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.
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 _________
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 :
Electrolysis of 600 mL aqueous solution of NaCl for 5 min changes the pH of the solution to 12. The current in Amperes used for the given electrolysis is ….. (Nearest integer).
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 \( 1.212 \, \text{V} \).
Let \( a \in \mathbb{R} \) and \( A \) be a matrix of order \( 3 \times 3 \) such that \( \det(A) = -4 \) and \[ A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} \] where \( I \) is the identity matrix of order \( 3 \times 3 \).
If \( \det\left( (a + 1) \cdot \text{adj}\left( (a - 1) A \right) \right) \) is \( 2^m 3^n \), \( m, n \in \{ 0, 1, 2, \dots, 20 \} \), then \( m + n \) is equal to:
Rate law for a reaction between $A$ and $B$ is given by $\mathrm{R}=\mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}}$. If concentration of A is doubled and concentration of B is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction $\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)$ is