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) |
The correct expression that represents the equivalent conductance at infinite dilution of is:
(Given that are the equivalent conductances at infinite dilution of the respective ions)
1.
2.
3.
4.
Molar conductivities at infinite dilution of
NaCl, HCl, and are 126.4, 425.9, and 91.0 S cm2 mol–1 respectively.
for will be:
1. | \(180.5~S~cm^2~mol^{-1}\) | 2. | \(290.8~S~cm^2~mol^{-1}\) |
3. | \(390.5~S~cm^2~mol^{-1}\) | 4. | \(425.5~S~cm^2~mol^{-1}\) |
The Gibb's energy for the decomposition of \(\mathrm{A l_{2} O_{3}}\) at \(\mathrm{500~ ^\circ C}\) is as follows:
2/3Al2O3 → 4/3Al + O2 ; ∆rG = + 960 k J mol–1
The potential difference needed for the electrolytic reduction of aluminium oxide (Al2O3) at \(\mathrm{500~ ^\circ C}\) is at least,
1. 3.0 V
2. 2.5 V
3. 5.0 V
4. 4.5 V
1. | –4.18 V and Yes | 2. | +0.33 V and Yes |
3. | +2.69 V and No | 4. | –2.69 V and No |
1. | 4.0 | 2. | 20.0 |
3. | 40.0 | 4. | 0.66 |
A hypothetical electrochemical cell is shown below.
A|A+(x M) || B+(y M)|B
The Emf measured is +0.20 V. The cell reaction is:
1. A+ + B → A + B+
2. A+ + e- → A ; B+ + e- → B
3. The cell reaction cannot be predicted.
4. A + B+ → A+ + B
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 |
1. | 17.6 mg | 2. | 21.3 mg |
3. | 24.3 mg | 4. | 13.6 mg |
The efficiency of a fuel cell is given by:
1.
2.
3.
4.