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

The Correct Option is C

To determine in which galvanic cell the given reaction occurs, we need to analyze the components and chemical processes involved in the options.

The reaction is:

\(\frac{1}{2}H_{2(g)} + AgCl_{(s)} \rightarrow H^+_{(aq)} + Cl^-_{(aq)} + Ag_{(s)}\)

This represents a galvanic cell where hydrogen gas and silver chloride are involved. The hydrogen gas is oxidized to produce \(H^+\), and the \(AgCl\) is reduced to solid silver \(Ag\). This process would typically occur in a galvanic cell where:

1. Hydrogen gas \(H_2\) is used at one electrode, often composed of platinum (\(Pt \vert H_2 \vert H^+\)).
2. Silver chloride \(AgCl\) and its reduction to silver \(Ag\) forms the other electrode component.

Now, let's analyze each option:

• Option 1: \(Pt \vert H_{2(g)} \vert HCl_{(soln.)} \vert AgCl_{(s)} \vert Ag\)
• Option 2: \(Pt \vert H_{2(g)} \vert HCl_{(soln.)} \vert AgNO_{3(aq)} \vert Ag\)
• Option 3: \(Pt \vert H_{2(g)} \vert KCl_{(soln.)} \vert AgCl_{(s)} \vert Ag\)
• Option 4: \(Ag \vert AgCl_{(s)} \vert KCl_{(soln.)} \vert AgNO_{3(aq)} \vert Ag\)

The correct option should have the components that match the reactions required in the question. Here, Option 3 accurately represents the process:

1. \(Pt \vert H_{2(g)}\): This is where hydrogen gas is oxidized to \(H^+\).
2. \(KCl_{(soln.)} \vert AgCl_{(s)} \vert Ag\): This involves the incorporation of \(AgCl\) and the reduction to silver, which is the solid deposited at the electrode.

Conclusion: The galvanic cell corresponding to the reaction is \(Pt \vert H_{2(g)} \vert KCl_{(soln.)} \vert AgCl_{(s)} \vert Ag\), which is Option 3.

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.