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).

A square loop of sides \( a = 1 \, {m} \) is held normally in front of a point charge \( q = 1 \, {C} \). The flux of the electric field through the shaded region is \( \frac{5}{p} \times \frac{1}{\varepsilon_0} \, {Nm}^2/{C} \), where the value of \( p \) is: