Standard reduction potentials of the half-reactions are given below:
The strongest oxidizing and reducing agents, respectively, are:
1. and
2. and
3. and
4. and
A solution contains Fe2+, Fe3+ and I– ions. This solution was treated with iodine at 35°C. E° for Fe3+/Fe2+ is +0.77 V and E° for I2/2I– = 0.536 V.
The favourable redox reaction is:
1. Fe2+ will be oxidized to Fe3+.
2. I2 will be reduced to I–.
3. There will be no redox reaction.
4. I– will be oxidized to I2.
From the following, identify the reaction having the top position in the EMF series (standard reduction potential) according to their electrode potential at 298 K.
1. Mg2++ 2e–Mg(s)
2. Fe2+ + 2e– Fe(s)
3. Au3++ 3e–Au(s)
4. K++ le –K(s)
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. |
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 |
The formulas for the following compounds are:
(a) Mercury(II) chloride and (b) Thallium(I) sulphate
1. HgCl2, Tl2SO4
2. Hg2Cl2, Tl2SO4
3. HgCl2, TlSO4
4. HgCl2, Tl3SO4
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. oxidises F-
2. The oxidation number of F increases from -1 to zero
3. is a stronger oxidizing agent that
4. All of the above.
The oxidising agent and reducing agent in the given reaction are
1. Oxidising agent = ; Reducing agent =
2. Oxidising agent = ; Reducing agent =
3. Oxidising agent = ; Reducing agent =
4. None of the above
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 |
The ion is unstable in solution and undergoes disproportionation reaction to give 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. | \(\small{2 \mathrm{Mn}^{3+}{ }_{(\mathrm{aq})}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \rightarrow 2 \mathrm{MnO}_{2(\mathrm{s})}+2 \mathrm{Mn}^{2+}{ }_{(\mathrm{aq})}+4 \mathrm{H}^{+}{ }_{(\mathrm{aq})}}\) |