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
Subtopic:  Electrode & Electrode Potential |
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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 \(\text{E}_{\text {cell }}^{\circ}=+2.733 \text{ V}\)
Subtopic:  Electrode & Electrode Potential |
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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 |
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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. \(\text{HBrO}\)

Subtopic:  Electrode & Electrode Potential |
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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. \(\mathrm{E_{1} = E_{2}}\)

Subtopic:  Electrode & Electrode Potential | Nernst Equation |
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A button cell used in watches functions as following
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
Subtopic:  Electrode & Electrode Potential |
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Standard electrode potential of three metals X, Y and Z are -1.2 V, +0.5 V and -3.0 V respectively. The reducing power of these metals will be: 
1. Y > X > Z 2. Z > X > Y
3. X > Y > Z 4. Y > Z > X
Subtopic:  Electrode & Electrode Potential |
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Based on electrode potentials in the table below: 
Cu2+(aq) + e- → Cu+(aq) 0.15 V
Cu+(aq) + e- → Cu(s) 0.50 V

The value of \(E_{Cu^{2+}/Cu}^{o}\) will be:
1. 0.325 V 2. 0650 V
3. 0.150 V 4. 0.500 V
Subtopic:  Electrode & Electrode Potential |
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Standard electrode potential for Sn4+/Sn2+ couple is +0.15 V and that for Cr3+/Cr couple is -0.74. 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

Subtopic:  Electrode & Electrode Potential |
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Consider the following relations for emf of an electrochemical cell:

(a) emf of cell = (Oxidation potential of anode) – (Reduction potential of cathode)
(b) emf of cell = (Oxidation potential of anode) + (Reduction potential of cathode)
(c) emf of cell = (Reduction potential of anode) + (Reduction potential of cathode)
(d) emf of cell = (Oxidation potential of anode) – (Oxidation potential of cathode)

The correct relation among the given options is: 

1. (a) and (b) 2. (c) and (d)
3. (b) and (d) 4. (c) and (a)
Subtopic:  Electrode & Electrode Potential |
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