Q. No.\(\mathrm{Fe}_{\text {(aq) }}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe}(\mathrm{s}), \mathrm{E}^{\circ}=-0.44 \mathrm{~V} \)
\( \mathrm{Cr}_2 \mathrm{O}_7^{2-}{ }_{\text {(aq) }}+14 \mathrm{H}^{+}+6 e^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O},\)
\( \mathrm{E}^{\circ}=+1.33 \mathrm{~V}\)
| 1. | +1.77 V | 2. | +2.65 V |
| 3. | +0.01 V | 4. | +0.89 V |
\(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}, \)
\( \mathrm{E}_{\mathrm{Mn}^{2+}}^{\circ} / \mathrm{MnO}_{4}^{-}=-1.510 \mathrm{~V} \)
\( \frac{1}{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}, \)
\( \mathrm{E}_{\mathrm{O}_{2} / \mathrm{H}_{2} \mathrm{O}}^{\circ}=+1.223 \mathrm{~V}\)
Will the permanganate ion, \(\mathrm{MnO}_{4}^{-}\) , liberate \(\mathrm{O}_{2}\) from water in the presence of an acid?
1. No, because \(\mathrm{E}_{\text {cell }}^{\circ}=-2.733 \mathrm{~V}\)
2. Yes, because \(\mathrm{E}_{\text {cell }}^{\circ}=+0.287 \mathrm{~V}\)
3. No, because \(\mathrm{E}_{\text {cell }}^{\circ}=-0.287 \mathrm{~V}\)
4. Yes, because \(\mathrm{E}_{\text {cell }}^{\circ}=+2.733 \mathrm{~V}\)
Two half cell reactions are given below:
\(\begin{aligned} &\mathrm{Co}^{3+}+e^{-} \rightarrow \mathrm{Co}^{2+}, \mathrm{E}_{\mathrm{Co}^{2+} / \mathrm{Co}^{3+}}^{\circ}=-1.81 \mathrm{~V} \\ &2 \mathrm{Al}^{3+}+6 e^{-} \rightarrow 2 \mathrm{Al}(\mathrm{s}), \mathrm{E}_{\mathrm{Al} / \mathrm{Al}^{3+}}^{\circ}=+1.66 \mathrm{~V} \end{aligned} \)
The standard EMF of a cell with feasible redox reaction will be:
| 1. | +7.09 V | 2. | +0.15 V |
| 3. | +3.47 V | 4. | –3.47 V |
Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below:
Then the species undergoing disproportionation is:-
| 1. | \(\text{BrO}^-_3\) | 2. | \(\text{BrO}^-_4\) |
| 3. | \(\text{Br}_2\) | 4. | \(\text{HBrO}\) |
In the electrochemical cell:
the emf of this Daniel cell is E1. When the concentration of ZnSO4 is changed to 1.0 M and that of CuSO4 is changed to 0.01 M, the emf changes to E2. From the following, which one is the relationship between E1 and E2 ?
(Given, \(\frac{RT}{F}\) = 0.059)
1.
2.
3.
4.
Zn(s) + Ag2O(s) + H2O(l) \(\rightleftharpoons\) 2Ag(s) + Zn2+(aq) + 2OH–(aq)
If half-cell potentials are-
| Zn2+(aq) + 2e–→ Zn(s) | Eo = – 0.76 V |
| Ag2O(s) + H2O(l) + 2e– → 2Ag(s) + 2OH–(aq) | Eo = 0.34 V |
The cell potential will be:
| 1. | 0.42 V | 2. | 0.84 V |
| 3. | 1.34 V | 4. | 1.10 V |
| 1. | Y > X > Z | 2. | Z > X > Y |
| 3. | X > Y > Z | 4. | Y > Z > X |
| Cu2+(aq) + e- → Cu+(aq) | 0.15 V |
| Cu+(aq) + e- → Cu(s) | 0.50 V |
The value of \(E_{Cu^{2+}/Cu}^{o}\) will be:
| 1. | 0.325 V | 2. | 0650 V |
| 3. | 0.150 V | 4. | 0.500 V |
Standard electrode potential for Sn4+/Sn2+ couple is +0.15 V and that for Cr3+/Cr couple is -0.74. These two couples in their standard state are connected to make a cell. The cell potential will be:
1. +0.89 V
2. +0.18 V
3. +1.83 V
4. +1.199 V
Consider the following relations for emf of an electrochemical cell:
| (a) | emf of cell = (Oxidation potential of anode) – (Reduction potential of cathode) |
| (b) | emf of cell = (Oxidation potential of anode) + (Reduction potential of cathode) |
| (c) | emf of cell = (Reduction potential of anode) + (Reduction potential of cathode) |
| (d) | emf of cell = (Oxidation potential of anode) – (Oxidation potential of cathode) |
The correct relation among the given options is:
| 1. | (a) and (b) | 2. | (c) and (d) |
| 3. | (b) and (d) | 4. | (c) and (a) |
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