- 1(current)
Q. No.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-
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 |
87%
Level 1: 80%+
AIPMT - 2013
Hints
Links
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:
\(\text{Mn}^{2+}+2e^-\rightarrow \text{Mn},\ E^{0}= -1.18~ \text V \)
\(\text{Mn}^{2+}\rightarrow \text{Mn}^{3+}+e^-,\ E^{0}= -1.5~ \text V \)
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 |
54%
Level 3: 35%-60%
NEET - 2013
Hints
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 |
82%
Level 1: 80%+
AIPMT - 2011
Hints
Based on electrode potentials in the table below:
The value of \(E_{Cu^{2+}/Cu}^{o}\) will be:
| 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 |
63%
Level 2: 60%+
AIPMT - 2011
Hints
Links
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 |
91%
Level 1: 80%+
AIPMT - 2011
Hints
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) |
Which of the following combinations correctly represents the relation for the emf of the cell?
| 1. | (a) and (b) | 2. | (c) and (d) |
| 3. | (b) and (d) | 4. | (c) and (a) |
Subtopic: Electrode & Electrode Potential |
70%
Level 2: 60%+
AIPMT - 2010
Hints
Links
If = -0.441 V and = 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 |
74%
Level 2: 60%+
AIPMT - 2006
Hints
Links
- 1(current)
Q. No.
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