Q. No.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:
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. | \(\mathrm{E_{1} = E_{2}}\) |
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. | \(+61.75 \mathrm{{~kJ} {~mol}}^{-1}\) |
| Assertion (A): | In the equation, \(\Delta_{\mathrm{r}} \mathrm{G}=-\mathrm{nFE} _{\text {cell }}, \) value of \(\mathrm{\Delta_rG }\) depends on n. |
| Reason (R): | \(\mathrm{E_{cell} }\) is an intensive property and \(\mathrm{\Delta_rG }\) is an extensive property. |
| 1. | (A) is False but (R) is True. |
| 2. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
| 3. | Both (A) and (R) are True and (R) is not the correct explanation of (A). |
| 4. | (A) is True but (R) is False. |
For a cell involving one electron \(E_{cell}^{\ominus} = 0 . 59 V\) at 298 K. The equilibrium constant for the cell reaction is :
\(\mathrm{[Given~ that~ \frac {2.303 ~RT}{F} = 0.059 ~V~ at~ T = 298 K]}\)
| 1. | \(1 . 0 \times \left(10\right)^{30}\) | 2. | \(1 . 0 \times \left(10\right)^{2}\) |
| 3. | \(1 . 0 \times \left(10\right)^{5}\) | 4. | \(1 . 0 \times \left(10\right)^{10}\)
|
If the Eocell for a given reaction has a negative value, which of the following gives correct relationships for the values of ∆Go and Keq?
1.
2.
3.
4.
© 2025 GoodEd Technologies Pvt. Ltd.