- 1(current)
Q. No.| 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
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & Biology1. Y > X > Z
2. Z > X > Y
3. X > Y > Z
4. Y > Z > X
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & Biology| 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
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & BiologyStandard 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
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & BiologyConsider 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)
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & BiologyIf = -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
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & Biology- 1(current)
Q. No.
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