Rusting of iron is envisaged as setting up of an electrochemical cell because:
1. | Oxidation of iron takes place during rusting. |
2. | Reduction of iron takes place. |
3. | Iron shows disproportionation reaction. |
4. | Oxygen is getting oxidised. |
The cell in which the following reactions occurs:
2Fe3+(aq)+2I-(aq)→2Fe2+(aq)+I2(s) has \(E_{cell}^{o}\)= 0.236 V at 298 K.
The standard Gibbs energy of the cell reaction is −45.54 kJ mol−1 . The value of equilibrium constant of the cell reaction is
1. | 10.27 × 109 | 2. | 9.57 × 107 |
3. | 12.17 × 108 | 4. | 7.17 × 106 |
1. | Methane | 2. | Methanol |
3. | Sulphuric acid | 4. | Both 1 and 2 |
The metals among the following that can be extracted electrolytically is -
1. | K | 2. | Na |
3. | Al | 4. | All of the above |
The number of electrons flow through the wire when a current of 0.5 ampere flows through a metallic wire for 2 hours, is-
1. \(6.25×10^{22}\)
2. \(2.25×10^{22}\)
3 \(8.25×10^{22}\)
4. \(4.25×10^{22}\)
The molar conductivity of 0.025 mol L−1 methanoic acid is 46.1 S cm2 mol−1.
The dissociation constant of methanoic acid is-
Given λ°(H+)= 349.6 S cm2 mol−1 and λ°(HCOO−) = 54.6 S cm2 mol
1. \(1.27×10^{-4}mol ~L^{−1}\)
2. \(5.17×10^{-5}mol ~L^{−1}\)
3. \(3.67×10^{-4}mol ~L^{−1}\)
4. \(4.87×10^{-5}mol ~L^{−1}\)
Assertion (A): | The conductivity of a solution decreases with dilution. |
Reason (R): | Conductivity of a solution depends on the number of ions and the number of ions decreases when the solution is diluted. |
1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
2. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |
3. | (A) is True but (R) is False. |
4. | (A) is False but (R) is True. |
The cell in which the following reactions occurs:
2Fe3+(aq)+2I-(aq)→2Fe2+(aq)+I2(s) has \(E_{cell}^{o}\)= 0.236 V at 298 K.
The standard Gibbs energy of the cell reaction is -
1. 41.14 kJ mol−1
2. 45.54 kJ mol−1
3. −49.94 kJ mol−1
4. −45.54 kJ mol−1
The emf of the cell in which the following reaction takes place is:
Ni(s)+2Ag+(0.002 M) →Ni2+(0.160 M) + 2Ag(s)
( Given that \(E_{cell}^{o}\)= 1.05 V)
1. Ecell= 9.14 V
2. Ecell= 0.0914 V
3. Ecell= 0.00914 V
4. Ecell= 0.914 V