- 1(current)
Q. No.The correct value of cell potential in volt for the reaction that occurs when the following two half cells are connected, is:
\(\mathrm{{Fe}_{ {(aq) }}^{2+}+2 {e}^{-} \rightarrow {Fe}({s}), {E}^{\circ}=-0.44{~V} }\)
\( \mathrm{{Cr}_2 {O}_7^{2-}{ }_{ {(aq) }}+14 {H}^{+}+6 e^{-} \rightarrow 2 {Cr}^{3+}+7 {H}_2 {O}},\)
\( \mathrm{{E}^{\circ}=+1.33 {~V}}\)
1. +1.77 V
2. +2.65 V
3. +0.01 V
4. +0.89 V
\(\mathrm{{Fe}_{ {(aq) }}^{2+}+2 {e}^{-} \rightarrow {Fe}({s}), {E}^{\circ}=-0.44{~V} }\)
\( \mathrm{{Cr}_2 {O}_7^{2-}{ }_{ {(aq) }}+14 {H}^{+}+6 e^{-} \rightarrow 2 {Cr}^{3+}+7 {H}_2 {O}},\)
\( \mathrm{{E}^{\circ}=+1.33 {~V}}\)
1. +1.77 V
2. +2.65 V
3. +0.01 V
4. +0.89 V
Subtopic: Electrode & Electrode Potential |
72%
From NCERT
NEET - 2023
Please attempt this question first.
Hints
Please attempt this question first.
Given below are half-cell reactions:
\(\text{MnO}_{4}^{-}+8 \text{H}^{+}+5 \text{e}^{-} \rightarrow \text{Mn}^{2+}+4 \text{H}_{2} \text{O}, \)
\( \text{E}_{\text{Mn}^{2+}}^{\circ} / \text{MnO}_{4}^{-}=-1.510 \text{ V} \)
\( \frac{1}{2} \text{O}_{2}+2 \text{H}^{+}+2 \text{e}^{-} \rightarrow \text{H}_{2} \text{O}, \)
\( \text{E}_{\text{O}_{2} / \text{H}_{2} \text{O}}^{\circ}=+1.223 \text{ V}\)
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 \(\text{E}_{\text {cell }}^{\circ}=+2.733 \text{ V}\)
\(\text{MnO}_{4}^{-}+8 \text{H}^{+}+5 \text{e}^{-} \rightarrow \text{Mn}^{2+}+4 \text{H}_{2} \text{O}, \)
\( \text{E}_{\text{Mn}^{2+}}^{\circ} / \text{MnO}_{4}^{-}=-1.510 \text{ V} \)
\( \frac{1}{2} \text{O}_{2}+2 \text{H}^{+}+2 \text{e}^{-} \rightarrow \text{H}_{2} \text{O}, \)
\( \text{E}_{\text{O}_{2} / \text{H}_{2} \text{O}}^{\circ}=+1.223 \text{ V}\)
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 \(\text{E}_{\text {cell }}^{\circ}=+2.733 \text{ V}\)
Subtopic: Electrode & Electrode Potential |
From NCERT
NEET - 2022
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
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NEET 2025 - Target Batch
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 |
Subtopic: Electrode & Electrode Potential |
66%
From NCERT
NEET - 2022
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
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Q. No.
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