The reaction: (g) + 2AgCl(s) -> 2H^+(aq) + 2Cl^-(aq) + 2Ag(s) occurs in the galvanic cell — Electrochemistry Chemistry Question

💡 Solution & Explanation

The correct cell representation for the given reaction is determined by analyzing the oxidation and reduction half-cell reactions. \subsubsection*{Step 1 - Analyze the Overall Cell Reaction} The given overall cell reaction is: $$\ce{H2(g) + 2AgCl(s) -> 2H+(aq) + 2Cl^-(aq) + 2Ag(s)}$$ To determine the correct galvanic cell notation, we split this overall reaction into two individual half-reactions: oxidation (occurring at the anode) and reduction (occurring at the cathode). \subsubsection*{Step 2 - Identify the Anode Reaction (Oxidation)} In the overall cell reaction, hydrogen gas ($\ce{H2}$) is oxidized to hydrogen ions ($\ce{H+}$) by losing electrons: $$\ce{H2(g) -> 2H+(aq) + 2e-}$$ \begin{itemize} \item \textbf{Electrode:} Since hydrogen is in the gaseous state, it cannot act as an electrical conductor on its own. An inert metal, platinum ($\ce{Pt}$), is used to provide the electrical contact and the catalytic surface necessary for the reaction. \textbf{Representation:} The anode half-cell is represented on the left side of the cell notation as: $$\ce{Pt | H2(g) | H+(aq)}$$ \end{itemize} \subsubsection*{Step 3 - Identify the Cathode Reaction (Reduction)} Silver ions within the solid silver chloride ($\ce{AgCl}$) are reduced to metallic silver ($\ce{Ag}$) by gaining electrons: $$\ce{2AgCl(s) + 2e- -> 2Ag(s) + 2Cl^-(aq)}$$ This constitutes a metal-insoluble metal salt-anion electrode (specifically, the silver-silver chloride electrode). \begin{itemize} \item \textbf{Electrode:} Metallic silver ($\ce{Ag}$) coated with solid, sparingly soluble silver chloride ($\ce{AgCl}$). \item \textbf{Representation:} The reduction reaction takes place in the presence of chloride ions ($\ce{Cl-}$). Therefore, the cathode half-cell is represented on the right side of the cell notation as: $$\ce{Cl^-(aq) | AgCl(s) | Ag(s)}$$ \end{itemize} \subsubsection*{Step 4 - Construct the Complete Cell Notation} By IUPAC convention, the anode (oxidation half-cell) is written on the left and the cathode (reduction half-cell) is written on the right. Both electrodes are in contact with a common electrolyte solution containing both $\ce{H+(aq)}$ and $\ce{Cl^-(aq)}$ ions. This electrolyte is hydrochloric acid, represented as $\ce{HCl(aq)}$. Combining the anode and the cathode with their shared electrolyte, we obtain the complete cell representation: $$\ce{Pt | H2(g) | HCl(aq) | AgCl(s) | Ag(s)}$$ \subsubsection*{Step 5 - Analyze and Evaluate the Options} Let's evaluate each of the options printed in the original question: \begin{itemize} \item \textbf{Option (a):} $\ce{Ag | AgCl(s) | KCl(aq) | AgNO3(aq) | Ag}$ \\ This option lacks the hydrogen electrode altogether, meaning the oxidation of $\ce{H2}$ to $\ce{H+}$ cannot take place. Therefore, it is incorrect. \item \textbf{Option (b):} $\ce{Pt | H2(g) | HCl(aq) | AgNO3(aq) | Ag}$ \\ In this cell, the cathode would be a standard silver-silver ion electrode in contact with soluble silver nitrate ($\ce{AgNO3}$), which does not involve insoluble $\ce{AgCl(s)}$. The overall reaction for this notation would be $\ce{H2(g) + 2Ag+(aq) -> 2H+(aq) + 2Ag(s)}$, which lacks $\ce{Cl-}$ ions. Therefore, it is incorrect. \item \textbf{Option (c):} $\ce{Pt | H2(g) | HCl(aq) | AgCl(s) | Ag(s)}$ \\ This is the correct representation, matching our derived cell notation exactly. \item \textbf{Option (d):} $\ce{Pt | H2(g) | KCl(aq) | AgCl(s) | Ag(s)}$ \\ The electrolyte here is potassium chloride ($\ce{KCl}$), which lacks the hydrogen ions ($\ce{H+}$) required to establish the standard hydrogen electrode anode. Without an acidic medium providing $\ce{H+}$ ions, the reaction cannot be represented as shown. Therefore, it is incorrect. \end{itemize} In the original textbook options, the correct cell representation corresponds to **Option (c)**. If your specific test version has shuffled this correct cell representation ($\ce{Pt | H2(g) | HCl(aq) | AgCl(s) | Ag}$) to option **(a)**, then Option A is correct.