The standard electrode potential (E°) values of Al3+/ Al, Ag+ / Ag, K+ / K, and Cr3+ / Cr are –1.66 V, 0.80 V, –2.93 V, & –0.79 V respectively. The correct decreasing order of the reducing power of the metal is:
1. | Ag > Cr > Al > K | 2. | K > Al > Cr > Ag |
3. | K > Al > Ag > Cr | 4. | Al > K > Ag > Cr |
In the given balanced chemical reaction
\(\mathrm{IO}_3^{-}+\mathrm{aI}^{-}+\mathrm{bH}^{+} \rightarrow \mathrm{cH}_2 \mathrm{O}+\mathrm{dI}_2\)
The values of a, b, c, and d respectively are:
1. 5, 6, 3, 3
2. 5, 3, 6, 3
3. 3, 5, 3, 6
4. 5, 6, 5, 5
Balance the following equation and choose the quantity that is the sum of the coefficients of reactants and products :
1. 16
2. 13
3. 18
4. 12
Which of the following reactions does not involve disproportionation process?
1.
2.
3.
4.
Among the following the correct order of acidity is-
1. HClO < HClO2< HClO3< HClO4
2. HClO2< HClO < HClO3< HClO4
3. HClO4< HClO2< HClO < HClO3
4. HClO3< HClO4< HClO2< HClO
The change in oxidation number of chlorine when Cl2 gas reacts with hot and concentrated sodium hydroxide solution is:
1. Zero to +1 and Zero to –5
2. Zero to –1 and Zero to +5
3. Zero to –1 and Zero to +3
4. Zero to +1 and Zero to –3
For the redox reaction,
a
the correct stoichiometric coefficients of the reactants a, b, and c respectively for the balanced equation are:
1. 16, 5, 2
2. 2, 5, 16
3. 2, 16, 5
4. 5, 16, 2
What is the alteration in the oxidation state of carbon in the given reaction?
\(\mathrm{{CH_4}_{(g)} + 4{Cl_2}_{(g)} \rightarrow {CCl_4}_{(l)} + 4 HCl_{(g)}}\)
1. | 0 to +4 | 2. | –4 to +4 |
3. | 0 to –4 | 4. | +4 to +4 |
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. |
(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. | \(\mathrm{K}<\mathrm{Mg}<\mathrm{Cr}<\mathrm{Hg}<\mathrm{Ag}\) |