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 EFe2+/Feo = -0.441 V and  EFe3+/Fe2+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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