Question

The reaction;$\frac{1}{2}H_{2(g)} + AgCl_{(s)} \rightarrow H^+_{(aq)} + Cl^-_{(aq)} + Ag_{(s)}$ occurs in which of the following galvanic cell:

• A. Pt $\vert$ H$_{2(g)}$ $\vert$ HCl$_{(soln.)}$ $\vert$ AgCl$_{(s)}$ $\vert$ Ag
• B. Pt $\vert$ H$_{2(g)}$ $\vert$ HCl$_{(soln.)}$ $\vert$ AgNO$_{3(aq)}$ $\vert$ Ag
• C. Pt $\vert$ H$_{2(g)}$ $\vert$ KCl$_{(soln.)}$ $\vert$ AgCl$_{(s)}$ $\vert$ Ag
• D. Ag $\vert$ AgCl$_{(s)}$ $\vert$ KCl$_{(soln.)}$ $\vert$ AgNO$_{3(aq)}$ $\vert$ Ag

Answer 1

Answer: (C)

Pt $\vert$ H$_{2(g)}$ $\vert$ KCl$_{(soln.)}$ $\vert$ AgCl$_{(s)}$ $\vert$ Ag

Answer 2

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.