JEE Main · 2023 · Shift-IeasyEC-004

The reaction occurs in which of the given galvanic cell? {1}{2}{H_2(g)} + {AgCl(s)} {H^+(aq)} + {Cl^-(aq)} + {Ag(s)}

Electrochemistry · Class 12 · JEE Main Previous Year Question

Question

The reaction occurs in which of the given galvanic cell? 12H2(g)+AgCl(s)H+(aq)+Cl(aq)+Ag(s)\frac{1}{2}\mathrm{H_2(g)} + \mathrm{AgCl(s)} \rightleftharpoons \mathrm{H^+(aq)} + \mathrm{Cl^-(aq)} + \mathrm{Ag(s)}

Options
  1. a

    PtH2(g)HCl(soln)AgCl(s)Ag\mathrm{Pt|H_2(g)|HCl(sol^n)|AgCl(s)|Ag}

  2. b

    AgAgCl(s)KCl(soln)AgNO3Ag\mathrm{Ag|AgCl(s)|KCl(sol^n)|AgNO_3|Ag}

  3. c

    PtH2(g)HCl(soln)AgNO3(soln)Ag\mathrm{Pt|H_2(g)|HCl(sol^n)|AgNO_3(sol^n)|Ag}

  4. d

    PtH2(g)KCl(soln)AgCl(s)Ag\mathrm{Pt|H_2(g)|KCl(sol^n)|AgCl(s)|Ag}

Correct Answera

PtH2(g)HCl(soln)AgCl(s)Ag\mathrm{Pt|H_2(g)|HCl(sol^n)|AgCl(s)|Ag}

Detailed Solution

Strategy: Break down the given net cell reaction into its oxidation (anode) and reduction (cathode) half-cells. Then, construct the cell notation following the "Anode | Solution | Cathode" convention.

Step 1: Identify half-reactions

  • Anode (Oxidation): 12\ceH2(g)H+(aq)+e\frac{1}{2}\ce{H2(g) \rightarrow H^+(aq) + e^-}. This requires a Platinum (Pt) electrode and acidic medium.
  • Cathode (Reduction): \ceAgCl(s)+eAg(s)+Cl(aq)\ce{AgCl(s) + e^- \rightarrow Ag(s) + Cl^-(aq)}. This is a metal-insoluble salt electrode.

Step 2: Construct cell notation

  1. Anode Side: \cePtH2(g)H+(aq)\ce{Pt|H2(g)|H^+(aq)}
  2. Cathode Side: \ceCl(aq)AgCl(s)Ag\ce{Cl^-(aq)|AgCl(s)|Ag}
  3. Electrolytic bridge: Both half-reactions can occur in an environment containing \ceH+\ce{H^+} and \ceCl\ce{Cl^-} (i.e., \ceHCl\ce{HCl}).

Conclusion: The notation is \cePtH2(g)HCl(soln)AgCl(s)Ag\ce{Pt|H2(g)|HCl(sol^n)|AgCl(s)|Ag}.

Answer: (a) \cePtH2(g)HCl(soln)AgCl(s)Ag\boxed{\text{Answer: (a) } \ce{Pt|H2(g)|HCl(sol^n)|AgCl(s)|Ag}}

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