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Q. No.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.
\(\mathrm{K}<\mathrm{Mg}<\mathrm{Cr}<\mathrm{Hg}<\mathrm{Ag}\)
The correct arrangement of increasing order for reducing power of elements is:
Subtopic: Emf & Electrode Potential |
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Subtopic: Introduction to Redox and Oxidation Number |
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Subtopic: Introduction to Redox and Oxidation Number |
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Nitric acid reacts with PbO but does not react with PbO2 , because -
1. PbO is a base while PbO2 is a strong oxidizing reagent
2. PbO is a base while PbO2 is a weak oxidizing reagent
3. PbO is neutral while PbO2 is a strong oxidizing reagent
4. PbO is acid while PbO2 is a strong oxidizing reagent
Subtopic: Oxidizing & Reducing Agents |
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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 |
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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 |
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Subtopic: Oxidizing & Reducing Agents |
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| Assertion (A): | In the presentation \( \mathrm{E}_{\mathrm{Fe}^{3+}}^{\ominus} /_{ \mathrm{Fe}^{2+}}\) \( \text { and } \mathrm{E}_{\mathrm{Cu}^{2+}}^{\ominus} / _{\mathrm{Cu}^{\prime}} \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+} \) and\(\text { } \mathrm{Cu}^{2+} / \mathrm{Cu}\) are redox couples. |
| 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 |
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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 |
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Using the standard electrode potentials of the redox couples given below, which is the strongest oxidizing agent?
| \(E^{\ominus} values : \) | \(\left(Fe\right)^{3 +} / \left(Fe\right)^{2 +} = + 0 . 77 V\) |
| \(I_{2} (s)/ I^{-} = + 0 . 54 V\) | |
| \(\left(Cu\right)^{2 +} / Cu = + 0 . 34 V\) | |
| \(\left(Ag\right)^{+} / Ag = 0 . 80 V\) |
1. Fe3+
2. I2(s)
3. Cu2+
4. Ag+
Subtopic: Application of Electrode Potential |
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