The given reaction involves the oxidation of hydrogen gas to H$^+$ ions and the reduction of AgCl to Ag. This is consistent with the following electrode reactions:
Anode (oxidation): H$_2 \to$ 2H$^+ + 2e^-$
Cathode (reduction): AgCl + e$^- \to$ Ag + Cl$^-$ \end{itemize}
The correct galvanic cell setup corresponding to this reaction is: \[ \text{Pt} \vert \text{H}_2(\text{g}) \vert \text{HCl(sol$^n$)} \vert \text{AgCl(s)} \vert \text{Ag} \] Here, the HCl provides the H$^+$ ions required for the anode reaction, and AgCl serves as the source of Ag$^+$ for the cathode reaction.
Calculate the potential for half-cell containing 0.01 M K\(_2\)Cr\(_2\)O\(_7\)(aq), 0.01 M Cr\(^{3+}\)(aq), and 1.0 x 10\(^{-4}\) M H\(^+\)(aq).
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is: