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:

Updated On: Nov 23, 2024
  • Pt $\vert$ H$_{2(g)}$ $\vert$ HCl$_{(soln.)}$ $\vert$ AgCl$_{(s)}$ $\vert$ Ag
  • Pt $\vert$ H$_{2(g)}$ $\vert$ HCl$_{(soln.)}$ $\vert$ AgNO$_{3(aq)}$ $\vert$ Ag
  • Pt $\vert$ H$_{2(g)}$ $\vert$ KCl$_{(soln.)}$ $\vert$ AgCl$_{(s)}$ $\vert$ Ag
  • Ag $\vert$ AgCl$_{(s)}$ $\vert$ KCl$_{(soln.)}$ $\vert$ AgNO$_{3(aq)}$ $\vert$ Ag
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The Correct Option is C

Solution and Explanation

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.

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