Molten sodium chloride conducts electricity due to the presence of:
1. Free ions.
2. Free molecules.
3. Free electrons.
4. Atoms of sodium and chlorine.
Find the relationship between E₁ and E₂ for the following Daniell cell:
Zn | ZnSO₄ (0.01 M) || CuSO₄ (1.0 M) | Cu
The emf of this cell is E₁.
If the concentration of ZnSO₄ is changed to 1.0 M and that of CuSO₄ is changed to 0.01 M, the emf becomes E₂.
Given:
RT/F = 0.059 V
1. E1 = E2
2. E1 > E2
3. E1 < E2
4. E2 = 0 \(\neq\)E1
| 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 |
The change in reduction potential of a hydrogen electrode when its solution initially at pH = 0 is neutralised to pH = 7, is a/an:
| 1. | Increase by 0.059 V | 2. | Decrease by 0.059 V |
| 3. | Increase by 0.41 V | 4. | Decrease by 0.41 V |
The standard electrode potential for Sn4+/Sn2+ couple is +0.15 V and that for the Cr3+/Cr couple is -0.74 V. 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 |
|
Half-reaction |
Eº, V |
|
Ag+(aq) + e¯ → Ag(s) |
+0.80 |
|
2 Hg2+(aq) + 2e¯ → Hg22+(aq) |
+0.91 |
| 1. | –0.11 V | 2. | 0.69 V |
| 3. | 1.71 V | 4. | 2.51 V |
| 1. | Y > X > Z | 2. | Z > X > Y |
| 3. | X > Y > Z | 4. | Y > Z > X |
Find the maximum electrical work obtainable from the Daniell cell:
Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)
Given:
E°(Zn²⁺/Zn) = −0.76 V
E°(Cu²⁺/Cu) = +0.34 V
| \(E^o_{Zn^{2+}/Zn}\) | −0.76 V |
| \(E^o_{Cu^{2+}/Cu}\) | 0.34 V |
1. +106.15 kJ
2. -212.3 kJ
3. -424.6 kJ
4. +212.3 kJ
| Cu2+(aq) + 2e- → Cu(s) Eº = 0.34 V 2H+(aq) + 2e- → H2(g) Eº = 0.00 V Cr3+(aq) + 3e- → Cr(s) Eº = -0.73 V |
Given that
I2 + 2e¯ 2I¯; E° = 0.54V
Br2 + 2e¯ 2Br¯; E° = 1.09V
Predict which of the following is true?
| 1. | I¯ ions will be able to reduce bromine |
| 2. | Br¯ ion will be able to reduce iodine |
| 3. | Iodine will be able to reduce bromide ions |
| 4. | Bromine will be able to reduce iodide ions |