EΘ values of some redox couples are given below. On the basis of these values choose the correct option.

EΘ values: Br2/Br-=+1.90
Ag+/Ag(s)=+0.80
Cu2+/Cu(s)=+0.34; 
I2(s)/I-=+0.54

1. Cu will reduce Br- 2. Cu will reduce Ag
3. Cu will reduce I- 4. Cu will reduce Br2

Subtopic:  Emf & Electrode Potential |
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Using the standard electrode potential, find out the pair between which redox reaction is not feasible.

E values: Fe3+/Fe2+=+0.77; I2/I-=+0.54;
Cu2+/Cu=+0.34; Ag+/Ag=+0.80 V

1. Fe3+ and I-

2. Ag+ and Cu

3. Fe3+ and Cu

4. Ag and Fe3+

Subtopic:  Application of Electrode Potential |
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In the reactions given below, thiosulphate reacts differently with iodine than with bromine.

2S2O32-+I2S4O62-+2I-

S2O32-+2Br2+5H2O2SO42-+2Br-+10H+

Choose the statements among the following that best describe the above dual behaviour of thiosulphate.

1. Bromine is a stronger oxidant than iodine
2. Bromine is a weaker oxidant than iodine
3. Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reactions
4. Bromine undergoes oxidation and iodine undergoes reduction in these reactions

Subtopic:  Oxidizing & Reducing Agents |
 61%
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The incorrect statement regarding the rule to find the oxidation number among the following is-

1. The oxidation number of hydrogen is always +1.
2. The algebraic sum of all the oxidation numbers carried by elements in a compound is zero.
3. An element in its free or uncombined state has an oxidation number of zero.
4. Generally, in all its compounds, the oxidation number of fluorine is -1.

Subtopic:  Oxidizing & Reducing Agents |
 76%
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In which of the following compounds, an element exhibits two different oxidation state?

1. NH2OH

2. NH4NO3

3. N2H4

4. N3H

Subtopic:  Oxidizing & Reducing Agents |
 84%
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Which of the following arrangements represents the increasing oxidation number of the central atom?

1. CrO2-, ClO3-, CrO42-, MnO4-

2. ClO3-, CrO42-, MnO4-, CrO2-

3. CrO2-, ClO3-, MnO4-, CrO42-

4. CrO42-, MnO4-, CrO2-, ClO3-

Subtopic:  Oxidizing & Reducing Agents |
 88%
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The largest oxidation number exhibited by an element depends on its electronic configuration. With which of the following electronic configurations will the element exhibit the largest oxidation number?

1. 3d14s2

2. 3d34s2

3. 3d54s1

4. 3d54s2

Subtopic:  Oxidizing & Reducing Agents |
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Match Column I with Column II for the oxidation states of the central atoms.

Column I

Column II

A. Cr2O72-

1. +3

B. MnO4-

2. +4

C. VO3-

3. +5

D. FeF63-

4. +6

5. +7

Codes

Options:  A   B   C   D 
1.  4  5   3   1
2.  1  2  3  5
3.  5  4  3   2
4.  4  5   3  2

Subtopic:  Introduction to Redox and Oxidation Number |
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Match the items in Column I with relevant items in Column II.

Column I

Column II

A. Ions having a positive charge

1. +7

B. The sum of oxidation number of all atoms in a neutral molecule

2. -1

C. Oxidation number of hydrogen ion. (H+)

3. +1

D. Oxidation number of fluorine in NaF

4. 0

E. Ions having a negative charge

5. Cation

6. Anion

Codes

Options:  A   B   C   D   E 
1. 5 4 3 2  6
2. 1 2 3 5  4
3. 5 4 3 2 1
4. 4 5 3 2 1
 

     

Subtopic:  Introduction to Redox and Oxidation Number |
 93%
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Given below are two statements: 

Assertion (A): Among halogens, fluorine is the best oxidant.
Reason (R): Fluorine is the most electronegative atom.
 
1. Both (A) and (R) are true and (R) is the correct explanation of (A).
2. Both (A) and (R) are true but (R) is not the correct explanation of (A).
3. (A) is true but (R) is false.
4. (A) is false but (R) is true.

Subtopic:  Emf & Electrode Potential |
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