Which of the following statements are correct?
i) \( CCl_4 \) undergoes hydrolysis easily.
(ii) Diamond has directional covalent bonds.
(iii)Fullerene is the thermodynamically most stable allotrope of carbon.
(iv)Glass is a man-made silicate.
\( i, ii \) only
Step 1: Analyzing Each Statement
1. Statement (i): Incorrect. Carbon tetrachloride (\( CCl_4 \)) does not undergo hydrolysis easily because it lacks vacant \( d \)-orbitals, making it resistant to nucleophilic attack by water molecules. 2. Statement (ii): Correct. Diamond has a tetrahedral structure where each carbon atom is bonded to four other carbon atoms through strong directional covalent bonds, giving it high hardness and rigidity. 3. Statement (iii): Incorrect. Graphite, not fullerene, is the most thermodynamically stable allotrope of carbon due to its lower energy configuration. 4. Statement (iv): Correct. Glass is an amorphous, man-made silicate composed mainly of silica (\( SiO_2 \)), which is fused with other compounds like sodium oxide and calcium oxide.
Step 2: Evaluating the Given Options
- Option (1): Incorrect, as a statement (i) is incorrect.
- Option (2): Correct, as statements (ii) and (iv) are the only correct ones.
- Option (3): Incorrect, as a statement (iii) is incorrect.
- Option (4): Incorrect, as a statement (i) is incorrect.
Thus, the correct answer is
Option (2).
If \( E^\circ_{Fe^{2+}/Fe} = -0.441 \, \text{V} \) and \( E^\circ_{Fe^{3+}/Fe^{2+}} = 0.771 \, \text{V} \),
the standard emf of the cell reaction \( Fe(s) + 2Fe^{3+}(aq) \rightarrow 3Fe^{2+}(aq) \) is:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] For the reaction, \( Fe^{3+} \) is reduced to \( Fe^{2+} \) (reduction at the cathode), and \( Fe \) is oxidized to \( Fe^{2+} \) (oxidation at the anode). So: \[ E^\circ_{\text{cell}} = E^\circ_{Fe^{3+}/Fe^{2+}} - E^\circ_{Fe^{2+}/Fe} \] \[ E^\circ_{\text{cell}} = 0.771 \, \text{V} - (-0.441 \, \text{V}) = 0.771 + 0.441 = 1.212 \, \text{V} \] Hence, the standard emf of the cell reaction is \( 1.212 \, \text{V} \).
Consider the following
Statement-I: Kolbe's electrolysis of sodium propionate gives n-hexane as product.
Statement-II: In Kolbe's process, CO$_2$ is liberated at anode and H$_2$ is liberated at cathode.
O\(_2\) gas will be evolved as a product of electrolysis of:
(A) an aqueous solution of AgNO3 using silver electrodes.
(B) an aqueous solution of AgNO3 using platinum electrodes.
(C) a dilute solution of H2SO4 using platinum electrodes.
(D) a high concentration solution of H2SO4 using platinum electrodes.
Choose the correct answer from the options given below :
A solution of aluminium chloride is electrolyzed for 30 minutes using a current of 2A. The amount of the aluminium deposited at the cathode is _________
Match the following: