0.25V and E0Al3+/Al=1.66V
(Log 8 × 10-6 = -0.54) (All India 2012)
Answer:
Half cell reactions and overall cell reaction are

Electrochemistry Class 12 Important Questions With Answers

What is corrosion? Explain the electrochemical theory of rusting of iron and write the reactions involved in the rusting of iron. (Comptt. Delhi 2012)
Answer:
Corrosion: Corrosion is defined as the deterioration of a substance because of its reaction with its environment. Corrosion is an electrochemical phenomenon. At a particular spot of an object made of iron, oxidation takes place and that spot behaves as anode and the reaction is
At Anode : 2Fe → 2Fe+2 + 4e
Electrons released at anodic spot move through the metal and go to another spot on the metal and reduce oxygen in presence of H+. This spot behaves as cathode
At Cathode : O2 + 4H+ + 4e-
Overall reaction : 2Fe + O2 + 4H+ → 2Fe+2 + 2H2O

When a certain conductance cell was filled with 0.1 M KCl, it has a resistance of 85 ohms at 25°C. When the same cell was filled with an aqueous solution of 0.052 M unknown electrolyte, the resistance was 96 ohms. Calculate the molar conductance of the electrolyte at this concentration.
[Specific conductance of 0.1 M KCl = 1.29 × 10-2 ohm-1 cm-1] (Comptt. All India 2012)
Answer:
Cell contant = Conductivity × Resistance
G* = K × R
= 1.29 × 10-2-1 × 85
= 109.65 × 10-2-1
= 1.0965 cm-1

(a) How many coulombs are required to reduce 1 mole Cr2O72- to Cr3+?
(b) The conductivity of 0.001 M acetic acid is 4 × 10-5 S/m. Calculate the dissociation constant of acetic acid if Λ0m,for acetic acid is 390 S cm2 mol-1. (Comptt. All India 2012)
Answer:
(a) Cr2O7-2 + 14H+ + 6e → 2Cr-3 + 7H2O
∴ 6 Faraday of charge is required
(b) Conductivity (K) = 4 × 10-5 S cm-1
Concentration (C) = 0.001M

The cell in which the following reaction occurs :
2Fe3+ (aq) + 2I (aq) → 2Fe2+ (aq) + I2 (s) has E0cell = 0.236V at 298K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
(Antilog of 6.5 = 3.162 × 106; of 8.0 = 10 × 108; of 8.5 = 3.162 × 108) (Comptt. All India 2012)
Answer:
log KC = nE0 cell 0.0591=2×0.2360.0591 = 8
KC = antilog 8 = 1 × 108
ΔG° = -nFE0cell = -2 × 96500 × 0.236
= -45548 J/mol-1
= -45.548 kJ/mol-1

2011 Electrochemistry Class 12 Questions With Answer

Express the relation between conductivity and molar conductivity of a solution held in a cell. (Delhi 2011)
Answer: Λm = K/C= Conductivity /Concentration 

The chemistry of corrosion of iron is essentially an electrochemical phenomenon. Explain the reactions occurring during the corrosion of iron in the atmosphere. (Delhi 2011)
Answer:
The mechanism of corrosion is explained on the basis of electrochemical theory. By taking example of rusting of iron, we Refer tothe formation of small electrochemical cells on the surface of iron.
The redox reaction involves
At anode : Fe(S) → Fe2+ (aq) + 2e
At cathode : H2O + CO2 ⇌ H2CO3 (Carbonic acid)
H2CO3 ⇌2H+ + CO22-
H2O ⇌ H+ + OH
H+ + e → H
4H + O2 → 2H2O
Then net resultant Redox reaction is
2Fe(s) + O2 (g) + 4H+ → 2Fe2+ + 2H2O

Determine the values of equilibrium constant (Kc) and ΔG° for the following reaction :
Ni(s) + 2Ag+ (aq) → Ni2+ (aq) + 2Ag(s),
E° = 1.05 V
(1F = 96500 C mol-1) (Delhi 2011)
Answer:
According to the formula
ΔG° = -nFE° = – 2 × 96500 ×1.05
or ΔG° = -202650 J mol-1 = -202.65 KJ mol-1
Now ΔG° ⇒ -202650 J Mol-1
R = 8.314 J/Mol/K, T = 298 K

Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions CBSE 2011

CBSE 2010 Electrochemistry Class 12 Questions With Answer

[1] What is meant by ‘limiting molar conductivity’? (All India 2010)
Answer:
The molar conductivity of a solution at infinite dilution is called limiting molar conductivity and is represented by the symbol Λ0m.

[2] Given that the standard electrode potentials (E°) of metals are :
K+/K = -2.93 V, Ag+/Ag = + 0.80 V, Cu2+/Cu = + 0.34 V,
Mg2+/Mg = – 2.37 V, Cr3+/Cr = – 0.74 V, Fe2+/Fe = – 0.44 V.
Arrange these metals in increasing order of their reducing power. (All India 2010)
Answer:
Ag < Cu < Fe < Cr <  Mg < K
More negative the value of standard electrode potentials of metals is, more will be the reducing power.

[3] Express the relation among the cell constant, the resistance of the solution in the cell and the conductivity of the solution. How is the conductivity of a solution related to its molar conductivity? (All India 2010)
Answer: Specific conductance = 1/R×L/A = Conductance (C) × Cell constant
Molar conductance : (Λm) = k×1000/M. where k is Specific conductance and M is Molarity of the electrolyte solution.

[4] Same Question of CBSE 2009 Question no [1]

CBSE 2009 Electrochemistry Class 12 Questions With Answer

[1] Two half cell reactions of an electrochemical cell are given below :
MnO4(aq) + 8H+ (aq) + 5e → Mn2+ (aq) + 4H2O (I), E° = + 1.51 V
Sn2+ (aq) → 4 Sn4+ (aq) + 2e, E° = + 0.15 V
Construct the redox equation from the two half cell reactions and predict if this reaction favours formation of reactants or product shown in the equation. (All India 2009)
Answer:
The reactions can be represented at anode and at cathode in the following ways :
At anode (oxidation) :
Sn2+ → = Sn4+ (aq) + 2e ] × 5 E° = + 0.15 V
At cathode (reduction) :
MnO4(aq) + 8H+ (aq) + 5e → Mn2+ (aq) + 4H2O (I)] × 2 E° = + 1.51 V
The Net R × M = 2MnO4(aq) + 16H+ + 5Sn2+ → 2Mn2+ + 5Sn4+ + 8H2O
Now E°cell = E°cathode – E°anode
= 1.51 – 0.15 = + 1.36 V
∴ Positive value of E°cell favours formation of product.

[2] A copper-silver cell is set up. The copper ion concentration in it is 0.10 M. The concentration of silver ion is not known. The cell potential is measured 0.422 V. Determine the concentration of silver ion in the cell.
Given : E°Ag+/Ag = + 0.80 V, E° Cu2+/Cu = + 0.34 V. (All India 2009)
Answer:
The reaction takes place at anode and cathode in the following ways :
At anode (oxidation) :
Cu(s) → Cu2+(aq) + 2e
At cathode (reduction) : Ag+(aq) + e → Ag(s)

The complete cell reaction is
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
A copper-silver cell is set up. The copper ion concentration is 0.10 M. The Ecell = 0.422 V

 

[3] A voltaic cell is set up at 25°C with the following half-cells, Al3+ (0.001 M) and Ni2+ (0.50 M). Write the cell reaction when the cell generates an electric current and determine the cell potential. Given : E° Ni2+/Ni = – 0.25 V, E° Al3+/Al = – 1.66 V. (All India 2009)

Answer: Thus, aluminium electrode is anode and nickel electrode is cathode reaction. 
3Ni+2+2Al→3Ni+2Al+3
Ecello​=−0.25−(−1.66)=1.41 V
Ecell=1.41+0.005319=1.415 V

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