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 |
86%
From NCERT
AIPMT - 2013
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Consider the half-cell reduction reaction:
$$\text{Mn}^{2+}+2e^-\rightarrow \text{Mn},\ E^{0}= -1.18~ \text V$$
$$\text{Mn}^{2+}\rightarrow \text{Mn}^{3+}+e^-,\ E^{0}= -1.5~ \text V$$

The $$E^{0}$$ for the reaction $$\mathrm{3\ Mn^{2+}\rightarrow Mn^{0}+2Mn^{3+} }$$ and possibility of the forward reaction are respectively:

 1 –4.18 V and Yes 2 +0.33 V and Yes 3 +2.69 V and No 4 –2.69 V and No
Subtopic:  Electrode & Electrode Potential |
53%
From NCERT
NEET - 2013
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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 |
81%
From NCERT
AIPMT - 2011
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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 |
60%
From NCERT
AIPMT - 2011
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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 |
90%
From NCERT
AIPMT - 2011
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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 |
68%
From NCERT
AIPMT - 2010
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If ${\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}^{\mathrm{o}}$ = -0.441 V and  ${\mathrm{E}}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}^{\mathrm{o}}$ = 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
Subtopic:  Electrode & Electrode Potential |
71%
From NCERT
AIPMT - 2006
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