Q. No.The cell that will measure the standard electrode potential of a copper electrode is:
| 1. | Pt(s) | H2(g, \(1 \over 10\) bar) | H+(aq, 1M) || Cu2+(aq, 1M) | Cu |
| 2. | Pt(s) | H2(g, 1 bar) | H+(aq, 1M) || Cu2+ (aq, 2M) | Cu |
| 3. | Pt(s) | H2(g, 1 bar) | H+(aq, 1M)|| Cu2+ (aq, 1M) | Cu |
| 4. | Pt(s) | H2(g, \(1 \over 10\) bar) | H+(aq, \(1 \over 10\)M) || Cu2+(aq, 1M) | Cu |
For the cell, Mg | Mg2+ || Cu2+ | Cu
a. Mg acts as cathode
b. Cu acts as cathode
c. The cell reaction is \(Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu\)
d. Cu is the oxidising agent
The correct choice among the given is:
| 1. | (a, b) | 2. | (b, c) |
| 3. | (c, d) | 4. | (a, d) |
1. Velocity of \(K^+\) is greater than that of \(NO^-_3\).
2. Velocity of \(NO^-_3\) is greater than that of \(K^+\) .
3. Velocities of both \(K^+\) and \(NO^-_3\) are nearly the same.
4. \(KNO_3\) is highly soluble in water.
\(E^\circ_{Fe/Fe^{2+}}=0.44~V~\text{and}~E^\circ_{Sn/Sn^{2+}}=0.14~V. \)
The \(E^\circ \) for the cell: \(\mathrm{Fe}(\mathrm{~s})+\mathrm{Sn}^{2+}(a q) \rightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{Sn}(\mathrm{~s})\) will be:
| 1. | 0.58 V | 2. | –0.30 V |
| 3. | 0.30 V | 4. | –0.58 V |
The difference between the electrode potentials of two electrodes when no current is drawn through the cell is called:
1. Cell potential.
2. Cell emf.
3. Potential difference.
4. Cell voltage.
\(\text{MnO}_{4}^{-}+8 \text{H}^{+}+5 \text{e}^{-} \rightarrow \text{Mn}^{2+}+4 \text{H}_{2} \text{O}, \)
\( \text{E}_{\text{Mn}^{2+}}^{\circ} / \text{MnO}_{4}^{-}=-1.510 \text{ V} \)
\( \frac{1}{2} \text{O}_{2}+2 \text{H}^{+}+2 \text{e}^{-} \rightarrow \text{H}_{2} \text{O}, \)
\( \text{E}_{\text{O}_{2} / \text{H}_{2} \text{O}}^{\circ}=+1.223 \text{ V}\)
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 \(\text{E}_{\text {cell }}^{\circ}=+2.733 \text{ V}\)
\(\mathrm{{Fe}_{ {(aq) }}^{2+}+2 {e}^{-} \rightarrow {Fe}({s}), {E}^{\circ}=-0.44{~V} }\)
\( \mathrm{{Cr}_2 {O}_7^{2-}{ }_{ {(aq) }}+14 {H}^{+}+6 e^{-} \rightarrow 2 {Cr}^{3+}+7 {H}_2 {O}},\)
\( \mathrm{{E}^{\circ}=+1.33 {~V}}\)
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 |
The incorrect statement about an inert electrode in a cell is:
| 1. | It does not participate in the cell reaction. |
| 2. | It provides a surface either for oxidation or for the reduction reaction. |
| 3. | It provides a surface for the conduction of electrons. |
| 4. | It provides a surface for redox reaction. |
The most stable oxidized species among the following is:
\(E_{{\mathrm{Cr}_2 \mathrm{O}_7^2}/ \mathrm{Cr}^{3+}}^{o} =1.33 \mathrm{~V} ; E_{\mathrm{Cl}_2 / \mathrm{Cl}^{-}}^{o}=1.36 \mathrm{~V} \)
\( E_{\mathrm{MnO_{4}}^{-} / \mathrm{Mn}^{2+}}^{o}=1.51 \mathrm{~V} ; E_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{o}=-0.74 \mathrm{~V}\)
| 1. | Cr3+ | 2. | MnO4- |
| 3. | Cr2O72- | 4. | Mn2+ |
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