- 1(current)
Q. No.Find the standard electrode potential (E°) for the reduction of Fe³⁺ to Fe²⁺ at 298 K using the following data.
Fe³⁺ + e⁻ → Fe²⁺
Given:
E°(Fe³⁺/Fe) = −0.04 V
E°(Fe²⁺/Fe) = −0.44 V
Choose the correct option:
| 1. | +0.92 V | 2. | +0.40 V |
| 3. | +0.76 V | 4. | -0.48 V |
Subtopic: Electrode & Electrode Potential |
50%
Level 3: 35%-60%
NEET - 2026
Please attempt this question first.
Hints
Please attempt this question first.
The correct value of cell potential in volts 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 |
75%
Level 2: 60%+
NEET - 2023
Hints
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 |
Level 3: 35%-60%
NEET - 2022
Hints
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 |
69%
Level 2: 60%+
NEET - 2022
Hints
- 1(current)
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