Q. No.If = -0.441 V and = 0.771 V, the standard emf of the reaction:
Fe + 2Fe3+→ 3Fe2+ will be:
| 1. | 0.330 V | 2. | 1.653 V |
| 3. | 1.212 V | 4. | 0.111 V |
Which of the following pairs represents the potential difference known as electrode potential?
1. Electrode and the electrolyte.
2. Anode and Cathode.
3. Anode and Atmosphere.
4. Cathode and Atmosphere.
The standard electrode potential for Sn4+/Sn2+ couple is +0.15 V and that for the Cr3+/Cr couple is -0.74 V. 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 |
The standard reduction potential for Fe2+|Fe and Sn2+|Sn electrodes are -0.44 V and -0.14 V respectively. For the cell reaction,
Fe2+ + Sn → Fe + Sn2+, the standard Emf is :
1. +0.30 V
2. 0.58 V
3. +0.58 V
4. -0.30 V
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 |
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.
Consider the following relations for emf of an electrochemical cell:
| (a) | emf of a cell = (Oxidation potential of the anode) – (Reduction potential of the cathode) |
| (b) | emf of a cell = (Oxidation potential of the anode) + (Reduction potential of the cathode) |
| (c) | emf of a cell = (Reduction potential of the anode) + (Reduction potential of the cathode) |
| (d) | emf of a cell = (Oxidation potential of the anode) – (Oxidation potential of the 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) |
\(\text{MnO}_{4}^{-}+8 \text{H}^{+}+5 \text{e}^{-} \rightarrow \text{Mn}^{2+}+4 \text{H}_{2} \text{O}, \)
\( \text{E}_{\text{Mn}^{2+}}^{\circ} / \text{MnO}_{4}^{-}=-1.510 \text{ V} \)
\( \frac{1}{2} \text{O}_{2}+2 \text{H}^{+}+2 \text{e}^{-} \rightarrow \text{H}_{2} \text{O}, \)
\( \text{E}_{\text{O}_{2} / \text{H}_{2} \text{O}}^{\circ}=+1.223 \text{ V}\)
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 |
\(\mathrm{{Fe}_{ {(aq) }}^{2+}+2 {e}^{-} \rightarrow {Fe}({s}), {E}^{\circ}=-0.44{~V} }\)
\( \mathrm{{Cr}_2 {O}_7^{2-}{ }_{ {(aq) }}+14 {H}^{+}+6 e^{-} \rightarrow 2 {Cr}^{3+}+7 {H}_2 {O}},\)
\( \mathrm{{E}^{\circ}=+1.33 {~V}}\)
1. +1.77 V
2. +2.65 V
3. +0.01 V
4. +0.89 V
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