(a) \(E_{k^+/K}^o = - 2.93\ V\); \(E_{Ag^+/Ag}^o = 0.80\ V\)
(b) \(E_{Hg^{2+}/Hg}^o = 0.79\ V\); \(E_{Mg^{2+}/Mg}^o = - 2.37\ V\)  
(c) \(E_{Cr^{3+}/Cr}^o = -0.74\ V\)

Based on standard electrode potentials given above, the correct arrangement for increasing order of reducing power of
elements is: 

1. \(\mathrm{Ag}<\mathrm{Hg}<\mathrm{Cr}<\mathrm{Mg}<\mathrm{K} \)
2. \(\mathrm{Ag}>\mathrm{Cr}>\mathrm{Mg}>\mathrm{Hg}>\mathrm{K}\)
3. \(\mathrm{K}>\mathrm{Mg}<\mathrm{Cr}<\mathrm{Hg}>\mathrm{Ag} \)
4. \(\mathrm{K}<\mathrm{Mg}<\mathrm{Cr}<\mathrm{Hg}<\mathrm{Ag}\)
Subtopic:  Emf & Electrode Potential |
 75%
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The compound AgF2 (unstable) acts as a/ an:

1. Oxidising agent.

2. Reducing agent.

3. Both oxidising and reducing agent.

4. Neither oxidising and reducing agent.

Subtopic:  Introduction to Redox and Oxidation Number |
 52%
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Fluorine reacts with ice as per the following reaction

H2O(s) + F2(g) → HF(g) + HOF(g)

This reaction is a redox reaction because-

1. F2 is getting oxidized. 2. F2 is getting reduced.
3. Both (1) and (2) 

4. None of the above.
Subtopic:  Introduction to Redox and Oxidation Number |
 71%
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The oxidation number of sulphur and nitrogen in H2SO5 and NO3- are respectively-

1. +6, +5 2. -6, -6
3. +8, +6 4. -8, -6
Subtopic:  Introduction to Redox and Oxidation Number |
 77%
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The correct statement(s) about the given reaction is -

\(\mathrm{{X} eO_{6 (aq)}^{4 -} + 2 F^{- 1}_{(aq)} + 6 H^{+}_{(aq )} \rightarrow}~\mathrm{{X} eO_{3 (g)} + {F}_{2 (g)} + 3 H_{2} O_{(l)}}\)

1. XeO64- oxidises F-

2. The oxidation number of F increases from -1  to  zero

3. XeO64- is a stronger oxidizing agent that F-

4. All of the above.

Subtopic:  Introduction to Redox and Oxidation Number | Oxidizing & Reducing Agents |
 87%
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The oxidising agent and reducing agent in the given reaction are

5P4s+12H2Ol+12HO-aq
8PH3g+12HPO2-aq

1. Oxidising agent = P4; Reducing agent = P4

2. Oxidising agent = P4; Reducing agent = H2O

3. Oxidising agent = H2O; Reducing agent = P4

4. None of the above

Subtopic:  Redox Titration & Type of Redox |
 71%
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The correct statement about the given reaction is-

(CN)2(g) + 2OH-(aq) CN-(aq) + CNO-(aq) + H2O(l)

1. The reaction is an example of a disproportionation reaction.
2. Hydrogen atom gets oxidized.
3. Reaction occurs in acidic medium.
4. None of the above

Subtopic:  Emf & Electrode Potential |
 82%
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The Mn3+ ion is unstable in solution and undergoes disproportionation reaction to give Mn2+, MnO2 and H+ ion. The balanced ionic equation for the reaction is-

1. \(\small{2 \mathrm{Mn}^{3+}{ }_{(\mathrm{aq})}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \rightarrow \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{Mn}^{2+}{ }_{(\mathrm{aq})}+4 \mathrm{H}^{+}{ }_{(\mathrm{aq})}}\)
2. Mn3+(aq) + H2O(l) → MnO2(s) + 2Mn2+(aq) + 4H+(aq)
3. \(\small{5 \mathrm{Mn}^{3+}(\mathrm{aq})+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \rightarrow \mathrm{MnO}_{2(\mathrm{s})}+3 \mathrm{Mn}^{2+}(\mathrm{aq})+4 \mathrm{H}^{+}(\mathrm{aq})}\)
4. \(\small{2 \mathrm{Mn}^{3+}{ }_{(\mathrm{aq})}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \rightarrow 2 \mathrm{MnO}_{2(\mathrm{s})}+2 \mathrm{Mn}^{2+}{ }_{(\mathrm{aq})}+4 \mathrm{H}^{+}{ }_{(\mathrm{aq})}}\)

Subtopic:  Emf & Electrode Potential |
 81%
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Which element exhibits both positive and negative oxidation states?
1. Cs
2. Ne
3. I
4. F

Subtopic:  Introduction to Redox and Oxidation Number |
 64%
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The balanced equation for the reaction between chlorine and sulphur dioxide in water is:

1. Cl2(s) + SO2(aq) + 2H2O(I) →2Cl-(aq) + SO42-(aq) + 4H+(aq)
2. 3Cl2(s) + SO2(aq) + 2H2O(I) →Cl-(aq) + SO42-(aq) + 3H+(aq)
3. Cl2(s) + 3SO2(aq) + H2O(I) →Cl-(aq) + 2SO42-(aq) + 4H+(aq)
4. 2Cl2(s) + SO2(aq) + H2O(I) →2Cl-(aq) + SO42-(aq) + 4H+(aq)

Subtopic:  Balancing of Equations |
 85%
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