CBSE 12th Chemistry - Electrochemistry Model Question Paper | QB365

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CBSE 12th Chemistry - Electrochemistry Model Question Paper

Published on: 04/12/2019

Electrochemistry

Download CBSE Class 12th Standard CBSE Chemistry question papers, sample papers, important questions, and previous year solved papers in PDF format. Get free study materials, NCERT solutions, and exam preparation resources for Class 12th Standard CBSE Chemistry

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Std: 12th Standard CBSE - Chemistry

chapter: - Model Question paper

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Questions:

Generic 2

2 x 2 = 4

The molar conductivity of 1.5 M solution of an electrolyte is found to be 138.9 S cm² mol^-1 . Calculete the conductivity of this solution.

Ezpress the relation among cell constant, resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of solution related to its conductivity.

Generic 3

2 x 3 = 6

(i) 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.
(ii) Given the standard electrode potentials;
K⁺/K = -2.93V,Ag⁺/Ag = 0.80V, Hg²⁺/Hg = +0.79V.
Mg²⁺/Mg = -2.37V, Cr³⁺/Cr = -0.74V.
Arrange these metals in their increasing order of reducing power.

When a certain electrolytic cell was filled with 0.1 M KCl, it has resistance of 85 ohms at 25 ^oC. 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 = \(1.29\times { 10 }^{ -2 }{ ohm }^{ -1 }{ cm }^{ -1 }\)]

Generic 5

3 x 5 = 15

EMF of Daniell cell was found using different concentrations of Zn²⁺ ion and Cu² ion. A graph was then plotted between E_cell and \(\log { \frac { \left[ { Zn }^{ 2+ } \right] }{ \left[ { Cu }^{ 2+ } \right] } } \) . The plot was found to be linear with intercept on E_cell axis equal to 1.10 V. Calculate E_cell for Zn | Zn²⁺ (0.1 M) || Cu²⁺ (0.01 M) | Cu.

Using the standard electrode potentials, 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) Br₂(aq) and Fe²⁺(aq)

Calculate the standard cell potentials of galvanic cell in which the following reactions take place:
(i) \(2Cr(s)+{ 3Cd }^{ 2+ }(aq)\rightarrow { 2Cr }^{ 3+ }(aq)+3Cd(s)\)
(ii) \({ Fe }^{ 2+ }(aq)+{ Ag }^{ + }(aq)\rightarrow { Fe }^{ 3+ }(aq)+Ag(s)\)
Calculate the \({ \triangle }_{ r }{ G }^{ o }\) and equilibrium constant for the reactions.

Multiple Choice Question

5 x 1 = 5

One faraday of electricity is passed thorugh molten Al₂ O₃ , aqueous solution of CuSO₄ and molten NaCl taken in three different electrolytic cells connected in series. The mole ratio of Al, Cu and Na deposited at the respective cathode is

(a)

2 : 3 : 6

(b)

6 : 2 : 3

(c)

6 : 3 : 2

(d)

1 : 2 : 3

(e)

3 : 6 : 2

Using the data given below find out the strongest reducing agent.
\({ E }_{ { Cr }_{ 2 }{ O }_{ 7 }^{ 2- }/{ Cr }^{ 3+ } }^{ \circleddash }=1.33V\ ,\ { E }_{ { Cl }_{ 2 }/{ Cl }^{ - } }^{ \circleddash }=1.36V\)
\({ E }_{ { MnO }_{ 4 }^{ - }/{ Mn }^{ 2+ } }^{ \circleddash }=1.51V\ ,\ { E }_{ { Cr }^{ 3+ }/{ Cr } }^{ \circleddash }=-0.74V\)

(a)

Cl^-

(b)

(c)

Cr³⁺

(d)

Mn²⁺

10.

For the reduction of silver ions with copper metal, the standard cell potential was found to be + 0.46 V at 25°C. The value of standard Gibbs energy, \({ \Delta G }^{ ° }\) will be (F = 96500 C mol^-1)

(a)

- 98.0 kJ

(b)

- 89.0 kJ

(c)

- 89.0 J

(d)

- 44.5 kJ

11.

Which has the highest oxidizing power ?

(a)

I₂

(b)

Br₂

(c)

F₂

(d)

Cl₂

12.

The time required to liberate one gram equivalent of an element by passing one ampere current through its solution is

(a)

6.7 hrs

(b)

13.4 hrs

(c)

19.9 hrs

(d)

26.8 hrs

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Answers Key:

Generic 2

2 x 2 = 4

\(A_m={1000\times k\over M}\)
138.9 S cm² mol^-1=\(1000\times k\over 1.5\)
\(k={138.9\times 1.5\over 1000}={208.05\over 1000}={2.0805}\times 10^{-1}\ S\ cm^{-1}\)

\(k=\frac { 1 }{ R } \times \frac { l }{ a } \), where 'k' is conductivity of solution, 'R' is resistance, \(\frac { l }{ a } \) is cell constant. \({ \Lambda }_{ m }=\frac { 1000k }{ M } \), where 'k' is conductivity in S cm^-1, 'M' is molarity of solution. \({ \Lambda }_{ m }\) is molar conductivity in S cm²mol^-1.

Generic 3

2 x 3 = 6

(i) Mg, Al, Zn, Fe, Cu is decreasing order of their reactivity.
(ii) Ag, Hg, Cr, Mg, K is increasing order of their reducing power.

\(K=1.29\times { 10 }^{ -2 }{ ohm }^{ -1 }{ cm }^{ -1 }\)
\(K=\frac { 1 }{ R } \times \frac { 1 }{ a } \Rightarrow \frac { 1 }{ a } =K\times R=1.29\times { 10 }^{ -2 }\times 85=109.65\times { 10 }^{ -2 }=1.0965{ cm }^{ -1 }\)
\({ \wedge }_{ m }=\frac { 1000K }{ M } =\frac { 1000 }{ M } \times \frac { 1 }{ R } \times \frac { 1 }{ a } \)
\({ \wedge }_{ m }=\frac { 1000\times 1\times 1.0965 }{ 0.052\times 96 } =\frac { 1096.50 }{ 4.992 } =219.65 \ S \ { cm }^{ 2 }{ mol }^{ -1 }\)

Generic 5

3 x 5 = 15

For Daniell cell, Zn + Cu²⁺⇾ Zn²⁺+ Cu
\(E_{cell}=E^0_{cell}-{0.0591\over 2}log{[Zn^{2+}]\over [Cu^{2+}]}\)
It is the equation of straight line (y = C + mx).
Intercept = E^O_cell= 1·10 V (Given)
\(E_{cell}\equiv 1.10-{0.0591\over 2}log{0.1\over 0.01}=1.10-0.0295=1.0705V\)

(i) \(2Fe^{3+}+2I^-\longrightarrow2Fe^{2+}+I_2\)
\(E^0_{cell}=E^0_{I_2/ I^-}-E^0_{Fe^{3+}/Few^{2+}}=0.54 V - (0.77 V) = 0.54 V - 0.11 v = - 0.23 V\)
It is not feasible therefore E⁰_cell is -ve ΔG^o = +ve
(ii) 2Ag⁺(aq) + Cu(s) ⟶ Cu²⁺(aq) + Ag(s)
\(E^0_{cell} = E^0_{Ag^+/Ag} -E^0_{Cu^{2+}/Cu} = +0.80 V - 0.34 V = 0.46 V\)
It is feasible because E⁰_cell is +ve
∴ ΔG⁰ = - ve
(iii) 2Fe³⁺(aq) + 2Br^-(aq) ⟶ 2Fe²⁺(aq) + Br₂'
\(E^0_{cell} = E^0_{Fe^{3+}/Fe^{2+}}- E^0_{Br_2/Br^-} = 0.77 V - 1.09 V = - 0.32 V\)
Since E⁰_cell is -ve, ΔG^o = +ve therefore, reaction is non spontaneous.
(iv) Ag(s) + Fe³⁺(aq) ⟶ Fe²⁺(aq) + Ag⁺(aq)
\(E^0_{cell} = E^0 _{Fe^{3+}/Fe^{2+}}-E^0_{Ag^+/Ag} = 0.77 V - 0.80 V = - 0.03 V\)
since E⁰_cell is -ve, ΔG^o is +ve therefore, reaction is non spontaneous.
(v) Br₂ + 2Fe²⁺(aq) ⟶ 2Br^-(aq) + 2Fe³⁺(aq)
\(E^0_{cell} = E^0_{Br^2/Br^-} E^0_{Fe^{3+}/Fe^{2+}}=1.09 V - 0.77 V = 0.32 V\)
Since E⁰_cell is +ve, ΔG^o = -ve, therefore, reaction will be spontaneous.

(i) \(E^0_{cell}=E^0_{cathode}-E^0_{anode}= - 0·40 V - (- 0·74 V) = + 0·34 V\)
\(Δ_r G^0 - n FE^o_{cell}= - 6 mol \times 96500\ C\ mol.^{-1}\times 0·34 V\)
= -196860 CV mol^-1 = -196860 J mol^-1 = -196.86 k.J mol^-1
- Δ_rG^o = 2.303 RT log K
196860 = 2.303 x 8.314 x 298 log K or log K = 34.5014
K = Antilog 34.5014 = 3.192 x 10³⁴
(ii) E⁰_cell = + 0.80 V - 0.77 V = + 0.03 V.
Δ_r G⁰ = - nF E⁰_cell= - (1 mol) x (96500 C mol^-1) x (0·03 V)
= - 2895 CV mol^-1 = - 2895 J mol^-1
= - 2.895 k.J mol^-1
Δ_rG⁰= - 2.303 RT log K
- 2895 = - 2.303 x 8.314 x 298 x log K
or log K = 0.5074 or K = Antilog (0.5074) = 3.22.