Brine is a concentrated aqueous solution of \( \text{NaCl} \). During electrolysis of brine, the following reactions occur:
Water molecules are preferentially reduced over sodium ions because the reduction potential of water is higher than that of sodium ions.
\( 2\text{H}_2\text{O(l)} + 2e^- \rightarrow \text{H}_2\text{(g)} + 2\text{OH}^-\text{(aq)} \)
Chloride ions are oxidized to chlorine gas.
\( 2\text{Cl}^-\text{(aq)} \rightarrow \text{Cl}_2\text{(g)} + 2e^- \)
Combining the cathode and anode reactions:
\( 2\text{NaCl(aq)} + 2\text{H}_2\text{O(l)} \rightarrow \text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} + 2\text{NaOH(aq)} \)
The products obtained during the electrolysis of brine are hydrogen gas (\( \text{H}_2 \)), chlorine gas (\( \text{Cl}_2 \)), and sodium hydroxide (\( \text{NaOH} \)). \( \text{HCl} \) is not formed.
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.
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 _________
Two positively charged particles \(m_1\) and \(m_2\) have been accelerated across the same potential difference of 200 keV. Given mass of \(m_1 = 1 \,\text{amu}\) and \(m_2 = 4 \,\text{amu}\). The de Broglie wavelength of \(m_1\) will be \(x\) times that of \(m_2\). The value of \(x\) is _______ (nearest integer). 