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
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 |
\(\text{MnO}_{4}^{-}+8 \text{H}^{+}+5 \text{e}^{-} \rightarrow \text{Mn}^{2+}+4 \text{H}_{2} \text{O}, \)
\( \text{E}_{\text{Mn}^{2+}}^{\circ} / \text{MnO}_{4}^{-}=-1.510 \text{ V} \)
\( \frac{1}{2} \text{O}_{2}+2 \text{H}^{+}+2 \text{e}^{-} \rightarrow \text{H}_{2} \text{O}, \)
\( \text{E}_{\text{O}_{2} / \text{H}_{2} \text{O}}^{\circ}=+1.223 \text{ V}\)
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 \(\text{E}_{\text {cell }}^{\circ}=+2.733 \text{ V}\)
Consider the change in oxidation state of bromine corresponding to different emf values as shown in the diagram below:
The species undergoing disproportionation is:
| 1. | \(\text{BrO}^-_3\) | 2. | \(\text{BrO}^-_4\) |
| 3. | \(\text{Br}_2\) | 4. | \(\text{HBrO}\) |
Consider the given cell:
\(\mathrm{Z n \left|\right. Z n S O_{4} \left(\right. 0 . 01 M \left.\right) \left|\right. \left|\right. C u S O_{4} \left(\right. 1 . 0 M \left.\right) \left|\right. C u}\)
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. | \(\mathrm{E_{1} < E_{2}}\) | 2. | \(\mathrm{E_{1} > E_{2}}\) |
| 3. | \(\mathrm{E_{2} = 0 \neq E_{1}}\) | 4. | \(\mathrm{E_{1} = E_{2}}\) |
\(\text{Mn}^{2+}+2e^-\rightarrow \text{Mn},\ E^{0}= -1.18~ \text V \)
\(\text{Mn}^{2+}\rightarrow \text{Mn}^{3+}+e^-,\ E^{0}= -1.51~ \text V \)
The \(E^{0}\) for the reaction \(\mathrm{3\ Mn^{2+}\rightarrow Mn^{0}+2Mn^{3+} }\) and possibility of the forward reaction are respectively:
| 1. | –4.18 V and Yes | 2. | +0.33 V and Yes |
| 3. | +2.69 V and No | 4. | –2.69 V and No |
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 |
Find the emf of the cell in which the following reaction takes place at 298 K:
\(\mathrm{Ni}(\mathrm{s})+2 \mathrm{Ag}^{+}(0.001 \mathrm{M}) \rightarrow \mathrm{Ni}^{2+}(0.001 \mathrm{M})+2 \mathrm{Ag}(\mathrm{s}) \)
\(\small{\text { (Given that } \mathrm{E}_{\text {cell }}^{\circ}=1.05 \mathrm{~V}; \dfrac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059} )\)
| 1. | 1.05 V | 2. | 1.0385 V |
| 3. | 1.385 V | 4. | 0.9615 V |
The pressure of H2 required to make the potential of H2 - electrode zero in pure water at 298 K is:
| 1. | 10–12 atm | 2. | 10–10 atm |
| 3. | 10–4 atm | 4. | 10–14 atm |
\(\mathrm{{Zn}({s})+{Fe}^{2+}({aq}) \rightarrow {Zn}^{2+}({aq})+{Fe}({s})}\)
(Given : \(1 \mathrm{~F}=96487 \mathrm{C} mol^{-1}\))
| 1. | \(-61.75 \mathrm{{~kJ} {~mol}}^{-1}\) | 2. | \(+5.006 \mathrm{{~kJ} {~mol}}^{-1}\) |
| 3. | \(-5.006 \mathrm{{~kJ} {~mol}}^{-1}\) | 4. | \(+61.75 \mathrm{{~kJ} {~mol}}^{-1}\) |
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