The reaction involves:
H$_2$(g) oxidizing to H$^+$(aq) in the anodic half-cell:
\[ \frac{1}{2}\text{H}_2(\text{g}) \rightarrow \text{H}^+(\text{aq}) + e^-. \]
AgCl(s) reducing to Ag(s) and Cl$^-$(aq) in the cathodic half-cell:
\[ \text{AgCl(s)} + e^- \rightarrow \text{Ag(s)} + \text{Cl}^-(\text{aq}). \]
Thus, the complete galvanic cell setup for the reaction is: \[ \text{Pt|H}_2(\text{g})|\text{KCl(soln.)|AgCl(s)|Ag}. \]
Here: H$_2$(g) serves as the gas electrode for the oxidation at the anode. AgCl(s) is reduced at the cathode.
If the molar conductivity ($\Lambda_m$) of a 0.050 mol $L^{–1}$ solution of a monobasic weak acid is 90 S $cm^{2} mol^{–1}$, its extent (degree) of dissociation will be:
[Assume: $\Lambda^0$ = 349.6 S $cm^{2} mol^{–1}$ and $\Lambda^0_{\text{acid}}$ = 50.4 S$ cm^{2} mol^{–1}$]
Let \( S = \left\{ m \in \mathbb{Z} : A^m + A^m = 3I - A^{-6} \right\} \), where
\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \]Then \( n(S) \) is equal to ______.
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: