Q. No.\(\mathrm{{Fe}_{ {(aq) }}^{2+}+2 {e}^{-} \rightarrow {Fe}({s}), {E}^{\circ}=-0.44{~V} }\)
\( \mathrm{{Cr}_2 {O}_7^{2-}{ }_{ {(aq) }}+14 {H}^{+}+6 e^{-} \rightarrow 2 {Cr}^{3+}+7 {H}_2 {O}},\)
\( \mathrm{{E}^{\circ}=+1.33 {~V}}\)
1. +1.77 V
2. +2.65 V
3. +0.01 V
4. +0.89 V
\(\text{MnO}_{4}^{-}+8 \text{H}^{+}+5 \text{e}^{-} \rightarrow \text{Mn}^{2+}+4 \text{H}_{2} \text{O}, \)
\( \text{E}_{\text{Mn}^{2+}}^{\circ} / \text{MnO}_{4}^{-}=-1.510 \text{ V} \)
\( \frac{1}{2} \text{O}_{2}+2 \text{H}^{+}+2 \text{e}^{-} \rightarrow \text{H}_{2} \text{O}, \)
\( \text{E}_{\text{O}_{2} / \text{H}_{2} \text{O}}^{\circ}=+1.223 \text{ V}\)
Will the permanganate ion, \(\text{MnO}_{4}^{-}\) , liberate \(\text{O}_{2}\) from water in the presence of an acid?
1. No, because \(\text{E}_{\text {cell }}^{\circ}=-2.733 \text{ V}\)
2. Yes, because \(\text{E}_{\text {cell }}^{\circ}=+0.287 \text{ V}\)
3. No, because \(\text{E}_{\text {cell }}^{\circ}=-0.287 \text{ V}\)
4. Yes, because \(\text{E}_{\text {cell }}^{\circ}=+2.733 \text{ V}\)
Two half cell reactions are given below:
\(\begin{aligned} &\mathrm{{Co}^{3+}+e^{-} \rightarrow {Co}^{2+}, {E}_{{Co}^{2+} / {Co}^{3+}}^{\circ}=-1.81 {~V}} \\ &2 \mathrm{{Al}^{3+}+6 e^{-} \rightarrow 2 {Al}({s}), {E}_{{Al} / {Al}^{3+}}^{\circ}=+1.66 {~V}} \end{aligned} \)
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:
\(\mathrm{Z n \left|\right. Z n S O_{4} \left(\right. 0 . 01 M \left.\right) \left|\right. \left|\right. C u S O_{4} \left(\right. 1 . 0 M \left.\right) \left|\right. C u}, \)
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. \(\mathrm{E_{1} < E_{2}}\)
2. \(\mathrm{E_{1} > E_{2}}\)
3. \(\mathrm{E_{2} = 0 \neq E_{1}}\)
4. \(\mathrm{E_{1} = E_{2}}\)
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) |
Which of the following combinations correctly represents the relation for the emf of the cell?
| 1. | (a) and (b) | 2. | (c) and (d) |
| 3. | (b) and (d) | 4. | (c) and (a) |
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