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TN 12th Computer Applications மின்னணு தரவு பரிமாற்றம் Sample Question Papers Study Material - QB365 Set A

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TN 12th Computer Applications மின் - வணிக பாதுகாப்பு அமைப்புகள் Sample Question Papers Study Material - QB365 Set A

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TN 12th Computer Applications மின்னணு செலுத்தல் முறைகள் Sample Question Papers Study Material - QB365 Set A

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TN 12th Computer Applications மின் - வணிகம் Sample Question Papers Study Material - QB365 Set A

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TN 12th Computer Applications வலையமைப்பு வடமிடல் Sample Question Papers Study Material - QB365 Set A

12th Standard English Medium Chemistry Subject Electro Chemistry Book Back 1 Mark Questions with Solution Part - II

Published on: 01/06/2021

QB365 provides detailed and simple solution for every Book back Questions in class 12 Chemistry Subject. It will helps to get more idea about question pattern in every book back questions with solution.

Download Tamil Nadu 12th Standard Chemistry question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.

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Std: 12th Standard - Chemistry

chapter: - Book back Questions

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Questions:

Multiple Choice Question

5 x 1 = 5

A gas X at 1 atm is bubbled through a solution containing a mixture of 1MY^-and 1MZ^- at 25^oC. If the reduction potential of Z>Y>X, then_____.

(a)

Y will oxidize X and not Z

(b)

Y will oxidize Z and not X

(c)

Y will oxidize both X and Z

(d)

Y will reduce both X and Z

A certain current liberated 0.504gm of hydrogen in 2 hours. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution ______.

(a)

31.75

(b)

15.8

(c)

7.5

(d)

63.5

For the cell reaction
2Fe³⁺(aq) + 2l⁻(aq) \(\rightarrow\)2Fe²⁺ (aq) + 1₂ (aq)
E^o_cell= 0.24V = at 298K. The standard Gibbs energy (Δ, G^o) of the cell reactions is:

(a)

-46.32 KJ mol⁻¹

(b)

-23.16 KJ mol^-1

(c)

46.32 KJ mol⁻¹

(d)

23.16 KJ mol^-1

Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below:
\({ BrO }_{ 4 }^{ - }\overset { 1.82V }{ \longrightarrow } { BrO }_{ 3 }^{ - }\overset { 1.5V }{ \longrightarrow } HBrO\overset { 1.595v }{ \longrightarrow } { Br }_{ 2 }\overset { 1.0652V }{ \longrightarrow } Br^{ - }\)
Then the species undergoing disproportional is

(a)

Br₂

(b)

BrO₄^-

(c)

BrO^-₃

(d)

HBrO

In the electrochemical cell: Zn|ZnSO₄ _(0.01M)|| CuSO₄ _(1.0M)|Cu, the emf of this Daniel cell is E₁. When the concentration of ZnSO₄ is changed to 1.0M and that CuSO₄ changed to 0.01M, the emf changes to E₂. From the above, which one is the relationship between E₁ and E₂?

(a)

E₁ < E₂

(b)

E₁ > E₂

(c)

E₂ ≥ E₁

(d)

E₁= E₂

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Answers Key:

Multiple Choice Question

5 x 1 = 5

Z is tie strongest oxidising agent (High SRP)
X is the strongest reducing agent (Low SRP)

\(\frac{m_{1}}{m_{2}} = \frac{E_{1}}{E_{2}} \)
\(\frac{0.504}{m_{2}} = \frac{1.008}{31.77} \)
m₂ = 15.885 g

ΔG^o = -nFE^o_cell
= -2 x 96500 x 0.24
= - 46320 J mol^-1
= -46.32 KJ mol⁻¹

(E_cell)_A = -1.82 + 1.5 = -0.32V
(E_cell)_B = -1.5 + 1.595 = + 0.095V
(E_cell)_C = -1.595 + 1.0652 = -0.529V
The species undergoing disproportionation is HBrO

E_cell= E^o_cell- \(\frac{0.0591}{2} \log [\frac{Zn^{2+}}{Cu^{2+}}]\)
E_l= E^o_cell- \(\frac{0.0591}{2} \log [\frac{10^{-2}}{1}]\)
E_l= E^o_cell+ 0.0591 ...(1)
E₂= E^o_cell- \(\frac{0.0591}{2} \log [\frac{1}{10^{-2}}]\)
E₂= E^o_cell- 0.0591 ...(2)
E₁ > E₂
Zn_(s)| ⟶ Zn²⁺_(aq) + 2e^-
Cu²⁺_(aq) + 2e^-⟶ Cu_(s)
Zn_(s) + Cu²⁺_(aq) ⟶ Zn²⁺_(aq) + Cu_(s)