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Q. No.Subtopic: Introduction to Redox and Oxidation Number |
85%
Level 1: 80%+
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Subtopic: Introduction to Redox and Oxidation Number |
Level 3: 35%-60%
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Consider the standard electrode potentials given below:
| (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) | \(E_{Cr^{3+}/Cr}^o = -0.74\ V\) |
The correct arrangement of increasing order for 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 |
76%
Level 2: 60%+
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Why does the MnO₄²⁻ ion undergo disproportionation in an acidic medium, whereas the MnO₄⁻ ion does not?
| 1. | Due to manganese being in its highest oxidation state in MnO₄²⁻. |
| 2. | Due to manganese being in its highest oxidation state in MnO₄⁻ |
| 3. | Because the disproportionation reaction of MnO₄²⁻ is endothermic. |
| 4. | Because the disproportionation reaction of MnO₄²⁻ is exothermic. |
Subtopic: Introduction to Redox and Oxidation Number |
81%
Level 1: 80%+
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In the given reaction, what is the name of the species that bleaches the substances due to its oxidising action?
Cl2(g) + 2OH-(aq) → ClO-(aq) + Cl-(aq) + H2O(l)
1. ClO-
2. Cl2
3. Cl-
4. Both ClO- and Cl-
Subtopic: Oxidizing & Reducing Agents |
51%
Level 3: 35%-60%
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The oxidation number of the atom (in bold) in the following species is given. Identify, which one is incorrectly related?
1. \(\mathbf{Cu}_{2} O\) is \(- 1\)
2. \(\mathbf{Cl} O_{3}^{-}\) is \(+ 5\)
3. \(K_{2} \mathbf{Cr}_{2} O_{7}\) is \(+ 6\)
4. \(H \mathbf{Au} Cl_{4}\) is \(+ 3\)
Subtopic: Oxidizing & Reducing Agents |
85%
Level 1: 80%+
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Subtopic: Oxidizing & Reducing Agents |
86%
Level 1: 80%+
Hints
Given below are two statements:
| Assertion (A): | In the representation Fe³⁺/Fe²⁺, Ce⁴⁺/Ce³⁺ , and Cu²⁺/Cu, each pair represents a redox couple. |
| Reason (R): | A redox couple is the combination of the oxidised and reduced forms of a substance involved in an oxidation or reduction half-cell. |
| 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: Introduction to Redox and Oxidation Number |
74%
Level 2: 60%+
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Given below are two statements:
| Assertion (A): | The decomposition of hydrogen peroxide to form water and oxygen is an example of a disproportionation reaction. |
| Reason (R): | The oxygen of peroxide is in –1 oxidation state and it is converted to zero oxidation state in O2 and –2 oxidation state in H2O. |
| 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: Redox Titration & Type of Redox |
90%
Level 1: 80%+
Hints
Which species is the strongest oxidizing agent based on the given standard electrode potentials?
| \(E^{\ominus} values : \) | \(\left(Fe\right)^{3 +} / \left(Fe\right)^{2 +} = + 0 . 77 V\) |
| \(I_{2} (s)/ I^{-} = + 0 . 54 V\) | |
| \(\left(Cu\right)^{2 +} / Cu = + 0 . 34 V\) | |
| \(\left(Ag\right)^{+} / Ag = +0 . 80 V\) |
1. Fe3+
2. I2(s)
3. Cu2+
4. Ag+
Subtopic: Application of Electrode Potential |
83%
Level 1: 80%+
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